video_title
stringlengths 12
104
| Sentence
stringlengths 131
8.39k
|
|---|---|
Divergence theorem proof (part 4) Divergence theorem Multivariable Calculus Khan Academy.mp3
|
Sometimes when you're doing a large multi-part proof like this, it's easy to lose your bearings. We're trying to prove the divergence theorem. We started off just rewriting the flux across the surface and rewriting the triple integral of the divergence. We said, if we can prove that each of these components are equal to each other, we will have our proof done. What I said is that we're going to use the fact that this is a type I region to prove this part, the fact that it's a type II region to prove this part, and the fact that it's a type III region to prove this part. In particular, I'm going to show the fact that if it is a type I region, we can prove this, and you can use the exact same argument to prove the other two, in which case the divergence theorem would be correct. We're focused on this part right here, and in particular we're focused on evaluating the surface integral.
|
Divergence theorem proof (part 4) Divergence theorem Multivariable Calculus Khan Academy.mp3
|
We said, if we can prove that each of these components are equal to each other, we will have our proof done. What I said is that we're going to use the fact that this is a type I region to prove this part, the fact that it's a type II region to prove this part, and the fact that it's a type III region to prove this part. In particular, I'm going to show the fact that if it is a type I region, we can prove this, and you can use the exact same argument to prove the other two, in which case the divergence theorem would be correct. We're focused on this part right here, and in particular we're focused on evaluating the surface integral. To evaluate that surface integral, we broke up the entire surface into three surfaces, that upper bound on z, the lower bound on z, and kind of the side of, if you imagine, some type of a funky cylinder. These don't have to be flat tops and bottoms. For a type I region, you don't even have to have this side surface.
|
Divergence theorem proof (part 4) Divergence theorem Multivariable Calculus Khan Academy.mp3
|
We're focused on this part right here, and in particular we're focused on evaluating the surface integral. To evaluate that surface integral, we broke up the entire surface into three surfaces, that upper bound on z, the lower bound on z, and kind of the side of, if you imagine, some type of a funky cylinder. These don't have to be flat tops and bottoms. For a type I region, you don't even have to have this side surface. This is only for the case where the two surfaces don't touch each other within this domain right over here. We broke it up into three surfaces, but we said, look, the normal vector along the side right over here is never going to have a k component. When you take the dot product with the k unit vector, this is just going to cross out.
|
Divergence theorem proof (part 4) Divergence theorem Multivariable Calculus Khan Academy.mp3
|
For a type I region, you don't even have to have this side surface. This is only for the case where the two surfaces don't touch each other within this domain right over here. We broke it up into three surfaces, but we said, look, the normal vector along the side right over here is never going to have a k component. When you take the dot product with the k unit vector, this is just going to cross out. Our surface integral is simplified to the surface integral over S2 and the surface integral of S1. In the last video, we evaluated the surface integral of S2, or at least we turned it into a double integral over the domain. Now we're going to do the same exact thing with S1.
|
Divergence theorem proof (part 4) Divergence theorem Multivariable Calculus Khan Academy.mp3
|
When you take the dot product with the k unit vector, this is just going to cross out. Our surface integral is simplified to the surface integral over S2 and the surface integral of S1. In the last video, we evaluated the surface integral of S2, or at least we turned it into a double integral over the domain. Now we're going to do the same exact thing with S1. Let's just remind ourselves what we are concerned with. We want to re-express S1, and S1 and this integral, the surface integral, I should say we want to re-express the surface integral over S1. This is R of x, y, and z.
|
Divergence theorem proof (part 4) Divergence theorem Multivariable Calculus Khan Academy.mp3
|
Now we're going to do the same exact thing with S1. Let's just remind ourselves what we are concerned with. We want to re-express S1, and S1 and this integral, the surface integral, I should say we want to re-express the surface integral over S1. This is R of x, y, and z. Up here we just wrote R, but this is making it explicit that R is a function of x, y, and z times k dot n dS. The way we evaluate any surface integral is we want to make sure or we want to have a parameterization for our actual surface. Let's introduce a parameterization for a surface.
|
Divergence theorem proof (part 4) Divergence theorem Multivariable Calculus Khan Academy.mp3
|
This is R of x, y, and z. Up here we just wrote R, but this is making it explicit that R is a function of x, y, and z times k dot n dS. The way we evaluate any surface integral is we want to make sure or we want to have a parameterization for our actual surface. Let's introduce a parameterization for a surface. Let's say that S1, and I'll use the letter O, it's kind of a weird one to use, it looks like a zero, and we're already using it for a normal vector. Let's use E. The parameterization of our surface could be x times i plus y times j plus, and it's a type 2 region, our surface is a function of x and y, plus F1, which is a function of x, y, times k. That's the F1 right over here. That's our lower bound on our region.
|
Divergence theorem proof (part 4) Divergence theorem Multivariable Calculus Khan Academy.mp3
|
Let's introduce a parameterization for a surface. Let's say that S1, and I'll use the letter O, it's kind of a weird one to use, it looks like a zero, and we're already using it for a normal vector. Let's use E. The parameterization of our surface could be x times i plus y times j plus, and it's a type 2 region, our surface is a function of x and y, plus F1, which is a function of x, y, times k. That's the F1 right over here. That's our lower bound on our region. This is 4 for all of the x, y pairs that are a member of our domain in question. We have our parameterization, and now we can think about how we can write this right over here, the n dS. n times dS.
|
Divergence theorem proof (part 4) Divergence theorem Multivariable Calculus Khan Academy.mp3
|
That's our lower bound on our region. This is 4 for all of the x, y pairs that are a member of our domain in question. We have our parameterization, and now we can think about how we can write this right over here, the n dS. n times dS. We've done this multiple, multiple, multiple times, which is actually the same thing as dS. We've done this multiple times. This is equal to the cross product of the parameterization in one direction with respect to one parameter, and then that crossed with respect to the other parameter, and then that times dA.
|
Divergence theorem proof (part 4) Divergence theorem Multivariable Calculus Khan Academy.mp3
|
n times dS. We've done this multiple, multiple, multiple times, which is actually the same thing as dS. We've done this multiple times. This is equal to the cross product of the parameterization in one direction with respect to one parameter, and then that crossed with respect to the other parameter, and then that times dA. We want to make sure we get the order right. I'm going to claim that this is going to be the partial of our parameterization with respect to y crossed with the partial of our parameterization with respect to x, and then we have times dA. We need to make sure this has the right orientation because for this bottom surface, remember, we need to be pointed straight down, outward from the region.
|
Divergence theorem proof (part 4) Divergence theorem Multivariable Calculus Khan Academy.mp3
|
This is equal to the cross product of the parameterization in one direction with respect to one parameter, and then that crossed with respect to the other parameter, and then that times dA. We want to make sure we get the order right. I'm going to claim that this is going to be the partial of our parameterization with respect to y crossed with the partial of our parameterization with respect to x, and then we have times dA. We need to make sure this has the right orientation because for this bottom surface, remember, we need to be pointed straight down, outward from the region. If we're going in the y direction, the partial with respect to y is like that. The partial with respect to x is like that. If you use the right-hand rule, your thumb will point downward like that.
|
Divergence theorem proof (part 4) Divergence theorem Multivariable Calculus Khan Academy.mp3
|
We need to make sure this has the right orientation because for this bottom surface, remember, we need to be pointed straight down, outward from the region. If we're going in the y direction, the partial with respect to y is like that. The partial with respect to x is like that. If you use the right-hand rule, your thumb will point downward like that. I could draw my right hand. My index finger could go like that. My middle finger would bend like this.
|
Divergence theorem proof (part 4) Divergence theorem Multivariable Calculus Khan Academy.mp3
|
If you use the right-hand rule, your thumb will point downward like that. I could draw my right hand. My index finger could go like that. My middle finger would bend like this. I don't really care what my other two fingers do. Then my thumb would go downwards. This is the right ordering, which is a different ordering, or it's the opposite ordering as we did last time.
|
Divergence theorem proof (part 4) Divergence theorem Multivariable Calculus Khan Academy.mp3
|
My middle finger would bend like this. I don't really care what my other two fingers do. Then my thumb would go downwards. This is the right ordering, which is a different ordering, or it's the opposite ordering as we did last time. We'll see. We'll just get a negative value. Let's just work it through just to be a little bit more convincing.
|
Divergence theorem proof (part 4) Divergence theorem Multivariable Calculus Khan Academy.mp3
|
This is the right ordering, which is a different ordering, or it's the opposite ordering as we did last time. We'll see. We'll just get a negative value. Let's just work it through just to be a little bit more convincing. This business right over here is going to be equal to we're going to divide our i, j, and k unit vectors. It's going to be all of that times dA. The partial of E with respect to y is 0, 1, 0, 1, partial of F1 with respect to y.
|
Divergence theorem proof (part 4) Divergence theorem Multivariable Calculus Khan Academy.mp3
|
Let's just work it through just to be a little bit more convincing. This business right over here is going to be equal to we're going to divide our i, j, and k unit vectors. It's going to be all of that times dA. The partial of E with respect to y is 0, 1, 0, 1, partial of F1 with respect to y. Partial with respect to x is going to be 1, 0. Partial of F1 with respect to x. Then when you evaluate this whole thing, let me draw a little dotted line here, this is going to be equal to some business times i minus some other business times j.
|
Divergence theorem proof (part 4) Divergence theorem Multivariable Calculus Khan Academy.mp3
|
The partial of E with respect to y is 0, 1, 0, 1, partial of F1 with respect to y. Partial with respect to x is going to be 1, 0. Partial of F1 with respect to x. Then when you evaluate this whole thing, let me draw a little dotted line here, this is going to be equal to some business times i minus some other business times j. On our k component, you're going to have 0 times 0 minus 1 times k. Minus k and then all of that times dA. The reason why I didn't even worry about what these things are going to be is I'm going to have to take the drop product with k. This whole thing, all of this business right over here, I can now express in the xy domain. I can now write is going to be equal to the double integral.
|
Divergence theorem proof (part 4) Divergence theorem Multivariable Calculus Khan Academy.mp3
|
Then when you evaluate this whole thing, let me draw a little dotted line here, this is going to be equal to some business times i minus some other business times j. On our k component, you're going to have 0 times 0 minus 1 times k. Minus k and then all of that times dA. The reason why I didn't even worry about what these things are going to be is I'm going to have to take the drop product with k. This whole thing, all of this business right over here, I can now express in the xy domain. I can now write is going to be equal to the double integral. Let me do it in that same purple color because that was the original color for that surface integral. It's going to be the double integral over the domain, our parameters domain in the xy plane of R of x. Let me write this a little bit cleaner.
|
Divergence theorem proof (part 4) Divergence theorem Multivariable Calculus Khan Academy.mp3
|
I can now write is going to be equal to the double integral. Let me do it in that same purple color because that was the original color for that surface integral. It's going to be the double integral over the domain, our parameters domain in the xy plane of R of x. Let me write this a little bit cleaner. R of xy, instead of z, I'm going to write z over that surface is F1 of x and y so that we have everything in terms of our parameters times all of this business, k dotted this. What is k dotted this? The dot product of k and negative k is negative 1, so we're just going to be left with negative dA.
|
Divergence theorem proof (part 4) Divergence theorem Multivariable Calculus Khan Academy.mp3
|
Let me write this a little bit cleaner. R of xy, instead of z, I'm going to write z over that surface is F1 of x and y so that we have everything in terms of our parameters times all of this business, k dotted this. What is k dotted this? The dot product of k and negative k is negative 1, so we're just going to be left with negative dA. We'll put the dA out front here and we are left with a negative dA right over here. We have now expressed this surface integral as the sum of two double integrals, as the sum of this, the sum of that, and that right over there. Actually, let me just rewrite it so that we can make everything clear.
|
Divergence theorem proof (part 4) Divergence theorem Multivariable Calculus Khan Academy.mp3
|
The dot product of k and negative k is negative 1, so we're just going to be left with negative dA. We'll put the dA out front here and we are left with a negative dA right over here. We have now expressed this surface integral as the sum of two double integrals, as the sum of this, the sum of that, and that right over there. Actually, let me just rewrite it so that we can make everything clear. This surface integral over our entire surface of R times k dot n dS is equal to the double integral and I'll do this in a new color, is equal to the double integral in the domain D of this thing minus this thing dA. I'll write R of x, y, and F2 of x, y, and that's that, minus this, minus R of x, y, and F2 of x, sorry, F1, be careful here, F1 of x, y, that's this thing right over here, all of that times dA. Now, we just showed this is equal to this.
|
Curvature formula, part 5.mp3
|
We're looking at this formula and trying to understand why it corresponds to curvature, why it tells you how much a curve actually curves. And the first thing we did is we noticed that this numerator corresponds to a certain cross product, the cross product between the first derivative and the second derivative of the function parameterizing the curve. And the way we started to understand that is we say, well, the function parameterizing the curve, s of t, produces vectors whose tips trace out that curve itself. And now if you think about how one tip moves to the next, the direction that it needs to go for that tip to move to the next one, that's what the first derivative tells you. And when you treat that in an infinitesimal way, this is why you get tangent vectors along the curve. And there's an entire series of videos on that for the derivative of a position vector-valued function. And they explain why the first derivative of a parametric function gives you tangent vectors to that curve.
|
Curvature formula, part 5.mp3
|
And now if you think about how one tip moves to the next, the direction that it needs to go for that tip to move to the next one, that's what the first derivative tells you. And when you treat that in an infinitesimal way, this is why you get tangent vectors along the curve. And there's an entire series of videos on that for the derivative of a position vector-valued function. And they explain why the first derivative of a parametric function gives you tangent vectors to that curve. But if we draw all of those tangent vectors just in their own space, their own little s prime of t space, you get all of these vectors, and we're rooting them at the same point to be able to relate them more easily. The way that you move from the tip of one of those to the next one is given by the second derivative. It kind of plays the same role for the first derivative as the first derivative plays for the original parametric function.
|
Curvature formula, part 5.mp3
|
And they explain why the first derivative of a parametric function gives you tangent vectors to that curve. But if we draw all of those tangent vectors just in their own space, their own little s prime of t space, you get all of these vectors, and we're rooting them at the same point to be able to relate them more easily. The way that you move from the tip of one of those to the next one is given by the second derivative. It kind of plays the same role for the first derivative as the first derivative plays for the original parametric function. And specifically, if you have a circumstance where the tangent vectors are just turning, the only thing they're doing is purely turning around, which is when your curve is actually curving, this corresponds to a case when the second derivative function is pretty much perpendicular as a vector. The vector it produces is perpendicular to the first derivative vector. So this is loosely why the cross product is kind of a good measure of curving because it tells you how perpendicular these guys are.
|
Curvature formula, part 5.mp3
|
It kind of plays the same role for the first derivative as the first derivative plays for the original parametric function. And specifically, if you have a circumstance where the tangent vectors are just turning, the only thing they're doing is purely turning around, which is when your curve is actually curving, this corresponds to a case when the second derivative function is pretty much perpendicular as a vector. The vector it produces is perpendicular to the first derivative vector. So this is loosely why the cross product is kind of a good measure of curving because it tells you how perpendicular these guys are. But there's a bit of a catch. The original formula for curvature, the whole reason we're doing it with respect to arc length and not with respect to the parameter t, is that curvature doesn't really care about how you parametrize the function. If you imagine zipping along it really quickly so your first derivative vectors are all super long, it shouldn't matter compared to crawling along it like a turtle.
|
Curvature formula, part 5.mp3
|
So this is loosely why the cross product is kind of a good measure of curving because it tells you how perpendicular these guys are. But there's a bit of a catch. The original formula for curvature, the whole reason we're doing it with respect to arc length and not with respect to the parameter t, is that curvature doesn't really care about how you parametrize the function. If you imagine zipping along it really quickly so your first derivative vectors are all super long, it shouldn't matter compared to crawling along it like a turtle. The curvature should just always be the same. But this is a problem if you think back to the cross product that we're now looking at, where you're taking the cross product between the first derivative and the second derivative because if you were traveling along this curve twice as quickly, what that would mean is your first derivative vector, so I'll kind of draw it again over here, would be twice as long to indicate that you're going twice as fast. And similarly, your second derivative vector to kind of keep up with that changing rate would also be twice as long.
|
Curvature formula, part 5.mp3
|
If you imagine zipping along it really quickly so your first derivative vectors are all super long, it shouldn't matter compared to crawling along it like a turtle. The curvature should just always be the same. But this is a problem if you think back to the cross product that we're now looking at, where you're taking the cross product between the first derivative and the second derivative because if you were traveling along this curve twice as quickly, what that would mean is your first derivative vector, so I'll kind of draw it again over here, would be twice as long to indicate that you're going twice as fast. And similarly, your second derivative vector to kind of keep up with that changing rate would also be twice as long. And as a result, the parallelogram that they trace out, and I should actually, kind of going off screen here, the parallelogram that they trace out would be actually four times as big, right, because both of the vectors get scaled up. So the way that we really want to be thinking about this is not the tangent vector due to the derivative, but normalizing this. And this should kind of make sense because we're thinking in terms of unit tangent vectors for the curvature as a whole.
|
Curvature formula, part 5.mp3
|
And similarly, your second derivative vector to kind of keep up with that changing rate would also be twice as long. And as a result, the parallelogram that they trace out, and I should actually, kind of going off screen here, the parallelogram that they trace out would be actually four times as big, right, because both of the vectors get scaled up. So the way that we really want to be thinking about this is not the tangent vector due to the derivative, but normalizing this. And this should kind of make sense because we're thinking in terms of unit tangent vectors for the curvature as a whole. So if you imagine instead kind of cutting off the vector to make sure that it's got a unit length, a length of one, and what that means is you're taking the derivative vector and dividing it by its own magnitude, by the magnitude of that derivative vector, and then similarly we'll want to scale everything else down. So you're taking this and kind of scaling it down by what you need. The resulting parallelogram they trace out is a more pure measurement of how perpendicular they are without caring about how long they are.
|
Curvature formula, part 5.mp3
|
And this should kind of make sense because we're thinking in terms of unit tangent vectors for the curvature as a whole. So if you imagine instead kind of cutting off the vector to make sure that it's got a unit length, a length of one, and what that means is you're taking the derivative vector and dividing it by its own magnitude, by the magnitude of that derivative vector, and then similarly we'll want to scale everything else down. So you're taking this and kind of scaling it down by what you need. The resulting parallelogram they trace out is a more pure measurement of how perpendicular they are without caring about how long they are. And what this guy would be, by the way, then, this is the second derivative vector, not normalized with respect to itself, but we're still dividing, you know, the thing we're dividing by as we scale everything down is still just the size of that first derivative vector. So this cross product, if we take the cross product between S prime normalized, S double prime, no, no, no, sorry, S prime itself, and I should be saying vectors for all of these, these are all vectors. If we take the cross product between that and S double prime scaled down by that same value that's still S prime, so it's not normalized, this is just scaled down by S prime, this here is a more pure measurement of how perpendicular the second derivative vector is to the first.
|
Curvature formula, part 5.mp3
|
The resulting parallelogram they trace out is a more pure measurement of how perpendicular they are without caring about how long they are. And what this guy would be, by the way, then, this is the second derivative vector, not normalized with respect to itself, but we're still dividing, you know, the thing we're dividing by as we scale everything down is still just the size of that first derivative vector. So this cross product, if we take the cross product between S prime normalized, S double prime, no, no, no, sorry, S prime itself, and I should be saying vectors for all of these, these are all vectors. If we take the cross product between that and S double prime scaled down by that same value that's still S prime, so it's not normalized, this is just scaled down by S prime, this here is a more pure measurement of how perpendicular the second derivative vector is to the first. And the reason we don't really care about the second derivative being normalized is if it was the case that, you know, the second derivative was really, really strong and wasn't necessarily a unit vector, that's fine. That's just telling us that the tangent vector turns much more quickly and the curvature should be higher. And in fact, it turns out that this whole expression is the derivative of the unit tangent vector, T, that unit tangent vector that I've talked about a lot, with respect to the parameter T. So whatever the parameter of your original function is.
|
Curvature formula, part 5.mp3
|
If we take the cross product between that and S double prime scaled down by that same value that's still S prime, so it's not normalized, this is just scaled down by S prime, this here is a more pure measurement of how perpendicular the second derivative vector is to the first. And the reason we don't really care about the second derivative being normalized is if it was the case that, you know, the second derivative was really, really strong and wasn't necessarily a unit vector, that's fine. That's just telling us that the tangent vector turns much more quickly and the curvature should be higher. And in fact, it turns out that this whole expression is the derivative of the unit tangent vector, T, that unit tangent vector that I've talked about a lot, with respect to the parameter T. So whatever the parameter of your original function is. And now if you think back to the, I'm not sure if it was the last video or the one before that, but I talked about how when you take, when we're looking for this derivative of the tangent vector with respect to arc length, the way that you compute this is to first take its derivative with respect to the parameter, which is something we can actually do because everything is expressed in terms of that parameter, and then dividing it by the, basically the change in arc length with respect to that parameter, which is the size of that first derivative function. So if this whole thing is the derivative of the tangent vector with respect to T, what that means is when we take this and we divide that whole thing by the derivative, by S prime, that should give us curvature. And in fact, that's just worth writing on its own here.
|
Curvature formula, part 5.mp3
|
And in fact, it turns out that this whole expression is the derivative of the unit tangent vector, T, that unit tangent vector that I've talked about a lot, with respect to the parameter T. So whatever the parameter of your original function is. And now if you think back to the, I'm not sure if it was the last video or the one before that, but I talked about how when you take, when we're looking for this derivative of the tangent vector with respect to arc length, the way that you compute this is to first take its derivative with respect to the parameter, which is something we can actually do because everything is expressed in terms of that parameter, and then dividing it by the, basically the change in arc length with respect to that parameter, which is the size of that first derivative function. So if this whole thing is the derivative of the tangent vector with respect to T, what that means is when we take this and we divide that whole thing by the derivative, by S prime, that should give us curvature. And in fact, that's just worth writing on its own here. That's curvature. Curvature is equal to, and what I'm gonna do is I'm gonna take, since we see this three different times, we see the magnitude of S prime here, magnitude of S prime here, magnitude of S prime here, since we see that three times, I'm gonna go over here and I'm gonna put that on the denominator, but cubing it. So S prime, that derivative vector, the magnitude of that derivative vector cubed.
|
Curvature formula, part 5.mp3
|
And in fact, that's just worth writing on its own here. That's curvature. Curvature is equal to, and what I'm gonna do is I'm gonna take, since we see this three different times, we see the magnitude of S prime here, magnitude of S prime here, magnitude of S prime here, since we see that three times, I'm gonna go over here and I'm gonna put that on the denominator, but cubing it. So S prime, that derivative vector, the magnitude of that derivative vector cubed. And then on the top, we still have S prime, that vector S prime cross product with S double prime. That vector. And this is, I mean, you could think of this as yet another formula for curvature.
|
Curvature formula, part 5.mp3
|
So S prime, that derivative vector, the magnitude of that derivative vector cubed. And then on the top, we still have S prime, that vector S prime cross product with S double prime. That vector. And this is, I mean, you could think of this as yet another formula for curvature. I think I've given you like four at this point. Or you could think of it as just kind of the same thing. And if we look back up to our original one that I was trying to justify, this is just the spelled out version of it because what is this bottom component here?
|
Curvature formula, part 5.mp3
|
And this is, I mean, you could think of this as yet another formula for curvature. I think I've given you like four at this point. Or you could think of it as just kind of the same thing. And if we look back up to our original one that I was trying to justify, this is just the spelled out version of it because what is this bottom component here? If we take X prime squared plus Y prime squared, and if we think of the square root of that, so kind of taking it to the 1 1⁄2 power, that would be the magnitude of the derivative. And I kind of showed that in some of the previous videos. And what we're doing is we're cubing that.
|
Curvature formula, part 5.mp3
|
And if we look back up to our original one that I was trying to justify, this is just the spelled out version of it because what is this bottom component here? If we take X prime squared plus Y prime squared, and if we think of the square root of that, so kind of taking it to the 1 1⁄2 power, that would be the magnitude of the derivative. And I kind of showed that in some of the previous videos. And what we're doing is we're cubing that. So that whole formula that's the very explicit, you know, in terms of X and Ys, what's going on, is really just expressing this idea. You take the cross product between the first derivative and the second derivative, and then because you're normalizing it, you know, normalizing with respect to the first derivative, you want to scale down the second derivative by that same amount just so that the parallelogram we're thinking kind of shrinks and everything stays in proportion. And then once again, we're dividing about that S prime basically because curvature is supposed to be with respect to S and not with respect to T. So that's a way of kind of getting a correction factor for how wrong you're going to be if you just think in terms of the parameter T instead of steps in terms of the arc length.
|
Curvature formula, part 5.mp3
|
And what we're doing is we're cubing that. So that whole formula that's the very explicit, you know, in terms of X and Ys, what's going on, is really just expressing this idea. You take the cross product between the first derivative and the second derivative, and then because you're normalizing it, you know, normalizing with respect to the first derivative, you want to scale down the second derivative by that same amount just so that the parallelogram we're thinking kind of shrinks and everything stays in proportion. And then once again, we're dividing about that S prime basically because curvature is supposed to be with respect to S and not with respect to T. So that's a way of kind of getting a correction factor for how wrong you're going to be if you just think in terms of the parameter T instead of steps in terms of the arc length. So hopefully this makes the original formula a little bit less random-seeming, you know, in this two-dimensional case. And it also gives another strong conceptual tool for understanding yet another way that you can think about how much a curve itself actually curves. And I think I've probably done enough videos here going through all the different formulas for what curvature should be, and then in the next one or two I'll go through some specific examples just to see what it looks like to compute that.
|
Directional derivative.mp3
|
And that's a way to extend the idea of a partial derivative. And partial derivatives, if you'll remember, have to do with functions with some kind of multivariable input, and I'll just use two inputs because that's the easiest to think about. And it could be some single variable output. It could also deal with vector variable outputs. We haven't gotten to that yet. So we'll just think about a single variable ordinary real number output that's, you know, an expression of x and y. And the partial derivative, one of the ways that I said you could think about it is to take a look at the input space, your x and y plane.
|
Directional derivative.mp3
|
It could also deal with vector variable outputs. We haven't gotten to that yet. So we'll just think about a single variable ordinary real number output that's, you know, an expression of x and y. And the partial derivative, one of the ways that I said you could think about it is to take a look at the input space, your x and y plane. So this would be the x axis, this is y. And, you know, vaguely in your mind, you're thinking that somehow this outputs to a line. This outputs to just the real numbers.
|
Directional derivative.mp3
|
And the partial derivative, one of the ways that I said you could think about it is to take a look at the input space, your x and y plane. So this would be the x axis, this is y. And, you know, vaguely in your mind, you're thinking that somehow this outputs to a line. This outputs to just the real numbers. And maybe you're thinking about a transformation that takes it there, or maybe you're just thinking, okay, this is the input space, that's the output. And when you take the partial derivative at some kind of point, so I'll write it out like partial derivative of f with respect to x at a point like one, two, you think about that point, you know, one, y is equal to two. And if you're taking it with respect to x, you think about just nudging it a little bit in that x direction.
|
Directional derivative.mp3
|
This outputs to just the real numbers. And maybe you're thinking about a transformation that takes it there, or maybe you're just thinking, okay, this is the input space, that's the output. And when you take the partial derivative at some kind of point, so I'll write it out like partial derivative of f with respect to x at a point like one, two, you think about that point, you know, one, y is equal to two. And if you're taking it with respect to x, you think about just nudging it a little bit in that x direction. And you see what the resulting nudge is in the output space. And the ratio between the size of that resulting nudge and the original one, the ratio between, you know, partial f and partial x, is the value that you want. And when you did it with respect to y, you know, you were thinking about traveling in a different direction.
|
Directional derivative.mp3
|
And if you're taking it with respect to x, you think about just nudging it a little bit in that x direction. And you see what the resulting nudge is in the output space. And the ratio between the size of that resulting nudge and the original one, the ratio between, you know, partial f and partial x, is the value that you want. And when you did it with respect to y, you know, you were thinking about traveling in a different direction. Maybe you nudge it straight up, and you're wondering, okay, how does that influence the output? And the question here, with directional derivatives, what if you have some vector, v, and I'll give a little vector hat on top of it, that, you know, I don't know, let's say it's negative one, two, is the vector. So you'd be thinking about that as a step of negative one in the x direction, and then two more in the y direction.
|
Directional derivative.mp3
|
And when you did it with respect to y, you know, you were thinking about traveling in a different direction. Maybe you nudge it straight up, and you're wondering, okay, how does that influence the output? And the question here, with directional derivatives, what if you have some vector, v, and I'll give a little vector hat on top of it, that, you know, I don't know, let's say it's negative one, two, is the vector. So you'd be thinking about that as a step of negative one in the x direction, and then two more in the y direction. So it's gonna be something that ends up there. This is your vector, v, at least if you're thinking of v as stemming from the original point. And you're wondering, what does a nudge in that direction do to the function itself?
|
Directional derivative.mp3
|
So you'd be thinking about that as a step of negative one in the x direction, and then two more in the y direction. So it's gonna be something that ends up there. This is your vector, v, at least if you're thinking of v as stemming from the original point. And you're wondering, what does a nudge in that direction do to the function itself? And remember, with these original, you know, nudges in the x direction, nudges in the y, you're not really thinking of it as, you know, this is kind of a large step. You're really thinking of it as something itty, itty, bitty, bitty, bitty. You know, it's not that, but it's really something very, very small.
|
Directional derivative.mp3
|
And you're wondering, what does a nudge in that direction do to the function itself? And remember, with these original, you know, nudges in the x direction, nudges in the y, you're not really thinking of it as, you know, this is kind of a large step. You're really thinking of it as something itty, itty, bitty, bitty, bitty. You know, it's not that, but it's really something very, very small. And formally, you'd be thinking about the limit, as this gets really, really, really small, approaching zero, and this gets really, really small, approaching zero, what does the ratio of the two approach? And similarly, with the y, you're not thinking of it as something, this is pretty sizable, but it would be something really, really small. And the directional derivative is similar.
|
Directional derivative.mp3
|
You know, it's not that, but it's really something very, very small. And formally, you'd be thinking about the limit, as this gets really, really, really small, approaching zero, and this gets really, really small, approaching zero, what does the ratio of the two approach? And similarly, with the y, you're not thinking of it as something, this is pretty sizable, but it would be something really, really small. And the directional derivative is similar. You're not thinking of the actual vector actually taking a step along that, but you'd be thinking of taking a step along, say, h, multiplied by that vector. And h might represent some really, really small numbers. I know maybe this here is like 0.001.
|
Directional derivative.mp3
|
And the directional derivative is similar. You're not thinking of the actual vector actually taking a step along that, but you'd be thinking of taking a step along, say, h, multiplied by that vector. And h might represent some really, really small numbers. I know maybe this here is like 0.001. And when you're doing this formally, you'd just be thinking the limit as h goes to zero. So the directional derivative is saying, when you take a slight nudge in the direction of that vector, what is the resulting change to the output? And one way to think about this is you say, well, that slight nudge of the vector, if we actually expand things out, and we look at the definition itself, it'll be negative h, negative one times that component, and then two h here.
|
Directional derivative.mp3
|
I know maybe this here is like 0.001. And when you're doing this formally, you'd just be thinking the limit as h goes to zero. So the directional derivative is saying, when you take a slight nudge in the direction of that vector, what is the resulting change to the output? And one way to think about this is you say, well, that slight nudge of the vector, if we actually expand things out, and we look at the definition itself, it'll be negative h, negative one times that component, and then two h here. So it's kind of like you took negative one nudge in the x direction, and then two nudges in the y direction. You know, so for whatever your nudge in the v direction there, you take a negative one step by x, and then two of them up by y. So when we actually write this out, the notation, by the way, is you take that same nabla from the gradient, but then you put the vector down here.
|
Directional derivative.mp3
|
And one way to think about this is you say, well, that slight nudge of the vector, if we actually expand things out, and we look at the definition itself, it'll be negative h, negative one times that component, and then two h here. So it's kind of like you took negative one nudge in the x direction, and then two nudges in the y direction. You know, so for whatever your nudge in the v direction there, you take a negative one step by x, and then two of them up by y. So when we actually write this out, the notation, by the way, is you take that same nabla from the gradient, but then you put the vector down here. So this is the directional derivative in the direction of v. And there's a whole bunch of other notations, too. You know, I think there's like derivative of f with respect to that vector is one way people think about it. Some people just write like partial with a little subscript vector.
|
Directional derivative.mp3
|
So when we actually write this out, the notation, by the way, is you take that same nabla from the gradient, but then you put the vector down here. So this is the directional derivative in the direction of v. And there's a whole bunch of other notations, too. You know, I think there's like derivative of f with respect to that vector is one way people think about it. Some people just write like partial with a little subscript vector. There's a whole bunch of different notations, but this is the one I like. You think that nabla with a little f down there, with a little v for your vector of f, and it's still a function of x and y. And the reason I like this is it's indicative of how you end up calculating it, which I'll talk about at the end of the video.
|
Directional derivative.mp3
|
Some people just write like partial with a little subscript vector. There's a whole bunch of different notations, but this is the one I like. You think that nabla with a little f down there, with a little v for your vector of f, and it's still a function of x and y. And the reason I like this is it's indicative of how you end up calculating it, which I'll talk about at the end of the video. And for this particular example, a good guess that you might have is to say, well, we take a negative step in the x direction, so you think of it as whatever the change that's caused by such a step in the x direction, you do the negative of that, and then it's two steps in the y direction, so whatever the change caused by a tiny step in the y direction, let's just take two of those. Two times partial f, partial y. And this is actually how you calculate it.
|
Directional derivative.mp3
|
And the reason I like this is it's indicative of how you end up calculating it, which I'll talk about at the end of the video. And for this particular example, a good guess that you might have is to say, well, we take a negative step in the x direction, so you think of it as whatever the change that's caused by such a step in the x direction, you do the negative of that, and then it's two steps in the y direction, so whatever the change caused by a tiny step in the y direction, let's just take two of those. Two times partial f, partial y. And this is actually how you calculate it. And if I was gonna be more general, you know, let's say we've got a vector w. I'm gonna keep it abstract and just call it a and b as its components, rather than the specific numbers. You would say that the directional derivative in the direction of w, whatever that is, of f, is equal to a times the partial derivative of f with respect to x plus b times the partial derivative of f with respect to y. And this is it.
|
Directional derivative.mp3
|
And this is actually how you calculate it. And if I was gonna be more general, you know, let's say we've got a vector w. I'm gonna keep it abstract and just call it a and b as its components, rather than the specific numbers. You would say that the directional derivative in the direction of w, whatever that is, of f, is equal to a times the partial derivative of f with respect to x plus b times the partial derivative of f with respect to y. And this is it. This is the formula that you would use for the directional derivative. And again, the way that you're thinking about this is you're really saying, you know, you take a little nudge that's a in the x direction and b in the y direction, so this should kind of make sense. And sometimes you see this written not with respect to the partial derivatives themselves and the actual components a and b, but with respect to the gradient.
|
Directional derivative.mp3
|
And this is it. This is the formula that you would use for the directional derivative. And again, the way that you're thinking about this is you're really saying, you know, you take a little nudge that's a in the x direction and b in the y direction, so this should kind of make sense. And sometimes you see this written not with respect to the partial derivatives themselves and the actual components a and b, but with respect to the gradient. And this is because it makes it much more compact, more general if you're dealing with other dimensions. So I'll just write it over here. If you look at this expression, it looks like a dot product.
|
Directional derivative.mp3
|
And sometimes you see this written not with respect to the partial derivatives themselves and the actual components a and b, but with respect to the gradient. And this is because it makes it much more compact, more general if you're dealing with other dimensions. So I'll just write it over here. If you look at this expression, it looks like a dot product. If you take the dot product of the vectors, a, b, and the one that has the partial derivatives in it, so what's lined up with a is the partial derivative with respect to x, partial f, partial x, and what's lined up with b is the partial derivative with respect to y. And you look at this and you say, hey, a, b, I mean, that's just the original vector, right? That's w. That's the vector w. And then you're dotting this with, well, partial derivative with respect to x in one component, the other partial derivative in the other component.
|
Directional derivative.mp3
|
If you look at this expression, it looks like a dot product. If you take the dot product of the vectors, a, b, and the one that has the partial derivatives in it, so what's lined up with a is the partial derivative with respect to x, partial f, partial x, and what's lined up with b is the partial derivative with respect to y. And you look at this and you say, hey, a, b, I mean, that's just the original vector, right? That's w. That's the vector w. And then you're dotting this with, well, partial derivative with respect to x in one component, the other partial derivative in the other component. That's just the gradient. That is the gradient of f. And here, you know, it's nabla without that little w at the bottom. And this is why we use this notation, because it's so suggestive of the way that you ultimately calculate it.
|
Directional derivative.mp3
|
That's w. That's the vector w. And then you're dotting this with, well, partial derivative with respect to x in one component, the other partial derivative in the other component. That's just the gradient. That is the gradient of f. And here, you know, it's nabla without that little w at the bottom. And this is why we use this notation, because it's so suggestive of the way that you ultimately calculate it. So this is really what you'll see in a textbook or see as the compact way of writing it. And you can see how this is more flexible for dimensions. So if we were talking about something that has like a five-dimensional input and the vector, the direction you move, has five different components, this is flexible.
|
Directional derivative.mp3
|
And this is why we use this notation, because it's so suggestive of the way that you ultimately calculate it. So this is really what you'll see in a textbook or see as the compact way of writing it. And you can see how this is more flexible for dimensions. So if we were talking about something that has like a five-dimensional input and the vector, the direction you move, has five different components, this is flexible. When you expand it, the gradient would have five components and the vector itself would have five components. So this is the directional derivative and how you calculate it. And the way you interpret, you're thinking of moving along that vector by a tiny nudge, by a tiny, you know, little value multiplied by that vector, and saying, how does that change the output and what's the ratio of the resulting change?
|
Divergence theorem proof (part 2) Divergence theorem Multivariable Calculus Khan Academy.mp3
|
In the last video, we said if we can just prove that each of these parts are equal to each other, we essentially have proven that that is equal to that. Because here in yellow is another way of writing the flux across the surface. And here in green is another way of writing the triple integral over our region of the divergence of f. What I'm going to do in this video and probably the next video is prove that these two are equivalent to each other. And I'm going to prove it using the fact that our original region is a simple solid region, or in particular, is a type 1 region. And then that's essentially going to be it because you can use the exact same argument and the fact that it is a type 2 region to prove this. And the exact same argument and the fact that it is a type 3 region to prove that. So I'm going to assume it's a type 1, which I can.
|
Divergence theorem proof (part 2) Divergence theorem Multivariable Calculus Khan Academy.mp3
|
And I'm going to prove it using the fact that our original region is a simple solid region, or in particular, is a type 1 region. And then that's essentially going to be it because you can use the exact same argument and the fact that it is a type 2 region to prove this. And the exact same argument and the fact that it is a type 3 region to prove that. So I'm going to assume it's a type 1, which I can. It's a type 1, 2, and 3 region. So given the fact that it's type 1, I'm now going to prove this relationship right over here. And then I'll leave it up to you to do the exact same argument with a type 2 and a type 3 region.
|
Divergence theorem proof (part 2) Divergence theorem Multivariable Calculus Khan Academy.mp3
|
So I'm going to assume it's a type 1, which I can. It's a type 1, 2, and 3 region. So given the fact that it's type 1, I'm now going to prove this relationship right over here. And then I'll leave it up to you to do the exact same argument with a type 2 and a type 3 region. So let's get going. So type 1 region, just to remind ourselves, a type 1 region is a region that is equal to the set of all x, y's, and z's, such that the xy pairs are a member of a domain in the xy plane. And z is bounded by two functions.
|
Divergence theorem proof (part 2) Divergence theorem Multivariable Calculus Khan Academy.mp3
|
And then I'll leave it up to you to do the exact same argument with a type 2 and a type 3 region. So let's get going. So type 1 region, just to remind ourselves, a type 1 region is a region that is equal to the set of all x, y's, and z's, such that the xy pairs are a member of a domain in the xy plane. And z is bounded by two functions. z's lower bound is f1 of x and y. And that's going to be less than or equal to z. And z's upper bound, we can call it f2 of x, y.
|
Divergence theorem proof (part 2) Divergence theorem Multivariable Calculus Khan Academy.mp3
|
And z is bounded by two functions. z's lower bound is f1 of x and y. And that's going to be less than or equal to z. And z's upper bound, we can call it f2 of x, y. And then let me close the set notation right over here. And let me just draw a general version of a type 1 region. So let me draw my x, y, and z axes.
|
Divergence theorem proof (part 2) Divergence theorem Multivariable Calculus Khan Academy.mp3
|
And z's upper bound, we can call it f2 of x, y. And then let me close the set notation right over here. And let me just draw a general version of a type 1 region. So let me draw my x, y, and z axes. So this is my z-axis. This is my x-axis. And there is my y-axis.
|
Divergence theorem proof (part 2) Divergence theorem Multivariable Calculus Khan Academy.mp3
|
So let me draw my x, y, and z axes. So this is my z-axis. This is my x-axis. And there is my y-axis. And so we might have a region D here. So our region, I'll draw it as a little circle right over here. This is our region D. And for any x, y in our region D, you can evaluate the function f. You can figure out an f1, I should say.
|
Divergence theorem proof (part 2) Divergence theorem Multivariable Calculus Khan Academy.mp3
|
And there is my y-axis. And so we might have a region D here. So our region, I'll draw it as a little circle right over here. This is our region D. And for any x, y in our region D, you can evaluate the function f. You can figure out an f1, I should say. So this might define an f1, which we can kind of imagine as a surface or a little thing that's at the bottom of a cylinder if you want. So every x, y there, when you evaluate or when you figure out what the corresponding f1 of those points in this domain would be, you might get a surface that looks something like this. It doesn't have to be flat.
|
Divergence theorem proof (part 2) Divergence theorem Multivariable Calculus Khan Academy.mp3
|
This is our region D. And for any x, y in our region D, you can evaluate the function f. You can figure out an f1, I should say. So this might define an f1, which we can kind of imagine as a surface or a little thing that's at the bottom of a cylinder if you want. So every x, y there, when you evaluate or when you figure out what the corresponding f1 of those points in this domain would be, you might get a surface that looks something like this. It doesn't have to be flat. But hopefully this gives the idea. It doesn't have to be completely flat. It can be curved or whatever else.
|
Divergence theorem proof (part 2) Divergence theorem Multivariable Calculus Khan Academy.mp3
|
It doesn't have to be flat. But hopefully this gives the idea. It doesn't have to be completely flat. It can be curved or whatever else. But this just shows that every x, y, when you evaluate it right over here, it gets associated with a point, this lower bound surface right over here. And I'll draw a dotted line to show that's only for the x, y's in this domain. And then we have an upper bound surface that might be up here.
|
Divergence theorem proof (part 2) Divergence theorem Multivariable Calculus Khan Academy.mp3
|
It can be curved or whatever else. But this just shows that every x, y, when you evaluate it right over here, it gets associated with a point, this lower bound surface right over here. And I'll draw a dotted line to show that's only for the x, y's in this domain. And then we have an upper bound surface that might be up here. Give me any x, y. When I evaluate f2, I get this surface up here. And once again, they don't have to look the same.
|
Divergence theorem proof (part 2) Divergence theorem Multivariable Calculus Khan Academy.mp3
|
And then we have an upper bound surface that might be up here. Give me any x, y. When I evaluate f2, I get this surface up here. And once again, they don't have to look the same. This could be like a dome, or it could be slanted, or who knows what it might be. But this will give you the general idea. And then z fills up the region.
|
Divergence theorem proof (part 2) Divergence theorem Multivariable Calculus Khan Academy.mp3
|
And once again, they don't have to look the same. This could be like a dome, or it could be slanted, or who knows what it might be. But this will give you the general idea. And then z fills up the region. Remember, the region isn't just the surface of the figure. It's the entire volume inside of it. So when z varies between that surface and that surface, for any given x, y in our domain, we fill up the entire region.
|
Divergence theorem proof (part 2) Divergence theorem Multivariable Calculus Khan Academy.mp3
|
And then z fills up the region. Remember, the region isn't just the surface of the figure. It's the entire volume inside of it. So when z varies between that surface and that surface, for any given x, y in our domain, we fill up the entire region. And this way I drew it, it looks like a cylinder. But it doesn't have to be a cylinder like this. And these two surfaces might touch each other, in which case there would be no side of the cylinder.
|
Divergence theorem proof (part 2) Divergence theorem Multivariable Calculus Khan Academy.mp3
|
So when z varies between that surface and that surface, for any given x, y in our domain, we fill up the entire region. And this way I drew it, it looks like a cylinder. But it doesn't have to be a cylinder like this. And these two surfaces might touch each other, in which case there would be no side of the cylinder. They could be lumpier than this. They might be inclined in some way. But hopefully this is a good generalization of a type I region.
|
Divergence theorem proof (part 2) Divergence theorem Multivariable Calculus Khan Academy.mp3
|
And these two surfaces might touch each other, in which case there would be no side of the cylinder. They could be lumpier than this. They might be inclined in some way. But hopefully this is a good generalization of a type I region. Now, a type I region, you can kind of think of it, it can be broken up into three parts. It can be broken up into surface, or the surfaces of a type I region, I should say, can be broken up into three parts. It can be broken up into, let's call that surface one.
|
Divergence theorem proof (part 2) Divergence theorem Multivariable Calculus Khan Academy.mp3
|
But hopefully this is a good generalization of a type I region. Now, a type I region, you can kind of think of it, it can be broken up into three parts. It can be broken up into surface, or the surfaces of a type I region, I should say, can be broken up into three parts. It can be broken up into, let's call that surface one. Let's call this right over here surface two, the top of the cylinder, or whatever kind of lumpy top it might be. And let's call the side, if these two surfaces don't touch each other, let's call that surface three. There might not necessarily even be a surface three if these two touch each other in the case of a sphere.
|
Divergence theorem proof (part 2) Divergence theorem Multivariable Calculus Khan Academy.mp3
|
It can be broken up into, let's call that surface one. Let's call this right over here surface two, the top of the cylinder, or whatever kind of lumpy top it might be. And let's call the side, if these two surfaces don't touch each other, let's call that surface three. There might not necessarily even be a surface three if these two touch each other in the case of a sphere. But let's just assume that there actually is a surface three. So if we're evaluating the surface integral, we'll think about this triple integral in a second. But let's think about how we can rewrite this surface integral right over here.
|
Divergence theorem proof (part 2) Divergence theorem Multivariable Calculus Khan Academy.mp3
|
There might not necessarily even be a surface three if these two touch each other in the case of a sphere. But let's just assume that there actually is a surface three. So if we're evaluating the surface integral, we'll think about this triple integral in a second. But let's think about how we can rewrite this surface integral right over here. Instead of this entire surface is s1 plus s2 plus s3. So we can essentially break this up into three separate surface integrals. So let's do that.
|
Divergence theorem proof (part 2) Divergence theorem Multivariable Calculus Khan Academy.mp3
|
But let's think about how we can rewrite this surface integral right over here. Instead of this entire surface is s1 plus s2 plus s3. So we can essentially break this up into three separate surface integrals. So let's do that. So remember, we're just focusing on this part right over here. So the surface integral of r times k dot n, the dot product of k and n, ds can be rewritten as the surface integral over s2 of r times k dot n, ds, plus the surface integral over s1 of r times k dot n, ds, plus the surface integral over surface three of the same thing, r times k dot n, ds. Now, the way I've drawn it, and this is actually the case, s3 is if those surfaces don't touch each other.
|
Divergence theorem proof (part 2) Divergence theorem Multivariable Calculus Khan Academy.mp3
|
So let's do that. So remember, we're just focusing on this part right over here. So the surface integral of r times k dot n, the dot product of k and n, ds can be rewritten as the surface integral over s2 of r times k dot n, ds, plus the surface integral over s1 of r times k dot n, ds, plus the surface integral over surface three of the same thing, r times k dot n, ds. Now, the way I've drawn it, and this is actually the case, s3 is if those surfaces don't touch each other. And for a type one situation right over here, the normal vector at any given point on this side of the cylinder for this type one region, if there is this in-between region, there always isn't. In a sphere, there wouldn't be the surface, in which case this would be 0. But if there is this surface in a type one region, the one that essentially connects the boundaries of the top and the bottom, then the normal vector will never have a k component.
|
Divergence theorem proof (part 2) Divergence theorem Multivariable Calculus Khan Academy.mp3
|
Now, the way I've drawn it, and this is actually the case, s3 is if those surfaces don't touch each other. And for a type one situation right over here, the normal vector at any given point on this side of the cylinder for this type one region, if there is this in-between region, there always isn't. In a sphere, there wouldn't be the surface, in which case this would be 0. But if there is this surface in a type one region, the one that essentially connects the boundaries of the top and the bottom, then the normal vector will never have a k component. The normal vector will always be pointing flat out. It will only have an i and j component. So if you take this normal vector right over here, does not have a k component, and you are dotting it with a k vector, then the dot product of two things that are orthogonal, the k vector, goes like that, you're going to get 0.
|
Divergence theorem proof (part 2) Divergence theorem Multivariable Calculus Khan Academy.mp3
|
But if there is this surface in a type one region, the one that essentially connects the boundaries of the top and the bottom, then the normal vector will never have a k component. The normal vector will always be pointing flat out. It will only have an i and j component. So if you take this normal vector right over here, does not have a k component, and you are dotting it with a k vector, then the dot product of two things that are orthogonal, the k vector, goes like that, you're going to get 0. So this thing is going to be 0, because k dot n is going to be 0 in this situation for this surface. k dot n is going to be equal to 0. So this part right over here simplifies to this right over here.
|
Example of closed line integral of conservative field Multivariable Calculus Khan Academy.mp3
|
Let's see if we can apply some of our new tools to solve some line integrals. So let's say we have a line integral along a closed curve. I'm going to define the path in a second. Of x squared plus y squared times dx plus 2xy times dy. And then our curve C is going to be defined by the parameterization x is equal to cosine of t and y is equal to sine of t. And this is valid for t between 0 and 2pi. So this is essentially a circle, a unit circle in the xy plane. We know how to solve these, but let's see if we can use some of our discoveries in the last couple of videos to maybe simplify this process.
|
Example of closed line integral of conservative field Multivariable Calculus Khan Academy.mp3
|
Of x squared plus y squared times dx plus 2xy times dy. And then our curve C is going to be defined by the parameterization x is equal to cosine of t and y is equal to sine of t. And this is valid for t between 0 and 2pi. So this is essentially a circle, a unit circle in the xy plane. We know how to solve these, but let's see if we can use some of our discoveries in the last couple of videos to maybe simplify this process. So the first thing you might say is, hey, this looks like a line integral, but you have a dx and dy. I don't see a dot dr here. It's not clear to me that this is some type of even a vector line integral.
|
Example of closed line integral of conservative field Multivariable Calculus Khan Academy.mp3
|
We know how to solve these, but let's see if we can use some of our discoveries in the last couple of videos to maybe simplify this process. So the first thing you might say is, hey, this looks like a line integral, but you have a dx and dy. I don't see a dot dr here. It's not clear to me that this is some type of even a vector line integral. I don't see any vectors. What I want to do first, and the reason why I wanted to show you this example, is just to show you that this is just another form of writing, really, a vector line integral. And to show you that, you just have to realize if I have some r of t, this is our curve, and I won't even write these functions in there.
|
Example of closed line integral of conservative field Multivariable Calculus Khan Academy.mp3
|
It's not clear to me that this is some type of even a vector line integral. I don't see any vectors. What I want to do first, and the reason why I wanted to show you this example, is just to show you that this is just another form of writing, really, a vector line integral. And to show you that, you just have to realize if I have some r of t, this is our curve, and I won't even write these functions in there. I'm just going to write it's x of t times i plus y of t times j. We've seen several videos now that we can write dr dt as being equal to dx dt times i plus dy dt times j. We've seen this multiple times.
|
Example of closed line integral of conservative field Multivariable Calculus Khan Academy.mp3
|
And to show you that, you just have to realize if I have some r of t, this is our curve, and I won't even write these functions in there. I'm just going to write it's x of t times i plus y of t times j. We've seen several videos now that we can write dr dt as being equal to dx dt times i plus dy dt times j. We've seen this multiple times. And we've seen multiple times we want to get the differential dr. We could just multiply everything times dt. And normally, I just put a dt here and a dt there and get rid of this dt. But if you multiply everything times dt, if you view this as the differentials as actual numbers you can multiply, normally you can treat them like that, then you just get rid of all of the dt.
|
Example of closed line integral of conservative field Multivariable Calculus Khan Academy.mp3
|
We've seen this multiple times. And we've seen multiple times we want to get the differential dr. We could just multiply everything times dt. And normally, I just put a dt here and a dt there and get rid of this dt. But if you multiply everything times dt, if you view this as the differentials as actual numbers you can multiply, normally you can treat them like that, then you just get rid of all of the dt. So dr, you can imagine, is equal to dx times the unit vector i plus dy times the unit vector j. So put that aside, and you might already see a pattern here. So if we define our vector field f of x, y as being equal to x squared plus y squared i plus 2xyj, what is this thing?
|
Example of closed line integral of conservative field Multivariable Calculus Khan Academy.mp3
|
But if you multiply everything times dt, if you view this as the differentials as actual numbers you can multiply, normally you can treat them like that, then you just get rid of all of the dt. So dr, you can imagine, is equal to dx times the unit vector i plus dy times the unit vector j. So put that aside, and you might already see a pattern here. So if we define our vector field f of x, y as being equal to x squared plus y squared i plus 2xyj, what is this thing? What is this thing over here? What is f dot dr going to be? Dot products, you just multiply the corresponding components of our vectors and then add them up.
|
Example of closed line integral of conservative field Multivariable Calculus Khan Academy.mp3
|
So if we define our vector field f of x, y as being equal to x squared plus y squared i plus 2xyj, what is this thing? What is this thing over here? What is f dot dr going to be? Dot products, you just multiply the corresponding components of our vectors and then add them up. So it's going to be, if you take this f and dot it with that dr, you're going to get the i component, x squared plus y squared times that dx plus the j component, 2xy times that dy. That's the dot product. Notice, this thing right here, that thing right here, is identical to that thing right there.
|
Example of closed line integral of conservative field Multivariable Calculus Khan Academy.mp3
|
Dot products, you just multiply the corresponding components of our vectors and then add them up. So it's going to be, if you take this f and dot it with that dr, you're going to get the i component, x squared plus y squared times that dx plus the j component, 2xy times that dy. That's the dot product. Notice, this thing right here, that thing right here, is identical to that thing right there. So our line integral, just to put it in a form that we're familiar with, this is the same exact thing as the line integral over this curve C, this closed curve C, of this f, maybe I'll write it in that magenta color, or that actually is more of a purple or pink color, f dot this dr. That's what this line integral is, it's just a different way of writing. So now that you've seen it, in the future if you see it in this differential form, you'll immediately know, okay, there's one vector field, that this is its x component, this is its y component, dotting with dr. This is the x component of dr, the i component, and this is the y component, or the j component of the dr.
|
Example of closed line integral of conservative field Multivariable Calculus Khan Academy.mp3
|
Notice, this thing right here, that thing right here, is identical to that thing right there. So our line integral, just to put it in a form that we're familiar with, this is the same exact thing as the line integral over this curve C, this closed curve C, of this f, maybe I'll write it in that magenta color, or that actually is more of a purple or pink color, f dot this dr. That's what this line integral is, it's just a different way of writing. So now that you've seen it, in the future if you see it in this differential form, you'll immediately know, okay, there's one vector field, that this is its x component, this is its y component, dotting with dr. This is the x component of dr, the i component, and this is the y component, or the j component of the dr. So you immediately know what the vector field is that we're taking a line integral of. This is the x, that's the y. Let's ask ourselves a question, is f conservative?
|
Example of closed line integral of conservative field Multivariable Calculus Khan Academy.mp3
|
This is the x component of dr, the i component, and this is the y component, or the j component of the dr. So you immediately know what the vector field is that we're taking a line integral of. This is the x, that's the y. Let's ask ourselves a question, is f conservative? So is f equal to the gradient of some scalar field, we'll call it capital F. Is this the case? So let's assume it is, and see if we can solve for a scalar field whose gradient really is f. Then we know that f is conservative. And then if f is conservative, and this is the whole reason we want to do it, that means that any closed loop, any line integral over a closed curve of f, is going to be equal to 0.
|
Example of closed line integral of conservative field Multivariable Calculus Khan Academy.mp3
|
Let's ask ourselves a question, is f conservative? So is f equal to the gradient of some scalar field, we'll call it capital F. Is this the case? So let's assume it is, and see if we can solve for a scalar field whose gradient really is f. Then we know that f is conservative. And then if f is conservative, and this is the whole reason we want to do it, that means that any closed loop, any line integral over a closed curve of f, is going to be equal to 0. And we'd be done. So if we can show this, then the answer to this question, or this question, is going to be 0. We don't even have to mess with the cosine of t's and the sine of t's and all of that, and actually we don't even have to take anti-derivatives.
|
Example of closed line integral of conservative field Multivariable Calculus Khan Academy.mp3
|
And then if f is conservative, and this is the whole reason we want to do it, that means that any closed loop, any line integral over a closed curve of f, is going to be equal to 0. And we'd be done. So if we can show this, then the answer to this question, or this question, is going to be 0. We don't even have to mess with the cosine of t's and the sine of t's and all of that, and actually we don't even have to take anti-derivatives. So let's see if we can find an f whose gradient is equal to that right there. So in order for f's gradient to be that, that means that the partial derivative of our capital F with respect to x has got to be equal to that right there. It's got to be equal to x squared plus y squared.
|
Example of closed line integral of conservative field Multivariable Calculus Khan Academy.mp3
|
We don't even have to mess with the cosine of t's and the sine of t's and all of that, and actually we don't even have to take anti-derivatives. So let's see if we can find an f whose gradient is equal to that right there. So in order for f's gradient to be that, that means that the partial derivative of our capital F with respect to x has got to be equal to that right there. It's got to be equal to x squared plus y squared. And that also tells us that the partial derivative of capital F with respect to y has got to be equal to 2xy. And just as a review, if I have the gradient of any function of any scalar field is equal to the partial of f with respect to x times i plus the partial of capital F with respect to y times j. So that's why I'm just pattern matching.
|
Example of closed line integral of conservative field Multivariable Calculus Khan Academy.mp3
|
It's got to be equal to x squared plus y squared. And that also tells us that the partial derivative of capital F with respect to y has got to be equal to 2xy. And just as a review, if I have the gradient of any function of any scalar field is equal to the partial of f with respect to x times i plus the partial of capital F with respect to y times j. So that's why I'm just pattern matching. I'm just saying, well, gee, if this is the gradient of that, then this must be that, which I wrote down right here. And this must be that, which I wrote down here. So let's see if I can find an f that satisfies both of these constraints.
|
Example of closed line integral of conservative field Multivariable Calculus Khan Academy.mp3
|
So that's why I'm just pattern matching. I'm just saying, well, gee, if this is the gradient of that, then this must be that, which I wrote down right here. And this must be that, which I wrote down here. So let's see if I can find an f that satisfies both of these constraints. So we can just take the anti-derivative with respect to x on both sides. And if we take the anti-derivative with respect to x on both sides, remember, you just treat y like a constant or y squared like a constant. It's just a number.
|
Example of closed line integral of conservative field Multivariable Calculus Khan Academy.mp3
|
So let's see if I can find an f that satisfies both of these constraints. So we can just take the anti-derivative with respect to x on both sides. And if we take the anti-derivative with respect to x on both sides, remember, you just treat y like a constant or y squared like a constant. It's just a number. So then we could say that f is equal to the anti-derivative of x squared is x to the third over 3. And then the anti-derivative of y squared, remember, with respect to x. So you just treat it like a number.
|
Example of closed line integral of conservative field Multivariable Calculus Khan Academy.mp3
|
It's just a number. So then we could say that f is equal to the anti-derivative of x squared is x to the third over 3. And then the anti-derivative of y squared, remember, with respect to x. So you just treat it like a number. That could just be the number k. Or this could be the number 5. So this is just going to be that times x. So plus x times y squared.
|
Example of closed line integral of conservative field Multivariable Calculus Khan Academy.mp3
|
So you just treat it like a number. That could just be the number k. Or this could be the number 5. So this is just going to be that times x. So plus x times y squared. And then there could be some function of y here. So plus some, I don't know, I'll call it g of y. Because there could have been some function of y here.
|
End of preview. Expand
in Data Studio
No dataset card yet
- Downloads last month
- 39