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Optimization box volume (Part 2) Applications of derivatives AP Calculus AB Khan Academy.mp3 | All right, so first I'll try when I add the radical. So I'm going to get 200, 200 plus the square root of 40,000. I could have just written that, I could have just written that as 200 squared, but that's fine, 40,000 minus four times 12, times 600, times 600. And I get 305, which I then need to divide by 24. Which I'll... |
Optimization box volume (Part 2) Applications of derivatives AP Calculus AB Khan Academy.mp3 | And I get 305, which I then need to divide by 24. Which I'll divide by 24. And I get 12.74. So one of my possible x's. So it equals 12.74. And now let me do the situation where I subtract what I had in the radical sign. So let me get my calculator back. |
Optimization box volume (Part 2) Applications of derivatives AP Calculus AB Khan Academy.mp3 | So one of my possible x's. So it equals 12.74. And now let me do the situation where I subtract what I had in the radical sign. So let me get my calculator back. And so now let me do 200. I probably could have done this slightly more efficiently, but this is fine. Minus the square root of 40,000, one, two, three. |
Optimization box volume (Part 2) Applications of derivatives AP Calculus AB Khan Academy.mp3 | So let me get my calculator back. And so now let me do 200. I probably could have done this slightly more efficiently, but this is fine. Minus the square root of 40,000, one, two, three. Minus four times 12, times 600. And I get, that's just the numerator. And then I'm gonna divide that by 24. |
Optimization box volume (Part 2) Applications of derivatives AP Calculus AB Khan Academy.mp3 | Minus the square root of 40,000, one, two, three. Minus four times 12, times 600. And I get, that's just the numerator. And then I'm gonna divide that by 24. Divide that by 24. And I get 3.92. Did I do that right? |
Optimization box volume (Part 2) Applications of derivatives AP Calculus AB Khan Academy.mp3 | And then I'm gonna divide that by 24. Divide that by 24. And I get 3.92. Did I do that right? 200 minus 40,000 minus four times 12 times 600. All of that divided by 24. My previous answer divided by 24. |
Optimization box volume (Part 2) Applications of derivatives AP Calculus AB Khan Academy.mp3 | Did I do that right? 200 minus 40,000 minus four times 12 times 600. All of that divided by 24. My previous answer divided by 24. Gives me 3.92. So it's 12.74 or 3.92. Now which of these can I use? |
Optimization box volume (Part 2) Applications of derivatives AP Calculus AB Khan Academy.mp3 | My previous answer divided by 24. Gives me 3.92. So it's 12.74 or 3.92. Now which of these can I use? Well, 12, x equals 12.74 is outside of our, outside of our valid values for x. If x was equal to 12.74, we would cut past, we would completely cut past, the x's would start to overlap with each other. So x cannot be 12... |
Optimization box volume (Part 2) Applications of derivatives AP Calculus AB Khan Academy.mp3 | Now which of these can I use? Well, 12, x equals 12.74 is outside of our, outside of our valid values for x. If x was equal to 12.74, we would cut past, we would completely cut past, the x's would start to overlap with each other. So x cannot be 12.74. So we get a critical point at x is equal to 3.92. And you could loo... |
Optimization box volume (Part 2) Applications of derivatives AP Calculus AB Khan Academy.mp3 | So x cannot be 12.74. So we get a critical point at x is equal to 3.92. And you could look at the graph, and you could say, oh look, that looks like a maximum value. But if you didn't have the graph at your disposal, you can then do the second derivative test and say, hey, are we concave upwards or concave downwards wh... |
Optimization box volume (Part 2) Applications of derivatives AP Calculus AB Khan Academy.mp3 | But if you didn't have the graph at your disposal, you can then do the second derivative test and say, hey, are we concave upwards or concave downwards when x is equal to 3.92? Well, in order to do the second derivative test, you have to figure out what the second derivative is, so let's do that. V prime prime of x is ... |
Optimization box volume (Part 2) Applications of derivatives AP Calculus AB Khan Academy.mp3 | Minus, minus 200. And you can just look at inspection that this number right over here is less than four, so this thing right over here is going to be less than 100. You subtract, you subtract 200, so we can write the second derivative at 3.92 is going to be less than zero. You can figure out what the exact value is if... |
Optimization box volume (Part 2) Applications of derivatives AP Calculus AB Khan Academy.mp3 | You can figure out what the exact value is if you like. So because this is less than zero, we are concave downwards, concave downwards. Another way of saying it is that the slope is, the slope is decreasing the entire time, concave downwards. When the slope is decreasing the entire time, our shape looks like that. The ... |
Optimization box volume (Part 2) Applications of derivatives AP Calculus AB Khan Academy.mp3 | When the slope is decreasing the entire time, our shape looks like that. The slope could start off high, lower, lower, gets to zero, even lower, lower, lower. And we even saw that on the graph right over here. And since it's concave downwards, that implies that our critical point that sits where, during where the inter... |
Optimization box volume (Part 2) Applications of derivatives AP Calculus AB Khan Academy.mp3 | And since it's concave downwards, that implies that our critical point that sits where, during where the interval is concave downward, that critical point is a local maximum, is a maximum. So this is the x value at which our function attains our maximum. Now, what is that maximum value? Well, we could type that back in... |
Optimization box volume (Part 2) Applications of derivatives AP Calculus AB Khan Academy.mp3 | Well, we could type that back in into our original expression for volume to figure what that is. So let's figure out what the volume, when we get to 3.92, is equal to. What is our maximum volume? So get the calculator back out. And I could use, it's obviously roughly 3.92, I could use this exact value. Actually, let's,... |
Optimization box volume (Part 2) Applications of derivatives AP Calculus AB Khan Academy.mp3 | So get the calculator back out. And I could use, it's obviously roughly 3.92, I could use this exact value. Actually, let's, well, I'll just use 3.92 to get a rough sense of what our maximum value is, our maximum volume. So it'll be 3.92, 3.92. I'll just use this expression for the volume as a function of x. 3.92 times... |
Optimization box volume (Part 2) Applications of derivatives AP Calculus AB Khan Academy.mp3 | So it'll be 3.92, 3.92. I'll just use this expression for the volume as a function of x. 3.92 times 20 minus two times 3.92 times 30, 30 minus two times 3.92 gives us, and we deserve a drum roll now, gives us 1,056.3. So 1,056.3, which is a higher volume than we got when we just inspected it graphically. We probably co... |
Definite integral of piecewise function AP Calculus AB Khan Academy.mp3 | For x is greater than or equal to zero, f of x is cosine of pi x. And we want to evaluate the definite integral from negative one to one of f of x dx. And you might immediately say, well, which of these versions of f of x am I going to take the antiderivative from? Because from negative one to zero, I would think about... |
Definite integral of piecewise function AP Calculus AB Khan Academy.mp3 | Because from negative one to zero, I would think about x plus one, but then from zero to one, I would think about cosine pi x. And if you were thinking that, you're thinking in the right direction. And the way that we can make this a little bit more straightforward is to actually split up this definite integral. This i... |
Definite integral of piecewise function AP Calculus AB Khan Academy.mp3 | This is going to be equal to the definite integral from negative one to zero of f of x dx plus the integral from zero to one of f of x dx. Now, why was it useful for me to split it up this way? And in particular, to split it up, split the interval from negative one to one, split it into two intervals from negative one ... |
Definite integral of piecewise function AP Calculus AB Khan Academy.mp3 | Well, I did that because zero, x equals zero is where we switch, where f of x switches from being x plus one to cosine pi x. So if you look at the interval from negative one to zero, f of x is x plus one. So f of x here is x plus one. And then when you go from zero to one, f of x is cosine pi x. So cosine of pi x. And ... |
Definite integral of piecewise function AP Calculus AB Khan Academy.mp3 | And then when you go from zero to one, f of x is cosine pi x. So cosine of pi x. And so now we just have to evaluate each of these separately and add them together. So let's take the definite integral from negative one to zero of x plus one dx. Well, let's see, the antiderivative of x plus one is, antiderivative of x i... |
Definite integral of piecewise function AP Calculus AB Khan Academy.mp3 | So let's take the definite integral from negative one to zero of x plus one dx. Well, let's see, the antiderivative of x plus one is, antiderivative of x is x squared over two. I'm just incrementing the exponent and then dividing by that value. And then plus x, and you could view it as I'm doing the same thing. If this... |
Definite integral of piecewise function AP Calculus AB Khan Academy.mp3 | And then plus x, and you could view it as I'm doing the same thing. If this is x to the zero, it'll be x to the first, x to the first over one, which is just x. And I'm gonna evaluate that at zero and subtract from that it evaluated at one, sorry, it evaluated at negative one. And so this is going to be equal to, if I ... |
Definite integral of piecewise function AP Calculus AB Khan Academy.mp3 | And so this is going to be equal to, if I evaluate it at zero, let me do this in another color. If I evaluate it at zero, it's going to be zero squared over two, which is, well, I'll just write it zero squared over two plus zero. Well, all of that's just gonna be equal to zero, minus it evaluated at, it evaluated at ne... |
Definite integral of piecewise function AP Calculus AB Khan Academy.mp3 | So minus negative one squared, negative one squared over two plus negative one. So negative one squared is just one. So it's 1 1 2 plus negative one, 1 1 2 plus negative one is, or 1 1 2 minus one is negative 1 1 2. So all of that is negative 1 1 2. But then we're subtracting negative 1 1 2. Zero minus negative 1 1 2 i... |
Definite integral of piecewise function AP Calculus AB Khan Academy.mp3 | So all of that is negative 1 1 2. But then we're subtracting negative 1 1 2. Zero minus negative 1 1 2 is going to be equal to positive 1 1 2. So this is going to be equal to positive 1 1 2. So this first part right over here is positive 1 1 2. And now let's evaluate the integral from zero to one of cosine pi, I don't ... |
Definite integral of piecewise function AP Calculus AB Khan Academy.mp3 | So this is going to be equal to positive 1 1 2. So this first part right over here is positive 1 1 2. And now let's evaluate the integral from zero to one of cosine pi, I don't need that first parenthesis, of cosine of pi x dx. What is this equal to? Now, if we were just trying to find the antiderivative of cosine of x... |
Definite integral of piecewise function AP Calculus AB Khan Academy.mp3 | What is this equal to? Now, if we were just trying to find the antiderivative of cosine of x, it's pretty straightforward. We know that the derivative with respect to x of sine of x is equal to cosine of x. Cosine of x. But that's not what we have here. We have cosine of pi x. So there is a technique here, you could ca... |
Definite integral of piecewise function AP Calculus AB Khan Academy.mp3 | But that's not what we have here. We have cosine of pi x. So there is a technique here, you could call it u substitution. You could say u is equal to pi x. If you don't know how to do that, you could still try to think about, think this through, where we could say, all right, well, maybe it involves sine of pi x someho... |
Definite integral of piecewise function AP Calculus AB Khan Academy.mp3 | You could say u is equal to pi x. If you don't know how to do that, you could still try to think about, think this through, where we could say, all right, well, maybe it involves sine of pi x somehow. So the derivative with respect to x of sine of pi x would be what? Well, we would use the chain rule. It would be the d... |
Definite integral of piecewise function AP Calculus AB Khan Academy.mp3 | Well, we would use the chain rule. It would be the derivative of the outside function with respect to the inside, or sine of pi x with respect to pi x, which would be cosine of pi x, and then times the derivative of the inside function with respect to x, so it would be times pi. Or you could say the derivative of sine ... |
Definite integral of piecewise function AP Calculus AB Khan Academy.mp3 | Now, we almost have that here, except we just need a pi. So what if we were to throw a pi right over here, but so we don't change the value, we also multiply by one over pi? So if you divide and multiply by the same number, you're not changing its value. One over pi times pi is just equal to one. But this is useful. Th... |
Definite integral of piecewise function AP Calculus AB Khan Academy.mp3 | One over pi times pi is just equal to one. But this is useful. This is useful because we now know that pi cosine pi x is the derivative of sine pi x. So this is all going to be equal to, this is equal to one. Let me take that one over pi. So this is equal to one over pi times, now we're going to evaluate, so the antide... |
Definite integral of piecewise function AP Calculus AB Khan Academy.mp3 | So this is all going to be equal to, this is equal to one. Let me take that one over pi. So this is equal to one over pi times, now we're going to evaluate, so the antiderivative here we just said is sine, sine of pi x, and we're going to evaluate that at one and at zero. So this is going to be equal to one over pi, on... |
Definite integral of piecewise function AP Calculus AB Khan Academy.mp3 | So this is going to be equal to one over pi, one over pi, not pi, my hand is not listening to my mouth, one over pi times sine of pi, sine of pi minus sine of, minus sine of pi times zero, which is just zero. Well sine of pi, that's zero. Sine of zero is zero. So you're going to have one over pi times zero minus zero. ... |
Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3 | Where we left off in the last video, we'd actually set up our definite integral to figure out the volume of this figure. So now we just have to evaluate it. And really the hardest part is going to be simplifying that and that right over there. So let's get to it. So what is this thing squared? Well, we're going to have... |
Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3 | So let's get to it. So what is this thing squared? Well, we're going to have to do a little bit of polynomial multiplication here. So I'm going to go into that same color I had. So we're going to have 4 minus x squared plus 2x times 4 minus. Actually, let me write that in the order of the terms or the degree of the ter... |
Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3 | So I'm going to go into that same color I had. So we're going to have 4 minus x squared plus 2x times 4 minus. Actually, let me write that in the order of the terms or the degree of the terms. It's negative x squared plus 2x plus 4. I just switched the order of these things. Times negative x squared plus 2x plus 4. So ... |
Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3 | It's negative x squared plus 2x plus 4. I just switched the order of these things. Times negative x squared plus 2x plus 4. So we're just going to multiply these two things. I shouldn't even write a multiplication symbol. It looks too much like an x. So 4 times 4 is 16. |
Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3 | So we're just going to multiply these two things. I shouldn't even write a multiplication symbol. It looks too much like an x. So 4 times 4 is 16. 4 times 2x is 8x. 4 times negative x squared is negative 4x squared. 2x times 4 is 8x. |
Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3 | So 4 times 4 is 16. 4 times 2x is 8x. 4 times negative x squared is negative 4x squared. 2x times 4 is 8x. 2x times 2x is 4x squared. And then 2x times negative x squared is negative 2x to the third power. And now we just have to multiply negative x times all of this. |
Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3 | 2x times 4 is 8x. 2x times 2x is 4x squared. And then 2x times negative x squared is negative 2x to the third power. And now we just have to multiply negative x times all of this. So negative x squared times 4 is negative 4x squared. Negative x squared times 2x is negative 2x to the third power. And then negative x squ... |
Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3 | And now we just have to multiply negative x times all of this. So negative x squared times 4 is negative 4x squared. Negative x squared times 2x is negative 2x to the third power. And then negative x squared times negative x squared is positive x to the fourth. So it's going to be positive x to the fourth. And now we j... |
Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3 | And then negative x squared times negative x squared is positive x to the fourth. So it's going to be positive x to the fourth. And now we just have to add up all of these terms. And we get x to the fourth. Add these two, minus 4x to the third. And then this cancels with this, but we still have that. So minus 4x square... |
Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3 | And we get x to the fourth. Add these two, minus 4x to the third. And then this cancels with this, but we still have that. So minus 4x squared. You add these two right over here, you get plus 16x. And then you have plus 16. So that's this expanded out. |
Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3 | So minus 4x squared. You add these two right over here, you get plus 16x. And then you have plus 16. So that's this expanded out. And now if we want to x minus 4, or 4 minus x times 4 minus x. So if we just have 4 minus x times 4 minus x. We could actually do it this way as well. |
Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3 | So that's this expanded out. And now if we want to x minus 4, or 4 minus x times 4 minus x. So if we just have 4 minus x times 4 minus x. We could actually do it this way as well. But that's just going to be 4 times 4, which is 16. Plus 4 times negative x, which is negative 4x. Negative x times 4, another negative 4x. |
Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3 | We could actually do it this way as well. But that's just going to be 4 times 4, which is 16. Plus 4 times negative x, which is negative 4x. Negative x times 4, another negative 4x. And then negative x times negative x is equal to plus x squared. So if we were to swap the order, we get x squared minus 8x plus 16. But w... |
Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3 | Negative x times 4, another negative 4x. And then negative x times negative x is equal to plus x squared. So if we were to swap the order, we get x squared minus 8x plus 16. But we're going to subtract this. So if you have the negative sign out there, we're going to subtract this business. So let's just do it right ove... |
Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3 | But we're going to subtract this. So if you have the negative sign out there, we're going to subtract this business. So let's just do it right over here, since we already have it set up. So we have this minus this. So we have this minus, we're going to subtract this. Or we could add the negative of it. So we'll put neg... |
Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3 | So we have this minus this. So we have this minus, we're going to subtract this. Or we could add the negative of it. So we'll put negative x squared plus 8x minus 16. So I'm just going to add the negative of this. And we get, and I'll do this in a new color, let's see, we get x to the fourth minus 4x to the third power... |
Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3 | So we'll put negative x squared plus 8x minus 16. So I'm just going to add the negative of this. And we get, and I'll do this in a new color, let's see, we get x to the fourth minus 4x to the third power minus 5x squared plus 24x. And then these cancel out. So that's what we are left with. And so that's the inside of o... |
Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3 | And then these cancel out. So that's what we are left with. And so that's the inside of our integral. So we're going to take the integral of this thing, just so I don't have to keep rewriting this thing, from 0 until, if I remember correctly, 3. Yep, from 0 to 3 of this dx. And then we had a pi out front here. We had a... |
Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3 | So we're going to take the integral of this thing, just so I don't have to keep rewriting this thing, from 0 until, if I remember correctly, 3. Yep, from 0 to 3 of this dx. And then we had a pi out front here. We had a pi out front here. So I'll just take that out of the integral. So times pi. Well, now we just have to... |
Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3 | We had a pi out front here. So I'll just take that out of the integral. So times pi. Well, now we just have to take the antiderivative. And this is going to be equal to pi times, antiderivative of x to the fourth is x to the fifth over 5. Antiderivative of 4x to the third is actually x to the fourth. So this is going t... |
Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3 | Well, now we just have to take the antiderivative. And this is going to be equal to pi times, antiderivative of x to the fourth is x to the fifth over 5. Antiderivative of 4x to the third is actually x to the fourth. So this is going to be minus x to the fourth. You can verify that. Derivative is 4x. And then you decre... |
Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3 | So this is going to be minus x to the fourth. You can verify that. Derivative is 4x. And then you decrement the exponent 4x to the third. So that works out. And then the antiderivative of this is negative 5 thirds x to the third. Just incremented the exponent and divided by that. |
Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3 | And then you decrement the exponent 4x to the third. So that works out. And then the antiderivative of this is negative 5 thirds x to the third. Just incremented the exponent and divided by that. And then you have plus 24x squared over 2, or 12x squared. And we're going to evaluate that. I like to match colors for my o... |
Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3 | Just incremented the exponent and divided by that. And then you have plus 24x squared over 2, or 12x squared. And we're going to evaluate that. I like to match colors for my opening and closing parentheses. We're going to evaluate that. Actually, let me do it as brackets. We're going to evaluate that from 0 to 3. |
Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3 | I like to match colors for my opening and closing parentheses. We're going to evaluate that. Actually, let me do it as brackets. We're going to evaluate that from 0 to 3. So this is going to be equal to pi times, let's evaluate all this business at 3. So we're going to get 3 to the fifth power. So let's see. |
Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3 | We're going to evaluate that from 0 to 3. So this is going to be equal to pi times, let's evaluate all this business at 3. So we're going to get 3 to the fifth power. So let's see. 3 to the third is equal to 27. 3 to the fourth is equal to 81. 3 to the fifth is going to be equal to, this times 3 is 243. |
Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3 | So let's see. 3 to the third is equal to 27. 3 to the fourth is equal to 81. 3 to the fifth is going to be equal to, this times 3 is 243. So it's going to be 243. It's going to be some hairy math. 243 over 5 minus, well, 3 to the fourth is 81. |
Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3 | 3 to the fifth is going to be equal to, this times 3 is 243. So it's going to be 243. It's going to be some hairy math. 243 over 5 minus, well, 3 to the fourth is 81. Let's see. We're going to have 3 to the third. So it's going to be minus 5 over 3 times 3 to the third power. |
Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3 | 243 over 5 minus, well, 3 to the fourth is 81. Let's see. We're going to have 3 to the third. So it's going to be minus 5 over 3 times 3 to the third power. So times 27. Well, 27 divided by 3 is just 9. 9 times 5 is 45. |
Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3 | So it's going to be minus 5 over 3 times 3 to the third power. So times 27. Well, 27 divided by 3 is just 9. 9 times 5 is 45. So let's just simplify that. So this is going to be equal to 45. Did I do that right? |
Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3 | 9 times 5 is 45. So let's just simplify that. So this is going to be equal to 45. Did I do that right? Yeah, it's essentially going to be like 3 squared times 5, because you're dividing by 3 here. So it's going to have 45. And then finally, 3 squared is 9. |
Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3 | Did I do that right? Yeah, it's essentially going to be like 3 squared times 5, because you're dividing by 3 here. So it's going to have 45. And then finally, 3 squared is 9. 9 times 12 is 108. So plus 108. These problems that involve hairy arithmetic are always the most stressful for me, but I'll try not to get too st... |
Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3 | And then finally, 3 squared is 9. 9 times 12 is 108. So plus 108. These problems that involve hairy arithmetic are always the most stressful for me, but I'll try not to get too stressed. And then we're going to subtract out this whole thing evaluated at 0. But lucky for us, that's pretty simple. This evaluated at 0, 0,... |
Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3 | These problems that involve hairy arithmetic are always the most stressful for me, but I'll try not to get too stressed. And then we're going to subtract out this whole thing evaluated at 0. But lucky for us, that's pretty simple. This evaluated at 0, 0, 0, 0, 0. So we're going to subtract out 0, which simplifies thing... |
Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3 | This evaluated at 0, 0, 0, 0, 0. So we're going to subtract out 0, which simplifies things a good bit. So now we just have to do some hairy fraction arithmetic. So let's do it. So what I'm going to do first is simplify all of this part. And then I'm going to worry about putting it over a denominator of 5. So let's see ... |
Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3 | So let's do it. So what I'm going to do first is simplify all of this part. And then I'm going to worry about putting it over a denominator of 5. So let's see what we have. We have negative 81 minus 45. So these two right over here become negative 126. And then negative 126 plus 108, well, that's just going to be the s... |
Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3 | So let's see what we have. We have negative 81 minus 45. So these two right over here become negative 126. And then negative 126 plus 108, well, that's just going to be the same thing as negative 26 plus 8, which is going to be negative 18. So this whole thing simplifies to negative 18. Did I do that right? So this is ... |
Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3 | And then negative 126 plus 108, well, that's just going to be the same thing as negative 26 plus 8, which is going to be negative 18. So this whole thing simplifies to negative 18. Did I do that right? So this is negative 126. And then negative 126, yes, it will be negative 18. So now we just have to write negative 18 ... |
Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3 | So this is negative 126. And then negative 126, yes, it will be negative 18. So now we just have to write negative 18 over 5. Negative 18 over 5 is the same thing as negative, let's see, 5 times 10 is 50, plus 40. So that's going to be negative 90 over 5. So this whole thing has simplified to equal to pi times 243 over... |
Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3 | Negative 18 over 5 is the same thing as negative, let's see, 5 times 10 is 50, plus 40. So that's going to be negative 90 over 5. So this whole thing has simplified to equal to pi times 243 over 5 minus 90 over 5, which is equal to pi. I deserve a drum roll now. This was some hairy mathematics. So 243 minus 90 is going... |
Verifying solutions to differential equations AP Calculus AB Khan Academy.mp3 | So let's write down a differential equation. The derivative of y with respect to x is equal to four y over x. And what we'll see in this video is the solution to a differential equation isn't a value or a set of values. It's a function or a set of functions. But before we go about actually trying to solve this or figur... |
Verifying solutions to differential equations AP Calculus AB Khan Academy.mp3 | It's a function or a set of functions. But before we go about actually trying to solve this or figure out all of the solutions, let's test whether certain equations, certain functions are solutions to this differential equation. So for example, if I have y is equal to four x, is this a solution to this differential equ... |
Verifying solutions to differential equations AP Calculus AB Khan Academy.mp3 | Pause the video and see if you can figure it out. Well, to see if this is a solution, what we have to do is figure out the derivative of y with respect to x, and see is that truly equal to four times y over x. And I'm gonna try to express everything in terms of x to see if I really have an equality there. So first, let... |
Verifying solutions to differential equations AP Calculus AB Khan Academy.mp3 | So first, let's figure out the derivative of y with respect to x. Well, that's just going to be equal to four. We've seen that many times before. And so what we need to test is, is four, the derivative of y with respect to x, equal to four times, I could write y, but instead of y, let's write four x. I'm gonna put ever... |
Verifying solutions to differential equations AP Calculus AB Khan Academy.mp3 | And so what we need to test is, is four, the derivative of y with respect to x, equal to four times, I could write y, but instead of y, let's write four x. I'm gonna put everything in terms of x. So y is equal to four x. So instead of four y, I could write four times four x. All of that over x. Is this true? Well, that... |
Verifying solutions to differential equations AP Calculus AB Khan Academy.mp3 | All of that over x. Is this true? Well, that x cancels with that, and I'm gonna get four is equal to 16, which it clearly is not. And so this is not a solution. Not a solution to our differential equation. Let's look at another equation. What about y is equal to x to the fourth power? |
Verifying solutions to differential equations AP Calculus AB Khan Academy.mp3 | And so this is not a solution. Not a solution to our differential equation. Let's look at another equation. What about y is equal to x to the fourth power? Pause this video and see if this is a solution to our original differential equation. Well, we're going to do the same thing. What's the derivative of y with respec... |
Verifying solutions to differential equations AP Calculus AB Khan Academy.mp3 | What about y is equal to x to the fourth power? Pause this video and see if this is a solution to our original differential equation. Well, we're going to do the same thing. What's the derivative of y with respect to x? This is equal to, just using the power rule, four x to the third power. And so what we have to test ... |
Verifying solutions to differential equations AP Calculus AB Khan Academy.mp3 | What's the derivative of y with respect to x? This is equal to, just using the power rule, four x to the third power. And so what we have to test is, is four x to the third power, that's the derivative of y with respect to x, equal to four times y. Instead of writing a y, I'm gonna write it all in terms of x. So is tha... |
Verifying solutions to differential equations AP Calculus AB Khan Academy.mp3 | Instead of writing a y, I'm gonna write it all in terms of x. So is that equal to four times x to the fourth, because x to the fourth is the same thing as y, divided by x. And so let's see, x to the fourth divided by x, that is going to be x to the third. And so you will indeed get four x to the third is equal to four ... |
Verifying solutions to differential equations AP Calculus AB Khan Academy.mp3 | And so you will indeed get four x to the third is equal to four x to the third. So check, this is a solution. So is a solution. It's not necessarily the only solution, but it is a solution to that differential equation. Let's look at another differential equation. Let's say that I had, and I'm gonna write it with diffe... |
Verifying solutions to differential equations AP Calculus AB Khan Academy.mp3 | It's not necessarily the only solution, but it is a solution to that differential equation. Let's look at another differential equation. Let's say that I had, and I'm gonna write it with different notation, f prime of x is equal to f of x minus x. And the first function that I wanna test, let's say I have f of x is equ... |
Verifying solutions to differential equations AP Calculus AB Khan Academy.mp3 | And the first function that I wanna test, let's say I have f of x is equal to two x. Is this a solution to this differential equation? Pause the video again and see if you can figure it out. Well, to figure that out, you have to say, well, what is f prime of x? F prime of x is just going to be equal to two. And then te... |
Verifying solutions to differential equations AP Calculus AB Khan Academy.mp3 | Well, to figure that out, you have to say, well, what is f prime of x? F prime of x is just going to be equal to two. And then test the equality. Is two, is f prime of x equal to f of x, which is two x, minus x, minus x. And so let's see, we are going to get two is equal to x. So you might be tempted to say, oh, hey, I... |
Verifying solutions to differential equations AP Calculus AB Khan Academy.mp3 | Is two, is f prime of x equal to f of x, which is two x, minus x, minus x. And so let's see, we are going to get two is equal to x. So you might be tempted to say, oh, hey, I just solved for x or something like that. But this would tell you that this is not a solution because this needs to be true for any x that is in ... |
Verifying solutions to differential equations AP Calculus AB Khan Academy.mp3 | But this would tell you that this is not a solution because this needs to be true for any x that is in the domain of this function. And so this is, I'll just put an x there, or I'll put an incorrect there to say not, not a, not a solution. Just to be clear again, this needs, in order for a function to be a solution of ... |
Verifying solutions to differential equations AP Calculus AB Khan Academy.mp3 | Let's look at another one. Let's say that we have f of x is equal to x plus one. Pause the video and see, is this a solution to our differential equation? Well, same drill. f prime of x is going to be equal to one. And so we have to see, is f prime of x, which is equal to one, is it equal to f of x, which is x plus one... |
Verifying solutions to differential equations AP Calculus AB Khan Academy.mp3 | Well, same drill. f prime of x is going to be equal to one. And so we have to see, is f prime of x, which is equal to one, is it equal to f of x, which is x plus one, x plus one, minus x? And so here, you see no matter what x is, this equation is going to be true. So this is a solution, is a solution. Let's do a few mo... |
Verifying solutions to differential equations AP Calculus AB Khan Academy.mp3 | And so here, you see no matter what x is, this equation is going to be true. So this is a solution, is a solution. Let's do a few more of these. Let me scroll down a little bit so I have a little bit more, a little bit more space, but make sure we see our original differential equation. Let's test whether, I'm gonna do... |
Verifying solutions to differential equations AP Calculus AB Khan Academy.mp3 | Let me scroll down a little bit so I have a little bit more, a little bit more space, but make sure we see our original differential equation. Let's test whether, I'm gonna do it in a red color. Let's test whether f of x equals e to the x plus x plus one is a solution to this differential equation. Pause the video agai... |
Verifying solutions to differential equations AP Calculus AB Khan Academy.mp3 | Pause the video again and see if you can figure it out. All right, well, let's figure out the derivative here. f prime of x is going to be equal to, derivative of e to the x with respect to x is e to the x, which I always find amazing, and so, and then plus one, and the derivative of this with respect to x is just zero... |
Verifying solutions to differential equations AP Calculus AB Khan Academy.mp3 | And then let's substitute this into our original differential equation. So f prime of x is e to the x plus one. Is that equal to f of x, which is e to the x plus x plus one minus x, minus x? And if that x cancels out with that x, it is indeed. They are indeed equal. So this is also a solution. So this, this is a soluti... |
Volume with cross sections squares and rectangles (no graph) AP Calculus AB Khan Academy.mp3 | Cross sections of the solid perpendicular to the x-axis are rectangles whose height is x. Express the volume of the solid with a definite integral. So pause this video and see if you can have a go at that. All right, now what's interesting about this is they've just given us the equations for the graphs, but we haven't... |
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