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Approximating limits using tables Limits and continuity AP Calculus AB Khan Academy.mp3 | 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. So is this equal to 1.8? |
Washer method rotating around vertical line (not y-axis), part 2 AP Calculus AB Khan Academy.mp3 | Using the, I guess we could call it the washer method or the ring method, we were able to come up with a definite integral for the volume of this solid of revolution right over here. So this is equal to the volume. 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. |
Washer method rotating around vertical line (not y-axis), part 2 AP Calculus AB Khan Academy.mp3 | 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. And then negative y squared squared is plus y to the fourth. |
Washer method rotating around vertical line (not y-axis), part 2 AP Calculus AB Khan Academy.mp3 | 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. And then all of that dy. |
Washer method rotating around vertical line (not y-axis), part 2 AP Calculus AB Khan Academy.mp3 | 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. And so this is going to be equal to pi times the definite integral from 0 to 1. |
Washer method rotating around vertical line (not y-axis), part 2 AP Calculus AB Khan Academy.mp3 | 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. So that cancels with that. |
Washer method rotating around vertical line (not y-axis), part 2 AP Calculus AB Khan Academy.mp3 | 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. So we have a y to the fourth. |
Washer method rotating around vertical line (not y-axis), part 2 AP Calculus AB Khan Academy.mp3 | 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. And so the next highest degree term right here is this negative 4y squared. |
Washer method rotating around vertical line (not y-axis), part 2 AP Calculus AB Khan Academy.mp3 | 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. We have negative 4y squared. |
Washer method rotating around vertical line (not y-axis), part 2 AP Calculus AB Khan Academy.mp3 | 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. So it's a negative y. |
Washer method rotating around vertical line (not y-axis), part 2 AP Calculus AB Khan Academy.mp3 | 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. So this is going to end up being a positive 4 square roots of y. |
Washer method rotating around vertical line (not y-axis), part 2 AP Calculus AB Khan Academy.mp3 | 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. Now we're ready to take the antiderivative. |
Washer method rotating around vertical line (not y-axis), part 2 AP Calculus AB Khan Academy.mp3 | 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. Antiderivative of the negative y is negative y squared over 2. |
Washer method rotating around vertical line (not y-axis), part 2 AP Calculus AB Khan Academy.mp3 | 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. Multiply by 2 thirds. |
Washer method rotating around vertical line (not y-axis), part 2 AP Calculus AB Khan Academy.mp3 | 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. Yep, that all works out. |
Washer method rotating around vertical line (not y-axis), part 2 AP Calculus AB Khan Academy.mp3 | 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. So this is all going to be equal to pi times evaluating all this business at 1. |
Washer method rotating around vertical line (not y-axis), part 2 AP Calculus AB Khan Academy.mp3 | 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. So whenever you evaluate at 1, it's just going to be. |
Washer method rotating around vertical line (not y-axis), part 2 AP Calculus AB Khan Academy.mp3 | 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. Plus 8 thirds. |
Washer method rotating around vertical line (not y-axis), part 2 AP Calculus AB Khan Academy.mp3 | 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? Let's see. |
Washer method rotating around vertical line (not y-axis), part 2 AP Calculus AB Khan Academy.mp3 | 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. So we could rewrite this as equal to pi. |
Washer method rotating around vertical line (not y-axis), part 2 AP Calculus AB Khan Academy.mp3 | 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. 4 thirds is 40 over 30. |
Washer method rotating around vertical line (not y-axis), part 2 AP Calculus AB Khan Academy.mp3 | 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. Well, actually, let me make that other shade of green. |
Washer method rotating around vertical line (not y-axis), part 2 AP Calculus AB Khan Academy.mp3 | 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. And then finally, 8 thirds is the same thing as 80 over 30. |
Washer method rotating around vertical line (not y-axis), part 2 AP Calculus AB Khan Academy.mp3 | 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. So let's see. |
Washer method rotating around vertical line (not y-axis), part 2 AP Calculus AB Khan Academy.mp3 | 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. So 80 minus 40 gets us 40. |
Washer method rotating around vertical line (not y-axis), part 2 AP Calculus AB Khan Academy.mp3 | 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. So this is equal to 31 pi over 30. |
Washer method rotating around vertical line (not y-axis), part 2 AP Calculus AB Khan Academy.mp3 | 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. Let's see. |
Washer method rotating around vertical line (not y-axis), part 2 AP Calculus AB Khan Academy.mp3 | 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. I'm going to do it one more time. |
Washer method rotating around vertical line (not y-axis), part 2 AP Calculus AB Khan Academy.mp3 | 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. 46 minus 10 is 36. |
Derivative as a concept Derivatives introduction AP Calculus AB Khan Academy.mp3 | You are likely already familiar with the idea of a slope of a line. If you're not, I encourage you to review it on Khan Academy. 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. |
Derivative as a concept Derivatives introduction AP Calculus AB Khan Academy.mp3 | 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. And I could also calculate the change in y. |
Derivative as a concept Derivatives introduction AP Calculus AB Khan Academy.mp3 | 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. So slope is equal to the rate of change of our vertical variable over the rate of change of our horizontal variable. |
Derivative as a concept Derivatives introduction AP Calculus AB Khan Academy.mp3 | 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. 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. |
Derivative as a concept Derivatives introduction AP Calculus AB Khan Academy.mp3 | 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. So for example, here's a curve where the rate of change of y with respect to x is constantly changing. |
Derivative as a concept Derivatives introduction AP Calculus AB Khan Academy.mp3 | 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. So it'd be the slope of this line, of the secant line. |
Derivative as a concept Derivatives introduction AP Calculus AB Khan Academy.mp3 | 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. So even when we take the slopes between two points on the line, the secant lines, you can see that those slopes are changing. |
Derivative as a concept Derivatives introduction AP Calculus AB Khan Academy.mp3 | 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? So for example, how fast is y changing with respect to x exactly at that point, exactly when x is equal to that value? |
Derivative as a concept Derivatives introduction AP Calculus AB Khan Academy.mp3 | 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? And we can calculate the slope of that line. |
Derivative as a concept Derivatives introduction AP Calculus AB Khan Academy.mp3 | 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. If we know the slope of this, well, then we could say that that's the instantaneous rate of change at that point. |
Derivative as a concept Derivatives introduction AP Calculus AB Khan Academy.mp3 | 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. 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. |
Derivative as a concept Derivatives introduction AP Calculus AB Khan Academy.mp3 | 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. And this idea is a central idea of differential calculus, and it's known as a derivative. |
Derivative as a concept Derivatives introduction AP Calculus AB Khan Academy.mp3 | 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? One way is known as Leibniz's notation, and Leibniz is one of the fathers of calculus along with Isaac Newton. |
Derivative as a concept Derivatives introduction AP Calculus AB Khan Academy.mp3 | 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? Because it really comes from this idea of a slope, which is change in y over change in x. |
Derivative as a concept Derivatives introduction AP Calculus AB Khan Academy.mp3 | 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. But then let's get even closer. |
Derivative as a concept Derivatives introduction AP Calculus AB Khan Academy.mp3 | 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. 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? |
Derivative as a concept Derivatives introduction AP Calculus AB Khan Academy.mp3 | 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. Now there's other notations. |
Derivative as a concept Derivatives introduction AP Calculus AB Khan Academy.mp3 | 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. It's saying f prime is representing the derivative. |
Derivative as a concept Derivatives introduction AP Calculus AB Khan Academy.mp3 | 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. If you input an x into f prime, you're getting the slope of the tangent line at that point. |
Derivative as a concept Derivatives introduction AP Calculus AB Khan Academy.mp3 | 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. You might also see y prime. |
Derivative as a concept Derivatives introduction AP Calculus AB Khan Academy.mp3 | 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. 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. |
Mean value theorem example polynomial Existence theorems AP Calculus AB Khan Academy.mp3 | 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. |
Mean value theorem example polynomial Existence theorems AP Calculus AB Khan Academy.mp3 | 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. |
Mean value theorem example polynomial Existence theorems AP Calculus AB Khan Academy.mp3 | 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. |
Mean value theorem example polynomial Existence theorems AP Calculus AB Khan Academy.mp3 | 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. |
Mean value theorem example polynomial Existence theorems AP Calculus AB Khan Academy.mp3 | 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. |
Mean value theorem example polynomial Existence theorems AP Calculus AB Khan Academy.mp3 | 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. |
Mean value theorem example polynomial Existence theorems AP Calculus AB Khan Academy.mp3 | 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. |
Mean value theorem example polynomial Existence theorems AP Calculus AB Khan Academy.mp3 | 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. |
Mean value theorem example polynomial Existence theorems AP Calculus AB Khan Academy.mp3 | 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. |
Mean value theorem example polynomial Existence theorems AP Calculus AB Khan Academy.mp3 | 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. |
Mean value theorem example polynomial Existence theorems AP Calculus AB Khan Academy.mp3 | 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. |
Mean value theorem example polynomial Existence theorems AP Calculus AB Khan Academy.mp3 | 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. |
Mean value theorem example polynomial Existence theorems AP Calculus AB Khan Academy.mp3 | 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. |
Mean value theorem example polynomial Existence theorems AP Calculus AB Khan Academy.mp3 | 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. |
Mean value theorem example polynomial Existence theorems AP Calculus AB Khan Academy.mp3 | 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 1⁄2 or 7 halves. So it makes sense. So this right over here is our c. And it's equal to 7 halves. |
Analyzing related rates problems equations (Pythagoras) AP Calculus AB Khan Academy.mp3 | 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. |
Analyzing related rates problems equations (Pythagoras) AP Calculus AB Khan Academy.mp3 | 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. |
Analyzing related rates problems equations (Pythagoras) AP Calculus AB Khan Academy.mp3 | 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. |
Analyzing related rates problems equations (Pythagoras) AP Calculus AB Khan Academy.mp3 | 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. |
Analyzing related rates problems equations (Pythagoras) AP Calculus AB Khan Academy.mp3 | 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? |
Analyzing related rates problems equations (Pythagoras) AP Calculus AB Khan Academy.mp3 | 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. |
Analyzing related rates problems equations (Pythagoras) AP Calculus AB Khan Academy.mp3 | 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. |
Derivative of e_____cos(e_) Advanced derivatives AP Calculus AB Khan Academy.mp3 | 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. |
Derivative of e_____cos(e_) Advanced derivatives AP Calculus AB Khan Academy.mp3 | 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. |
Derivative of e_____cos(e_) Advanced derivatives AP Calculus AB Khan Academy.mp3 | 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. |
Derivative of e_____cos(e_) Advanced derivatives AP Calculus AB Khan Academy.mp3 | 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. |
Derivative of e_____cos(e_) Advanced derivatives AP Calculus AB Khan Academy.mp3 | 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. |
Derivative of e_____cos(e_) Advanced derivatives AP Calculus AB Khan Academy.mp3 | 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. |
Derivative of e_____cos(e_) Advanced derivatives AP Calculus AB Khan Academy.mp3 | 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. |
Derivative of e_____cos(e_) Advanced derivatives AP Calculus AB Khan Academy.mp3 | 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. |
Derivative of e_____cos(e_) Advanced derivatives AP Calculus AB Khan Academy.mp3 | 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. |
Derivative of e_____cos(e_) Advanced derivatives AP Calculus AB Khan Academy.mp3 | 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. |
Derivative of e_____cos(e_) Advanced derivatives AP Calculus AB Khan Academy.mp3 | 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. |
Derivative of e_____cos(e_) Advanced derivatives AP Calculus AB Khan Academy.mp3 | 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. |
Derivative of e_____cos(e_) Advanced derivatives AP Calculus AB Khan Academy.mp3 | 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. |
Derivative of e_____cos(e_) Advanced derivatives AP Calculus AB Khan Academy.mp3 | 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. |
Derivative of e_____cos(e_) Advanced derivatives AP Calculus AB Khan Academy.mp3 | 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. |
Derivative of e_____cos(e_) Advanced derivatives AP Calculus AB Khan Academy.mp3 | 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. |
Derivative of e_____cos(e_) Advanced derivatives AP Calculus AB Khan Academy.mp3 | 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. |
Derivative of e_____cos(e_) Advanced derivatives AP Calculus AB Khan Academy.mp3 | 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. |
Derivative of e_____cos(e_) Advanced derivatives AP Calculus AB Khan Academy.mp3 | 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. |
Derivative of e_____cos(e_) Advanced derivatives AP Calculus AB Khan Academy.mp3 | 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. |
Derivative of e_____cos(e_) Advanced derivatives AP Calculus AB Khan Academy.mp3 | 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. |
Derivative of e_____cos(e_) Advanced derivatives AP Calculus AB Khan Academy.mp3 | 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. |
Derivative of e_____cos(e_) Advanced derivatives AP Calculus AB Khan Academy.mp3 | 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. |
Derivative of e_____cos(e_) Advanced derivatives AP Calculus AB Khan Academy.mp3 | 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. |
Inflection points (graphical) AP Calculus AB Khan Academy.mp3 | 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. |
Inflection points (graphical) AP Calculus AB Khan Academy.mp3 | 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. |
Inflection points (graphical) AP Calculus AB Khan Academy.mp3 | 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. |
Inflection points (graphical) AP Calculus AB Khan Academy.mp3 | 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. |
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