video_title stringlengths 24 104 | Sentence stringlengths 126 3.25k |
|---|---|
Related rates balloon Applications of derivatives AP Calculus AB Khan Academy.mp3 | 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. ... |
Related rates balloon Applications of derivatives AP Calculus AB Khan Academy.mp3 | 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, I'm gonna take the derivative with respect to t on the left, w... |
Related rates balloon Applications of derivatives AP Calculus AB Khan Academy.mp3 | So let's write it like that. And now, let's take the derivative with respect to t. So d, d t, I'm 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 c... |
Related rates balloon Applications of derivatives AP Calculus AB Khan Academy.mp3 | 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 d t. Once again this is just d, the derivative, the, sorry, the derivative of th... |
Related rates balloon Applications of derivatives AP Calculus AB Khan Academy.mp3 | 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 1 over 500 dh dt. So 1 over 500 dh dt. We're literally saying it's just 1 over 500 times the derivative of h with re... |
Related rates balloon Applications of derivatives AP Calculus AB Khan Academy.mp3 | We're literally saying it's just 1 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 r... |
Related rates balloon Applications of derivatives AP Calculus AB Khan Academy.mp3 | 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 4. Secant squared of pi over 4. |
Related rates balloon Applications of derivatives AP Calculus AB Khan Academy.mp3 | So we get secant squared of theta. So we get secant squared. Right now our theta is pi over 4. Secant squared of pi over 4. Let me write those colors in to show you that I'm putting these values in. Secant squared of pi over 4 times d theta dt. Well, that is just 0.2. |
Related rates balloon Applications of derivatives AP Calculus AB Khan Academy.mp3 | Secant squared of pi over 4. Let me write those colors in to show you that I'm putting these values in. Secant squared of pi over 4 times d theta dt. Well, that is just 0.2. So times 0.2. And then this is going to be equal to 1 over 500. And we want to make sure, since this is in radians per minute, we're going to get ... |
Related rates balloon Applications of derivatives AP Calculus AB Khan Academy.mp3 | Well, that is just 0.2. So times 0.2. And then this is going to be equal to 1 over 500. And we want to make sure, since this is in radians per minute, we're going to get meters per minute. And this is meters right over here. We're going to get meters per minute right over here. We just want to make sure we know what ou... |
Related rates balloon Applications of derivatives AP Calculus AB Khan Academy.mp3 | And we want to make sure, since this is in radians per minute, we're going to get meters per minute. And this is meters right over here. We're going to get meters per minute right over here. We just want to make sure we know what our units are doing. I haven't written the units here to save some space. But we get 1 ove... |
Related rates balloon Applications of derivatives AP Calculus AB Khan Academy.mp3 | We just want to make sure we know what our units are doing. I haven't written the units here to save some space. But we get 1 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 secant squared of pi over 4... |
Related rates balloon Applications of derivatives AP Calculus AB Khan Academy.mp3 | 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 secant squared of pi over 4. That is 1 over cosine squared of pi over 4. Cosine of pi over 4. Let me write this over here. Cosine of pi over 4 is square root of 2 over 2. Cosi... |
Related rates balloon Applications of derivatives AP Calculus AB Khan Academy.mp3 | Cosine of pi over 4. Let me write this over here. Cosine of pi over 4 is square root of 2 over 2. Cosine squared of pi over 4 is going to be equal to 2 over 4, which is equal to 1 half. And so secant squared of pi over 4 is just 1 over that is equal to 2. So this is going to be equal to, let me rewrite this. So the sec... |
Related rates balloon Applications of derivatives AP Calculus AB Khan Academy.mp3 | Cosine squared of pi over 4 is going to be equal to 2 over 4, which is equal to 1 half. And so secant squared of pi over 4 is just 1 over that is equal to 2. So this is going to be equal to, let me rewrite this. So the secant squared of pi over 4, let me erase this right over here. Secant squared of pi over 4, all of t... |
Related rates balloon Applications of derivatives AP Calculus AB Khan Academy.mp3 | So the secant squared of pi over 4, let me erase this right over here. Secant squared of pi over 4, all of this business right over here, simplifies to 2. So times 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 0.4, which is equal to, let me make sure I get th... |
Related rates balloon Applications of derivatives AP Calculus AB Khan Academy.mp3 | So what is this going to be? This is going to be 500 times 0.4. So this is equal to 500 times 0.4, which is equal to, let me make sure I get this right. This would be with two 0's and one behind the decimal. Yep, there you go. It would be 200. So that rate at which our height is changing with respect to time right at t... |
Introduction to infinite limits Limits and continuity AP Calculus AB Khan Academy.mp3 | This is y is equal to one over x squared. 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 un... |
Introduction to infinite limits Limits and continuity AP Calculus AB Khan Academy.mp3 | 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, t... |
Introduction to infinite limits Limits and continuity AP Calculus AB Khan Academy.mp3 | 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 notati... |
Introduction to infinite limits Limits and continuity AP Calculus AB Khan Academy.mp3 | 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... |
Introduction to infinite limits Limits and continuity AP Calculus AB Khan Academy.mp3 | 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... |
Introduction to infinite limits Limits and continuity AP Calculus AB Khan Academy.mp3 | 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... |
Introduction to infinite limits Limits and continuity AP Calculus AB Khan Academy.mp3 | 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 ... |
Introduction to infinite limits Limits and continuity AP Calculus AB Khan Academy.mp3 | 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... |
Introduction to infinite limits Limits and continuity AP Calculus AB Khan Academy.mp3 | 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... |
Introduction to infinite limits Limits and continuity AP Calculus AB Khan Academy.mp3 | 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... |
Introduction to infinite limits Limits and continuity AP Calculus AB Khan Academy.mp3 | 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 negativ... |
Introduction to infinite limits Limits and continuity AP Calculus AB Khan Academy.mp3 | 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, the... |
Introduction to infinite limits Limits and continuity AP Calculus AB Khan Academy.mp3 | 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 g... |
Unbounded limits Limits and continuity AP Calculus AB Khan Academy.mp3 | 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 unb... |
Unbounded limits Limits and continuity AP Calculus AB Khan Academy.mp3 | 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'... |
Unbounded limits Limits and continuity AP Calculus AB Khan Academy.mp3 | 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... |
Unbounded limits Limits and continuity AP Calculus AB Khan Academy.mp3 | 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 o... |
Unbounded limits Limits and continuity AP Calculus AB Khan Academy.mp3 | 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... |
Worked examples Definite integral properties 1 AP Calculus AB Khan Academy.mp3 | And we're given the graph of f of x and of y equals f of x, and the area between f of x and the x-axis over different intervals. 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... |
Worked examples Definite integral properties 1 AP Calculus AB Khan Academy.mp3 | 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. So we want to go from seven to fou... |
Worked examples Definite integral properties 1 AP Calculus AB Khan Academy.mp3 | 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. But the key realization is... |
Worked examples Definite integral properties 1 AP Calculus AB Khan Academy.mp3 | 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 fou... |
Worked examples Definite integral properties 1 AP Calculus AB Khan Academy.mp3 | 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 thi... |
Worked examples Definite integral properties 1 AP Calculus AB Khan Academy.mp3 | 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 t... |
2015 AP Calculus AB 5a AP Calculus AB solved exams AP Calculus AB Khan Academy.mp3 | So you have a horizontal tangent right over, a horizontal tangent right over there. Let me draw that a little bit neater. Right over there, a horizontal tangent right over there, and a horizontal tangent right over there. All right, the areas of the regions bounded by the x-axis in the graph of f prime on the intervals... |
2015 AP Calculus AB 5a AP Calculus AB solved exams AP Calculus AB Khan Academy.mp3 | All right, the areas of the regions bounded by the x-axis in the graph of f prime on the intervals negative two to one, closed intervals from negative two to one, so this region right over here, and the region from one to four, so this region right over there, they tell us the areas are nine and 12 respectively. So tha... |
2015 AP Calculus AB 5a AP Calculus AB solved exams AP Calculus AB Khan Academy.mp3 | Give a reason for your answer. All x-coordinates at which f has a relative maximum. So you might say, oh, look, this looks like a relative maximum over here, but this isn't f, this is the graph of f prime. So let's think about what needs to, we don't have a graph of f in front of us. So let's think about what needs to ... |
2015 AP Calculus AB 5a AP Calculus AB solved exams AP Calculus AB Khan Academy.mp3 | So let's think about what needs to, we don't have a graph of f in front of us. So let's think about what needs to be true for f to have a relative maximum at a point. So let's, we are probably familiar with what relative maxima look like. They look like a little lump like that. They could also actually look like that, ... |
2015 AP Calculus AB 5a AP Calculus AB solved exams AP Calculus AB Khan Academy.mp3 | They look like a little lump like that. They could also actually look like that, but since this is a differentiable function over the interval, we're probably not dealing with a relative maximum that looks like that. And so what do we know about a relative maximum point? So let's say, let's say that's our relative maxi... |
2015 AP Calculus AB 5a AP Calculus AB solved exams AP Calculus AB Khan Academy.mp3 | So let's say, let's say that's our relative maximum. Well, as we approach our relative maximum for values for, as we have x-values that are approaching the x-value of our relative maximum point, as we approach it from values below that x-value, we see that we have a positive slope. Our function needs to be increasing. ... |
2015 AP Calculus AB 5a AP Calculus AB solved exams AP Calculus AB Khan Academy.mp3 | So over here, over here, we see f is increasing, going into the relative maximum point. f is increasing, which means that the derivative of f, the derivative of f must be greater than zero. And then after we pass that maximum point, after we pass that maximum point, we see that our function needs to be decreasing. Let ... |
2015 AP Calculus AB 5a AP Calculus AB solved exams AP Calculus AB Khan Academy.mp3 | Let me just add another color. We see that our function is decreasing right over here. So f decreasing, decreasing, which means that f-prime of x needs to be less than zero. So our relative maximum point should be, should happen at an x-value. It should happen at an x-value where our first derivative transitions from b... |
2015 AP Calculus AB 5a AP Calculus AB solved exams AP Calculus AB Khan Academy.mp3 | So our relative maximum point should be, should happen at an x-value. It should happen at an x-value where our first derivative transitions from being greater than zero to being less than zero. So what x-values, let me say this. So we have f has relative, let me just write it shorthand, relative maximum at x-values whe... |
2015 AP Calculus AB 5a AP Calculus AB solved exams AP Calculus AB Khan Academy.mp3 | So we have f has relative, let me just write it shorthand, relative maximum at x-values where f-prime transitions, transitions, transitions from positive, positive, to negative, to, let me write this a little bit neater, to negative, to negative. And where do we see f-prime transitioning from positive to negative? Well... |
2015 AP Calculus AB 5a AP Calculus AB solved exams AP Calculus AB Khan Academy.mp3 | We see right here f-prime is positive, positive, positive, and then it goes negative, negative, negative. So we see f-prime is positive over here. And then right when we hit x equals negative two, f-prime becomes negative. f-prime becomes negative. So we know that the function itself, not f-prime, f must be increasing ... |
2015 AP Calculus AB 5a AP Calculus AB solved exams AP Calculus AB Khan Academy.mp3 | f-prime becomes negative. So we know that the function itself, not f-prime, f must be increasing here because f-prime is positive. And then our function f is decreasing here because f-prime is negative. And so this happens at x equals two. So let me write that down. This happens at x equals two. This happens, happens a... |
2015 AP Calculus AB BC 3cd AP Calculus AB solved exams AP Calculus AB Khan Academy.mp3 | For zero is less than or equal to t is less than or equal to 10, Bob's velocity is modeled by b of t is equal to t to the third minus 60 squared plus 300, where t is measured in minutes and b of t is measured in meters per minute. Find Bob's acceleration at time t equals 5. Well, acceleration, this is a velocity functi... |
2015 AP Calculus AB BC 3cd AP Calculus AB solved exams AP Calculus AB Khan Academy.mp3 | What is the rate of change of velocity with respect to time? That's acceleration. So we really just want to evaluate. This Bob's acceleration at t equals 5, that's going to be b prime of 5. So let's first figure out what b prime of t is. b prime of t is equal to, we'll take the derivative here, it's pretty straightforw... |
2015 AP Calculus AB BC 3cd AP Calculus AB solved exams AP Calculus AB Khan Academy.mp3 | This Bob's acceleration at t equals 5, that's going to be b prime of 5. So let's first figure out what b prime of t is. b prime of t is equal to, we'll take the derivative here, it's pretty straightforward, just use the power rule. So it's going to be 3t squared minus 12t, 2 times negative 6 is 12, or negative 12, and ... |
2015 AP Calculus AB BC 3cd AP Calculus AB solved exams AP Calculus AB Khan Academy.mp3 | So it's going to be 3t squared minus 12t, 2 times negative 6 is 12, or negative 12, and then the derivative of 300, 300 doesn't change with respect to time, so it's just a zero. And so b prime of 5 is going to be equal to 3 times 5 squared minus 12 times 5, which is equal to 75 minus 60, which is equal to 15. And the u... |
2015 AP Calculus AB BC 3cd AP Calculus AB solved exams AP Calculus AB Khan Academy.mp3 | So we could write as meters, let me write it out, meters per minute per minute, which is the same thing as meters per minute, meters per minute squared. Alright, let's do the next part. Based on the model, based on the model b from part c, find Bob's average velocity during the interval from zero is less than or equal ... |
2015 AP Calculus AB BC 3cd AP Calculus AB solved exams AP Calculus AB Khan Academy.mp3 | And if the notion of average velocity or average value of a function is completely foreign to you, I encourage you to watch the videos on Khan Academy on finding the average of a function. But straight, just to kind of cut to the chase, the average velocity, the average velocity is going to be the area under the veloci... |
2015 AP Calculus AB BC 3cd AP Calculus AB solved exams AP Calculus AB Khan Academy.mp3 | So it's going to be divided by, well, you're going from zero to 10, so 10 minus zero is going to be equal to 10. And if you wanted the intuition here, it's like, well, if you know the area of something, and if you wanted to find its average height, you could just divide by its width, and that's what we're doing here. I... |
2015 AP Calculus AB BC 3cd AP Calculus AB solved exams AP Calculus AB Khan Academy.mp3 | That's, I guess, a very high level intuition for where this expression came from. And so this is going to be equal to 1 tenth times the integral from zero to 10, and b of t is t to the third power minus 6t squared plus 300 dt. And so this is going to be equal to 1 over 10. Take the antiderivative here, so this is going... |
2015 AP Calculus AB BC 3cd AP Calculus AB solved exams AP Calculus AB Khan Academy.mp3 | Take the antiderivative here, so this is going to be t to the fourth over 4, t to the fourth over 4. And then this is going to be, if we increase the exponent here by 1, it's t to the third, and then you divide by 3. So it's negative 6 divided by 3 is negative 2t to the third, and then plus 300t, 300t, and I'm going to... |
2015 AP Calculus AB BC 3cd AP Calculus AB solved exams AP Calculus AB Khan Academy.mp3 | I am going to evaluate it at 10, and subtract from that, and evaluate it at 0. And so this is going to be equal to 1 tenth, that same 1 tenth there. And when you evaluate all of this at 10, what are we going to get? Let's see, 10 to the fourth power is 10,000 divided by 4 is 2,500. And then minus 2 times 10 to the thir... |
2015 AP Calculus AB BC 3cd AP Calculus AB solved exams AP Calculus AB Khan Academy.mp3 | Let's see, 10 to the fourth power is 10,000 divided by 4 is 2,500. And then minus 2 times 10 to the third, so it's 2 times 1,000. So minus 2,000. And then 300 times 10, well that's plus 3,000. And then you subtract all of this evaluated at 0, which is just going to be 0. So this is going to be equal to 2,500 minus 2,00... |
2015 AP Calculus AB BC 3cd AP Calculus AB solved exams AP Calculus AB Khan Academy.mp3 | And then 300 times 10, well that's plus 3,000. And then you subtract all of this evaluated at 0, which is just going to be 0. So this is going to be equal to 2,500 minus 2,000 is 500, plus 3,000. This all simplifies to 3,500, or 3,500. And then you divide it by 10. This is going to be 350, and it's an average velocity,... |
Tangents of polynomials Derivative rules AP Calculus AB Khan Academy.mp3 | 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 d... |
Tangents of polynomials Derivative rules AP Calculus AB Khan Academy.mp3 | 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, ... |
Tangents of polynomials Derivative rules AP Calculus AB Khan Academy.mp3 | 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... |
Tangents of polynomials Derivative rules AP Calculus AB Khan Academy.mp3 | 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 equ... |
Tangents of polynomials Derivative rules AP Calculus AB Khan Academy.mp3 | 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. It is negative nine. |
Tangents of polynomials Derivative rules AP Calculus AB Khan Academy.mp3 | 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. And now let's evaluate what the derivative is at one. |
Tangents of polynomials Derivative rules AP Calculus AB Khan Academy.mp3 | 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. Well here it's just a polynomial. |
Tangents of polynomials Derivative rules AP Calculus AB Khan Academy.mp3 | 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. We bring the three out front. |
Tangents of polynomials Derivative rules AP Calculus AB Khan Academy.mp3 | 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. So you bring the two times the six to get 12. |
Tangents of polynomials Derivative rules AP Calculus AB Khan Academy.mp3 | 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 o... |
Tangents of polynomials Derivative rules AP Calculus AB Khan Academy.mp3 | 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. ... |
Tangents of polynomials Derivative rules AP Calculus AB Khan Academy.mp3 | 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 e... |
Tangents of polynomials Derivative rules AP Calculus AB Khan Academy.mp3 | 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 negati... |
Tangents of polynomials Derivative rules AP Calculus AB Khan Academy.mp3 | 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 th... |
Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy (2).mp3 | 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 ... |
Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy (2).mp3 | 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... |
Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy (2).mp3 | 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 ab... |
Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy (2).mp3 | 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. |
Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy (2).mp3 | 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. |
Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy (2).mp3 | 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 he... |
Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy (2).mp3 | 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 p... |
Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy (2).mp3 | 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... |
Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy (2).mp3 | 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 equ... |
Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy (2).mp3 | 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 t... |
Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy (2).mp3 | 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 i... |
Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy (2).mp3 | 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 i... |
Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy (2).mp3 | 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,... |
Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy (2).mp3 | 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 righ... |
Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy (2).mp3 | 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 1⁄2 times base t... |
Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy (2).mp3 | The area here is gonna be 1 1⁄2 times base times height. So the area, and once again, this is this entire area, that's going to be 1 1⁄2 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 t... |
Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy (2).mp3 | 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 l... |
Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy (2).mp3 | 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 ... |
Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy (2).mp3 | 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 ... |
Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy (2).mp3 | 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 t... |
Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy (2).mp3 | 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 abso... |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.