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That's what we're doing. But what if associated with each of those little parallelograms we had some value, where that value is defined by some third function, f of x, y, z. So every parallelogram, it's super small, it's around a point, you could kind of say it's maybe the center of it, it doesn't have to be the center. It may be the center of it is some point in three-dimensional space. And if you use some other function, f of x, y, and z, you'll get the value of that point. And what we want to do is figure out what happens if for every one of those parallelograms we were to multiply it times the value of the function at that point. So we could write it this way.
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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It may be the center of it is some point in three-dimensional space. And if you use some other function, f of x, y, and z, you'll get the value of that point. And what we want to do is figure out what happens if for every one of those parallelograms we were to multiply it times the value of the function at that point. So we could write it this way. So this is where you can imagine the function is just 1. We're just multiplying each of the parallelograms by 1. But we could imagine we're multiplying each of the little parallelograms by f of x, y, and z d sigma.
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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So we could write it this way. So this is where you can imagine the function is just 1. We're just multiplying each of the parallelograms by 1. But we could imagine we're multiplying each of the little parallelograms by f of x, y, and z d sigma. That's going to be the exact same thing. Where this is each of the little parallelograms, we're just going to multiply it by f of x, y, and z there. So we're going to integrate it over the area, over that region of f of x, y, and z, and then times the magnitude of the partial of r with respect to s, crossed with the partial of r with respect to t, ds dt.
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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But we could imagine we're multiplying each of the little parallelograms by f of x, y, and z d sigma. That's going to be the exact same thing. Where this is each of the little parallelograms, we're just going to multiply it by f of x, y, and z there. So we're going to integrate it over the area, over that region of f of x, y, and z, and then times the magnitude of the partial of r with respect to s, crossed with the partial of r with respect to t, ds dt. And of course, we're integrating with respect to s and t. Hopefully, we can express this function in terms of s and t. And we should be able to, because we have a parametrization there. Wherever we see an x there, it's really x is a function of s and t. y is a function of s and t. z is a function of s and t. And this might look super convoluted and hard. And the visualizations for this, of why you'd want to do this, it has applications in physics.
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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So we're going to integrate it over the area, over that region of f of x, y, and z, and then times the magnitude of the partial of r with respect to s, crossed with the partial of r with respect to t, ds dt. And of course, we're integrating with respect to s and t. Hopefully, we can express this function in terms of s and t. And we should be able to, because we have a parametrization there. Wherever we see an x there, it's really x is a function of s and t. y is a function of s and t. z is a function of s and t. And this might look super convoluted and hard. And the visualizations for this, of why you'd want to do this, it has applications in physics. It's a little hard to visualize. It's easier just to visualize the straight up surface area. But we're going to see in the next few videos that it's a little hairy to calculate these problems, but they're not too hard to do.
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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If I have some region, so this is my region right over here. We'll call it r. And let's call the boundary of my region, let's call that c. And if I have some vector field in this region, so let me draw a vector field like this. If I draw a vector field just like that, our two-dimensional divergence theorem, which we really derived from Green's theorem, told us that the flux across our boundary of this region, so let me write that out. The flux across the boundary. So the flux is essentially going to be the vector field. It's going to be our vector field f dotted with the normal outward-facing vector. So the normal vector at any point is this outward-facing vector.
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3D divergence theorem intuition Divergence theorem Multivariable Calculus Khan Academy.mp3
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The flux across the boundary. So the flux is essentially going to be the vector field. It's going to be our vector field f dotted with the normal outward-facing vector. So the normal vector at any point is this outward-facing vector. So our vector field dotted with the normal-facing vector at our boundary times our little chunk of the boundary. If we were to sum them all up over the entire boundary, that's the same thing as summing up over the entire region. So let's sum up over the entire region.
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3D divergence theorem intuition Divergence theorem Multivariable Calculus Khan Academy.mp3
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So the normal vector at any point is this outward-facing vector. So our vector field dotted with the normal-facing vector at our boundary times our little chunk of the boundary. If we were to sum them all up over the entire boundary, that's the same thing as summing up over the entire region. So let's sum up over the entire region. So summing up over this entire region, each little chunk of area, dA, we could call that dx dy if we wanted to, if we're dealing in the xy domain right over here, but each little chunk of area times the divergence of f, which is really saying how much is that vector field pulling apart. So it's times the divergence of f. And hopefully it made intuitive sense. The way that I drew this vector field right over here, you see everything's kind of coming out.
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3D divergence theorem intuition Divergence theorem Multivariable Calculus Khan Academy.mp3
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So let's sum up over the entire region. So summing up over this entire region, each little chunk of area, dA, we could call that dx dy if we wanted to, if we're dealing in the xy domain right over here, but each little chunk of area times the divergence of f, which is really saying how much is that vector field pulling apart. So it's times the divergence of f. And hopefully it made intuitive sense. The way that I drew this vector field right over here, you see everything's kind of coming out. We could almost call this a source right here, where the vector field seems like it's popping out of there. This has positive divergence right over here. And so because of this, you actually see that the vector field at the boundary is actually going in the direction of the normal vector, pretty close to the direction of the normal vector.
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3D divergence theorem intuition Divergence theorem Multivariable Calculus Khan Academy.mp3
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The way that I drew this vector field right over here, you see everything's kind of coming out. We could almost call this a source right here, where the vector field seems like it's popping out of there. This has positive divergence right over here. And so because of this, you actually see that the vector field at the boundary is actually going in the direction of the normal vector, pretty close to the direction of the normal vector. So it makes sense. You have positive divergence, and this is going to be a positive value. The vector field is going, for the most part, in the direction of the normal vector.
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3D divergence theorem intuition Divergence theorem Multivariable Calculus Khan Academy.mp3
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And so because of this, you actually see that the vector field at the boundary is actually going in the direction of the normal vector, pretty close to the direction of the normal vector. So it makes sense. You have positive divergence, and this is going to be a positive value. The vector field is going, for the most part, in the direction of the normal vector. So the larger this is, the larger that is. So hopefully some intuitive sense. If you had another vector field, so let me draw another region, that looked like this.
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3D divergence theorem intuition Divergence theorem Multivariable Calculus Khan Academy.mp3
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The vector field is going, for the most part, in the direction of the normal vector. So the larger this is, the larger that is. So hopefully some intuitive sense. If you had another vector field, so let me draw another region, that looked like this. So I could draw a couple of situations. So one where there's very limited divergence, maybe it's just a constant vector. The vector field doesn't really change as you go in any given direction.
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3D divergence theorem intuition Divergence theorem Multivariable Calculus Khan Academy.mp3
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If you had another vector field, so let me draw another region, that looked like this. So I could draw a couple of situations. So one where there's very limited divergence, maybe it's just a constant vector. The vector field doesn't really change as you go in any given direction. Over here, you'll get positive fluxes. I don't know what the plural of flux is. You'll get positive fluxes because the vector field seems to be going in roughly the same direction as our normal vector.
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3D divergence theorem intuition Divergence theorem Multivariable Calculus Khan Academy.mp3
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The vector field doesn't really change as you go in any given direction. Over here, you'll get positive fluxes. I don't know what the plural of flux is. You'll get positive fluxes because the vector field seems to be going in roughly the same direction as our normal vector. But here you'll get a negative flux. So stuff is coming in here. If you imagine your vector field is essentially some type of mass density times volume, and we've thought about that before, this is showing how much stuff is coming in, and then stuff is coming out.
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3D divergence theorem intuition Divergence theorem Multivariable Calculus Khan Academy.mp3
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You'll get positive fluxes because the vector field seems to be going in roughly the same direction as our normal vector. But here you'll get a negative flux. So stuff is coming in here. If you imagine your vector field is essentially some type of mass density times volume, and we've thought about that before, this is showing how much stuff is coming in, and then stuff is coming out. So your net flux will be close to 0. Stuff is coming in and stuff is coming out. Here you're just saying, hey, stuff is constantly coming out of this surface.
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3D divergence theorem intuition Divergence theorem Multivariable Calculus Khan Academy.mp3
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If you imagine your vector field is essentially some type of mass density times volume, and we've thought about that before, this is showing how much stuff is coming in, and then stuff is coming out. So your net flux will be close to 0. Stuff is coming in and stuff is coming out. Here you're just saying, hey, stuff is constantly coming out of this surface. So hopefully this gives you a sense that here you have very low divergence and you would have a low flux, total aggregate flux going through your boundary. Here you have a high divergence and you would have a high aggregate flux. I could draw another situation.
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3D divergence theorem intuition Divergence theorem Multivariable Calculus Khan Academy.mp3
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Here you're just saying, hey, stuff is constantly coming out of this surface. So hopefully this gives you a sense that here you have very low divergence and you would have a low flux, total aggregate flux going through your boundary. Here you have a high divergence and you would have a high aggregate flux. I could draw another situation. So this is my region R. And let's say that we have negative divergence, or we could even call it convergence. Convergence isn't an actual technical term, but you can imagine if the vector field is converging within R, well, the divergence is going to be negative in this situation. It's actually converging, which is the opposite of diverging.
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3D divergence theorem intuition Divergence theorem Multivariable Calculus Khan Academy.mp3
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I could draw another situation. So this is my region R. And let's say that we have negative divergence, or we could even call it convergence. Convergence isn't an actual technical term, but you can imagine if the vector field is converging within R, well, the divergence is going to be negative in this situation. It's actually converging, which is the opposite of diverging. So the divergence is negative in this situation. And also the flux across the boundary is going to be negative, because as we see here, the way I drew it, across most of this boundary, the vector field is going in the opposite direction. It's going in the opposite direction as our normal vector at any point.
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3D divergence theorem intuition Divergence theorem Multivariable Calculus Khan Academy.mp3
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It's actually converging, which is the opposite of diverging. So the divergence is negative in this situation. And also the flux across the boundary is going to be negative, because as we see here, the way I drew it, across most of this boundary, the vector field is going in the opposite direction. It's going in the opposite direction as our normal vector at any point. So hopefully this gives you a sense of why there's this connection between the divergence over the region and the flux across the boundary. Well, now we're just going to extend this to three dimensions. And it's the exact same reasoning.
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3D divergence theorem intuition Divergence theorem Multivariable Calculus Khan Academy.mp3
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It's going in the opposite direction as our normal vector at any point. So hopefully this gives you a sense of why there's this connection between the divergence over the region and the flux across the boundary. Well, now we're just going to extend this to three dimensions. And it's the exact same reasoning. If we have a, and I'll define it a little bit more precisely in future videos, a simple solid region. So let me just draw it. And I'm going to try to draw it in three dimensions.
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3D divergence theorem intuition Divergence theorem Multivariable Calculus Khan Academy.mp3
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And it's the exact same reasoning. If we have a, and I'll define it a little bit more precisely in future videos, a simple solid region. So let me just draw it. And I'm going to try to draw it in three dimensions. So let's say it looks something like that. And one way to think about it is this is going to be a region that doesn't bend back on itself. If you have a region that bends back on itself, and we'll think about it in multiple ways, but out of all of the volumes of three dimensions that you can imagine, these are the ones that don't bend back on themselves.
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3D divergence theorem intuition Divergence theorem Multivariable Calculus Khan Academy.mp3
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And I'm going to try to draw it in three dimensions. So let's say it looks something like that. And one way to think about it is this is going to be a region that doesn't bend back on itself. If you have a region that bends back on itself, and we'll think about it in multiple ways, but out of all of the volumes of three dimensions that you can imagine, these are the ones that don't bend back on themselves. And there are some that you might not be able to imagine that would also not make the case. But even if you had ones that bent back on themselves, you could separate them out into other ones that don't. So here's just a simple solid region over here.
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3D divergence theorem intuition Divergence theorem Multivariable Calculus Khan Academy.mp3
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If you have a region that bends back on itself, and we'll think about it in multiple ways, but out of all of the volumes of three dimensions that you can imagine, these are the ones that don't bend back on themselves. And there are some that you might not be able to imagine that would also not make the case. But even if you had ones that bent back on themselves, you could separate them out into other ones that don't. So here's just a simple solid region over here. I'll make it look three dimensional. So maybe if it was transparent, you would see it like that. And then you see the front of it like that.
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3D divergence theorem intuition Divergence theorem Multivariable Calculus Khan Academy.mp3
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So here's just a simple solid region over here. I'll make it look three dimensional. So maybe if it was transparent, you would see it like that. And then you see the front of it like that. So it's this kind of elliptical, circular, blob-looking thing. So that could be the back of it. And then if you go to the front, it could look something like that.
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3D divergence theorem intuition Divergence theorem Multivariable Calculus Khan Academy.mp3
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And then you see the front of it like that. So it's this kind of elliptical, circular, blob-looking thing. So that could be the back of it. And then if you go to the front, it could look something like that. So this is our simple solid region. I'll call it R still, but we're dealing with a three dimensional. We are now dealing with a three dimensional region.
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3D divergence theorem intuition Divergence theorem Multivariable Calculus Khan Academy.mp3
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And then if you go to the front, it could look something like that. So this is our simple solid region. I'll call it R still, but we're dealing with a three dimensional. We are now dealing with a three dimensional region. And now the boundary of this is no longer a line. We're now in three dimensions. The boundary is a surface.
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3D divergence theorem intuition Divergence theorem Multivariable Calculus Khan Academy.mp3
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We are now dealing with a three dimensional region. And now the boundary of this is no longer a line. We're now in three dimensions. The boundary is a surface. So I'll call that S. S is the boundary of R. And now let's throw on a vector field here. Now this is a vector field in three dimensions. And now let's imagine that we actually have positive divergence of our vector field within this region right over here.
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3D divergence theorem intuition Divergence theorem Multivariable Calculus Khan Academy.mp3
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The boundary is a surface. So I'll call that S. S is the boundary of R. And now let's throw on a vector field here. Now this is a vector field in three dimensions. And now let's imagine that we actually have positive divergence of our vector field within this region right over here. So we have positive divergence. So you could imagine that it's kind of the vector field within the region. It's a source of the vector field.
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3D divergence theorem intuition Divergence theorem Multivariable Calculus Khan Academy.mp3
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And now let's imagine that we actually have positive divergence of our vector field within this region right over here. So we have positive divergence. So you could imagine that it's kind of the vector field within the region. It's a source of the vector field. Or the vector field is diverging out of the vector field. That's just the case I drew right over here. And the other thing we want to say about vector field S, it's oriented in a way that its normal vector is outward facing.
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3D divergence theorem intuition Divergence theorem Multivariable Calculus Khan Academy.mp3
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It's a source of the vector field. Or the vector field is diverging out of the vector field. That's just the case I drew right over here. And the other thing we want to say about vector field S, it's oriented in a way that its normal vector is outward facing. So outward normal vector. So it's oriented so that the surface, the normal vector, is like that. The other option is that you have an inward facing normal vector.
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3D divergence theorem intuition Divergence theorem Multivariable Calculus Khan Academy.mp3
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And the other thing we want to say about vector field S, it's oriented in a way that its normal vector is outward facing. So outward normal vector. So it's oriented so that the surface, the normal vector, is like that. The other option is that you have an inward facing normal vector. But we're assuming it's an outward facing n. Well then we just extrapolate this to three dimensions. We essentially say the flux across the surface. So the flux across the surface, you would take your vector field, dot it with the normal vector at the surface, and then multiply that times a little chunk of surface.
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3D divergence theorem intuition Divergence theorem Multivariable Calculus Khan Academy.mp3
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The other option is that you have an inward facing normal vector. But we're assuming it's an outward facing n. Well then we just extrapolate this to three dimensions. We essentially say the flux across the surface. So the flux across the surface, you would take your vector field, dot it with the normal vector at the surface, and then multiply that times a little chunk of surface. So multiply that times a little chunk of surface. And then sum it up along the whole surface. So sum it up.
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3D divergence theorem intuition Divergence theorem Multivariable Calculus Khan Academy.mp3
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So the flux across the surface, you would take your vector field, dot it with the normal vector at the surface, and then multiply that times a little chunk of surface. So multiply that times a little chunk of surface. And then sum it up along the whole surface. So sum it up. So it's going to be a surface integral. So this is flux across the surface. Is going to be equal to, if we were to sum up the divergence of f, if we were to sum up across the whole volume.
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3D divergence theorem intuition Divergence theorem Multivariable Calculus Khan Academy.mp3
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So sum it up. So it's going to be a surface integral. So this is flux across the surface. Is going to be equal to, if we were to sum up the divergence of f, if we were to sum up across the whole volume. So now if we're summing up things on every little chunk of volume over here in three dimensions, we're going to have to take integrals along each dimension. So it's going to be a triple integral over the region of the divergence of f. So we're going to say, what is the divergence of f at each point? And then multiply it times the volume of that little chunk.
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3D divergence theorem intuition Divergence theorem Multivariable Calculus Khan Academy.mp3
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Is going to be equal to, if we were to sum up the divergence of f, if we were to sum up across the whole volume. So now if we're summing up things on every little chunk of volume over here in three dimensions, we're going to have to take integrals along each dimension. So it's going to be a triple integral over the region of the divergence of f. So we're going to say, what is the divergence of f at each point? And then multiply it times the volume of that little chunk. And multiply it times the volume of that chunk, in the sense of how much is it totally diverging in that volume. And then you sum it up. That should be equal to the flux.
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3D divergence theorem intuition Divergence theorem Multivariable Calculus Khan Academy.mp3
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And then multiply it times the volume of that little chunk. And multiply it times the volume of that chunk, in the sense of how much is it totally diverging in that volume. And then you sum it up. That should be equal to the flux. It's completely analogous to what here. Here we had a flux across the line. We had essentially a two-dimensional, or I guess we could say it's a one-dimensional boundary.
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3D divergence theorem intuition Divergence theorem Multivariable Calculus Khan Academy.mp3
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That should be equal to the flux. It's completely analogous to what here. Here we had a flux across the line. We had essentially a two-dimensional, or I guess we could say it's a one-dimensional boundary. So flux across the curve. And here we have the flux across the surface. Here we were summing the divergence in the region.
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3D divergence theorem intuition Divergence theorem Multivariable Calculus Khan Academy.mp3
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We had essentially a two-dimensional, or I guess we could say it's a one-dimensional boundary. So flux across the curve. And here we have the flux across the surface. Here we were summing the divergence in the region. Here we're summing it in the volume. But it's the exact same logic. If you had a vector field like this that was fairly constant going through the surface, on one side you would have a negative flux.
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3D divergence theorem intuition Divergence theorem Multivariable Calculus Khan Academy.mp3
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Here we were summing the divergence in the region. Here we're summing it in the volume. But it's the exact same logic. If you had a vector field like this that was fairly constant going through the surface, on one side you would have a negative flux. On the other side you would have a positive flux. And they would roughly cancel out. And that makes sense because there would be no diverging going on.
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3D divergence theorem intuition Divergence theorem Multivariable Calculus Khan Academy.mp3
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If you had a vector field like this that was fairly constant going through the surface, on one side you would have a negative flux. On the other side you would have a positive flux. And they would roughly cancel out. And that makes sense because there would be no diverging going on. If you had a converging vector field where it's coming in, the flux would be negative because it's going in the opposite direction of the normal vector. And so the divergence would be negative as well. Because essentially the vector field would be converging.
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3D divergence theorem intuition Divergence theorem Multivariable Calculus Khan Academy.mp3
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And that makes sense because there would be no diverging going on. If you had a converging vector field where it's coming in, the flux would be negative because it's going in the opposite direction of the normal vector. And so the divergence would be negative as well. Because essentially the vector field would be converging. So hopefully this gives you an intuition of what the divergence theorem is actually saying something very, very, very, very almost common sense or intuitive. And now in the next few videos, we can do some worked examples just so you feel comfortable computing or manipulating these integrals. And then we'll do a couple of proof videos where we actually prove the divergence theorem.
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3D divergence theorem intuition Divergence theorem Multivariable Calculus Khan Academy.mp3
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It's a scalar valued function. The question is, how do we take the derivative of an expression like this? And there's a certain method called a partial derivative which is very similar to ordinary derivatives and I kinda wanna show how they're secretly the same thing. So to do that, let me just remind ourselves of how we interpret the notation for ordinary derivatives. So if you have something like f of x is equal to x squared, and let's say you wanna take its derivative, and I'll use the Leibniz notation here, df dx, and let's evaluate it at two, let's say. I really like this notation because it's suggestive of what's going on. If we sketch out a graph, so you know, this axis represents our output, this over here represents our input, and x squared has a certain parabolic shape to it, something like that.
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Partial derivatives, introduction.mp3
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So to do that, let me just remind ourselves of how we interpret the notation for ordinary derivatives. So if you have something like f of x is equal to x squared, and let's say you wanna take its derivative, and I'll use the Leibniz notation here, df dx, and let's evaluate it at two, let's say. I really like this notation because it's suggestive of what's going on. If we sketch out a graph, so you know, this axis represents our output, this over here represents our input, and x squared has a certain parabolic shape to it, something like that. Then we go to the input, x equals one, two. This little dx here, I like to interpret as just a little nudge in the x direction, and it's kinda the size of that nudge. And then df, df, is the resulting change in the output after you make that initial little nudge.
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Partial derivatives, introduction.mp3
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If we sketch out a graph, so you know, this axis represents our output, this over here represents our input, and x squared has a certain parabolic shape to it, something like that. Then we go to the input, x equals one, two. This little dx here, I like to interpret as just a little nudge in the x direction, and it's kinda the size of that nudge. And then df, df, is the resulting change in the output after you make that initial little nudge. So it's this resulting change. And when you're thinking in terms of graphs, this is slope, you kind of have this rise over run for your ratio between the tiny change of the output that's caused by a tiny change in the input, and of course this is dependent on where you start. Over here we have x equals two.
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Partial derivatives, introduction.mp3
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And then df, df, is the resulting change in the output after you make that initial little nudge. So it's this resulting change. And when you're thinking in terms of graphs, this is slope, you kind of have this rise over run for your ratio between the tiny change of the output that's caused by a tiny change in the input, and of course this is dependent on where you start. Over here we have x equals two. But you could also think about this without graphs if you really wanted to. You might just think about, you know, your input space is just a number line, and your output space also is just a number line, the output of f over here. And really you're just thinking of somehow mapping numbers from here onto the second line.
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Partial derivatives, introduction.mp3
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Over here we have x equals two. But you could also think about this without graphs if you really wanted to. You might just think about, you know, your input space is just a number line, and your output space also is just a number line, the output of f over here. And really you're just thinking of somehow mapping numbers from here onto the second line. And in that case, your initial nudge, your initial little dx, would be some nudge on that number line. And you're wondering how that influences the function itself. So maybe that causes a nudge that's, you know, four times as big.
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Partial derivatives, introduction.mp3
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And really you're just thinking of somehow mapping numbers from here onto the second line. And in that case, your initial nudge, your initial little dx, would be some nudge on that number line. And you're wondering how that influences the function itself. So maybe that causes a nudge that's, you know, four times as big. And that would mean your derivative is four at that point. So the reason that I'm talking about this is because over in the multivariable world, we can pretty much do the same thing. You know, you could write df dx, and interpret that as saying, hey, how does a tiny change in the input in the x direction influence the output?
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Partial derivatives, introduction.mp3
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So maybe that causes a nudge that's, you know, four times as big. And that would mean your derivative is four at that point. So the reason that I'm talking about this is because over in the multivariable world, we can pretty much do the same thing. You know, you could write df dx, and interpret that as saying, hey, how does a tiny change in the input in the x direction influence the output? But this time the way that you might visualize it, you'd be thinking of your input space, here I'll draw it down here, as the xy plane. So this time this is not gonna be graphing the function, this is every point on the plane is an input. And let's say you were evaluating this at a point like one, two.
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Partial derivatives, introduction.mp3
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You know, you could write df dx, and interpret that as saying, hey, how does a tiny change in the input in the x direction influence the output? But this time the way that you might visualize it, you'd be thinking of your input space, here I'll draw it down here, as the xy plane. So this time this is not gonna be graphing the function, this is every point on the plane is an input. And let's say you were evaluating this at a point like one, two. Okay? So in that case, so you'd go over to the input that's one, and then two. And then you'd say, okay, so this tiny nudge in the input, this tiny change dx, how does that influence the output?
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Partial derivatives, introduction.mp3
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And let's say you were evaluating this at a point like one, two. Okay? So in that case, so you'd go over to the input that's one, and then two. And then you'd say, okay, so this tiny nudge in the input, this tiny change dx, how does that influence the output? And in this case, the output, I mean, it's still just a number. So maybe we go off to the side here, and we draw just like a number line as our output. And somehow we're thinking about the function as mapping points on the plane to the number line.
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Partial derivatives, introduction.mp3
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And then you'd say, okay, so this tiny nudge in the input, this tiny change dx, how does that influence the output? And in this case, the output, I mean, it's still just a number. So maybe we go off to the side here, and we draw just like a number line as our output. And somehow we're thinking about the function as mapping points on the plane to the number line. So you'd say, okay, that's your dx, how much does it change the output? And maybe this time it changes it negatively, it depends on your function. And that would be your df.
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Partial derivatives, introduction.mp3
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And somehow we're thinking about the function as mapping points on the plane to the number line. So you'd say, okay, that's your dx, how much does it change the output? And maybe this time it changes it negatively, it depends on your function. And that would be your df. And you can also do this with the y variable, right? There's no reason that you can't say df dy, and evaluate at that same point, one, two, and interpret totally the same way. Except this time your dy would be a change in the y direction.
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Partial derivatives, introduction.mp3
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And that would be your df. And you can also do this with the y variable, right? There's no reason that you can't say df dy, and evaluate at that same point, one, two, and interpret totally the same way. Except this time your dy would be a change in the y direction. So maybe I should really emphasize here that that dx is a change in the x direction here, and that dy is a change in the y direction. And maybe when you change your f according to y, it does something different, right? Maybe the output increases, and it increases by a lot, it's more sensitive to y.
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Partial derivatives, introduction.mp3
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Except this time your dy would be a change in the y direction. So maybe I should really emphasize here that that dx is a change in the x direction here, and that dy is a change in the y direction. And maybe when you change your f according to y, it does something different, right? Maybe the output increases, and it increases by a lot, it's more sensitive to y. Again, it depends on the function. And I'll show you how you can compute something like this in just a moment here. But first, there's kind of an annoying thing associated with partial derivatives, where we don't write them with ds, and dx, df.
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Partial derivatives, introduction.mp3
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Maybe the output increases, and it increases by a lot, it's more sensitive to y. Again, it depends on the function. And I'll show you how you can compute something like this in just a moment here. But first, there's kind of an annoying thing associated with partial derivatives, where we don't write them with ds, and dx, df. And a lot of people came up with this new notation, mostly just to emphasize to the reader of your equation that it's a multivariable function involved. And what you do is you say, you write a d, but it's got kind of a curl at the top. It's this new symbol, and people will often read it as partial, so you might read like partial f, partial y.
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Partial derivatives, introduction.mp3
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But first, there's kind of an annoying thing associated with partial derivatives, where we don't write them with ds, and dx, df. And a lot of people came up with this new notation, mostly just to emphasize to the reader of your equation that it's a multivariable function involved. And what you do is you say, you write a d, but it's got kind of a curl at the top. It's this new symbol, and people will often read it as partial, so you might read like partial f, partial y. If you're wondering, by the way, why we call these partial derivatives, it's sort of like this doesn't tell the full story of how f changes, because it only cares about the x direction, neither does this, this only cares about the y direction, each one is only a small part of the story. So let's actually evaluate something like this. I'm gonna go ahead and clear the board over here.
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Partial derivatives, introduction.mp3
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It's this new symbol, and people will often read it as partial, so you might read like partial f, partial y. If you're wondering, by the way, why we call these partial derivatives, it's sort of like this doesn't tell the full story of how f changes, because it only cares about the x direction, neither does this, this only cares about the y direction, each one is only a small part of the story. So let's actually evaluate something like this. I'm gonna go ahead and clear the board over here. I think the one-dimensional analogy is something we probably have already. So, ah, little remnants. So if you're actually evaluating something like this, here, I'll write it again up here.
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Partial derivatives, introduction.mp3
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I'm gonna go ahead and clear the board over here. I think the one-dimensional analogy is something we probably have already. So, ah, little remnants. So if you're actually evaluating something like this, here, I'll write it again up here. Partial derivative of f with respect to x, and we're doing it at one, two. It only cares about movement in the x direction, so it's treating y as a constant. It doesn't even care about the fact that y changes.
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Partial derivatives, introduction.mp3
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So if you're actually evaluating something like this, here, I'll write it again up here. Partial derivative of f with respect to x, and we're doing it at one, two. It only cares about movement in the x direction, so it's treating y as a constant. It doesn't even care about the fact that y changes. As far as it's concerned, y is always equal to two. So we can just plug that in ahead of time. So I'm gonna say partial partial x, this is another way you might write it, put the expression in here, and I'll say x squared, but instead of writing y, I'm just gonna plug in that constant ahead of time, because when you're only moving in the x direction, this is kind of how the multivariable function sees the world, and I'll just keep a little note that we're evaluating this whole thing at x equals one.
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Partial derivatives, introduction.mp3
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It doesn't even care about the fact that y changes. As far as it's concerned, y is always equal to two. So we can just plug that in ahead of time. So I'm gonna say partial partial x, this is another way you might write it, put the expression in here, and I'll say x squared, but instead of writing y, I'm just gonna plug in that constant ahead of time, because when you're only moving in the x direction, this is kind of how the multivariable function sees the world, and I'll just keep a little note that we're evaluating this whole thing at x equals one. And here, this is actually just an ordinary derivative. This is an expression that's an x. You're asking how it changes as you shift around x, and you know how to do this.
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Partial derivatives, introduction.mp3
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So I'm gonna say partial partial x, this is another way you might write it, put the expression in here, and I'll say x squared, but instead of writing y, I'm just gonna plug in that constant ahead of time, because when you're only moving in the x direction, this is kind of how the multivariable function sees the world, and I'll just keep a little note that we're evaluating this whole thing at x equals one. And here, this is actually just an ordinary derivative. This is an expression that's an x. You're asking how it changes as you shift around x, and you know how to do this. This is just taking the derivative, the derivative of x squared times two is gonna be four x, because x squared goes to two x, and then the derivative of a constant, sine of two, is just a constant, is zero. And of course, we're evaluating this at x equals one, so your overall answer is gonna be four. And just for practice, let's also do that with the derivative with respect to y.
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Partial derivatives, introduction.mp3
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You're asking how it changes as you shift around x, and you know how to do this. This is just taking the derivative, the derivative of x squared times two is gonna be four x, because x squared goes to two x, and then the derivative of a constant, sine of two, is just a constant, is zero. And of course, we're evaluating this at x equals one, so your overall answer is gonna be four. And just for practice, let's also do that with the derivative with respect to y. So we look over here, I'm gonna write the same thing. You're taking the partial derivative of f with respect to y. We're evaluating it at the same point, one, two.
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Partial derivatives, introduction.mp3
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And just for practice, let's also do that with the derivative with respect to y. So we look over here, I'm gonna write the same thing. You're taking the partial derivative of f with respect to y. We're evaluating it at the same point, one, two. This time, it doesn't care about movement in the x direction, so as far as it's concerned, that x just stays constant at one, so we'd write one squared times y plus sine of y. Sine of y. And you're saying, oh, I'm keeping track of this at y equals two, so that's kind of your evaluating at y equals two.
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Partial derivatives, introduction.mp3
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We're evaluating it at the same point, one, two. This time, it doesn't care about movement in the x direction, so as far as it's concerned, that x just stays constant at one, so we'd write one squared times y plus sine of y. Sine of y. And you're saying, oh, I'm keeping track of this at y equals two, so that's kind of your evaluating at y equals two. When you take the derivative, this is just one times y, so the derivative is one. This over here, the derivative is cosine, cosine of y. Again, we're evaluating this whole thing at y equals two, so your overall answer, it would be one plus cosine of two.
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Partial derivatives, introduction.mp3
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And you're saying, oh, I'm keeping track of this at y equals two, so that's kind of your evaluating at y equals two. When you take the derivative, this is just one times y, so the derivative is one. This over here, the derivative is cosine, cosine of y. Again, we're evaluating this whole thing at y equals two, so your overall answer, it would be one plus cosine of two. I'm not sure what the value of cosine of two is off the top of my head, but that would be your answer. And this is a partial derivative at a point, but a lot of times, you're not asked to just compute it at a point. What you want is a general formula that tells you, hey, plug in any point x, y, and it should spit out the answer.
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Partial derivatives, introduction.mp3
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Again, we're evaluating this whole thing at y equals two, so your overall answer, it would be one plus cosine of two. I'm not sure what the value of cosine of two is off the top of my head, but that would be your answer. And this is a partial derivative at a point, but a lot of times, you're not asked to just compute it at a point. What you want is a general formula that tells you, hey, plug in any point x, y, and it should spit out the answer. So let me just kind of go over how you would do that. It's actually very similar, but this time, instead of plugging in the constant ahead of time, we just have to pretend that it's a constant. So let me make a little bit of space for ourselves here.
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Partial derivatives, introduction.mp3
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What you want is a general formula that tells you, hey, plug in any point x, y, and it should spit out the answer. So let me just kind of go over how you would do that. It's actually very similar, but this time, instead of plugging in the constant ahead of time, we just have to pretend that it's a constant. So let me make a little bit of space for ourselves here. Really. We don't need any of this anymore. I'm gonna leave the partial partial f, partial partial y.
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Partial derivatives, introduction.mp3
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So let me make a little bit of space for ourselves here. Really. We don't need any of this anymore. I'm gonna leave the partial partial f, partial partial y. And we want this as a more general function of x and y. Well, we kind of do the same thing. We're gonna say that this is a derivative with respect to x, and I'm using partials just to kind of emphasize that it's a partial derivative.
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Partial derivatives, introduction.mp3
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I'm gonna leave the partial partial f, partial partial y. And we want this as a more general function of x and y. Well, we kind of do the same thing. We're gonna say that this is a derivative with respect to x, and I'm using partials just to kind of emphasize that it's a partial derivative. But now we'd write x squared, and then kind of emphasize that it's a constant value of y, plus the sine, and again, I'll say y. And here, I'm writing the variable y, but we have to pretend like it's a constant. Now, you're pretending that you plug in two or something like that, and you still just take the derivative.
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Partial derivatives, introduction.mp3
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We're gonna say that this is a derivative with respect to x, and I'm using partials just to kind of emphasize that it's a partial derivative. But now we'd write x squared, and then kind of emphasize that it's a constant value of y, plus the sine, and again, I'll say y. And here, I'm writing the variable y, but we have to pretend like it's a constant. Now, you're pretending that you plug in two or something like that, and you still just take the derivative. So in this case, the derivative of x squared times a constant is just two x times that constant, two x times that constant. And over here, the derivative of a constant is always zero. So that's just always gonna be zero.
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Partial derivatives, introduction.mp3
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Now, you're pretending that you plug in two or something like that, and you still just take the derivative. So in this case, the derivative of x squared times a constant is just two x times that constant, two x times that constant. And over here, the derivative of a constant is always zero. So that's just always gonna be zero. So this is your partial derivative as a more general formula. If you plugged in one, two to this, you'd get what we had before. And similarly, if you're doing this with partial f, partial y, we write down all of the same things.
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Partial derivatives, introduction.mp3
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So that's just always gonna be zero. So this is your partial derivative as a more general formula. If you plugged in one, two to this, you'd get what we had before. And similarly, if you're doing this with partial f, partial y, we write down all of the same things. Now, you're taking it with respect to y, and I'm just gonna copy this formula here, actually. But this time, we're considering all of the x's to be constants. So in this case, when you take the derivative with respect to y of some kind of constant, you know, constant squared is a constant, times y, it's just gonna equal that constant.
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Partial derivatives, introduction.mp3
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And similarly, if you're doing this with partial f, partial y, we write down all of the same things. Now, you're taking it with respect to y, and I'm just gonna copy this formula here, actually. But this time, we're considering all of the x's to be constants. So in this case, when you take the derivative with respect to y of some kind of constant, you know, constant squared is a constant, times y, it's just gonna equal that constant. So this is gonna be x squared. And over here, you're taking the derivative of sine of y. There's no x's in there.
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Partial derivatives, introduction.mp3
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So in this case, when you take the derivative with respect to y of some kind of constant, you know, constant squared is a constant, times y, it's just gonna equal that constant. So this is gonna be x squared. And over here, you're taking the derivative of sine of y. There's no x's in there. So that remains the sine of y. If you're being clear. And now, this is a more general formula.
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Partial derivatives, introduction.mp3
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There's no x's in there. So that remains the sine of y. If you're being clear. And now, this is a more general formula. If you plugged in one, two, you would get one. Oh, sorry, that's cosine of y. Cosine of y. Because we're taking a derivative.
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Partial derivatives, introduction.mp3
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And now, this is a more general formula. If you plugged in one, two, you would get one. Oh, sorry, that's cosine of y. Cosine of y. Because we're taking a derivative. So if you plugged in one, two, you know, you would get one plus the cosine of one, which is what we had before. So this, this is really what you'll see for how to compute a partial derivative. You pretend that one of the variables is constant, and you take an ordinary derivative.
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Partial derivatives, introduction.mp3
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Because we're taking a derivative. So if you plugged in one, two, you know, you would get one plus the cosine of one, which is what we had before. So this, this is really what you'll see for how to compute a partial derivative. You pretend that one of the variables is constant, and you take an ordinary derivative. And in the back of your mind, you're thinking this is because you're just moving in one direction for the input, and you're seeing how that influences things, and then, you know, you might move in one direction for another input, and see how that influences things. In the next video, I'll show you what this means in terms of graphs and slopes. But it's important to understand that graphs and slopes are not the only way to understand derivatives, because as soon as you start thinking about vector-valued functions, or functions with inputs of higher dimensions than just two, you can no longer think in terms of graphs.
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Partial derivatives, introduction.mp3
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Or if you prefer my kindergartener drawing skills, you can just look at the helix over here while we work it through. And the part where we left off, we have this tangent vector function, this unit tangent vector function for our curve, so at every given value t, whatever point that corresponds to on the curve, this function is going to give us the vector that is of unit length and tangent to the curve. And the ultimate goal for curvature is to find the derivative of that unit tangent vector with respect to arc length, and specifically we want its magnitude. And what that typically requires, and I talked about this in the videos on the formula, is you take its derivative with respect to the parameter t, because that's the thing you can actually do, and that might not correspond to unit length. If you nudge the parameter t, it might not nudge you a corresponding length on the curve, but you correct that by dividing out by the derivative of the parameterization function with respect to t. And that's actually arc length, the magnitude of the derivative of the parameterization with respect to t. Boy, is that a hard word. So with our given function, let's go ahead and start doing that. And I think first, given that we have this kind of fractions of fractions form, I'm just going to start by writing it a little bit more simply.
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Curvature of a helix, part 2.mp3
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And what that typically requires, and I talked about this in the videos on the formula, is you take its derivative with respect to the parameter t, because that's the thing you can actually do, and that might not correspond to unit length. If you nudge the parameter t, it might not nudge you a corresponding length on the curve, but you correct that by dividing out by the derivative of the parameterization function with respect to t. And that's actually arc length, the magnitude of the derivative of the parameterization with respect to t. Boy, is that a hard word. So with our given function, let's go ahead and start doing that. And I think first, given that we have this kind of fractions of fractions form, I'm just going to start by writing it a little bit more simply. So our unit tangent vector function, that first component where we're dividing by the square root of 26 divided by 5, instead I'm just going to write that as negative 5 times the sine of t, so I'm just kind of moving that 5 up into the numerator, divided by the root of 26. And I'll do the same thing for cosine, where we move that 5 up into the numerator. Cosine of t divided by the root of 26.
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Curvature of a helix, part 2.mp3
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And I think first, given that we have this kind of fractions of fractions form, I'm just going to start by writing it a little bit more simply. So our unit tangent vector function, that first component where we're dividing by the square root of 26 divided by 5, instead I'm just going to write that as negative 5 times the sine of t, so I'm just kind of moving that 5 up into the numerator, divided by the root of 26. And I'll do the same thing for cosine, where we move that 5 up into the numerator. Cosine of t divided by the root of 26. And then that last part, 1 5th divided by the square root of 26 5ths, gives us just 1 divided by the square root of 26. All right, so the first step in using our curvature formula is going to be to take the derivative of this guy, right? We need the derivative of the tangent vector function.
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Curvature of a helix, part 2.mp3
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Cosine of t divided by the root of 26. And then that last part, 1 5th divided by the square root of 26 5ths, gives us just 1 divided by the square root of 26. All right, so the first step in using our curvature formula is going to be to take the derivative of this guy, right? We need the derivative of the tangent vector function. So we go ahead and start doing that. We see that d big T, d little t, so tangent vector function parameter, is equal to, and we just take the derivative of each component. So negative 5 sine goes to negative 5 cosine, and we divide out by that constant, square root of 26.
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Curvature of a helix, part 2.mp3
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We need the derivative of the tangent vector function. So we go ahead and start doing that. We see that d big T, d little t, so tangent vector function parameter, is equal to, and we just take the derivative of each component. So negative 5 sine goes to negative 5 cosine, and we divide out by that constant, square root of 26. Similarly, 5 cosine goes to negative 5 sine, since the derivative of cosine is negative sine. Negative 5 sine divided by 26, or square root of 26. And that last component is just a constant, so derivative is nothing, it's zero.
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Curvature of a helix, part 2.mp3
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So negative 5 sine goes to negative 5 cosine, and we divide out by that constant, square root of 26. Similarly, 5 cosine goes to negative 5 sine, since the derivative of cosine is negative sine. Negative 5 sine divided by 26, or square root of 26. And that last component is just a constant, so derivative is nothing, it's zero. Next step, we're going to take the magnitude of that vector that we just found. So we're trying to find the magnitude of the derivative of the tangent vector function. So we say, all right, magnitude of what we just found, d big T, d little t, involves, so this magnitude will be the square root, make the little tick there, square root of the sum of the squares of these guys.
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Curvature of a helix, part 2.mp3
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And that last component is just a constant, so derivative is nothing, it's zero. Next step, we're going to take the magnitude of that vector that we just found. So we're trying to find the magnitude of the derivative of the tangent vector function. So we say, all right, magnitude of what we just found, d big T, d little t, involves, so this magnitude will be the square root, make the little tick there, square root of the sum of the squares of these guys. So the square of that first component is going to be 25 multiplied by cosine squared, cosine squared of T, all divided by 26, right, because the square root of the square, the square of the square root of 26 is 26. And then we add to that 25 times sine squared, sine squared of T, also divided by 26. And from that, we can factor out, factor out 25 over 26 inside that radical, because both terms involve multiplying by 25 and dividing by 26, and what we're left with is a nice and friendly cosine sine pair.
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Curvature of a helix, part 2.mp3
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So we say, all right, magnitude of what we just found, d big T, d little t, involves, so this magnitude will be the square root, make the little tick there, square root of the sum of the squares of these guys. So the square of that first component is going to be 25 multiplied by cosine squared, cosine squared of T, all divided by 26, right, because the square root of the square, the square of the square root of 26 is 26. And then we add to that 25 times sine squared, sine squared of T, also divided by 26. And from that, we can factor out, factor out 25 over 26 inside that radical, because both terms involve multiplying by 25 and dividing by 26, and what we're left with is a nice and friendly cosine sine pair. The reason we love things involving circles, this always happens, nice cancellation. This just becomes one, so what we're left with on the whole is root 25 over 26, pretty nice, root 25 over 26. So for our curvature equation, we go up and we can start plugging that in.
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Curvature of a helix, part 2.mp3
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And from that, we can factor out, factor out 25 over 26 inside that radical, because both terms involve multiplying by 25 and dividing by 26, and what we're left with is a nice and friendly cosine sine pair. The reason we love things involving circles, this always happens, nice cancellation. This just becomes one, so what we're left with on the whole is root 25 over 26, pretty nice, root 25 over 26. So for our curvature equation, we go up and we can start plugging that in. So we just found that numerator and we found that it was the square root of 25 divided by 26, the entire thing, 25 over 26. And we already found the magnitude of the derivative itself, that's one of the things we needed to do to find the tangent vector. That's where this 26 over 5 came from.
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Curvature of a helix, part 2.mp3
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So for our curvature equation, we go up and we can start plugging that in. So we just found that numerator and we found that it was the square root of 25 divided by 26, the entire thing, 25 over 26. And we already found the magnitude of the derivative itself, that's one of the things we needed to do to find the tangent vector. That's where this 26 over 5 came from. I deleted it from where we did last video to make room, but if you look at the last video, you can see where we got that square root of 26 divided by 5, and I'll actually write that as square root of 26 divided by 25. That's how we originally found it, I'm just putting the 5 back under the radical. And it's tempting if you aren't looking too closely to think these guys cancel out, but it's actually they're the opposite of each other, right?
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Curvature of a helix, part 2.mp3
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That's where this 26 over 5 came from. I deleted it from where we did last video to make room, but if you look at the last video, you can see where we got that square root of 26 divided by 5, and I'll actually write that as square root of 26 divided by 25. That's how we originally found it, I'm just putting the 5 back under the radical. And it's tempting if you aren't looking too closely to think these guys cancel out, but it's actually they're the opposite of each other, right? One is 25 over 26, the other is 26 over 25. So if we put everything under the radical there, I'm going to say equals the square root, and we're going to have 25 over 26 divided by 26 over 25. And when we flip that bottom and multiply, what we get is 25 squared divided by 26 squared.
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Curvature of a helix, part 2.mp3
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And it's tempting if you aren't looking too closely to think these guys cancel out, but it's actually they're the opposite of each other, right? One is 25 over 26, the other is 26 over 25. So if we put everything under the radical there, I'm going to say equals the square root, and we're going to have 25 over 26 divided by 26 over 25. And when we flip that bottom and multiply, what we get is 25 squared divided by 26 squared. And the root of that whole thing just gives us 25 over 26. And that is our curvature. That right there is the answer, that's the curvature.
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Curvature of a helix, part 2.mp3
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And when we flip that bottom and multiply, what we get is 25 squared divided by 26 squared. And the root of that whole thing just gives us 25 over 26. And that is our curvature. That right there is the answer, that's the curvature. So it's a little bit greater than 1. No, no, sorry, so it's a little bit less than 1, which means that you're curving a little bit less than you would if it was a circle with radius 1, which kind of makes sense if we look at the image here, because if the helix were completely flattened out, if you imagine squishing this down onto the xy-plane, you'd just be going around a circle with radius 1. But by kind of pulling that spring and pulling it so that there's a z component, you're making it a little bit more straight.
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Curvature of a helix, part 2.mp3
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That right there is the answer, that's the curvature. So it's a little bit greater than 1. No, no, sorry, so it's a little bit less than 1, which means that you're curving a little bit less than you would if it was a circle with radius 1, which kind of makes sense if we look at the image here, because if the helix were completely flattened out, if you imagine squishing this down onto the xy-plane, you'd just be going around a circle with radius 1. But by kind of pulling that spring and pulling it so that there's a z component, you're making it a little bit more straight. So the curvature should go down a little bit because it's becoming a little bit more straight. The radius of curvature will go up. So that's the curvature of a helix.
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Curvature of a helix, part 2.mp3
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But by kind of pulling that spring and pulling it so that there's a z component, you're making it a little bit more straight. So the curvature should go down a little bit because it's becoming a little bit more straight. The radius of curvature will go up. So that's the curvature of a helix. And that's a pretty good example of how you can find the curvature by walking through directly the idea of finding dT, dS, and, you know, getting that unit tangent vector, getting that little unit arc length. And in the next example, I think I'll go through one where you just use the formula, where it's something a little bit more complicated than thinking about this, and you turn to the formula itself. See you then.
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Curvature of a helix, part 2.mp3
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So let's say surface 2 can be represented by the vector position function. I'll just call it t, t for 2. So t, t, which is a vector, it's going to be a function of x and y. Those are going to be our parameters. And it's going to be equal to, and we can do this once again, because we're dealing with our surface is a function of x and y. So it's going to be equal to x times i plus y times j plus f2 of x, y times k for all the x, y's that are a member of our domain. Now with that out of the way, we can re-express what k dot n ds is.
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Divergence theorem proof (part 3) Divergence theorem Multivariable Calculus Khan Academy.mp3
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Those are going to be our parameters. And it's going to be equal to, and we can do this once again, because we're dealing with our surface is a function of x and y. So it's going to be equal to x times i plus y times j plus f2 of x, y times k for all the x, y's that are a member of our domain. Now with that out of the way, we can re-express what k dot n ds is. So let me write this over here. k dot n ds is equal to, and we could put parentheses here that we're going to take this dot product. That's at least how I like to think about it.
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Divergence theorem proof (part 3) Divergence theorem Multivariable Calculus Khan Academy.mp3
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Now with that out of the way, we can re-express what k dot n ds is. So let me write this over here. k dot n ds is equal to, and we could put parentheses here that we're going to take this dot product. That's at least how I like to think about it. This is the exact same thing as k dotted with the cross product of the partial of t with respect to x crossed with the partial of t with respect to y times a little chunk of our area, times da, a little chunk of our area in the x, y domain. We've done this multiple times as we evaluated surface integrals, and we got the intuition for why this works. So we're essentially just evaluating the surface integral.
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Divergence theorem proof (part 3) Divergence theorem Multivariable Calculus Khan Academy.mp3
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That's at least how I like to think about it. This is the exact same thing as k dotted with the cross product of the partial of t with respect to x crossed with the partial of t with respect to y times a little chunk of our area, times da, a little chunk of our area in the x, y domain. We've done this multiple times as we evaluated surface integrals, and we got the intuition for why this works. So we're essentially just evaluating the surface integral. And the one thing we want to make sure is make sure this has the right orientation. Because remember, in order for the divergence theorem to be true, the way we've defined it is all the normal vectors have to be outward facing. So for this top surface, the normal vector has to be pointing straight up.
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Divergence theorem proof (part 3) Divergence theorem Multivariable Calculus Khan Academy.mp3
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So we're essentially just evaluating the surface integral. And the one thing we want to make sure is make sure this has the right orientation. Because remember, in order for the divergence theorem to be true, the way we've defined it is all the normal vectors have to be outward facing. So for this top surface, the normal vector has to be pointing straight up. Or not necessarily straight up, at least upwards. If this is a curve, it wouldn't be necessarily straight up, but it needs to be kind of outward facing like that. On the sides, it would be outward facing like that.
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Divergence theorem proof (part 3) Divergence theorem Multivariable Calculus Khan Academy.mp3
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So for this top surface, the normal vector has to be pointing straight up. Or not necessarily straight up, at least upwards. If this is a curve, it wouldn't be necessarily straight up, but it needs to be kind of outward facing like that. On the sides, it would be outward facing like that. And down here, it would be outward facing, going in the general downwards direction. So let's just make sure that this is upwards facing. If we're changing with respect to x, we're going in this direction.
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Divergence theorem proof (part 3) Divergence theorem Multivariable Calculus Khan Academy.mp3
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On the sides, it would be outward facing like that. And down here, it would be outward facing, going in the general downwards direction. So let's just make sure that this is upwards facing. If we're changing with respect to x, we're going in this direction. Changing with respect to y, we're going in that direction. Take the right hand rule with the cross product. Index finger there, middle finger there, your left thumb, or your right thumb, I should say, will go straight up.
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Divergence theorem proof (part 3) Divergence theorem Multivariable Calculus Khan Academy.mp3
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If we're changing with respect to x, we're going in this direction. Changing with respect to y, we're going in that direction. Take the right hand rule with the cross product. Index finger there, middle finger there, your left thumb, or your right thumb, I should say, will go straight up. So it goes in the right direction. So this would be an upward pointing vector. So we got the right orientation for our surface.
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Divergence theorem proof (part 3) Divergence theorem Multivariable Calculus Khan Academy.mp3
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