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Fancy algebra to find a limit and make a function continuous Differential Calculus Khan Academy.mp3 | 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, s... |
Fancy algebra to find a limit and make a function continuous Differential Calculus Khan Academy.mp3 | 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, altho... |
Fancy algebra to find a limit and make a function continuous Differential Calculus Khan Academy.mp3 | 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... |
Fancy algebra to find a limit and make a function continuous Differential Calculus Khan Academy.mp3 | 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 ... |
Fancy algebra to find a limit and make a function continuous Differential Calculus Khan Academy.mp3 | 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 fi... |
Fancy algebra to find a limit and make a function continuous Differential Calculus Khan Academy.mp3 | 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 ... |
Fancy algebra to find a limit and make a function continuous Differential Calculus Khan Academy.mp3 | 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 ... |
Finding definite integrals using area formulas AP Calculus AB Khan Academy.mp3 | 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,... |
Finding definite integrals using area formulas AP Calculus AB Khan Academy.mp3 | 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 ... |
Finding definite integrals using area formulas AP Calculus AB Khan Academy.mp3 | 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 def... |
Finding definite integrals using area formulas AP Calculus AB Khan Academy.mp3 | 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 ... |
Finding definite integrals using area formulas AP Calculus AB Khan Academy.mp3 | 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, jus... |
Finding definite integrals using area formulas AP Calculus AB Khan Academy.mp3 | 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 c... |
Finding definite integrals using area formulas AP Calculus AB Khan Academy.mp3 | 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 1⁄2. So this has an ar... |
Finding definite integrals using area formulas AP Calculus AB Khan Academy.mp3 | So if you split it up like this, this triangle right over here has an area of one times two times 1 1⁄2. 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 ... |
Finding definite integrals using area formulas AP Calculus AB Khan Academy.mp3 | 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 1⁄2. Remember, the area of a triangle is 1 1⁄2 base times height, so it's one. So if you add up those areas, one plus two plus one is four.... |
Finding definite integrals using area formulas AP Calculus AB Khan Academy.mp3 | So it's one times two times 1 1⁄2. Remember, the area of a triangle is 1 1⁄2 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 t... |
Finding definite integrals using area formulas AP Calculus AB Khan Academy.mp3 | 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. So we'r... |
Finding definite integrals using area formulas AP Calculus AB Khan Academy.mp3 | 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? Well, just the formula for the area of a triangle, base times h... |
Finding definite integrals using area formulas AP Calculus AB Khan Academy.mp3 | 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 1⁄2. So, or you could say 1 1⁄2 times our base, which is a length of, see, we have a base of three right over here, we go from ... |
Finding definite integrals using area formulas AP Calculus AB Khan Academy.mp3 | Well, just the formula for the area of a triangle, base times height times 1 1⁄2. So, or you could say 1 1⁄2 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 1⁄2 times three times our height, which is one, two, three, four, times four. Well, this is just g... |
Finding definite integrals using area formulas AP Calculus AB Khan Academy.mp3 | 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 ci... |
Finding definite integrals using area formulas AP Calculus AB Khan Academy.mp3 | 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 ... |
Analyzing functions for discontinuities (discontinuity example) AP Calculus AB Khan Academy.mp3 | So we've got this function f of x that is piecewise continuous, it's defined over several intervals here. For x being, or for zero less than x, being less than or equal to two, f of x is natural log of x. 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 wan... |
Analyzing functions for discontinuities (discontinuity example) AP Calculus AB Khan Academy.mp3 | 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... |
Analyzing functions for discontinuities (discontinuity example) AP Calculus AB Khan Academy.mp3 | 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 ... |
Analyzing functions for discontinuities (discontinuity example) AP Calculus AB Khan Academy.mp3 | 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 ... |
Analyzing functions for discontinuities (discontinuity example) AP Calculus AB Khan Academy.mp3 | 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 ... |
Analyzing functions for discontinuities (discontinuity example) AP Calculus AB Khan Academy.mp3 | 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'... |
Analyzing functions for discontinuities (discontinuity example) AP Calculus AB Khan Academy.mp3 | 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 ... |
Analyzing functions for discontinuities (discontinuity example) AP Calculus AB Khan Academy.mp3 | 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 approa... |
Analyzing functions for discontinuities (discontinuity example) AP Calculus AB Khan Academy.mp3 | 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... |
Analyzing functions for discontinuities (discontinuity example) AP Calculus AB Khan Academy.mp3 | 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 ... |
Analyzing functions for discontinuities (discontinuity example) AP Calculus AB Khan Academy.mp3 | 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. But the thing that might jump out at you is that these are two d... |
Analyzing functions for discontinuities (discontinuity example) AP Calculus AB Khan Academy.mp3 | 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... |
Analyzing functions for discontinuities (discontinuity example) AP Calculus AB Khan Academy.mp3 | 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 ... |
Proof d dx(sqrt(x)) Taking derivatives Differential Calculus Khan Academy.mp3 | 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. An... |
Proof d dx(sqrt(x)) Taking derivatives Differential Calculus Khan Academy.mp3 | 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. So what I'm going to do, right now when I look at that, there's not much simp... |
Proof d dx(sqrt(x)) Taking derivatives Differential Calculus Khan Academy.mp3 | 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 fractio... |
Proof d dx(sqrt(x)) Taking derivatives Differential Calculus Khan Academy.mp3 | 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. So I said the square root of x plus delta x minus square root of x, all ... |
Proof d dx(sqrt(x)) Taking derivatives Differential Calculus Khan Academy.mp3 | 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 p... |
Proof d dx(sqrt(x)) Taking derivatives Differential Calculus Khan Academy.mp3 | 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 numb... |
Proof d dx(sqrt(x)) Taking derivatives Differential Calculus Khan Academy.mp3 | 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. Well, if you view this as a minus b times a plus b, right? Let me do a little aside here. |
Proof d dx(sqrt(x)) Taking derivatives Differential Calculus Khan Academy.mp3 | 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? So this is a plus b times a minus b. So it's going to be eq... |
Proof d dx(sqrt(x)) Taking derivatives Differential Calculus Khan Academy.mp3 | 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? These are my a's. Well, it's just going to be x plus delta x... |
Proof d dx(sqrt(x)) Taking derivatives Differential Calculus Khan Academy.mp3 | 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. And then what's b squared? So minus square root of x is b in this analogy. |
Proof d dx(sqrt(x)) Taking derivatives Differential Calculus Khan Academy.mp3 | 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. 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... |
Proof d dx(sqrt(x)) Taking derivatives Differential Calculus Khan Academy.mp3 | 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. So those cancel out. So we have delta x minus x. |
Proof d dx(sqrt(x)) Taking derivatives Differential Calculus Khan Academy.mp3 | 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. 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. |
Proof d dx(sqrt(x)) Taking derivatives Differential Calculus Khan Academy.mp3 | 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. And so this equals the limit. I'll write smaller because I'm running out of space. |
Proof d dx(sqrt(x)) Taking derivatives Differential Calculus Khan Academy.mp3 | 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. And of course, we can only do this assuming that delta, well, we're dividing by delta x to begin with. So we know i... |
Proof d dx(sqrt(x)) Taking derivatives Differential Calculus Khan Academy.mp3 | 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. So we get square root of x plus delta x plus the square root of x. And now ... |
Proof d dx(sqrt(x)) Taking derivatives Differential Calculus Khan Academy.mp3 | 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. That's what it's approaching. So that that equals 1 over the square root of x. Delta x is 0, so we can... |
Proof d dx(sqrt(x)) Taking derivatives Differential Calculus Khan Academy.mp3 | 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. And then this is, of course, just a square root of x here plu... |
Proof d dx(sqrt(x)) Taking derivatives Differential Calculus Khan Academy.mp3 | 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. So we just proved... |
Proof d dx(sqrt(x)) Taking derivatives Differential Calculus Khan Academy.mp3 | 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... |
Proof d dx(sqrt(x)) Taking derivatives Differential Calculus Khan Academy.mp3 | 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. Square root of x. And it's hopefully not t... |
Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3 | So this right over here is a mouse. 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 abov... |
Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3 | 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. Let me draw that. S... |
Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3 | 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. And right now, it is going 20 feet per second. ... |
Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3 | 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. Ligh... |
Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3 | 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. And as the owl goes further and further down, the shadow is going to mo... |
Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3 | 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 ... |
Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3 | 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 t... |
Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3 | 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 tha... |
Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3 | 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 th... |
Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3 | 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 sha... |
Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3 | 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 g... |
Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3 | 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. So let's call that y. Right at this moment, it's equal to 15. |
Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3 | 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. 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? |
Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3 | 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 cha... |
Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3 | 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... |
Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3 | 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... |
Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3 | 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. They both share this angle. So i... |
Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3 | 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. So we know that the ratio of x to y must b... |
Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3 | 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 ... |
Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3 | 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 multipl... |
Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3 | 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. And on the left-hand side, it is 20x. And then on the right-hand ... |
Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3 | 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. 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 ... |
Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3 | 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... |
Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3 | 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 ... |
Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3 | 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. So the derivative of y with respect to t is just dy dt. And then... |
Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3 | So the derivative of y with respect to t is just 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 dt. And there you have it. You have your relationship between dx dt, d... |
Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3 | 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. |
Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3 | 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. |
Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3 | 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. |
Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3 | 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 ... |
Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3 | 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 coul... |
Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3 | 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? |
Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3 | 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. |
Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3 | 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. |
Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3 | 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. |
Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3 | 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 act... |
Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3 | 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 ... |
Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3 | 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. |
Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3 | 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. |
Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3 | 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. |
Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3 | 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. |
Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3 | 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 sid... |
Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3 | 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. |
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