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It's going to be minus 1 third plus minus 1 times 0 squared. So that's just going to be a 0. In both cases, the y is 0, so this term's going to disappear. So we can ignore that. And then we have minus F of 1, 0. So you put a 1 here, 1 to the third over 3, that is 1 third plus 1 times 0 squared, that's just 0. So this is going to be equal to minus 1 third minus 1 third is equal to minus 2 thirds.
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Second example of line integral of conservative vector field Multivariable Calculus Khan Academy.mp3
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So we can ignore that. And then we have minus F of 1, 0. So you put a 1 here, 1 to the third over 3, that is 1 third plus 1 times 0 squared, that's just 0. So this is going to be equal to minus 1 third minus 1 third is equal to minus 2 thirds. And we're done. And once again, because this is a conservative vector field and it's path independent, we really didn't have to mess with the cosines of t's and sines of t's when we actually took our antiderivative. We just had to find the potential function and evaluate it at the two endpoints to get the answer of our line integral, minus 2 thirds.
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Second example of line integral of conservative vector field Multivariable Calculus Khan Academy.mp3
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I'll write ds in magenta. So first let's conceptualize what this is even saying, and then let's manipulate it a little bit to see if we can come up with an interesting conclusion. We actually will manipulate it, we'll use Green's theorem, and we're going to actually come up with a two-dimensional version of the divergence theorem, which all sounds very complicated, but hopefully we can get a little bit of an intuition for it as to why it is actually a little bit of common sense. So first let's just think about this. Let me draw a coordinate plane here. So let me do it in white. So this right over here, that's our y-axis, that over there is our x-axis.
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2D divergence theorem Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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So first let's just think about this. Let me draw a coordinate plane here. So let me do it in white. So this right over here, that's our y-axis, that over there is our x-axis. Let me draw ourselves my curve. So my curve might look something like, I'll do it in the blue color. So my curve might look something like this, my contour, going in the positive counterclockwise direction, just like that.
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2D divergence theorem Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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So this right over here, that's our y-axis, that over there is our x-axis. Let me draw ourselves my curve. So my curve might look something like, I'll do it in the blue color. So my curve might look something like this, my contour, going in the positive counterclockwise direction, just like that. And now we have our vector field. And just a reminder, we've seen this multiple times. My vector field will associate a vector with any point on the xy-plane, and it can be defined as some function of x and y, actually I'll call that p, some function of x and y times the i unit vector, so it says what the i component of the vector field is for any x and y point, and then what the j component, or what we multiply the j component by, or the vertical component by, for any x and y point.
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2D divergence theorem Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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So my curve might look something like this, my contour, going in the positive counterclockwise direction, just like that. And now we have our vector field. And just a reminder, we've seen this multiple times. My vector field will associate a vector with any point on the xy-plane, and it can be defined as some function of x and y, actually I'll call that p, some function of x and y times the i unit vector, so it says what the i component of the vector field is for any x and y point, and then what the j component, or what we multiply the j component by, or the vertical component by, for any x and y point. So some function of x and y times i, plus some other scalar function of x and y times j. And so if you give me any point, there will be an associated vector with it. Any point, there's an associated vector, depending on how we define this function.
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2D divergence theorem Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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My vector field will associate a vector with any point on the xy-plane, and it can be defined as some function of x and y, actually I'll call that p, some function of x and y times the i unit vector, so it says what the i component of the vector field is for any x and y point, and then what the j component, or what we multiply the j component by, or the vertical component by, for any x and y point. So some function of x and y times i, plus some other scalar function of x and y times j. And so if you give me any point, there will be an associated vector with it. Any point, there's an associated vector, depending on how we define this function. But this expression right over here, we're taking a line integral. We care specifically about the points along this curve, along this contour right over here. And so let's think about what this is actually, what this piece right over here is actually telling us before we sum up all of these infinitesimally small pieces.
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2D divergence theorem Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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Any point, there's an associated vector, depending on how we define this function. But this expression right over here, we're taking a line integral. We care specifically about the points along this curve, along this contour right over here. And so let's think about what this is actually, what this piece right over here is actually telling us before we sum up all of these infinitesimally small pieces. So if we just take f dot n, so let's just think about a point on this curve. So a point on this curve, maybe this point right over here. So associated with that point, there is a vector.
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2D divergence theorem Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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And so let's think about what this is actually, what this piece right over here is actually telling us before we sum up all of these infinitesimally small pieces. So if we just take f dot n, so let's just think about a point on this curve. So a point on this curve, maybe this point right over here. So associated with that point, there is a vector. And that's what the vector field does. So f might look something like that. Right over there.
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2D divergence theorem Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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So associated with that point, there is a vector. And that's what the vector field does. So f might look something like that. Right over there. So that might be f at that point. And then we're going to dot it with the unit normal vector at that point. So the unit normal vector might look something like that.
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2D divergence theorem Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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Right over there. So that might be f at that point. And then we're going to dot it with the unit normal vector at that point. So the unit normal vector might look something like that. That would be n hat at that point. This is the vector field at that point. When you take the dot product, you get a scalar quantity.
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2D divergence theorem Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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So the unit normal vector might look something like that. That would be n hat at that point. This is the vector field at that point. When you take the dot product, you get a scalar quantity. You essentially just get a number, and you might remember it, and there are several videos where we go into detail about this. But that tells you how much those two vectors go together. It's essentially, if they're completely orthogonal to each other, you're going to get zero.
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2D divergence theorem Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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When you take the dot product, you get a scalar quantity. You essentially just get a number, and you might remember it, and there are several videos where we go into detail about this. But that tells you how much those two vectors go together. It's essentially, if they're completely orthogonal to each other, you're going to get zero. And if they go completely in the same direction, it's essentially you're just going to multiply their magnitudes times each other. And since you have a unit normal vector here, what this is essentially going to give you is the magnitude of the vector field f that goes in the normal direction. So think of it this way.
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2D divergence theorem Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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It's essentially, if they're completely orthogonal to each other, you're going to get zero. And if they go completely in the same direction, it's essentially you're just going to multiply their magnitudes times each other. And since you have a unit normal vector here, what this is essentially going to give you is the magnitude of the vector field f that goes in the normal direction. So think of it this way. So let's think about the component of this that goes in the normal direction. It might look something like that. This would be the component that goes in the tangential direction.
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2D divergence theorem Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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So think of it this way. So let's think about the component of this that goes in the normal direction. It might look something like that. This would be the component that goes in the tangential direction. So this expression right over here is going to give us the magnitude of this vector. Let me write this down. This right over here is the magnitude of the component of f in the normal direction, or in the same direction as that unit normal vector.
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2D divergence theorem Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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This would be the component that goes in the tangential direction. So this expression right over here is going to give us the magnitude of this vector. Let me write this down. This right over here is the magnitude of the component of f in the normal direction, or in the same direction as that unit normal vector. And then we're going to multiply that times a very infinitely small length of our contour, of our curve, right around that point. So we're going to multiply that thing times this right over here. And so you might say, well, OK, I kind of get what that is saying, but how could this ever be physically relevant?
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2D divergence theorem Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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This right over here is the magnitude of the component of f in the normal direction, or in the same direction as that unit normal vector. And then we're going to multiply that times a very infinitely small length of our contour, of our curve, right around that point. So we're going to multiply that thing times this right over here. And so you might say, well, OK, I kind of get what that is saying, but how could this ever be physically relevant? Or what could be the intuition for what this expression is even measuring? And to think about that, I always visualize this is in two dimensions. You will later do things like this in three dimensions.
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2D divergence theorem Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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And so you might say, well, OK, I kind of get what that is saying, but how could this ever be physically relevant? Or what could be the intuition for what this expression is even measuring? And to think about that, I always visualize this is in two dimensions. You will later do things like this in three dimensions. Imagine this is a two-dimensional universe, and we're studying gases. And so you have all these gas particles in a two-dimensional universe, so they only can have kind of an x and y coordinate. And this vector field is essentially telling you the velocity at any point there.
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2D divergence theorem Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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You will later do things like this in three dimensions. Imagine this is a two-dimensional universe, and we're studying gases. And so you have all these gas particles in a two-dimensional universe, so they only can have kind of an x and y coordinate. And this vector field is essentially telling you the velocity at any point there. So this is the velocity of the particles at this point. This is the velocity of the particles at that point. That is the velocity of the particles at that point.
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2D divergence theorem Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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And this vector field is essentially telling you the velocity at any point there. So this is the velocity of the particles at this point. This is the velocity of the particles at that point. That is the velocity of the particles at that point. And so when you take f right on this curve, that's the velocity of the particles at that point. They're going in that direction. When you dot it with n, it tells you essentially the speed going straight out, right at that point.
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2D divergence theorem Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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That is the velocity of the particles at that point. And so when you take f right on this curve, that's the velocity of the particles at that point. They're going in that direction. When you dot it with n, it tells you essentially the speed going straight out, right at that point. And when you multiply that times ds, you're essentially saying at any given moment, how fast or at any given, at that point right over there on the curve, how fast are the particles exiting the curve? And so if you were to sum up all of them, which is essentially what this integral is doing, with this line integral, it's essentially saying how fast are the particles exiting this contour? Or even entering the contour if you get a negative number.
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2D divergence theorem Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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When you dot it with n, it tells you essentially the speed going straight out, right at that point. And when you multiply that times ds, you're essentially saying at any given moment, how fast or at any given, at that point right over there on the curve, how fast are the particles exiting the curve? And so if you were to sum up all of them, which is essentially what this integral is doing, with this line integral, it's essentially saying how fast are the particles exiting this contour? Or even entering the contour if you get a negative number. But since we're taking the unit vector that goes, the unit normal vector that is outward pointing, it's saying how fast are they exiting this thing? If you get a negative number, that means there might be some net entrance. So this whole expression is, if you take that analogy, it doesn't have to have that physical representation.
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2D divergence theorem Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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Or even entering the contour if you get a negative number. But since we're taking the unit vector that goes, the unit normal vector that is outward pointing, it's saying how fast are they exiting this thing? If you get a negative number, that means there might be some net entrance. So this whole expression is, if you take that analogy, it doesn't have to have that physical representation. How fast are particles, our two-dimensional gas particle, exiting the contour? And in the future, you can do it in three dimensions where you have a surface and you can say how fast are things exiting that surface? And so let's start, now that we have hopefully a decent conceptual understanding of what this could represent, let's play around with it a little bit, because we know how to define a normal vector.
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2D divergence theorem Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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So this whole expression is, if you take that analogy, it doesn't have to have that physical representation. How fast are particles, our two-dimensional gas particle, exiting the contour? And in the future, you can do it in three dimensions where you have a surface and you can say how fast are things exiting that surface? And so let's start, now that we have hopefully a decent conceptual understanding of what this could represent, let's play around with it a little bit, because we know how to define a normal vector. So let's rewrite it using what we know about how to construct a normal vector. So if we rewrite it, our integral becomes this. We have our vector field F dotted with the normal vector.
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2D divergence theorem Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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And so let's start, now that we have hopefully a decent conceptual understanding of what this could represent, let's play around with it a little bit, because we know how to define a normal vector. So let's rewrite it using what we know about how to construct a normal vector. So if we rewrite it, our integral becomes this. We have our vector field F dotted with the normal vector. The normal vector we can write this way. A normal vector we saw was dy times i minus dx times j. And then we had to divide it by its magnitude in order to make it a unit normal vector.
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2D divergence theorem Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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We have our vector field F dotted with the normal vector. The normal vector we can write this way. A normal vector we saw was dy times i minus dx times j. And then we had to divide it by its magnitude in order to make it a unit normal vector. The magnitude was this right over here, dx squared plus dy squared, which is the same thing as ds, which is the exact same thing as that little mini arc length, that infinitely small length of our contour. So we're going to divide it by ds. And then we're going to multiply it times ds.
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2D divergence theorem Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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And then we had to divide it by its magnitude in order to make it a unit normal vector. The magnitude was this right over here, dx squared plus dy squared, which is the same thing as ds, which is the exact same thing as that little mini arc length, that infinitely small length of our contour. So we're going to divide it by ds. And then we're going to multiply it times ds. And ds is just a scalar quantity, and so we can actually even multiply this thing times ds before taking the dot product or vice versa. But these two things are going to cancel out. And so we're essentially left with F dot this thing right over here.
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2D divergence theorem Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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And then we're going to multiply it times ds. And ds is just a scalar quantity, and so we can actually even multiply this thing times ds before taking the dot product or vice versa. But these two things are going to cancel out. And so we're essentially left with F dot this thing right over here. But we have F defined right over here, so let's take the dot product. So I'll just write the line integral symbol again. We're going in the counterclockwise direction.
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2D divergence theorem Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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And so we're essentially left with F dot this thing right over here. But we have F defined right over here, so let's take the dot product. So I'll just write the line integral symbol again. We're going in the counterclockwise direction. And when we evaluate, and now let's pick a color that I have not used. Well, I've used many colors. So I'll do yellow again.
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2D divergence theorem Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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We're going in the counterclockwise direction. And when we evaluate, and now let's pick a color that I have not used. Well, I've used many colors. So I'll do yellow again. So now let's evaluate F dot this business. So dot product, fairly straightforward. You take the product of the x components, or essentially the magnitude of the x components.
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2D divergence theorem Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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So I'll do yellow again. So now let's evaluate F dot this business. So dot product, fairly straightforward. You take the product of the x components, or essentially the magnitude of the x components. So it's going to be P of xy times dy. And plus the product of the magnitudes of the y components, or the j components. So it's going to be plus Q of xy times minus dx, or times negative dx.
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2D divergence theorem Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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You take the product of the x components, or essentially the magnitude of the x components. So it's going to be P of xy times dy. And plus the product of the magnitudes of the y components, or the j components. So it's going to be plus Q of xy times minus dx, or times negative dx. Well, that's going to give us negative Q of xy times dx. So this is kind of an interesting statement. We've seen something like not too different than this before, and when we saw even just the definition of Green's Theorem.
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2D divergence theorem Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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So it's going to be plus Q of xy times minus dx, or times negative dx. Well, that's going to give us negative Q of xy times dx. So this is kind of an interesting statement. We've seen something like not too different than this before, and when we saw even just the definition of Green's Theorem. Let me rewrite it here just so we remember. So the definition, when we learned about Green's Theorem, it told us if we're taking a line integral over this contour, and there's multiple ways to write it, but one way that we often see it, and we've explored it already in our videos, is if we were to say if you have m times dx plus n times dy, this is equal to, and I'm just restating Green's Theorem right over here, this is equal to the double integral over the region that this contour surrounds of, and whatever function you have here times dy, you take the partial of that with respect to x. So you take the partial of this with respect to x, the partial of n with respect to x, and from that you subtract whatever was on the dx side, so the partial of m with respect to y, and we could say times dx dy, or you could say da, for the infinitesimally small little chunk of area.
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2D divergence theorem Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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We've seen something like not too different than this before, and when we saw even just the definition of Green's Theorem. Let me rewrite it here just so we remember. So the definition, when we learned about Green's Theorem, it told us if we're taking a line integral over this contour, and there's multiple ways to write it, but one way that we often see it, and we've explored it already in our videos, is if we were to say if you have m times dx plus n times dy, this is equal to, and I'm just restating Green's Theorem right over here, this is equal to the double integral over the region that this contour surrounds of, and whatever function you have here times dy, you take the partial of that with respect to x. So you take the partial of this with respect to x, the partial of n with respect to x, and from that you subtract whatever was on the dx side, so the partial of m with respect to y, and we could say times dx dy, or you could say da, for the infinitesimally small little chunk of area. So I'll just write, I could write da here. I'll write dx dy, or it could be dy dx. Actually, let me just write da here, since we're speaking in generalizations.
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2D divergence theorem Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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So you take the partial of this with respect to x, the partial of n with respect to x, and from that you subtract whatever was on the dx side, so the partial of m with respect to y, and we could say times dx dy, or you could say da, for the infinitesimally small little chunk of area. So I'll just write, I could write da here. I'll write dx dy, or it could be dy dx. Actually, let me just write da here, since we're speaking in generalizations. Where da is an infinitely small little chunk of area. So this right over here, this is just a restatement of Green's Theorem. So we already know this.
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2D divergence theorem Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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Actually, let me just write da here, since we're speaking in generalizations. Where da is an infinitely small little chunk of area. So this right over here, this is just a restatement of Green's Theorem. So we already know this. This is a restatement of Green's Theorem, and how can we apply it here? Well, it's the same thing. We have a little bit of sign differences, but we can apply Green's Theorem right over here.
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2D divergence theorem Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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So we already know this. This is a restatement of Green's Theorem, and how can we apply it here? Well, it's the same thing. We have a little bit of sign differences, but we can apply Green's Theorem right over here. This is going to be equal to the double integral over the region that this contour surrounds, and then what we want to do is we want to look at whatever is the function that's being multiplied times the dy, and in this case, this is the function that's being multiplied times the dy, and we want to take the partial of that with respect to x. So we're going to take the partial of p with respect to x. And then from that, we are going to subtract the other function, whatever is being multiplied times dx.
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2D divergence theorem Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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We have a little bit of sign differences, but we can apply Green's Theorem right over here. This is going to be equal to the double integral over the region that this contour surrounds, and then what we want to do is we want to look at whatever is the function that's being multiplied times the dy, and in this case, this is the function that's being multiplied times the dy, and we want to take the partial of that with respect to x. So we're going to take the partial of p with respect to x. And then from that, we are going to subtract the other function, whatever is being multiplied times dx. We're going to take the partial of that with respect to y. So here, we're going to take the partial of this whole thing with respect to y, but we have a negative out here. So it's going to be minus partial of q with respect to y, and then we have da, and obviously, these two negatives, subtracting a negative just gives you a positive.
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2D divergence theorem Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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And then from that, we are going to subtract the other function, whatever is being multiplied times dx. We're going to take the partial of that with respect to y. So here, we're going to take the partial of this whole thing with respect to y, but we have a negative out here. So it's going to be minus partial of q with respect to y, and then we have da, and obviously, these two negatives, subtracting a negative just gives you a positive. So then this is going to be equal to the double integral over the region, and maybe you might already see where this is going, maybe getting a little bit excited, the partial of p with respect to x plus the partial of q with respect to y, da. Now, what is this telling me, Sal? Well, look at this thing right over here.
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2D divergence theorem Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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So it's going to be minus partial of q with respect to y, and then we have da, and obviously, these two negatives, subtracting a negative just gives you a positive. So then this is going to be equal to the double integral over the region, and maybe you might already see where this is going, maybe getting a little bit excited, the partial of p with respect to x plus the partial of q with respect to y, da. Now, what is this telling me, Sal? Well, look at this thing right over here. p was originally the function that is telling us the magnitude in the x direction. q was telling us the magnitude in the y direction. We're taking the partial of this with respect to x.
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2D divergence theorem Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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Well, look at this thing right over here. p was originally the function that is telling us the magnitude in the x direction. q was telling us the magnitude in the y direction. We're taking the partial of this with respect to x. We're taking the partial of this with respect to y, and we're summing them. This is essentially, or this is exactly, the divergence of f. And if that doesn't make any sense, go watch the video on divergence. This right over here is the divergence of f. This is the divergence, by definition, really, this is the divergence of our vector field f. And so we have a very interesting thing.
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2D divergence theorem Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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We're taking the partial of this with respect to x. We're taking the partial of this with respect to y, and we're summing them. This is essentially, or this is exactly, the divergence of f. And if that doesn't make any sense, go watch the video on divergence. This right over here is the divergence of f. This is the divergence, by definition, really, this is the divergence of our vector field f. And so we have a very interesting thing. This thing that we saw, this original expression that we started studying, which is essentially saying, what's the speed at which the particles are exiting this surface? We now get it in terms of this little expression, and we'll interpret it in an intuitive way in a little bit. So this is equal to the divergence of f times dA over the whole region, so the double integral.
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2D divergence theorem Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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This right over here is the divergence of f. This is the divergence, by definition, really, this is the divergence of our vector field f. And so we have a very interesting thing. This thing that we saw, this original expression that we started studying, which is essentially saying, what's the speed at which the particles are exiting this surface? We now get it in terms of this little expression, and we'll interpret it in an intuitive way in a little bit. So this is equal to the divergence of f times dA over the whole region, so the double integral. So we're summing up the divergences of f times the infinitely small little chunk of area, and we're summing them up over the whole region. Now, why does that make intuitive sense? And for you to realize why it makes intuitive sense, you just have to remind yourself what the divergence is.
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2D divergence theorem Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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So this is equal to the divergence of f times dA over the whole region, so the double integral. So we're summing up the divergences of f times the infinitely small little chunk of area, and we're summing them up over the whole region. Now, why does that make intuitive sense? And for you to realize why it makes intuitive sense, you just have to remind yourself what the divergence is. Divergence is a measure of whether things are expanding or diverging or kind of contracting. If you have a point over here where around that the particles are kind of moving away from each other, you would have a positive divergence here. If you have a point, sometimes called a sink, where all of the particles are kind of condensing or converging, you would have a negative divergence here.
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2D divergence theorem Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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And for you to realize why it makes intuitive sense, you just have to remind yourself what the divergence is. Divergence is a measure of whether things are expanding or diverging or kind of contracting. If you have a point over here where around that the particles are kind of moving away from each other, you would have a positive divergence here. If you have a point, sometimes called a sink, where all of the particles are kind of condensing or converging, you would have a negative divergence here. So this should make a lot of sense. You take any little infinitely small area in this, and then you multiply that times the divergence there. So the more divergence here, you're going to get a larger number, and then you sum them up across the entire region.
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2D divergence theorem Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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If you have a point, sometimes called a sink, where all of the particles are kind of condensing or converging, you would have a negative divergence here. So this should make a lot of sense. You take any little infinitely small area in this, and then you multiply that times the divergence there. So the more divergence here, you're going to get a larger number, and then you sum them up across the entire region. So that makes a lot of sense. The more diverging that's going on through your region, obviously the more stuff is going to be exiting your boundary over here. So it actually makes complete sense, or hopefully it makes a little bit of complete sense, why if you were to see how fast are things exiting the surface, it's really the two-dimensional flux, how fast are things exiting the surface, and you take the sum over all of them, that's going to be the same thing as summing all of the divergences over this area that the contour is surrounding.
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2D divergence theorem Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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So the more divergence here, you're going to get a larger number, and then you sum them up across the entire region. So that makes a lot of sense. The more diverging that's going on through your region, obviously the more stuff is going to be exiting your boundary over here. So it actually makes complete sense, or hopefully it makes a little bit of complete sense, why if you were to see how fast are things exiting the surface, it's really the two-dimensional flux, how fast are things exiting the surface, and you take the sum over all of them, that's going to be the same thing as summing all of the divergences over this area that the contour is surrounding. So hopefully that makes a little bit of sense to you, and it's actually another way of kind of thinking about Green's theorem. We also have just explored what we've just said, this expression, that the divergence summed over this region over here is the same thing as the F dot N over the contour. That essentially is the two-dimensional divergence theorem.
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2D divergence theorem Line integrals and Green's theorem Multivariable Calculus Khan Academy.mp3
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So the first type of region, and it's appropriately named, we will call a type one region. Type one region. At first I'll give a formal definition. Hopefully the formal definition makes some intuitive sense. But then I'll draw a couple of type one regions, and then I'll show you what would not be a type one region, because sometimes that's the more important question. So type one region, maybe a type one region R, is the set, and these little curly brackets mean set, is the set of all x, y's, and z's. It's the set of all points in three dimensions, such that the x and y's are part of some domain.
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Type I regions in three dimensions Divergence theorem Multivariable Calculus Khan Academy.mp3
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Hopefully the formal definition makes some intuitive sense. But then I'll draw a couple of type one regions, and then I'll show you what would not be a type one region, because sometimes that's the more important question. So type one region, maybe a type one region R, is the set, and these little curly brackets mean set, is the set of all x, y's, and z's. It's the set of all points in three dimensions, such that the x and y's are part of some domain. So the x and y's are part of some domain, are a member. That's what this little symbol represents. Are a member of some domain, and z can essentially vary between two functions of x and y.
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Type I regions in three dimensions Divergence theorem Multivariable Calculus Khan Academy.mp3
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It's the set of all points in three dimensions, such that the x and y's are part of some domain. So the x and y's are part of some domain, are a member. That's what this little symbol represents. Are a member of some domain, and z can essentially vary between two functions of x and y. So let me write it over here. So f1 of x, y is kind of the lower bound on z. So this is going to be less than or equal to z, which is less than or equal to another function of x and y, which is going to be less than or equal to f2 of x and y.
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Type I regions in three dimensions Divergence theorem Multivariable Calculus Khan Academy.mp3
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Are a member of some domain, and z can essentially vary between two functions of x and y. So let me write it over here. So f1 of x, y is kind of the lower bound on z. So this is going to be less than or equal to z, which is less than or equal to another function of x and y, which is going to be less than or equal to f2 of x and y. And let me close the curly brackets to show that this was all a set. This is a set of x, y's and z's, and right here we are defining that set. So what would be a reasonable type one region?
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Type I regions in three dimensions Divergence theorem Multivariable Calculus Khan Academy.mp3
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So this is going to be less than or equal to z, which is less than or equal to another function of x and y, which is going to be less than or equal to f2 of x and y. And let me close the curly brackets to show that this was all a set. This is a set of x, y's and z's, and right here we are defining that set. So what would be a reasonable type one region? Well, a very simple type one region is a sphere. So let me draw a sphere right over here. So in a sphere, where it intersects the x, y plane, that's essentially this domain D right over here.
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Type I regions in three dimensions Divergence theorem Multivariable Calculus Khan Academy.mp3
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So what would be a reasonable type one region? Well, a very simple type one region is a sphere. So let me draw a sphere right over here. So in a sphere, where it intersects the x, y plane, that's essentially this domain D right over here. So I'll do it in blue. So let me draw my best attempt at drawing that domain. So this is the domain D right over here for a sphere.
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Type I regions in three dimensions Divergence theorem Multivariable Calculus Khan Academy.mp3
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So in a sphere, where it intersects the x, y plane, that's essentially this domain D right over here. So I'll do it in blue. So let me draw my best attempt at drawing that domain. So this is the domain D right over here for a sphere. This is a sphere centered at zero, but you can make the same argument for a sphere anywhere else. So that is my domain. And then f1 of x, y, which is the lower bound of z, will be the bottom half of the sphere.
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Type I regions in three dimensions Divergence theorem Multivariable Calculus Khan Academy.mp3
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So this is the domain D right over here for a sphere. This is a sphere centered at zero, but you can make the same argument for a sphere anywhere else. So that is my domain. And then f1 of x, y, which is the lower bound of z, will be the bottom half of the sphere. So you really can't see it well right over here, but these contours right over here would be on the bottom half. And I can even color in this part right over here. The bottom surface of our sphere would be f1 of x, y, and f2 of x, y, as you can imagine, will be the top half of the sphere, the top hemisphere.
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Type I regions in three dimensions Divergence theorem Multivariable Calculus Khan Academy.mp3
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And then f1 of x, y, which is the lower bound of z, will be the bottom half of the sphere. So you really can't see it well right over here, but these contours right over here would be on the bottom half. And I can even color in this part right over here. The bottom surface of our sphere would be f1 of x, y, and f2 of x, y, as you can imagine, will be the top half of the sphere, the top hemisphere. So it will look something like that. So this thing that I'm drawing right over here is definitely a type one region. As we'll see, this could be a type one, type two, or type three region, but it's definitely a type one region.
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Type I regions in three dimensions Divergence theorem Multivariable Calculus Khan Academy.mp3
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The bottom surface of our sphere would be f1 of x, y, and f2 of x, y, as you can imagine, will be the top half of the sphere, the top hemisphere. So it will look something like that. So this thing that I'm drawing right over here is definitely a type one region. As we'll see, this could be a type one, type two, or type three region, but it's definitely a type one region. Another example of a type one region, and actually this might even be more obvious. Let me draw some axes again, and let me draw some type of a cylinder. And just to make it clear that our domain, where the x, y plane does not have to be inside of our region, let's imagine a cylinder that is below, that is, well actually I'll draw it above, that is above the x, y plane.
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Type I regions in three dimensions Divergence theorem Multivariable Calculus Khan Academy.mp3
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As we'll see, this could be a type one, type two, or type three region, but it's definitely a type one region. Another example of a type one region, and actually this might even be more obvious. Let me draw some axes again, and let me draw some type of a cylinder. And just to make it clear that our domain, where the x, y plane does not have to be inside of our region, let's imagine a cylinder that is below, that is, well actually I'll draw it above, that is above the x, y plane. So this is the bottom of the cylinder, it's right over here. And once again, it doesn't have to be centered around the z axis, but I'll do it that way just for this video. Actually I can draw it a little bit better than that.
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Type I regions in three dimensions Divergence theorem Multivariable Calculus Khan Academy.mp3
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And just to make it clear that our domain, where the x, y plane does not have to be inside of our region, let's imagine a cylinder that is below, that is, well actually I'll draw it above, that is above the x, y plane. So this is the bottom of the cylinder, it's right over here. And once again, it doesn't have to be centered around the z axis, but I'll do it that way just for this video. Actually I can draw it a little bit better than that. So this is the bottom surface of our cylinder, and then the top surface of our cylinder might be right over here. And these things actually don't even have to be flat, they could actually be curvy in some way. And in this situation, so in this cylinder, let me draw it a little bit neater, in this cylinder right over here, our domain, all of the values that the x and y's can take on, so our domain is going to be this region right over here in the x, y plane.
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Type I regions in three dimensions Divergence theorem Multivariable Calculus Khan Academy.mp3
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Actually I can draw it a little bit better than that. So this is the bottom surface of our cylinder, and then the top surface of our cylinder might be right over here. And these things actually don't even have to be flat, they could actually be curvy in some way. And in this situation, so in this cylinder, let me draw it a little bit neater, in this cylinder right over here, our domain, all of the values that the x and y's can take on, so our domain is going to be this region right over here in the x, y plane. And then for each of those x, y pairs, f1 of x, y defines the bottom boundary of our region, so f1 of x, y is going to be this right over here. So you give me any of these x, y's in this domain D, and then you evaluate the function at those points, and it will correspond to this surface right over here. And then f2 of x, y, once again, give me any one of those x, y points in our domain, and you evaluate f2 at those points, and it will give you this surface up here.
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Type I regions in three dimensions Divergence theorem Multivariable Calculus Khan Academy.mp3
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And in this situation, so in this cylinder, let me draw it a little bit neater, in this cylinder right over here, our domain, all of the values that the x and y's can take on, so our domain is going to be this region right over here in the x, y plane. And then for each of those x, y pairs, f1 of x, y defines the bottom boundary of our region, so f1 of x, y is going to be this right over here. So you give me any of these x, y's in this domain D, and then you evaluate the function at those points, and it will correspond to this surface right over here. And then f2 of x, y, once again, give me any one of those x, y points in our domain, and you evaluate f2 at those points, and it will give you this surface up here. And we're saying that z will take on all the values in between, and so it is really this whole solid, it's really this entire solid area. Likewise, over here, z could take on any value between this magenta surface and this green surface, so it would essentially fill up our entire volume, so it would become a solid region. Now, you might be wondering, what would not be a type I region?
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Type I regions in three dimensions Divergence theorem Multivariable Calculus Khan Academy.mp3
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And then f2 of x, y, once again, give me any one of those x, y points in our domain, and you evaluate f2 at those points, and it will give you this surface up here. And we're saying that z will take on all the values in between, and so it is really this whole solid, it's really this entire solid area. Likewise, over here, z could take on any value between this magenta surface and this green surface, so it would essentially fill up our entire volume, so it would become a solid region. Now, you might be wondering, what would not be a type I region? So let's think about that. So it would essentially be something that we could not define in this way, and I'll try my best to draw it, but you can imagine a shape that does something funky like this. So there's like one big, I guess you could imagine a sideways dumbbell.
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Type I regions in three dimensions Divergence theorem Multivariable Calculus Khan Academy.mp3
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Now, you might be wondering, what would not be a type I region? So let's think about that. So it would essentially be something that we could not define in this way, and I'll try my best to draw it, but you can imagine a shape that does something funky like this. So there's like one big, I guess you could imagine a sideways dumbbell. So a sideways dumbbell, I'll maybe curve it out a little bit. So maybe it's, so this is kind of the top of the dumbbell, and then it, or an hourglass, I guess you could say, or a dumbbell, it would look something like that. So I'm trying my best to draw it.
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Type I regions in three dimensions Divergence theorem Multivariable Calculus Khan Academy.mp3
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So there's like one big, I guess you could imagine a sideways dumbbell. So a sideways dumbbell, I'll maybe curve it out a little bit. So maybe it's, so this is kind of the top of the dumbbell, and then it, or an hourglass, I guess you could say, or a dumbbell, it would look something like that. So I'm trying my best to draw it. It would look something like that, and the reason why this is not definable in this way, it becomes obvious if you kind of look at a cross-section of it. There's no way to define only two functions, that's a lower bound and an upper bound, in terms of z. So even if you say, hey, maybe my domain, my domain will be all of the xy values that can be taken on.
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Type I regions in three dimensions Divergence theorem Multivariable Calculus Khan Academy.mp3
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So I'm trying my best to draw it. It would look something like that, and the reason why this is not definable in this way, it becomes obvious if you kind of look at a cross-section of it. There's no way to define only two functions, that's a lower bound and an upper bound, in terms of z. So even if you say, hey, maybe my domain, my domain will be all of the xy values that can be taken on. Let me see how well I can draw this. So you say my xy values, let me try to draw this whole thing a little bit better, a better attempt. So you might say, okay, for something like a dumbbell, let me clear out that part as well.
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Type I regions in three dimensions Divergence theorem Multivariable Calculus Khan Academy.mp3
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So even if you say, hey, maybe my domain, my domain will be all of the xy values that can be taken on. Let me see how well I can draw this. So you say my xy values, let me try to draw this whole thing a little bit better, a better attempt. So you might say, okay, for something like a dumbbell, let me clear out that part as well. For something like a dumbbell, let me erase that. So for something like a dumbbell, maybe my domain is right over here. So these are all the xy values that you can take on.
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Type I regions in three dimensions Divergence theorem Multivariable Calculus Khan Academy.mp3
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So you might say, okay, for something like a dumbbell, let me clear out that part as well. For something like a dumbbell, let me erase that. So for something like a dumbbell, maybe my domain is right over here. So these are all the xy values that you can take on. But in order to have a dumbbell shape, for any one xy, z is going to take on, there's not just an upper and a lower bound, and z doesn't take on all values in between. Let me just draw it more clearly. So our dumbbell, maybe it's centered on the z axis, this is the middle of our dumbbell, and then it comes out like that, and then up here, the z axis, so it looks like that.
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Type I regions in three dimensions Divergence theorem Multivariable Calculus Khan Academy.mp3
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So these are all the xy values that you can take on. But in order to have a dumbbell shape, for any one xy, z is going to take on, there's not just an upper and a lower bound, and z doesn't take on all values in between. Let me just draw it more clearly. So our dumbbell, maybe it's centered on the z axis, this is the middle of our dumbbell, and then it comes out like that, and then up here, the z axis, so it looks like that. And then it goes below the xy plane, and it does kind of a similar thing. It goes below the xy plane and looks something like that. So notice, for any given xy, what would be, if you attempted to make it a type I region, you would say, well, maybe this is the top surface, and maybe you would say down here is the bottom surface.
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Type I regions in three dimensions Divergence theorem Multivariable Calculus Khan Academy.mp3
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So our dumbbell, maybe it's centered on the z axis, this is the middle of our dumbbell, and then it comes out like that, and then up here, the z axis, so it looks like that. And then it goes below the xy plane, and it does kind of a similar thing. It goes below the xy plane and looks something like that. So notice, for any given xy, what would be, if you attempted to make it a type I region, you would say, well, maybe this is the top surface, and maybe you would say down here is the bottom surface. Down here is the bottom surface. But notice, z can't take on every value in between. You kind of have to break this up if you wanted to be able to do something like that.
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Type I regions in three dimensions Divergence theorem Multivariable Calculus Khan Academy.mp3
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So notice, for any given xy, what would be, if you attempted to make it a type I region, you would say, well, maybe this is the top surface, and maybe you would say down here is the bottom surface. Down here is the bottom surface. But notice, z can't take on every value in between. You kind of have to break this up if you wanted to be able to do something like that. You would have to break this up into two separate regions, where this would be the bottom region, and then this right over here would be another top region. So this dumbbell shape itself is not a type I region, but you could actually break it up into two separate type I regions. So hopefully that helps out.
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Type I regions in three dimensions Divergence theorem Multivariable Calculus Khan Academy.mp3
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You kind of have to break this up if you wanted to be able to do something like that. You would have to break this up into two separate regions, where this would be the bottom region, and then this right over here would be another top region. So this dumbbell shape itself is not a type I region, but you could actually break it up into two separate type I regions. So hopefully that helps out. And actually, another way to think about it, and this might be an easier way, if we were to look at it from this direction, and if we were to just think about the zy, if we were to just think about what's happening on the zy plane, so that's z, and this is y right over here, our dumbbell shape would look something like this. My best attempt to draw our dumbbell shape. So if you get a given x or y, maybe x is even zero, and you're sitting right here on the y-axis, notice z is not, even up here, cannot be a function of just y.
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Type I regions in three dimensions Divergence theorem Multivariable Calculus Khan Academy.mp3
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So hopefully that helps out. And actually, another way to think about it, and this might be an easier way, if we were to look at it from this direction, and if we were to just think about the zy, if we were to just think about what's happening on the zy plane, so that's z, and this is y right over here, our dumbbell shape would look something like this. My best attempt to draw our dumbbell shape. So if you get a given x or y, maybe x is even zero, and you're sitting right here on the y-axis, notice z is not, even up here, cannot be a function of just y. On this top part, there's two possible z values that we need to take on for that given y. Two possible z values for that given y. So you can't define it simply in terms of just one lower bound function and one upper bound function.
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Type I regions in three dimensions Divergence theorem Multivariable Calculus Khan Academy.mp3
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So if you take this guy, how do you compute the curl of that vector valued function? So what you do, as I mentioned in the last video, is you imagine taking this del operator and taking the cross product between that and your vector valued function. And what that means when you expand it is that del operator, you just kind of fill it with partial differential operators, you could say, but really it's just the symbol partial partial x, partial partial y, partial partial z. And these are things that are just waiting to take in some kind of function. So we're gonna take the cross product between that and the function that we have defined here. So let me just actually copy it over. Copy it over here.
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3d curl computation example.mp3
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And these are things that are just waiting to take in some kind of function. So we're gonna take the cross product between that and the function that we have defined here. So let me just actually copy it over. Copy it over here. And a little residue. And to compute this cross product, we take a certain determinant. So I'm gonna write over here determinant, and it's gonna be of a three by three matrix, but really it's kind of like a quote unquote matrix because each component has something funky.
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3d curl computation example.mp3
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Copy it over here. And a little residue. And to compute this cross product, we take a certain determinant. So I'm gonna write over here determinant, and it's gonna be of a three by three matrix, but really it's kind of like a quote unquote matrix because each component has something funky. So the top row, just like we would have with any other cross product that we're computing, is gonna have i, j, and k, these unit vectors in three dimensional space. And the second row here is gonna have all of these partial differential operators since that's the first vector in our cross product. So that's partial partial x, partial partial y, and again, all of these are just kind of waiting to be given a function that they can take the derivative of.
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3d curl computation example.mp3
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So I'm gonna write over here determinant, and it's gonna be of a three by three matrix, but really it's kind of like a quote unquote matrix because each component has something funky. So the top row, just like we would have with any other cross product that we're computing, is gonna have i, j, and k, these unit vectors in three dimensional space. And the second row here is gonna have all of these partial differential operators since that's the first vector in our cross product. So that's partial partial x, partial partial y, and again, all of these are just kind of waiting to be given a function that they can take the derivative of. And then that third row is gonna be the functions that we have, so the first component here is x, y. The second component is cosine of z, cosine of z. And then that final component is z squared plus y, z squared plus y.
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3d curl computation example.mp3
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So that's partial partial x, partial partial y, and again, all of these are just kind of waiting to be given a function that they can take the derivative of. And then that third row is gonna be the functions that we have, so the first component here is x, y. The second component is cosine of z, cosine of z. And then that final component is z squared plus y, z squared plus y. So I'll give some room here, maybe make it more visible. So this is the determinant we need to compute, and this is gonna be broken up into three different parts. The first one, we take this top part, i, and multiply it by the determinant of this sub matrix.
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3d curl computation example.mp3
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And then that final component is z squared plus y, z squared plus y. So I'll give some room here, maybe make it more visible. So this is the determinant we need to compute, and this is gonna be broken up into three different parts. The first one, we take this top part, i, and multiply it by the determinant of this sub matrix. So when we do that, this sub-determinant, we're taking partial derivative with respect to y of z squared plus y. Now as far as y is concerned, z looks like a constant. So z squared is a constant, and the partial derivative of this entire guy is just one.
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3d curl computation example.mp3
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The first one, we take this top part, i, and multiply it by the determinant of this sub matrix. So when we do that, this sub-determinant, we're taking partial derivative with respect to y of z squared plus y. Now as far as y is concerned, z looks like a constant. So z squared is a constant, and the partial derivative of this entire guy is just one. So that'll look like one. And then we're subtracting off the partial derivative with respect to z of cosine of z. And that just looks the same as a derivative of cosine z, which is negative sine.
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3d curl computation example.mp3
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So z squared is a constant, and the partial derivative of this entire guy is just one. So that'll look like one. And then we're subtracting off the partial derivative with respect to z of cosine of z. And that just looks the same as a derivative of cosine z, which is negative sine. So that's negative sine of z. So that's the first part. And then as the next part, we're gonna take j, but we're subtracting, because you're always kind of thinking plus, minus, plus when you're doing these determinants.
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3d curl computation example.mp3
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And that just looks the same as a derivative of cosine z, which is negative sine. So that's negative sine of z. So that's the first part. And then as the next part, we're gonna take j, but we're subtracting, because you're always kind of thinking plus, minus, plus when you're doing these determinants. So we're gonna subtract off j, multiply it by its own little sub-determinant, and this time the sub-determinant is gonna involve the two columns that it's not part of. So you're imagining this first column and this second column as being part of a matrix. So the first thing you do is you take this partial derivative with respect to x of z squared plus y.
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3d curl computation example.mp3
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And then as the next part, we're gonna take j, but we're subtracting, because you're always kind of thinking plus, minus, plus when you're doing these determinants. So we're gonna subtract off j, multiply it by its own little sub-determinant, and this time the sub-determinant is gonna involve the two columns that it's not part of. So you're imagining this first column and this second column as being part of a matrix. So the first thing you do is you take this partial derivative with respect to x of z squared plus y. Well, no x's show up there, right? That's z squared and y each look like constants as far as x is concerned. So that's zero.
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3d curl computation example.mp3
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So the first thing you do is you take this partial derivative with respect to x of z squared plus y. Well, no x's show up there, right? That's z squared and y each look like constants as far as x is concerned. So that's zero. Then we take the partial with respect to z of x times y. And again, there's no z that shows up there, so that's also zero. So we're kind of subtracting off zero.
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3d curl computation example.mp3
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So that's zero. Then we take the partial with respect to z of x times y. And again, there's no z that shows up there, so that's also zero. So we're kind of subtracting off zero. And then finally, we're adding this last component. We're gonna add that last component, k, multiplied by the determinant of this sub-matrix of the columns that it's not part of. So this involves partial derivative with respect to x of cosine z.
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3d curl computation example.mp3
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So we're kind of subtracting off zero. And then finally, we're adding this last component. We're gonna add that last component, k, multiplied by the determinant of this sub-matrix of the columns that it's not part of. So this involves partial derivative with respect to x of cosine z. Well, no x's show up there, so that's just zero. So that's just a zero. And then we're subtracting off the partial with respect to y of x times y.
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3d curl computation example.mp3
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So this involves partial derivative with respect to x of cosine z. Well, no x's show up there, so that's just zero. So that's just a zero. And then we're subtracting off the partial with respect to y of x times y. Well, x looks like a constant, y looks like the variable, so that partial derivative is just x. So we're subtracting off x, which means if we simplify this, so the curl of our vector field, curl of our vector field as a whole, as this function of x, y, and z, is equal to, and that first component, the i component, we've got one minus negative sine of z, so minus minus sine of z, that's one plus sine of z. And then the j component, we're subtracting off, but at zero.
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3d curl computation example.mp3
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And then we're subtracting off the partial with respect to y of x times y. Well, x looks like a constant, y looks like the variable, so that partial derivative is just x. So we're subtracting off x, which means if we simplify this, so the curl of our vector field, curl of our vector field as a whole, as this function of x, y, and z, is equal to, and that first component, the i component, we've got one minus negative sine of z, so minus minus sine of z, that's one plus sine of z. And then the j component, we're subtracting off, but at zero. Usually if you were subtracting off, you'd have to make sure to remember to flip those, but both of those are zero, so the entire j component here, or the y component of the output is zero. And then finally we're adding, the k component is zero minus x, so that entire thing is just negative x. And that's the curl of the function.
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3d curl computation example.mp3
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Hey everyone. So here and in the next few videos, I'm gonna be talking about tangent planes, tangent planes of graphs. And I'll specify that this is tangent planes of graphs and not of some other thing, because in different contexts of multivariable calculus, you might be taking a tangent plane of, say, a parametric surface or something like that, but here I'm just focused on graphs. So in the single variable world, a common problem that people like to ask in calculus is you have some sort of curve, and you wanna find, at a given point, what the tangent line to that curve is, what the tangent line is. And you'll find the equation for that tangent line, and this gives you various information, kind of how to, let's say you wanted to approximate the function around that point, and it turns out to be a nice, simple approximation. And in the multivariable world, it's actually pretty similar. In terms of geometric intuition, it's almost identical.
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What is a tangent plane.mp3
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So in the single variable world, a common problem that people like to ask in calculus is you have some sort of curve, and you wanna find, at a given point, what the tangent line to that curve is, what the tangent line is. And you'll find the equation for that tangent line, and this gives you various information, kind of how to, let's say you wanted to approximate the function around that point, and it turns out to be a nice, simple approximation. And in the multivariable world, it's actually pretty similar. In terms of geometric intuition, it's almost identical. You'll have some kind of graph of a function, like the one that I have here, and then instead of having a tangent line, because a line is a very one-dimensional thing, and here it's a very two-dimensional surface, instead you'll have some kind of tangent plane. So this is something where it's just gonna barely be kissing the graph, in the same way that the tangent line just barely kisses the function graph in the one-dimensional circumstance. And it could be at various different points, rather than just being at that point.
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What is a tangent plane.mp3
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In terms of geometric intuition, it's almost identical. You'll have some kind of graph of a function, like the one that I have here, and then instead of having a tangent line, because a line is a very one-dimensional thing, and here it's a very two-dimensional surface, instead you'll have some kind of tangent plane. So this is something where it's just gonna barely be kissing the graph, in the same way that the tangent line just barely kisses the function graph in the one-dimensional circumstance. And it could be at various different points, rather than just being at that point. You could kind of move it around, and say that, okay, it'll just barely be kissing the graph of this function, but at different points. And usually the way that a problem like this would be framed, if you're trying to find such a tangent plane, is first, you think about the specified input that you want. So in the same way that over in the single variable world, what you might do is say, okay, what is the input value here?
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What is a tangent plane.mp3
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And it could be at various different points, rather than just being at that point. You could kind of move it around, and say that, okay, it'll just barely be kissing the graph of this function, but at different points. And usually the way that a problem like this would be framed, if you're trying to find such a tangent plane, is first, you think about the specified input that you want. So in the same way that over in the single variable world, what you might do is say, okay, what is the input value here? Maybe you'd name it like X sub O, and then you're gonna find the graph of the function that corresponds to kind of just kissing the graph at that input point. Over here in the multivariable world, kind of move things about, you'll choose some kind of input point, like this little red dot, and that could be at various different spots. It doesn't have to be where I put it.
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What is a tangent plane.mp3
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So in the same way that over in the single variable world, what you might do is say, okay, what is the input value here? Maybe you'd name it like X sub O, and then you're gonna find the graph of the function that corresponds to kind of just kissing the graph at that input point. Over here in the multivariable world, kind of move things about, you'll choose some kind of input point, like this little red dot, and that could be at various different spots. It doesn't have to be where I put it. You could imagine putting it somewhere else. But once you decide on what input point you want, you see where that is on the graph. So you kind of go and say, oh, that input point corresponds to such and such a height.
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What is a tangent plane.mp3
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It doesn't have to be where I put it. You could imagine putting it somewhere else. But once you decide on what input point you want, you see where that is on the graph. So you kind of go and say, oh, that input point corresponds to such and such a height. So in this case, it actually looks like the graph is about zero at that point, so the output of the function would be zero. And what you want is a plane that's tangent right at that point. So you'll draw some kind of plane that's tangent right at that point.
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What is a tangent plane.mp3
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So you kind of go and say, oh, that input point corresponds to such and such a height. So in this case, it actually looks like the graph is about zero at that point, so the output of the function would be zero. And what you want is a plane that's tangent right at that point. So you'll draw some kind of plane that's tangent right at that point. And if we think about what this input point corresponds to, it's not X sub O, a single variable input like we have in the single variable world, but instead that red dot that you're seeing is gonna correspond to some kind of input pair, X sub O and Y sub O. So the ultimate goal over here in our multivariable circumstance is gonna be to find some kind of new function, so I'll write it down here, some kind of new function that I'll call L for linear that's gonna take in X and Y, and we want the graph of that function to be this plane. And you might specify that this is dependent on the original function that you have, and maybe you also specify that it's dependent on this input point in some way, but the basic idea is we're gonna be looking for a function whose graph is this plane tangent at a given point.
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What is a tangent plane.mp3
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So you'll draw some kind of plane that's tangent right at that point. And if we think about what this input point corresponds to, it's not X sub O, a single variable input like we have in the single variable world, but instead that red dot that you're seeing is gonna correspond to some kind of input pair, X sub O and Y sub O. So the ultimate goal over here in our multivariable circumstance is gonna be to find some kind of new function, so I'll write it down here, some kind of new function that I'll call L for linear that's gonna take in X and Y, and we want the graph of that function to be this plane. And you might specify that this is dependent on the original function that you have, and maybe you also specify that it's dependent on this input point in some way, but the basic idea is we're gonna be looking for a function whose graph is this plane tangent at a given point. And in the next couple videos, I'm gonna talk through how you actually compute that. It might seem a little intimidating at first because how do you control a plane in three dimensions like this? But it's actually very similar to the single variable circumstance, and you just kind of take it one step at a time.
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What is a tangent plane.mp3
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And that means that it's representing some kind of function that has a two-dimensional input and a one-dimensional output. So that might look something like f of x, y equals, and then just some expression that has a bunch of x's and y's in it. And graphs are great, but they're kind of clunky to draw. I mean, certainly you can't just scribble it down. It typically requires some kind of graphing software. And when you take a static image of it, it's not always clear what's going on. So here, I'm going to describe a way that you can represent these functions in these graphs two-dimensionally, just by scribbling down on a two-dimensional piece of paper.
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Contour plots Multivariable calculus Khan Academy.mp3
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I mean, certainly you can't just scribble it down. It typically requires some kind of graphing software. And when you take a static image of it, it's not always clear what's going on. So here, I'm going to describe a way that you can represent these functions in these graphs two-dimensionally, just by scribbling down on a two-dimensional piece of paper. And this is a very common way that you'll see if you're reading a textbook or if someone is drawing on a blackboard. It's known as a contour plot. And the idea of a contour plot is that we're going to take this graph and slice it a bunch of times.
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Contour plots Multivariable calculus Khan Academy.mp3
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So here, I'm going to describe a way that you can represent these functions in these graphs two-dimensionally, just by scribbling down on a two-dimensional piece of paper. And this is a very common way that you'll see if you're reading a textbook or if someone is drawing on a blackboard. It's known as a contour plot. And the idea of a contour plot is that we're going to take this graph and slice it a bunch of times. So I'm going to slice it with various planes that are all parallel to the xy-plane. And let's think for a moment about what these guys represent. So the bottom one here represents the value z is equal to negative 2.
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Contour plots Multivariable calculus Khan Academy.mp3
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And the idea of a contour plot is that we're going to take this graph and slice it a bunch of times. So I'm going to slice it with various planes that are all parallel to the xy-plane. And let's think for a moment about what these guys represent. So the bottom one here represents the value z is equal to negative 2. So this is the z-axis over here. And when we fix that to be negative 2 and let x and y run freely, we get this whole plane. And if you let z increase, keep it constant, but let it increase by 1 to negative 1, we get a new plane still parallel to the xy-plane.
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Contour plots Multivariable calculus Khan Academy.mp3
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So the bottom one here represents the value z is equal to negative 2. So this is the z-axis over here. And when we fix that to be negative 2 and let x and y run freely, we get this whole plane. And if you let z increase, keep it constant, but let it increase by 1 to negative 1, we get a new plane still parallel to the xy-plane. And it's got a, but its distance from the xy-plane is negative 1. And the rest of these guys, they're all still values, constant values of z. Now in terms of our graph, what that means is that these represent constant values of the graph itself.
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Contour plots Multivariable calculus Khan Academy.mp3
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