Sentence
stringlengths 131
8.39k
| video_title
stringlengths 12
104
|
|---|---|
And then I'm not going to draw this surface exactly. I'm just trying to give you a sense of what the volume of the figure we're trying to calculate is. So the top of, if this is just some arbitrary surface, let me do it in a different color. So this is the top. This line is going vertical in the z direction. Actually, I could draw it like this, like it's a curve. And then this curve back here is going to be like a wall.
|
Double integrals 5 Double and triple integrals Multivariable Calculus Khan Academy.mp3
|
So this is the top. This line is going vertical in the z direction. Actually, I could draw it like this, like it's a curve. And then this curve back here is going to be like a wall. That's going to be like a wall. And maybe I'll paint this side of the wall, just so you can see what it kind of looks like. Trying my best.
|
Double integrals 5 Double and triple integrals Multivariable Calculus Khan Academy.mp3
|
And then this curve back here is going to be like a wall. That's going to be like a wall. And maybe I'll paint this side of the wall, just so you can see what it kind of looks like. Trying my best. I think you get an idea. Let me make it a little darker. This is actually more of an exercise in art than in math, in many ways.
|
Double integrals 5 Double and triple integrals Multivariable Calculus Khan Academy.mp3
|
Trying my best. I think you get an idea. Let me make it a little darker. This is actually more of an exercise in art than in math, in many ways. You get the idea. And then the boundary here is like this. And this top isn't flat.
|
Double integrals 5 Double and triple integrals Multivariable Calculus Khan Academy.mp3
|
This is actually more of an exercise in art than in math, in many ways. You get the idea. And then the boundary here is like this. And this top isn't flat. It could be a curved surface. I do it a little like that, but it's a curved surface. And we know, in the example we're about to do, that this surface right here, z is equal to x squared.
|
Double integrals 5 Double and triple integrals Multivariable Calculus Khan Academy.mp3
|
And this top isn't flat. It could be a curved surface. I do it a little like that, but it's a curved surface. And we know, in the example we're about to do, that this surface right here, z is equal to x squared. So we want to figure out the volume under this. So how do we do it? Well, let's think about it.
|
Double integrals 5 Double and triple integrals Multivariable Calculus Khan Academy.mp3
|
And we know, in the example we're about to do, that this surface right here, z is equal to x squared. So we want to figure out the volume under this. So how do we do it? Well, let's think about it. We could actually use the intuition that I just gave you. We're essentially just going to take a dA, which is a little small square down here. And that little area, that's the same thing as a dx.
|
Double integrals 5 Double and triple integrals Multivariable Calculus Khan Academy.mp3
|
Well, let's think about it. We could actually use the intuition that I just gave you. We're essentially just going to take a dA, which is a little small square down here. And that little area, that's the same thing as a dx. Let me use a darker color. As a dx times a dy. And then we just have to multiply it times f of xy, which is this, for each area.
|
Double integrals 5 Double and triple integrals Multivariable Calculus Khan Academy.mp3
|
And that little area, that's the same thing as a dx. Let me use a darker color. As a dx times a dy. And then we just have to multiply it times f of xy, which is this, for each area. And then sum them all up. And then we could take a sum in the x direction first or the y direction first. Now before doing that, just to make sure that you have the intuition, because the boundaries are the hard part, let me just draw our xy plane.
|
Double integrals 5 Double and triple integrals Multivariable Calculus Khan Academy.mp3
|
And then we just have to multiply it times f of xy, which is this, for each area. And then sum them all up. And then we could take a sum in the x direction first or the y direction first. Now before doing that, just to make sure that you have the intuition, because the boundaries are the hard part, let me just draw our xy plane. So let me rotate up like that. I'm just going to draw our xy plane, because that's what matters. Because the hard part here is just figuring out our bounds of integration.
|
Double integrals 5 Double and triple integrals Multivariable Calculus Khan Academy.mp3
|
Now before doing that, just to make sure that you have the intuition, because the boundaries are the hard part, let me just draw our xy plane. So let me rotate up like that. I'm just going to draw our xy plane, because that's what matters. Because the hard part here is just figuring out our bounds of integration. So the curve is just y is equal to x squared. So it'll look something like that. y is equal to x squared.
|
Double integrals 5 Double and triple integrals Multivariable Calculus Khan Academy.mp3
|
Because the hard part here is just figuring out our bounds of integration. So the curve is just y is equal to x squared. So it'll look something like that. y is equal to x squared. This is the point y is equal to 1. This is y-axis. This is the x-axis.
|
Double integrals 5 Double and triple integrals Multivariable Calculus Khan Academy.mp3
|
y is equal to x squared. This is the point y is equal to 1. This is y-axis. This is the x-axis. This is the point x is equal to 1. x is equal to 1. There you go. That's not an x, that's a 1.
|
Double integrals 5 Double and triple integrals Multivariable Calculus Khan Academy.mp3
|
This is the x-axis. This is the point x is equal to 1. x is equal to 1. There you go. That's not an x, that's a 1. This is the x-axis. Anyway, so we want to figure out how do we sum up this dx times dy, or this da, along this domain. So let's draw it.
|
Double integrals 5 Double and triple integrals Multivariable Calculus Khan Academy.mp3
|
That's not an x, that's a 1. This is the x-axis. Anyway, so we want to figure out how do we sum up this dx times dy, or this da, along this domain. So let's draw it. Let's visually draw it. And it doesn't hurt to do this when you actually have to do the problem, because this frankly is the hard part. A lot of calculus teachers will just have you set up the integral and then say, OK, well the rest is easy, or the rest is calc 1.
|
Double integrals 5 Double and triple integrals Multivariable Calculus Khan Academy.mp3
|
So let's draw it. Let's visually draw it. And it doesn't hurt to do this when you actually have to do the problem, because this frankly is the hard part. A lot of calculus teachers will just have you set up the integral and then say, OK, well the rest is easy, or the rest is calc 1. So this area here is the same thing as this area here. So its base is dx, and its height is dy. And then you can imagine that we're looking at this thing from above.
|
Double integrals 5 Double and triple integrals Multivariable Calculus Khan Academy.mp3
|
A lot of calculus teachers will just have you set up the integral and then say, OK, well the rest is easy, or the rest is calc 1. So this area here is the same thing as this area here. So its base is dx, and its height is dy. And then you can imagine that we're looking at this thing from above. So the surface is up here someplace, and we're looking straight down on it, and so this is just this area. So let's say we wanted to take the integral with respect to x first. So we want to sum up the volume above this column, first of all, is this area times dx dy.
|
Double integrals 5 Double and triple integrals Multivariable Calculus Khan Academy.mp3
|
And then you can imagine that we're looking at this thing from above. So the surface is up here someplace, and we're looking straight down on it, and so this is just this area. So let's say we wanted to take the integral with respect to x first. So we want to sum up the volume above this column, first of all, is this area times dx dy. So let's write the volume above that column. It's going to be the value of the function, the height at that point, which is xy squared. So it's going to be xy squared times dx dy.
|
Double integrals 5 Double and triple integrals Multivariable Calculus Khan Academy.mp3
|
So we want to sum up the volume above this column, first of all, is this area times dx dy. So let's write the volume above that column. It's going to be the value of the function, the height at that point, which is xy squared. So it's going to be xy squared times dx dy. This expression gives us the volume above this area, or this column right here. And let's say we want to sum in the x direction first. So we want to sum that dx, sum one here, sum here, et cetera, et cetera.
|
Double integrals 5 Double and triple integrals Multivariable Calculus Khan Academy.mp3
|
So it's going to be xy squared times dx dy. This expression gives us the volume above this area, or this column right here. And let's say we want to sum in the x direction first. So we want to sum that dx, sum one here, sum here, et cetera, et cetera. So we're going to sum in the x direction. So my question to you is, what is our lower bound of integration? Well, we're kind of holding our y constant, right?
|
Double integrals 5 Double and triple integrals Multivariable Calculus Khan Academy.mp3
|
So we want to sum that dx, sum one here, sum here, et cetera, et cetera. So we're going to sum in the x direction. So my question to you is, what is our lower bound of integration? Well, we're kind of holding our y constant, right? And so if we go to the left, if we go lower and lower x's, we kind of bump into the curve here. So the lower bound of integration is actually the curve. And what is this curve if we were to write x as a function of y, right?
|
Double integrals 5 Double and triple integrals Multivariable Calculus Khan Academy.mp3
|
Well, we're kind of holding our y constant, right? And so if we go to the left, if we go lower and lower x's, we kind of bump into the curve here. So the lower bound of integration is actually the curve. And what is this curve if we were to write x as a function of y, right? This curve is y is equal to x squared, or x is equal to the square root of y. So if we're integrating with respect to x for a fixed y right here, we're integrating in the horizontal direction first. Our lower bound is x is equal to the square root of y. x is equal to the square root of y.
|
Double integrals 5 Double and triple integrals Multivariable Calculus Khan Academy.mp3
|
And what is this curve if we were to write x as a function of y, right? This curve is y is equal to x squared, or x is equal to the square root of y. So if we're integrating with respect to x for a fixed y right here, we're integrating in the horizontal direction first. Our lower bound is x is equal to the square root of y. x is equal to the square root of y. That's interesting. I think this is the first time you've probably seen a variable bound integral. But it makes sense, because for this row that we're adding up right here, the upper bound's easy.
|
Double integrals 5 Double and triple integrals Multivariable Calculus Khan Academy.mp3
|
Our lower bound is x is equal to the square root of y. x is equal to the square root of y. That's interesting. I think this is the first time you've probably seen a variable bound integral. But it makes sense, because for this row that we're adding up right here, the upper bound's easy. The upper bound is x is equal to 1. The upper bound is x is equal to 1. But the lower bound is x is equal to the square root of y, right?
|
Double integrals 5 Double and triple integrals Multivariable Calculus Khan Academy.mp3
|
But it makes sense, because for this row that we're adding up right here, the upper bound's easy. The upper bound is x is equal to 1. The upper bound is x is equal to 1. But the lower bound is x is equal to the square root of y, right? Because as you go back, like, oh, I bump into the curve. And what's the curve? Well, the curve is x is equal to the square root of y, because we don't know which y we picked.
|
Double integrals 5 Double and triple integrals Multivariable Calculus Khan Academy.mp3
|
But the lower bound is x is equal to the square root of y, right? Because as you go back, like, oh, I bump into the curve. And what's the curve? Well, the curve is x is equal to the square root of y, because we don't know which y we picked. Fair enough. So once we've figured out the volume, so that'll give us the volume above this rectangle right here. And then we want to add up the dy's, right?
|
Double integrals 5 Double and triple integrals Multivariable Calculus Khan Academy.mp3
|
Well, the curve is x is equal to the square root of y, because we don't know which y we picked. Fair enough. So once we've figured out the volume, so that'll give us the volume above this rectangle right here. And then we want to add up the dy's, right? And remember, there's a whole volume above what I'm drawing right here, right? I'm just drawing this part in the xy plane. So what we've done just now, this expression as it's written right now, figures out the volume above that rectangle.
|
Double integrals 5 Double and triple integrals Multivariable Calculus Khan Academy.mp3
|
And then we want to add up the dy's, right? And remember, there's a whole volume above what I'm drawing right here, right? I'm just drawing this part in the xy plane. So what we've done just now, this expression as it's written right now, figures out the volume above that rectangle. Now, if we want to figure out the entire volume of the solid, we integrate along the y-axis, or we add up all the dy's. This was a dy right here, not a dx. My dx's and dy's look too similar.
|
Double integrals 5 Double and triple integrals Multivariable Calculus Khan Academy.mp3
|
So what we've done just now, this expression as it's written right now, figures out the volume above that rectangle. Now, if we want to figure out the entire volume of the solid, we integrate along the y-axis, or we add up all the dy's. This was a dy right here, not a dx. My dx's and dy's look too similar. So now, what is the lower bound on the y-axis, if I'm summing up these rectangles? Well, the lower bound is y is equal to 0, right? So we're going to go from y is equal to 0 to what?
|
Double integrals 5 Double and triple integrals Multivariable Calculus Khan Academy.mp3
|
My dx's and dy's look too similar. So now, what is the lower bound on the y-axis, if I'm summing up these rectangles? Well, the lower bound is y is equal to 0, right? So we're going to go from y is equal to 0 to what? What is the upper bound? To y is equal to 1. And there you have it.
|
Double integrals 5 Double and triple integrals Multivariable Calculus Khan Academy.mp3
|
So we're going to go from y is equal to 0 to what? What is the upper bound? To y is equal to 1. And there you have it. Let me rewrite that integral. So the double integral is going to be from x is equal to the square root of y to x is equal to 1, xy squared, dx, dy. And then the y bounds, y goes from 0 to y to 1.
|
Double integrals 5 Double and triple integrals Multivariable Calculus Khan Academy.mp3
|
And there you have it. Let me rewrite that integral. So the double integral is going to be from x is equal to the square root of y to x is equal to 1, xy squared, dx, dy. And then the y bounds, y goes from 0 to y to 1. I've just realized I've run out of time. In the next video, we'll evaluate this, and then we'll do it in the other order. See you soon.
|
Double integrals 5 Double and triple integrals Multivariable Calculus Khan Academy.mp3
|
I'm just trying to draw a bit of an arbitrary path. And let's say we go in a counterclockwise direction, like that, along our path. And we could call this path, so we're in a counterclockwise direction, we could call that path C. And let's say we also have a vector field. And our vector field is going to be a little unusual, I'll call it P. P of xy, it only has an i component, or all of its vectors are only multiples of the i unit vector. So it's P, capital P of xy, times the unit vector i. There is no j component. So if you had to visualize this vector field, all of the vectors only go, they're all multiples of the i unit vector.
|
Green's theorem proof part 1 Multivariable Calculus Khan Academy.mp3
|
And our vector field is going to be a little unusual, I'll call it P. P of xy, it only has an i component, or all of its vectors are only multiples of the i unit vector. So it's P, capital P of xy, times the unit vector i. There is no j component. So if you had to visualize this vector field, all of the vectors only go, they're all multiples of the i unit vector. Or they could be negative multiples, so they could also go in that direction. But they don't go diagonal, or they don't go up, they all go left to right, or right to left. That's what this vector field would look like.
|
Green's theorem proof part 1 Multivariable Calculus Khan Academy.mp3
|
So if you had to visualize this vector field, all of the vectors only go, they're all multiples of the i unit vector. Or they could be negative multiples, so they could also go in that direction. But they don't go diagonal, or they don't go up, they all go left to right, or right to left. That's what this vector field would look like. Now what I'm interested in doing is figuring out the line integral over a closed loop, the closed loop C, or the closed path C, of P dot dr. Which is just our standard kind of way of solving for a line integral. And we've seen what dr is in the past.
|
Green's theorem proof part 1 Multivariable Calculus Khan Academy.mp3
|
That's what this vector field would look like. Now what I'm interested in doing is figuring out the line integral over a closed loop, the closed loop C, or the closed path C, of P dot dr. Which is just our standard kind of way of solving for a line integral. And we've seen what dr is in the past. dr is equal to dx times i, plus dy times the j unit vector. And you might say, isn't it dx dt times dt? Let me write that.
|
Green's theorem proof part 1 Multivariable Calculus Khan Academy.mp3
|
And we've seen what dr is in the past. dr is equal to dx times i, plus dy times the j unit vector. And you might say, isn't it dx dt times dt? Let me write that. Can't dr be written as dx dt times dt i, plus dy dt times dt j? And it could, but if you imagine these differentials could cancel out, and you'll just look for the dx and the dy. And we've seen that multiple times.
|
Green's theorem proof part 1 Multivariable Calculus Khan Academy.mp3
|
Let me write that. Can't dr be written as dx dt times dt i, plus dy dt times dt j? And it could, but if you imagine these differentials could cancel out, and you'll just look for the dx and the dy. And we've seen that multiple times. And I'm going to leave it in this form, because hopefully, if we're careful, we won't have to deal with the third parameter t. So let's just look at it in this form right here, with just the dx's and the dy's. So this integral can be rewritten as the line integral, the curve C, actually let me just do it over down here. The line integral over the path, or the curve C, of P dot dr.
|
Green's theorem proof part 1 Multivariable Calculus Khan Academy.mp3
|
And we've seen that multiple times. And I'm going to leave it in this form, because hopefully, if we're careful, we won't have to deal with the third parameter t. So let's just look at it in this form right here, with just the dx's and the dy's. So this integral can be rewritten as the line integral, the curve C, actually let me just do it over down here. The line integral over the path, or the curve C, of P dot dr. So we take the product of each of the coefficients, let's say the coefficient of the i component. So we get P, I'll do it in green, actually let me do it in that purple color. So we get P of xy times dx, plus, well there's no 0 times j times dy.
|
Green's theorem proof part 1 Multivariable Calculus Khan Academy.mp3
|
The line integral over the path, or the curve C, of P dot dr. So we take the product of each of the coefficients, let's say the coefficient of the i component. So we get P, I'll do it in green, actually let me do it in that purple color. So we get P of xy times dx, plus, well there's no 0 times j times dy. 0 times dy is just going to be 0. So this, our line integral simplified to this right here. This is equal to this original integral up here.
|
Green's theorem proof part 1 Multivariable Calculus Khan Academy.mp3
|
So we get P of xy times dx, plus, well there's no 0 times j times dy. 0 times dy is just going to be 0. So this, our line integral simplified to this right here. This is equal to this original integral up here. So we're literally just taking the line integral around this path. Now I said that if we play our cards right, we're not going to have to deal with the third variable t, that we might be able to just solve this integral only in terms of x. And so let's see if we can do that.
|
Green's theorem proof part 1 Multivariable Calculus Khan Academy.mp3
|
This is equal to this original integral up here. So we're literally just taking the line integral around this path. Now I said that if we play our cards right, we're not going to have to deal with the third variable t, that we might be able to just solve this integral only in terms of x. And so let's see if we can do that. So let's look at our minimum and maximum x points. That looks like our minimum x point. Let's call that a.
|
Green's theorem proof part 1 Multivariable Calculus Khan Academy.mp3
|
And so let's see if we can do that. So let's look at our minimum and maximum x points. That looks like our minimum x point. Let's call that a. And let's call that our maximum x point. Let's call that b. What we could do is we can break up this curve into two functions of x. Y is functions of x.
|
Green's theorem proof part 1 Multivariable Calculus Khan Academy.mp3
|
Let's call that a. And let's call that our maximum x point. Let's call that b. What we could do is we can break up this curve into two functions of x. Y is functions of x. So this bottom one right here we could call as y1 of x. This is just a standard curve. When we just start dealing with standard calculus, this is just, you can imagine this is f of x, and it's a function of x.
|
Green's theorem proof part 1 Multivariable Calculus Khan Academy.mp3
|
What we could do is we can break up this curve into two functions of x. Y is functions of x. So this bottom one right here we could call as y1 of x. This is just a standard curve. When we just start dealing with standard calculus, this is just, you can imagine this is f of x, and it's a function of x. And this is y2 of x. Just like that. So you can imagine two paths.
|
Green's theorem proof part 1 Multivariable Calculus Khan Academy.mp3
|
When we just start dealing with standard calculus, this is just, you can imagine this is f of x, and it's a function of x. And this is y2 of x. Just like that. So you can imagine two paths. One path defined by y1 of x. Let me do that in a different color, a little magenta. One path defined by y1 of x as we go from x is equal to a to x is equal to b.
|
Green's theorem proof part 1 Multivariable Calculus Khan Academy.mp3
|
So you can imagine two paths. One path defined by y1 of x. Let me do that in a different color, a little magenta. One path defined by y1 of x as we go from x is equal to a to x is equal to b. And then another path defined by y2 of x as we go from x is equal to b to x is equal to a. That is our curve. So what we could do is we could rewrite this integral, which is the same thing as that integral, as this is equal to the integral, we'll first do this first path, of x going from a to b of p of x.
|
Green's theorem proof part 1 Multivariable Calculus Khan Academy.mp3
|
One path defined by y1 of x as we go from x is equal to a to x is equal to b. And then another path defined by y2 of x as we go from x is equal to b to x is equal to a. That is our curve. So what we could do is we could rewrite this integral, which is the same thing as that integral, as this is equal to the integral, we'll first do this first path, of x going from a to b of p of x. And I could just say p of x and y, but we know along this path y is a function of x. It's y1 of x. So we say x and y1 of x.
|
Green's theorem proof part 1 Multivariable Calculus Khan Academy.mp3
|
So what we could do is we could rewrite this integral, which is the same thing as that integral, as this is equal to the integral, we'll first do this first path, of x going from a to b of p of x. And I could just say p of x and y, but we know along this path y is a function of x. It's y1 of x. So we say x and y1 of x. Wherever we see a y, we substitute it with y1 of x dx. So that covers that first path. I'll do it in the same color.
|
Green's theorem proof part 1 Multivariable Calculus Khan Academy.mp3
|
So we say x and y1 of x. Wherever we see a y, we substitute it with y1 of x dx. So that covers that first path. I'll do it in the same color. So this is, we could imagine this is c1. This is kind of the first half of our curve. Well, it's not exactly necessarily the half, but that takes us right from that point to that point.
|
Green's theorem proof part 1 Multivariable Calculus Khan Academy.mp3
|
I'll do it in the same color. So this is, we could imagine this is c1. This is kind of the first half of our curve. Well, it's not exactly necessarily the half, but that takes us right from that point to that point. And then we want to complete the circle. Maybe I'll do that. And that is going to be, I'll do that in yellow.
|
Green's theorem proof part 1 Multivariable Calculus Khan Academy.mp3
|
Well, it's not exactly necessarily the half, but that takes us right from that point to that point. And then we want to complete the circle. Maybe I'll do that. And that is going to be, I'll do that in yellow. That's going to be equal to, sorry, we're going to have to add these two, plus the integral from x is equal to b to x is equal to a of, and I'll do it in that same color, of p of x. And now y is going to be y2 of x. Wherever you see a y, you can substitute with y2 of x along this curve.
|
Green's theorem proof part 1 Multivariable Calculus Khan Academy.mp3
|
And that is going to be, I'll do that in yellow. That's going to be equal to, sorry, we're going to have to add these two, plus the integral from x is equal to b to x is equal to a of, and I'll do it in that same color, of p of x. And now y is going to be y2 of x. Wherever you see a y, you can substitute with y2 of x along this curve. y2 of x dx. This is already getting interesting, and you might already see where I'm going with this. So this is the curve c2.
|
Green's theorem proof part 1 Multivariable Calculus Khan Academy.mp3
|
Wherever you see a y, you can substitute with y2 of x along this curve. y2 of x dx. This is already getting interesting, and you might already see where I'm going with this. So this is the curve c2. I think you'd appreciate if you take the union of c1 and c2. We've got our whole curve. So let's see if we can simplify this integral a little bit.
|
Green's theorem proof part 1 Multivariable Calculus Khan Academy.mp3
|
So this is the curve c2. I think you'd appreciate if you take the union of c1 and c2. We've got our whole curve. So let's see if we can simplify this integral a little bit. Well, one thing we want to do, we might want to make their endpoints the same. So if you swap a and b here, it just turns the integral negative. So you make this into a b, that into an a, and then make that plus sign into a minus sign.
|
Green's theorem proof part 1 Multivariable Calculus Khan Academy.mp3
|
So let's see if we can simplify this integral a little bit. Well, one thing we want to do, we might want to make their endpoints the same. So if you swap a and b here, it just turns the integral negative. So you make this into a b, that into an a, and then make that plus sign into a minus sign. And now we can rewrite this whole thing as being equal to the integral from a to b of this thing, of p of x and y1 of x, minus this thing, minus p of x and y2 of x. And then all of that times dx. Maybe I'll write it in a third color.
|
Green's theorem proof part 1 Multivariable Calculus Khan Academy.mp3
|
So you make this into a b, that into an a, and then make that plus sign into a minus sign. And now we can rewrite this whole thing as being equal to the integral from a to b of this thing, of p of x and y1 of x, minus this thing, minus p of x and y2 of x. And then all of that times dx. Maybe I'll write it in a third color. All of that times dx. Now, I'm going to do something a little bit arbitrary, but I think you'll appreciate why I did this by the end of this video. And it's just a very simple operation.
|
Green's theorem proof part 1 Multivariable Calculus Khan Academy.mp3
|
Maybe I'll write it in a third color. All of that times dx. Now, I'm going to do something a little bit arbitrary, but I think you'll appreciate why I did this by the end of this video. And it's just a very simple operation. What I'm going to do is I'm going to swap these two. So I'm essentially going to multiply this whole thing by negative 1, or essentially multiply and divide by negative 1. So I can multiply this by negative 1, and then multiply the outside by negative 1, and I will not have changed the integral.
|
Green's theorem proof part 1 Multivariable Calculus Khan Academy.mp3
|
And it's just a very simple operation. What I'm going to do is I'm going to swap these two. So I'm essentially going to multiply this whole thing by negative 1, or essentially multiply and divide by negative 1. So I can multiply this by negative 1, and then multiply the outside by negative 1, and I will not have changed the integral. I'm multiplying by negative 1 twice. So if I swap these two things, if I multiply the inside times negative 1, so this is going to be equal to, we do the outside of the integral, a to b. If I multiply the inside, I'll do a dx out here.
|
Green's theorem proof part 1 Multivariable Calculus Khan Academy.mp3
|
So I can multiply this by negative 1, and then multiply the outside by negative 1, and I will not have changed the integral. I'm multiplying by negative 1 twice. So if I swap these two things, if I multiply the inside times negative 1, so this is going to be equal to, we do the outside of the integral, a to b. If I multiply the inside, I'll do a dx out here. If I multiply the inside of the integral by negative 1, these two guys switch. So this becomes p of x of y2 of x, and then you're going to have minus p of x and y1 of x. My handwriting is getting a little messy.
|
Green's theorem proof part 1 Multivariable Calculus Khan Academy.mp3
|
If I multiply the inside, I'll do a dx out here. If I multiply the inside of the integral by negative 1, these two guys switch. So this becomes p of x of y2 of x, and then you're going to have minus p of x and y1 of x. My handwriting is getting a little messy. But I can't just multiply just the inside by minus 1. I don't want to change the integral, so I multiply the inside by minus 1. Let me multiply the outside by minus 1, and since I multiply by minus 1 twice, these two things are equivalent.
|
Green's theorem proof part 1 Multivariable Calculus Khan Academy.mp3
|
My handwriting is getting a little messy. But I can't just multiply just the inside by minus 1. I don't want to change the integral, so I multiply the inside by minus 1. Let me multiply the outside by minus 1, and since I multiply by minus 1 twice, these two things are equivalent. You could say this is the negative of that. Either way, I think you appreciate that I haven't changed the integral at all numerically. I multiply the inside and the outside by minus 1.
|
Green's theorem proof part 1 Multivariable Calculus Khan Academy.mp3
|
Let me multiply the outside by minus 1, and since I multiply by minus 1 twice, these two things are equivalent. You could say this is the negative of that. Either way, I think you appreciate that I haven't changed the integral at all numerically. I multiply the inside and the outside by minus 1. Now the next step I'm going to do might look a little bit foreign to you, but I think you'll appreciate it. It might be obvious to you if you've recently done some double integrals. This thing can be rewritten as minus the integral from a to b of the function p of x, y, evaluated at y2 of x minus it evaluated.
|
Green's theorem proof part 1 Multivariable Calculus Khan Academy.mp3
|
I multiply the inside and the outside by minus 1. Now the next step I'm going to do might look a little bit foreign to you, but I think you'll appreciate it. It might be obvious to you if you've recently done some double integrals. This thing can be rewritten as minus the integral from a to b of the function p of x, y, evaluated at y2 of x minus it evaluated. Let me make it very clear. This is y is equal to y2 of x minus this function evaluated at y is equal to y1 of x, and then of course all of that times dx. This statement and what we saw right here and this statement right here are completely identical.
|
Green's theorem proof part 1 Multivariable Calculus Khan Academy.mp3
|
This thing can be rewritten as minus the integral from a to b of the function p of x, y, evaluated at y2 of x minus it evaluated. Let me make it very clear. This is y is equal to y2 of x minus this function evaluated at y is equal to y1 of x, and then of course all of that times dx. This statement and what we saw right here and this statement right here are completely identical. Then if we assume that a partial derivative of capital P with respect to y exists, hopefully you'll realize, and I'll focus on this a little bit because I don't want to confuse you in this step. Let me write the outside of the integral. This is going to be equal to, this is kind of a neat outcome, and we're starting to build up to a very neat outcome, which we'll probably have to take the next video to do.
|
Green's theorem proof part 1 Multivariable Calculus Khan Academy.mp3
|
This statement and what we saw right here and this statement right here are completely identical. Then if we assume that a partial derivative of capital P with respect to y exists, hopefully you'll realize, and I'll focus on this a little bit because I don't want to confuse you in this step. Let me write the outside of the integral. This is going to be equal to, this is kind of a neat outcome, and we're starting to build up to a very neat outcome, which we'll probably have to take the next video to do. Let me do the outside, dx. This right here, if we assume that capital P has a partial derivative, this right here is the exact same thing. This right here is the exact same thing as the partial derivative of P with respect to y, dy, the antiderivative of that, from y1 of x to y2 of x. I want to make you feel comfortable that these two things are equivalent.
|
Green's theorem proof part 1 Multivariable Calculus Khan Academy.mp3
|
This is going to be equal to, this is kind of a neat outcome, and we're starting to build up to a very neat outcome, which we'll probably have to take the next video to do. Let me do the outside, dx. This right here, if we assume that capital P has a partial derivative, this right here is the exact same thing. This right here is the exact same thing as the partial derivative of P with respect to y, dy, the antiderivative of that, from y1 of x to y2 of x. I want to make you feel comfortable that these two things are equivalent. To realize they're equivalent, you probably just have to start here and then go to that. We're used to seeing this. We're used to seeing a double integral like this.
|
Green's theorem proof part 1 Multivariable Calculus Khan Academy.mp3
|
This right here is the exact same thing as the partial derivative of P with respect to y, dy, the antiderivative of that, from y1 of x to y2 of x. I want to make you feel comfortable that these two things are equivalent. To realize they're equivalent, you probably just have to start here and then go to that. We're used to seeing this. We're used to seeing a double integral like this. Then the very first step, we say, okay, to solve this double integral, we start on the inside integral right there. We say, okay, let's take the antiderivative of this with respect to y. If you take the antiderivative of the partial of P with respect to y, you're going to end up with P. Since this is a definite integral, the boundaries are going to be in terms of x.
|
Green's theorem proof part 1 Multivariable Calculus Khan Academy.mp3
|
We're used to seeing a double integral like this. Then the very first step, we say, okay, to solve this double integral, we start on the inside integral right there. We say, okay, let's take the antiderivative of this with respect to y. If you take the antiderivative of the partial of P with respect to y, you're going to end up with P. Since this is a definite integral, the boundaries are going to be in terms of x. You're going to evaluate that from y is equal to y2 of x and you're going to subtract from that y is equal to y1 of x. Normally, we start with something like this and we go to something like this. This is kind of unusual that we started, we kind of solved, we started with the solution of the definite integral and then we slowly built back to the definite integral.
|
Green's theorem proof part 1 Multivariable Calculus Khan Academy.mp3
|
If you take the antiderivative of the partial of P with respect to y, you're going to end up with P. Since this is a definite integral, the boundaries are going to be in terms of x. You're going to evaluate that from y is equal to y2 of x and you're going to subtract from that y is equal to y1 of x. Normally, we start with something like this and we go to something like this. This is kind of unusual that we started, we kind of solved, we started with the solution of the definite integral and then we slowly built back to the definite integral. Hopefully, you realize that this is true, that this is just, we're kind of going in the reverse direction that we normally do. If you do realize that, then we've just established a pretty neat outcome because what is this right here? What is this right here?
|
Green's theorem proof part 1 Multivariable Calculus Khan Academy.mp3
|
This is kind of unusual that we started, we kind of solved, we started with the solution of the definite integral and then we slowly built back to the definite integral. Hopefully, you realize that this is true, that this is just, we're kind of going in the reverse direction that we normally do. If you do realize that, then we've just established a pretty neat outcome because what is this right here? What is this right here? Let me go back, let me see if I can fit everything. I have some function, I have some function and I'm assuming that the partial of P with respect to y exists, but I have some function defined over the xy plane. You can imagine if we're dealing in three dimensions now.
|
Green's theorem proof part 1 Multivariable Calculus Khan Academy.mp3
|
What is this right here? Let me go back, let me see if I can fit everything. I have some function, I have some function and I'm assuming that the partial of P with respect to y exists, but I have some function defined over the xy plane. You can imagine if we're dealing in three dimensions now. Let me draw a little bit neater. That's y, that's x, that's z, so that's y, that's x. This, you can imagine, is some surface.
|
Green's theorem proof part 1 Multivariable Calculus Khan Academy.mp3
|
You can imagine if we're dealing in three dimensions now. Let me draw a little bit neater. That's y, that's x, that's z, so that's y, that's x. This, you can imagine, is some surface. It just happens to be the partial of P with respect to x. It's some surface on the xy plane like that. What are we doing?
|
Green's theorem proof part 1 Multivariable Calculus Khan Academy.mp3
|
This, you can imagine, is some surface. It just happens to be the partial of P with respect to x. It's some surface on the xy plane like that. What are we doing? We're taking the double integral under that surface, around this region. The region's boundaries in terms of y are defined by y2 and y1 of x, by y2 and y1 of x. You literally have that curve.
|
Green's theorem proof part 1 Multivariable Calculus Khan Academy.mp3
|
What are we doing? We're taking the double integral under that surface, around this region. The region's boundaries in terms of y are defined by y2 and y1 of x, by y2 and y1 of x. You literally have that curve. That's y2 on top, y1 on the bottom. We're essentially taking the volume above this. If you imagine with the base, the whole floor of this is going to be the area inside of this curve.
|
Green's theorem proof part 1 Multivariable Calculus Khan Academy.mp3
|
You literally have that curve. That's y2 on top, y1 on the bottom. We're essentially taking the volume above this. If you imagine with the base, the whole floor of this is going to be the area inside of this curve. Then the height is going to be the function, partial of P with respect to y. The height is going to be this function, partial of P with respect to y. It's going to be a little hard for me to draw, but this is essentially some type of a volume if you want to visualize it that way.
|
Green's theorem proof part 1 Multivariable Calculus Khan Academy.mp3
|
If you imagine with the base, the whole floor of this is going to be the area inside of this curve. Then the height is going to be the function, partial of P with respect to y. The height is going to be this function, partial of P with respect to y. It's going to be a little hard for me to draw, but this is essentially some type of a volume if you want to visualize it that way. The really neat outcome here is, if you call this region R, we've just simplified this line integral. This was a special one. It only had an x component, the vector field.
|
Green's theorem proof part 1 Multivariable Calculus Khan Academy.mp3
|
It's going to be a little hard for me to draw, but this is essentially some type of a volume if you want to visualize it that way. The really neat outcome here is, if you call this region R, we've just simplified this line integral. This was a special one. It only had an x component, the vector field. We've just simplified this line integral to being equivalent to, and maybe I should write this line integral because that's what's the really neat outcome. We've just established that this thing right here, which is the same as our original one, so let me write that. The closed line integral around the curve C of P of x, y, dx.
|
Green's theorem proof part 1 Multivariable Calculus Khan Academy.mp3
|
It only had an x component, the vector field. We've just simplified this line integral to being equivalent to, and maybe I should write this line integral because that's what's the really neat outcome. We've just established that this thing right here, which is the same as our original one, so let me write that. The closed line integral around the curve C of P of x, y, dx. We've just established that that's the same thing as the double integral over the region R. This is the region R. Over the region R of the partial of P with respect to y. The partial of P with respect to y. We could write dy dx, or we could write da, or whatever you want to write, but this is the double integral over that region.
|
Green's theorem proof part 1 Multivariable Calculus Khan Academy.mp3
|
The closed line integral around the curve C of P of x, y, dx. We've just established that that's the same thing as the double integral over the region R. This is the region R. Over the region R of the partial of P with respect to y. The partial of P with respect to y. We could write dy dx, or we could write da, or whatever you want to write, but this is the double integral over that region. The neat thing here is, using a vector field that only had an x component, we were able to connect its line integral to the double integral over a region. Oh, and I forgot something very important. We had a negative sign out here, so this was a minus sign out here.
|
Green's theorem proof part 1 Multivariable Calculus Khan Academy.mp3
|
Hey everyone, so in the next couple videos I'm gonna be talking about a different sort of optimization problem, something called a constrained optimization problem. And an example of this is something where you might see, you might be asked to maximize some kind of multivariable function. And let's just say it was the function f of x, y is equal to x squared times y. But that's not all you're asked to do. You're subject to a certain constraint where you're only allowed values of x and y on a certain set. And I'm just gonna say the set of all values of x and y such that x squared plus y squared equals one. And this is something you might recognize as the unit circle, this particular constraint that I've put on here.
|
Constrained optimization introduction.mp3
|
But that's not all you're asked to do. You're subject to a certain constraint where you're only allowed values of x and y on a certain set. And I'm just gonna say the set of all values of x and y such that x squared plus y squared equals one. And this is something you might recognize as the unit circle, this particular constraint that I've put on here. This is the unit circle. So one way that you might think about a problem like this, you know, you're maximizing a certain two variable function, is to first think of the graph of that function. That's what I have pictured here, is the graph of f of x, y equals x squared times y.
|
Constrained optimization introduction.mp3
|
And this is something you might recognize as the unit circle, this particular constraint that I've put on here. This is the unit circle. So one way that you might think about a problem like this, you know, you're maximizing a certain two variable function, is to first think of the graph of that function. That's what I have pictured here, is the graph of f of x, y equals x squared times y. And now this constraint, x squared plus y squared, is basically just a subset of the x, y plane. So if we look at it head on here, and we look at the x, y plane, this circle represents all of the points x, y such that this holds. And what I've actually drawn here isn't the circle on the x, y plane, but I've projected it up onto the graph.
|
Constrained optimization introduction.mp3
|
That's what I have pictured here, is the graph of f of x, y equals x squared times y. And now this constraint, x squared plus y squared, is basically just a subset of the x, y plane. So if we look at it head on here, and we look at the x, y plane, this circle represents all of the points x, y such that this holds. And what I've actually drawn here isn't the circle on the x, y plane, but I've projected it up onto the graph. So this is showing you basically the values where this constraint holds, and also what they look like when graphed. So a way you can think about a problem like this is that you're looking on this circle, this kind of projected circle onto the graph, and looking for the highest points. And you might notice kind of here, there's sort of a peak on that wiggly circle, and over here there's another one.
|
Constrained optimization introduction.mp3
|
And what I've actually drawn here isn't the circle on the x, y plane, but I've projected it up onto the graph. So this is showing you basically the values where this constraint holds, and also what they look like when graphed. So a way you can think about a problem like this is that you're looking on this circle, this kind of projected circle onto the graph, and looking for the highest points. And you might notice kind of here, there's sort of a peak on that wiggly circle, and over here there's another one. And then the low points would be, you know, around that point and around over here. Now, this is good, and I think this is a nice way to sort of wrap your head around what this problem is asking. But there's actually a better way to visualize it in terms of finding the actual solution.
|
Constrained optimization introduction.mp3
|
And you might notice kind of here, there's sort of a peak on that wiggly circle, and over here there's another one. And then the low points would be, you know, around that point and around over here. Now, this is good, and I think this is a nice way to sort of wrap your head around what this problem is asking. But there's actually a better way to visualize it in terms of finding the actual solution. And that's to look only in the x, y plane, rather than trying to graph things, and just limit our perspective to the input space. So what I have here are the contour lines for f of x, y equals x squared plus y squared. And if you're unfamiliar with contour lines, or contour map, I have a video on that.
|
Constrained optimization introduction.mp3
|
But there's actually a better way to visualize it in terms of finding the actual solution. And that's to look only in the x, y plane, rather than trying to graph things, and just limit our perspective to the input space. So what I have here are the contour lines for f of x, y equals x squared plus y squared. And if you're unfamiliar with contour lines, or contour map, I have a video on that. You can go back and take a look. It's gonna be pretty crucial for the next couple videos to have a feel for that. But basically, each one of these lines represents a certain constant value for f. So for example, one of them might represent all of the values of x and y, where f of x, y is equal to, you know, two, right?
|
Constrained optimization introduction.mp3
|
And if you're unfamiliar with contour lines, or contour map, I have a video on that. You can go back and take a look. It's gonna be pretty crucial for the next couple videos to have a feel for that. But basically, each one of these lines represents a certain constant value for f. So for example, one of them might represent all of the values of x and y, where f of x, y is equal to, you know, two, right? So if you looked at all the values of x and y where this is true, you'd find yourself on one of these lines. And each line represents a different possible value for what this constant here actually is. So what I'm gonna do here is I'm actually gonna just zoom in on one particular contour line, right?
|
Constrained optimization introduction.mp3
|
But basically, each one of these lines represents a certain constant value for f. So for example, one of them might represent all of the values of x and y, where f of x, y is equal to, you know, two, right? So if you looked at all the values of x and y where this is true, you'd find yourself on one of these lines. And each line represents a different possible value for what this constant here actually is. So what I'm gonna do here is I'm actually gonna just zoom in on one particular contour line, right? So this here is something that I'm gonna vary, where I'm gonna be able to change what the constant we're setting f equal to is, and look at how the contour line changes as a result. So for example, if I put it around here-ish, what you're looking at is the contour line, the contour line for f of x, y equals 0.1. So all of the values on these two blue lines here tell you what values of x and y satisfy 0.1.
|
Constrained optimization introduction.mp3
|
So what I'm gonna do here is I'm actually gonna just zoom in on one particular contour line, right? So this here is something that I'm gonna vary, where I'm gonna be able to change what the constant we're setting f equal to is, and look at how the contour line changes as a result. So for example, if I put it around here-ish, what you're looking at is the contour line, the contour line for f of x, y equals 0.1. So all of the values on these two blue lines here tell you what values of x and y satisfy 0.1. But on the other hand, I could also shift this guy up, and maybe I'll shift it up, I'm gonna set it to where that constant is actually equal to one. So that would be kind of an alternative. I'll just kinda separate over here.
|
Constrained optimization introduction.mp3
|
So all of the values on these two blue lines here tell you what values of x and y satisfy 0.1. But on the other hand, I could also shift this guy up, and maybe I'll shift it up, I'm gonna set it to where that constant is actually equal to one. So that would be kind of an alternative. I'll just kinda separate over here. That would be the line where f of x, y is set equal to one itself. And the main thing I wanna highlight here is that at some values, like 0.1, this contour line intersects with the circle. It intersects with our constraint.
|
Constrained optimization introduction.mp3
|
I'll just kinda separate over here. That would be the line where f of x, y is set equal to one itself. And the main thing I wanna highlight here is that at some values, like 0.1, this contour line intersects with the circle. It intersects with our constraint. And let's just think about what that means. If there's a point, x and y, on that intersection there, that basically gives us a pair of numbers, x and y, such that this is true, that fact, that f of x, y equals 0.1, and also that x squared plus y squared equals one. So it means this is something that actually exists and is possible.
|
Constrained optimization introduction.mp3
|
It intersects with our constraint. And let's just think about what that means. If there's a point, x and y, on that intersection there, that basically gives us a pair of numbers, x and y, such that this is true, that fact, that f of x, y equals 0.1, and also that x squared plus y squared equals one. So it means this is something that actually exists and is possible. In fact, we can see that there's four different pairs of numbers where that's true, where they intersect here, where they intersect over here, and then the other two kind of symmetrically on that side. But on the other hand, if we look at this other world, where we shift up to the line f of x, y equals one, this never intersects with the constraint. So what that means is x, y, the pairs of numbers that satisfy this guy, are off the constraint.
|
Constrained optimization introduction.mp3
|
So it means this is something that actually exists and is possible. In fact, we can see that there's four different pairs of numbers where that's true, where they intersect here, where they intersect over here, and then the other two kind of symmetrically on that side. But on the other hand, if we look at this other world, where we shift up to the line f of x, y equals one, this never intersects with the constraint. So what that means is x, y, the pairs of numbers that satisfy this guy, are off the constraint. They're off of that circle, x squared plus y squared equals one. So what that tells us, as we try to maximize this function, subject to this constraint, is that we can never get as high as one. 0.1 would be achievable, and in fact, if we kind of go back to that and we look at 0.1, if I upped that value and changed it to the line where instead what you're looking at is 0.2, that's also possible because it intersects with the circle.
|
Constrained optimization introduction.mp3
|
So what that means is x, y, the pairs of numbers that satisfy this guy, are off the constraint. They're off of that circle, x squared plus y squared equals one. So what that tells us, as we try to maximize this function, subject to this constraint, is that we can never get as high as one. 0.1 would be achievable, and in fact, if we kind of go back to that and we look at 0.1, if I upped that value and changed it to the line where instead what you're looking at is 0.2, that's also possible because it intersects with the circle. In fact, you could play around with it and increase it a little bit more. And if I go to 0.3 instead, and I go over here and I say 0.3, that's also possible. And what we're basically trying to do is find the maximum value that we can put here, the maximum value so that if we look at the line that represents f of x, y equals that value, it still intersects with the circle.
|
Constrained optimization introduction.mp3
|
0.1 would be achievable, and in fact, if we kind of go back to that and we look at 0.1, if I upped that value and changed it to the line where instead what you're looking at is 0.2, that's also possible because it intersects with the circle. In fact, you could play around with it and increase it a little bit more. And if I go to 0.3 instead, and I go over here and I say 0.3, that's also possible. And what we're basically trying to do is find the maximum value that we can put here, the maximum value so that if we look at the line that represents f of x, y equals that value, it still intersects with the circle. And the key here, the key observation, is that that maximum value happens when these guys are tangent. And in the next video, I'll start going into the details of how we can use that observation, this notion of tangency, to solve the problem, to find the actual value of x and y that maximizes this subject to the constraint. But in the interim, I kind of want you to mull on that and think a little bit about how you might use that.
|
Constrained optimization introduction.mp3
|
And what we're basically trying to do is find the maximum value that we can put here, the maximum value so that if we look at the line that represents f of x, y equals that value, it still intersects with the circle. And the key here, the key observation, is that that maximum value happens when these guys are tangent. And in the next video, I'll start going into the details of how we can use that observation, this notion of tangency, to solve the problem, to find the actual value of x and y that maximizes this subject to the constraint. But in the interim, I kind of want you to mull on that and think a little bit about how you might use that. What does tangency mean here? How can you take advantage of certain other notions that we've learned about in multivariable calculus, like hint, hint, the gradient, to actually solve something like this? So with that, I will see you next video.
|
Constrained optimization introduction.mp3
|
In the last couple videos, I talked about this multivariable chain rule, and I gave some justification, and it might have been considered a little bit hand-wavy by some. I was doing a lot of things that looked kind of like taking a derivative with respect to t, and then multiplying that by an infinitesimal quantity dt, and thinking of canceling those out. And some people might say, ah, but this isn't really a fraction, that's a derivative, that's a differential operator, and you're treating it incorrectly. And while that's true, the intuitions underlying a lot of this actually matches with the formal argument pretty well. So what I wanna do here is just talk about what the formal argument behind the multivariable chain rule is. And just to remind ourselves of the setup, of where we are, you're thinking of v as a vector-valued function. So this is something that takes as an input t that lives on a number line, and then v maps this to some kind of high-dimensional space.
|
More formal treatment of multivariable chain rule.mp3
|
And while that's true, the intuitions underlying a lot of this actually matches with the formal argument pretty well. So what I wanna do here is just talk about what the formal argument behind the multivariable chain rule is. And just to remind ourselves of the setup, of where we are, you're thinking of v as a vector-valued function. So this is something that takes as an input t that lives on a number line, and then v maps this to some kind of high-dimensional space. In the simplest case, you might just think of that as a two-dimensional space, maybe it's three-dimensional space, or it could be 100-dimensional, you don't have to literally be visualizing it. And then f, our function f, somehow takes that 100-dimensional space, or two-dimensional, or three-dimensional, whatever it is, and then maps it onto the number line. So the overall effect of the composition function is to just take a real number to a real number.
|
More formal treatment of multivariable chain rule.mp3
|
So this is something that takes as an input t that lives on a number line, and then v maps this to some kind of high-dimensional space. In the simplest case, you might just think of that as a two-dimensional space, maybe it's three-dimensional space, or it could be 100-dimensional, you don't have to literally be visualizing it. And then f, our function f, somehow takes that 100-dimensional space, or two-dimensional, or three-dimensional, whatever it is, and then maps it onto the number line. So the overall effect of the composition function is to just take a real number to a real number. So it's a single-variable function, so that's why we're taking this ordinary derivative, rather than a partial derivative, or gradient, or anything like that. But because it goes through a multi-dimensional space, and you have this intermediary multivariable nature to it, that's why you have a gradient and a vector-valued derivative. So with the formal argument, the first thing you might do is just write out the formal definition of a derivative.
|
More formal treatment of multivariable chain rule.mp3
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.