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So we have our surface integral of the curl of f dot ds. Well, we've already figured out what our curl of f is here two videos ago. And we've almost figured out what ds is. ds is a cross product of these two vectors times da. The cross product of these two vectors is this right over here. So we could just write that ds is going to be equal to this thing times da. This is the cross product of the partial of r with respect to x and the partial of r with respect to y.
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Stokes' theorem proof part 3 Multivariable Calculus Khan Academy.mp3
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ds is a cross product of these two vectors times da. The cross product of these two vectors is this right over here. So we could just write that ds is going to be equal to this thing times da. This is the cross product of the partial of r with respect to x and the partial of r with respect to y. And then we have to multiply that times da right over there. So this expression is just going to be the dot product of the curl of f, which is this stuff up here, dotted with this stuff down here. And essentially, we're going to take the dot product of this vector and that vector and then multiply it times this.
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Stokes' theorem proof part 3 Multivariable Calculus Khan Academy.mp3
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This is the cross product of the partial of r with respect to x and the partial of r with respect to y. And then we have to multiply that times da right over there. So this expression is just going to be the dot product of the curl of f, which is this stuff up here, dotted with this stuff down here. And essentially, we're going to take the dot product of this vector and that vector and then multiply it times this. We can actually consider this to be a scalar value. So let's do that. So this is going to be equal to.
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Stokes' theorem proof part 3 Multivariable Calculus Khan Academy.mp3
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And essentially, we're going to take the dot product of this vector and that vector and then multiply it times this. We can actually consider this to be a scalar value. So let's do that. So this is going to be equal to. And when we do this, all of these, we're now going to start operating in the domain of our parameters. So it's going to turn from a surface integral into a double integral over that domain, over that region that we care about. This is the domain of our parameters, this region r. And this is how we've manipulated any of the surface integrals that we've come across so far.
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Stokes' theorem proof part 3 Multivariable Calculus Khan Academy.mp3
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So this is going to be equal to. And when we do this, all of these, we're now going to start operating in the domain of our parameters. So it's going to turn from a surface integral into a double integral over that domain, over that region that we care about. This is the domain of our parameters, this region r. And this is how we've manipulated any of the surface integrals that we've come across so far. We've turned them into double integrals over the domain of the parameters. So this is going to turn into a double integral over the domain of our parameters, which is the region r. It's the region r in the xy-plane right up here. And now we can take the dot product of the curl of f with the dotted ds, which is all of this business right over here.
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Stokes' theorem proof part 3 Multivariable Calculus Khan Academy.mp3
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This is the domain of our parameters, this region r. And this is how we've manipulated any of the surface integrals that we've come across so far. We've turned them into double integrals over the domain of the parameters. So this is going to turn into a double integral over the domain of our parameters, which is the region r. It's the region r in the xy-plane right up here. And now we can take the dot product of the curl of f with the dotted ds, which is all of this business right over here. Let me see if I can show both of them on the screen at the same time. There you go. So first, let's think about the x components.
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Stokes' theorem proof part 3 Multivariable Calculus Khan Academy.mp3
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And now we can take the dot product of the curl of f with the dotted ds, which is all of this business right over here. Let me see if I can show both of them on the screen at the same time. There you go. So first, let's think about the x components. So you have that right over there. And then you have this right over here. You multiply the two, the negative.
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Stokes' theorem proof part 3 Multivariable Calculus Khan Academy.mp3
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So first, let's think about the x components. So you have that right over there. And then you have this right over here. You multiply the two, the negative. We can swap the order right over there. You have the partial of z with respect to x times, and we're going to swap this order right over here, the partial of q with respect to z minus the partial of r with respect to y. Now let's think about the j component.
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Stokes' theorem proof part 3 Multivariable Calculus Khan Academy.mp3
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You multiply the two, the negative. We can swap the order right over there. You have the partial of z with respect to x times, and we're going to swap this order right over here, the partial of q with respect to z minus the partial of r with respect to y. Now let's think about the j component. You have negative z sub y times all of this up here, at least the stuff that's multiplied times the j component. This negative can cancel out with that negative. So you have plus z sub y, the partial of z with respect to y, times the partial of r with respect to x minus the partial of p with respect to z.
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Stokes' theorem proof part 3 Multivariable Calculus Khan Academy.mp3
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Now let's think about the j component. You have negative z sub y times all of this up here, at least the stuff that's multiplied times the j component. This negative can cancel out with that negative. So you have plus z sub y, the partial of z with respect to y, times the partial of r with respect to x minus the partial of p with respect to z. Let me make that clear. That is an r right over here. And then we have the k component.
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Stokes' theorem proof part 3 Multivariable Calculus Khan Academy.mp3
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So you have plus z sub y, the partial of z with respect to y, times the partial of r with respect to x minus the partial of p with respect to z. Let me make that clear. That is an r right over here. And then we have the k component. And the k component's actually the easiest because it's just one here. So it's just going to be 1 times, and I'll just do it in that same color, 1 times the partial of q with respect to x minus the partial of p with respect to y. And then finally, we just have this dA over here.
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Stokes' theorem proof part 3 Multivariable Calculus Khan Academy.mp3
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And then we have the k component. And the k component's actually the easiest because it's just one here. So it's just going to be 1 times, and I'll just do it in that same color, 1 times the partial of q with respect to x minus the partial of p with respect to y. And then finally, we just have this dA over here. And this dA is multiplied by everything. So we'll put some parentheses, and we'll write dA. So we're done.
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Stokes' theorem proof part 3 Multivariable Calculus Khan Academy.mp3
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And then finally, we just have this dA over here. And this dA is multiplied by everything. So we'll put some parentheses, and we'll write dA. So we're done. We've expressed our surface integral as a double integral over the domain of our parameters. And what we're going to do in the next few videos is do the same thing with this using Green's theorem. We're going to see that we get the exact same value.
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Stokes' theorem proof part 3 Multivariable Calculus Khan Academy.mp3
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They can be types one, type two, and type three, and for a lot of situations that aren't simple solid regions, you can break them up into simple solid regions, but let's just prove it for this case, for this case right over here. So let's just assume that our vector field F, our vector field F can be written as P, which is a function of X, Y, and Z, times I, plus Q, which is a function of X, Y, and Z, times J, plus R, which is a function of X, Y, and Z, times K. So let's think about what each of these sides of the equation would come out to be. Well, first of all, what is going to be F dot N? So let's think about that a little bit. F dot N, F dot N, F dot N, is going to be equal to this, this component right over here, times N's I component, plus this component right over here, times N's J component, plus this component here, times N's K component. So we could write it as, it could be written as P times, P times, or I'll just write P open parentheses, the dot product of I and N, I dotted N, I dotted N, and let me make sure I write I as a vector, as a unit vector. I want to be clear, what's going to happen right over here?
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Divergence theorem proof (part 1) Divergence theorem Multivariable Calculus Khan Academy.mp3
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So let's think about that a little bit. F dot N, F dot N, F dot N, is going to be equal to this, this component right over here, times N's I component, plus this component right over here, times N's J component, plus this component here, times N's K component. So we could write it as, it could be written as P times, P times, or I'll just write P open parentheses, the dot product of I and N, I dotted N, I dotted N, and let me make sure I write I as a vector, as a unit vector. I want to be clear, what's going to happen right over here? If you take the dot product of I and N, you're just going to get the I component, the scaling factor of the I component, of the N normal vector, and we're just going to multiply that times P. So that's essentially the product of the X components, or I guess you could say the magnitude of the X components. And then to that, we are going to add Q times J, J dotted with N. And once again, when you dot J with N, you get the magnitude of the J component of the normal vector right over there. And then times, or plus, I should say, plus R times K dotted with N. K dotted with N. This isn't how we normally see it, but I think it's reasonable to say that this is actually true.
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Divergence theorem proof (part 1) Divergence theorem Multivariable Calculus Khan Academy.mp3
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I want to be clear, what's going to happen right over here? If you take the dot product of I and N, you're just going to get the I component, the scaling factor of the I component, of the N normal vector, and we're just going to multiply that times P. So that's essentially the product of the X components, or I guess you could say the magnitude of the X components. And then to that, we are going to add Q times J, J dotted with N. And once again, when you dot J with N, you get the magnitude of the J component of the normal vector right over there. And then times, or plus, I should say, plus R times K dotted with N. K dotted with N. This isn't how we normally see it, but I think it's reasonable to say that this is actually true. This right over here is going to be equal to P times the magnitude of the I component of N's normal vector, which is exactly what we want in a dot product. This is the same thing for the J component. This is the same thing for the K component.
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Divergence theorem proof (part 1) Divergence theorem Multivariable Calculus Khan Academy.mp3
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And then times, or plus, I should say, plus R times K dotted with N. K dotted with N. This isn't how we normally see it, but I think it's reasonable to say that this is actually true. This right over here is going to be equal to P times the magnitude of the I component of N's normal vector, which is exactly what we want in a dot product. This is the same thing for the J component. This is the same thing for the K component. And you could try it out. Define N is equal to, I don't know, M times I plus N times J plus O times K or something like that, and you'll see that this actually does work out fine. So how can we simplify this expression right up here?
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Divergence theorem proof (part 1) Divergence theorem Multivariable Calculus Khan Academy.mp3
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This is the same thing for the K component. And you could try it out. Define N is equal to, I don't know, M times I plus N times J plus O times K or something like that, and you'll see that this actually does work out fine. So how can we simplify this expression right up here? Well, we can rewrite this as, we can rewrite the left-hand side as, so the surface integral of F, and let me write it multiple ways, F dot DS, which is equal to the surface integral, the surface integral of F dot N times the scalar DS, is equal to the double integral of the surface of all of this business right over here, is equal to the double integral over the surface of, let me just copy and paste that, of all of that business. So let me copy and then let me paste. So all of that business right over there.
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Divergence theorem proof (part 1) Divergence theorem Multivariable Calculus Khan Academy.mp3
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So how can we simplify this expression right up here? Well, we can rewrite this as, we can rewrite the left-hand side as, so the surface integral of F, and let me write it multiple ways, F dot DS, which is equal to the surface integral, the surface integral of F dot N times the scalar DS, is equal to the double integral of the surface of all of this business right over here, is equal to the double integral over the surface of, let me just copy and paste that, of all of that business. So let me copy and then let me paste. So all of that business right over there. I just noticed that I forgot to put the little unit vector symbol, the little caret, right over there. Put some parentheses, and then we are left with our DS. And then this, all of this can be rewritten as, this can be written as the surface integral of P times this business, and I'll just do it in the same color, of P times the dot product of I and N, DS, plus the surface integral of Q times the dot product of J and N, DS, plus the surface integral of R times the dot product of K and N, forgot a caret, K and N, DS.
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Divergence theorem proof (part 1) Divergence theorem Multivariable Calculus Khan Academy.mp3
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So all of that business right over there. I just noticed that I forgot to put the little unit vector symbol, the little caret, right over there. Put some parentheses, and then we are left with our DS. And then this, all of this can be rewritten as, this can be written as the surface integral of P times this business, and I'll just do it in the same color, of P times the dot product of I and N, DS, plus the surface integral of Q times the dot product of J and N, DS, plus the surface integral of R times the dot product of K and N, forgot a caret, K and N, DS. So I just broke it up, we were taking the integral of this sum, and so I just took it as, I just rewrote it as the sum of the integrals. So that's the left-hand side right over here. Now let's think about the right-hand side.
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Divergence theorem proof (part 1) Divergence theorem Multivariable Calculus Khan Academy.mp3
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And then this, all of this can be rewritten as, this can be written as the surface integral of P times this business, and I'll just do it in the same color, of P times the dot product of I and N, DS, plus the surface integral of Q times the dot product of J and N, DS, plus the surface integral of R times the dot product of K and N, forgot a caret, K and N, DS. So I just broke it up, we were taking the integral of this sum, and so I just took it as, I just rewrote it as the sum of the integrals. So that's the left-hand side right over here. Now let's think about the right-hand side. What is the divergence of F? And actually I'm gonna take some space up here. What is the divergence of F?
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Divergence theorem proof (part 1) Divergence theorem Multivariable Calculus Khan Academy.mp3
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Now let's think about the right-hand side. What is the divergence of F? And actually I'm gonna take some space up here. What is the divergence of F? Well the divergence of F, based on this expression of F, is just going to be, let me just write it over here real small, the divergence of F is going to be the partial of P with respect to, let me do this in a new color, just because I'm using the yellow too much. The divergence of F, the divergence of F is going to be the partial of P with respect to X, plus the partial of Q with respect to Y, plus the partial of R, plus the partial of R with respect to Z. So we can, this triple integral right over here could be written as the triple integral of the partial of P with respect to X plus the partial of Q with respect to Y, plus the partial of R with respect to Z.
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Divergence theorem proof (part 1) Divergence theorem Multivariable Calculus Khan Academy.mp3
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What is the divergence of F? Well the divergence of F, based on this expression of F, is just going to be, let me just write it over here real small, the divergence of F is going to be the partial of P with respect to, let me do this in a new color, just because I'm using the yellow too much. The divergence of F, the divergence of F is going to be the partial of P with respect to X, plus the partial of Q with respect to Y, plus the partial of R, plus the partial of R with respect to Z. So we can, this triple integral right over here could be written as the triple integral of the partial of P with respect to X plus the partial of Q with respect to Y, plus the partial of R with respect to Z. Well this thing once again, Instead of writing it as the triple integral of this sum, we could write it as the sum of triple integrals. So this thing right over here can be rewritten as the triple integral over our three dimensional region. Actually, let me copy and paste that so I don't have to keep rewriting it.
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Divergence theorem proof (part 1) Divergence theorem Multivariable Calculus Khan Academy.mp3
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So we can, this triple integral right over here could be written as the triple integral of the partial of P with respect to X plus the partial of Q with respect to Y, plus the partial of R with respect to Z. Well this thing once again, Instead of writing it as the triple integral of this sum, we could write it as the sum of triple integrals. So this thing right over here can be rewritten as the triple integral over our three dimensional region. Actually, let me copy and paste that so I don't have to keep rewriting it. So let me copy it. So it's going to be equal to the triple integral of the partial of p with respect to x dv plus, let me paste that again, triple integral of the partial of q with respect to y dv plus, once again, triple integral of the partial of r with respect to z dv. So we've essentially restated our divergence theorem.
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Divergence theorem proof (part 1) Divergence theorem Multivariable Calculus Khan Academy.mp3
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Actually, let me copy and paste that so I don't have to keep rewriting it. So let me copy it. So it's going to be equal to the triple integral of the partial of p with respect to x dv plus, let me paste that again, triple integral of the partial of q with respect to y dv plus, once again, triple integral of the partial of r with respect to z dv. So we've essentially restated our divergence theorem. This is our surface integral, and the divergence theorem says that this needs to be equal to this business right over here. We've just written it a different way. And so what I'm going to do in order to prove it is just show that each of these corresponding terms are equal to each other.
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Divergence theorem proof (part 1) Divergence theorem Multivariable Calculus Khan Academy.mp3
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So we've essentially restated our divergence theorem. This is our surface integral, and the divergence theorem says that this needs to be equal to this business right over here. We've just written it a different way. And so what I'm going to do in order to prove it is just show that each of these corresponding terms are equal to each other. That these are equal to each other. And in particular, we're going to focus the proof on this, and we're going to use the fact that our region is a type 1 region. It's a type 1, type 2, and type 3.
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Divergence theorem proof (part 1) Divergence theorem Multivariable Calculus Khan Academy.mp3
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So I've talked about the partial derivative and how you compute it, how you interpret it in terms of graphs, but what I'd like to do here is give its formal definition. So it's the kind of thing, just to remind you, that applies to a function that has a multivariable input. So x, y, and I'll just emphasize that it could actually be a number of other inputs. You could have 100 inputs or something like that. And as with a lot of things here, I think it's helpful to take a look at the one-dimensional analogy and think about how we define the derivative, just the ordinary derivative, when you have a function that's just one variable. You know, this would be just something simple, f of x, and you know, maybe you're thinking in the back of your mind that it's a function like f of x equals x squared. And the way to think about the definition of this is to just actually spell out how we interpret this df and dx and then slowly start to tighten it up into a formalization.
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Formal definition of partial derivatives.mp3
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You could have 100 inputs or something like that. And as with a lot of things here, I think it's helpful to take a look at the one-dimensional analogy and think about how we define the derivative, just the ordinary derivative, when you have a function that's just one variable. You know, this would be just something simple, f of x, and you know, maybe you're thinking in the back of your mind that it's a function like f of x equals x squared. And the way to think about the definition of this is to just actually spell out how we interpret this df and dx and then slowly start to tighten it up into a formalization. So you might be thinking of the graph of this function. You know, maybe it's some kind of curve. And when you think of evaluating it at some point, you know, let's say you're evaluating it at a point a, you're imagining dx here as representing a slight nudge, just a slight nudge in the input value.
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Formal definition of partial derivatives.mp3
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And the way to think about the definition of this is to just actually spell out how we interpret this df and dx and then slowly start to tighten it up into a formalization. So you might be thinking of the graph of this function. You know, maybe it's some kind of curve. And when you think of evaluating it at some point, you know, let's say you're evaluating it at a point a, you're imagining dx here as representing a slight nudge, just a slight nudge in the input value. So this is in the x direction. You've got your x-coordinate, your output, f of x, is what the y-axis represents here. And then you're thinking of df as being the resulting nudge here, the resulting change to the function.
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Formal definition of partial derivatives.mp3
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And when you think of evaluating it at some point, you know, let's say you're evaluating it at a point a, you're imagining dx here as representing a slight nudge, just a slight nudge in the input value. So this is in the x direction. You've got your x-coordinate, your output, f of x, is what the y-axis represents here. And then you're thinking of df as being the resulting nudge here, the resulting change to the function. So when we formalize this, we're gonna be thinking of a fraction that's gonna represent df over dx. And I'll leave myself some room. You can probably anticipate why if you know where this is going.
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Formal definition of partial derivatives.mp3
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And then you're thinking of df as being the resulting nudge here, the resulting change to the function. So when we formalize this, we're gonna be thinking of a fraction that's gonna represent df over dx. And I'll leave myself some room. You can probably anticipate why if you know where this is going. And instead of saying dx, I'll say h. So instead of thinking, you know, dx is that tiny nudge, you'll think h. And I'm not sure why h is used necessarily, but just having some kind of variable that you think of as getting small. Maybe all the other letters in the alphabet were taken. And now, when you actually say, what do we mean by the resulting change in f, we should be writing as well, where does it go after you nudge?
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Formal definition of partial derivatives.mp3
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You can probably anticipate why if you know where this is going. And instead of saying dx, I'll say h. So instead of thinking, you know, dx is that tiny nudge, you'll think h. And I'm not sure why h is used necessarily, but just having some kind of variable that you think of as getting small. Maybe all the other letters in the alphabet were taken. And now, when you actually say, what do we mean by the resulting change in f, we should be writing as well, where does it go after you nudge? So when you take, you know, from that input point plus that nudge, plus that little h, what's the difference between that and the original function just, or the original value of the function at that point? So this top part is really what's representing df. You know, this is what's representing df over here.
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Formal definition of partial derivatives.mp3
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And now, when you actually say, what do we mean by the resulting change in f, we should be writing as well, where does it go after you nudge? So when you take, you know, from that input point plus that nudge, plus that little h, what's the difference between that and the original function just, or the original value of the function at that point? So this top part is really what's representing df. You know, this is what's representing df over here. But you don't do this for any actual value of h. You don't do it for any specific nudge. The whole point, largely the whole point of calculus is that you're considering the limit as h goes to zero of this. And this is what makes concrete the idea of, you know, a tiny little nudge or a tiny little resulting change.
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Formal definition of partial derivatives.mp3
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You know, this is what's representing df over here. But you don't do this for any actual value of h. You don't do it for any specific nudge. The whole point, largely the whole point of calculus is that you're considering the limit as h goes to zero of this. And this is what makes concrete the idea of, you know, a tiny little nudge or a tiny little resulting change. It's not that it's any specific one. You're taking the limit. And, you know, you could get into the formal definition of a limit, but it gives you room for rigor as soon as you start writing something like this.
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Formal definition of partial derivatives.mp3
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And this is what makes concrete the idea of, you know, a tiny little nudge or a tiny little resulting change. It's not that it's any specific one. You're taking the limit. And, you know, you could get into the formal definition of a limit, but it gives you room for rigor as soon as you start writing something like this. Now, over in the multivariable world, very similar story. We can pretty much do the same thing and we're gonna look at our original fraction and just start to formalize what we think of each of these variables as representing. That partial x still is common to use the letter h just to represent a tiny nudge in the x direction.
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Formal definition of partial derivatives.mp3
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And, you know, you could get into the formal definition of a limit, but it gives you room for rigor as soon as you start writing something like this. Now, over in the multivariable world, very similar story. We can pretty much do the same thing and we're gonna look at our original fraction and just start to formalize what we think of each of these variables as representing. That partial x still is common to use the letter h just to represent a tiny nudge in the x direction. And now if we think about what is that nudge, and here, let me draw it out actually. The way that I kind of like to draw this out is you think of your entire input space as, you know, the xy-plane. If it was more variables, this would be a high-dimensional space.
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Formal definition of partial derivatives.mp3
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That partial x still is common to use the letter h just to represent a tiny nudge in the x direction. And now if we think about what is that nudge, and here, let me draw it out actually. The way that I kind of like to draw this out is you think of your entire input space as, you know, the xy-plane. If it was more variables, this would be a high-dimensional space. And you're thinking of some point, you know, maybe you're thinking of it as a, b. Or maybe I should specify that actually. We're, you know, we're doing this at a specific point, how you define it.
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Formal definition of partial derivatives.mp3
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If it was more variables, this would be a high-dimensional space. And you're thinking of some point, you know, maybe you're thinking of it as a, b. Or maybe I should specify that actually. We're, you know, we're doing this at a specific point, how you define it. We're doing this at a very specific point, a, b. And when you're thinking of your tiny little change in x, you'd be thinking, you know, a tiny little nudge in the x direction, tiny little shift there. And the entire function maps that input space, whatever it is, to the real number line.
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Formal definition of partial derivatives.mp3
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We're, you know, we're doing this at a specific point, how you define it. We're doing this at a very specific point, a, b. And when you're thinking of your tiny little change in x, you'd be thinking, you know, a tiny little nudge in the x direction, tiny little shift there. And the entire function maps that input space, whatever it is, to the real number line. This is your output space. And you're saying, hey, how does that tiny nudge influence the output? I've drawn this diagram a lot, just this loose sketch.
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Formal definition of partial derivatives.mp3
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And the entire function maps that input space, whatever it is, to the real number line. This is your output space. And you're saying, hey, how does that tiny nudge influence the output? I've drawn this diagram a lot, just this loose sketch. I think it's actually a pretty good model because once we start thinking of higher-dimensional outputs or things like that, it's pretty flexible. And you're thinking of this as your partial f, partial x. Sorry, you're changing the x direction.
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Formal definition of partial derivatives.mp3
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I've drawn this diagram a lot, just this loose sketch. I think it's actually a pretty good model because once we start thinking of higher-dimensional outputs or things like that, it's pretty flexible. And you're thinking of this as your partial f, partial x. Sorry, you're changing the x direction. And this is that resulting change to the function. But we go back up here and we say, well, what does that mean, right? If h represents that tiny change to your x value, well, then you have to evaluate the function at the point a, but plus that h. And you're adding it to the x value, that first component, just because this is the partial derivative with respect to x.
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Formal definition of partial derivatives.mp3
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Sorry, you're changing the x direction. And this is that resulting change to the function. But we go back up here and we say, well, what does that mean, right? If h represents that tiny change to your x value, well, then you have to evaluate the function at the point a, but plus that h. And you're adding it to the x value, that first component, just because this is the partial derivative with respect to x. And the point b, the point b just remains unchanged, right? So this is you evaluating it kind of at the new point. And you have to say, what's the difference between that and the old evaluation where it was just at a and b?
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Formal definition of partial derivatives.mp3
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If h represents that tiny change to your x value, well, then you have to evaluate the function at the point a, but plus that h. And you're adding it to the x value, that first component, just because this is the partial derivative with respect to x. And the point b, the point b just remains unchanged, right? So this is you evaluating it kind of at the new point. And you have to say, what's the difference between that and the old evaluation where it was just at a and b? And that's it. That's the formal definition of your partial derivative. Except, oh, I mean, the most important part, right?
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Formal definition of partial derivatives.mp3
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And you have to say, what's the difference between that and the old evaluation where it was just at a and b? And that's it. That's the formal definition of your partial derivative. Except, oh, I mean, the most important part, right? The most important part, given that this is calculus, is that we're not doing this for any specific value of h, but we're actually, actually, let me just move this guy, give a little bit of room here. Yes. But we're actually taking the limit here, limit as h goes to zero.
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Formal definition of partial derivatives.mp3
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Except, oh, I mean, the most important part, right? The most important part, given that this is calculus, is that we're not doing this for any specific value of h, but we're actually, actually, let me just move this guy, give a little bit of room here. Yes. But we're actually taking the limit here, limit as h goes to zero. And what this means is you're not considering any specific size of dx, any specific size of this. And clearly, this is h, considering the notation up here. But any size for that partial x.
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Formal definition of partial derivatives.mp3
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But we're actually taking the limit here, limit as h goes to zero. And what this means is you're not considering any specific size of dx, any specific size of this. And clearly, this is h, considering the notation up here. But any size for that partial x. You're imagining that nudge shrinking more and more and more and the resulting change shrinks more and more and more, and you're wondering what the ratio between them approaches. So that would be the partial derivative with respect to x. And just for practice, let's actually write out what the partial derivative with respect to y would be.
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Formal definition of partial derivatives.mp3
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But any size for that partial x. You're imagining that nudge shrinking more and more and more and the resulting change shrinks more and more and more, and you're wondering what the ratio between them approaches. So that would be the partial derivative with respect to x. And just for practice, let's actually write out what the partial derivative with respect to y would be. So I'll get rid of some of this one-dimensional analogy stuff here. Don't need that anymore. And let's just think about what the partial derivative with respect to a different variable would be.
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Formal definition of partial derivatives.mp3
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And just for practice, let's actually write out what the partial derivative with respect to y would be. So I'll get rid of some of this one-dimensional analogy stuff here. Don't need that anymore. And let's just think about what the partial derivative with respect to a different variable would be. So if we were doing it as partial derivative of f with respect to y, now we're nudging slightly in the other direction, right? We're nudging in the y direction. And you'd be thinking, okay, so we're still gonna divide something by that nudge.
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Formal definition of partial derivatives.mp3
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And let's just think about what the partial derivative with respect to a different variable would be. So if we were doing it as partial derivative of f with respect to y, now we're nudging slightly in the other direction, right? We're nudging in the y direction. And you'd be thinking, okay, so we're still gonna divide something by that nudge. And again, I'm just using the same variable. Maybe it would be clearer to write something like the change in y, or to go up here and write something like, you know, the change in x. And people will do that, but it's less common.
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Formal definition of partial derivatives.mp3
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And you'd be thinking, okay, so we're still gonna divide something by that nudge. And again, I'm just using the same variable. Maybe it would be clearer to write something like the change in y, or to go up here and write something like, you know, the change in x. And people will do that, but it's less common. I think people just kinda want the standard go-to limiting variable. But this time, when you're considering what is the resulting change. Oh, and again, I always forget to write in.
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Formal definition of partial derivatives.mp3
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And people will do that, but it's less common. I think people just kinda want the standard go-to limiting variable. But this time, when you're considering what is the resulting change. Oh, and again, I always forget to write in. We're evaluating this at a specific point, at a specific point, a, b. And as a result, maybe I'll give myself a little bit more room here. So we're taking this whole thing, dividing by h, but what is the resulting change in f?
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Formal definition of partial derivatives.mp3
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Oh, and again, I always forget to write in. We're evaluating this at a specific point, at a specific point, a, b. And as a result, maybe I'll give myself a little bit more room here. So we're taking this whole thing, dividing by h, but what is the resulting change in f? This time, you say f, the new value is still gonna be at a, but the change happens for that second variable. It's gonna be that b, b plus h. So you're adding that nudge to the y value. And as before, you subtract off.
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Formal definition of partial derivatives.mp3
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So we're taking this whole thing, dividing by h, but what is the resulting change in f? This time, you say f, the new value is still gonna be at a, but the change happens for that second variable. It's gonna be that b, b plus h. So you're adding that nudge to the y value. And as before, you subtract off. You see the difference between that and how you evaluate it at the original point. And again, the whole reason I moved this over and gave myself some room is because we're taking the limit as this h goes to zero. And the way that you're thinking about this is very similar.
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Formal definition of partial derivatives.mp3
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And as before, you subtract off. You see the difference between that and how you evaluate it at the original point. And again, the whole reason I moved this over and gave myself some room is because we're taking the limit as this h goes to zero. And the way that you're thinking about this is very similar. It's just that when you change the input by adding h to the y value, you're shifting it upwards. So again, this is the partial derivative, the formal definition of the partial derivative. Looks very similar to the formal definition of the derivative, but I just always think about this as spelling out what we mean by partial y and partial f and kind of spelling out why it is that the Leibniz came up with this notation in the first place.
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Formal definition of partial derivatives.mp3
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And the way that you're thinking about this is very similar. It's just that when you change the input by adding h to the y value, you're shifting it upwards. So again, this is the partial derivative, the formal definition of the partial derivative. Looks very similar to the formal definition of the derivative, but I just always think about this as spelling out what we mean by partial y and partial f and kind of spelling out why it is that the Leibniz came up with this notation in the first place. Well, I don't know if Leibniz came up with the partials, but the df dx portion, and this is good to keep in the back of your mind, especially as we introduce new notions, new types of multivariable derivatives, like the directional derivative. I think it helps clarify what's really going on in certain contexts. Great, see you next video.
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Formal definition of partial derivatives.mp3
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And as you can see, we can kind of decompose it into three separate surfaces. The first surface is its base, which is really the filled-in unit circle down here. The second surface, which we have in blue, you can kind of view it as the side of it. You can view it as it's kind of like the side of a cylinder, but the cylinder has been cut by a plane up here. The plane that cuts it is the plane z is equal to 1 minus x. And obviously, the plane itself goes well beyond the shape. But where that plane cuts the cylinder kind of defines this shape.
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Surface integral ex3 part 1 Parameterizing the outside surface Khan Academy.mp3
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You can view it as it's kind of like the side of a cylinder, but the cylinder has been cut by a plane up here. The plane that cuts it is the plane z is equal to 1 minus x. And obviously, the plane itself goes well beyond the shape. But where that plane cuts the cylinder kind of defines this shape. So the blue surface is above the boundary of the unit circle and below the plane. And then the third surface is the subset of the plane, of that purple plane of z equals 1 minus x, that overlaps, that kind of forms the top of this cylinder. And so we can rewrite this surface integral as the sum of three surface integrals.
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Surface integral ex3 part 1 Parameterizing the outside surface Khan Academy.mp3
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But where that plane cuts the cylinder kind of defines this shape. So the blue surface is above the boundary of the unit circle and below the plane. And then the third surface is the subset of the plane, of that purple plane of z equals 1 minus x, that overlaps, that kind of forms the top of this cylinder. And so we can rewrite this surface integral as the sum of three surface integrals. It's the surface integral of z over S1 plus the surface integral of z over S2 plus the surface integral of z over S3. And we can just tackle each of these independently. So let's start with surface 1.
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Surface integral ex3 part 1 Parameterizing the outside surface Khan Academy.mp3
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And so we can rewrite this surface integral as the sum of three surface integrals. It's the surface integral of z over S1 plus the surface integral of z over S2 plus the surface integral of z over S3. And we can just tackle each of these independently. So let's start with surface 1. And you might immediately want to start parameterizing things and all the rest, but there's actually a very fast way to handle this surface integral, especially because we're taking the surface integral of z. What value does z take on throughout this surface, throughout this filled in unit circle right over here? Well, that surface is on the xy plane.
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Surface integral ex3 part 1 Parameterizing the outside surface Khan Academy.mp3
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So let's start with surface 1. And you might immediately want to start parameterizing things and all the rest, but there's actually a very fast way to handle this surface integral, especially because we're taking the surface integral of z. What value does z take on throughout this surface, throughout this filled in unit circle right over here? Well, that surface is on the xy plane. When we're on the xy plane, z is equal to 0. So for this entire surface, z is going to be equal to 0. You're essentially integrating 0.
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Surface integral ex3 part 1 Parameterizing the outside surface Khan Academy.mp3
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Well, that surface is on the xy plane. When we're on the xy plane, z is equal to 0. So for this entire surface, z is going to be equal to 0. You're essentially integrating 0. 0 times dS is just going to be 0. So this whole thing is going to evaluate to 0. So that's about as simple as evaluating a surface integral can get.
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Surface integral ex3 part 1 Parameterizing the outside surface Khan Academy.mp3
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You're essentially integrating 0. 0 times dS is just going to be 0. So this whole thing is going to evaluate to 0. So that's about as simple as evaluating a surface integral can get. But it's always important to at least keep a lookout for things like that. And it'll keep you from going down this wild goose chase. That wouldn't be a wild goose chase.
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Surface integral ex3 part 1 Parameterizing the outside surface Khan Academy.mp3
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So that's about as simple as evaluating a surface integral can get. But it's always important to at least keep a lookout for things like that. And it'll keep you from going down this wild goose chase. That wouldn't be a wild goose chase. You would eventually get the answer anyway. But you don't have to waste so much time in parameterizing things and all the rest. Now let's tackle the other two surfaces.
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Surface integral ex3 part 1 Parameterizing the outside surface Khan Academy.mp3
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That wouldn't be a wild goose chase. You would eventually get the answer anyway. But you don't have to waste so much time in parameterizing things and all the rest. Now let's tackle the other two surfaces. And we'll focus on surface 2. So the x and y values, the valid x and y values, are the x and y values along the unit circle right over here. So we can really parameterize it the way that we would parameterize a traditional unit circle.
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Surface integral ex3 part 1 Parameterizing the outside surface Khan Academy.mp3
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Now let's tackle the other two surfaces. And we'll focus on surface 2. So the x and y values, the valid x and y values, are the x and y values along the unit circle right over here. So we can really parameterize it the way that we would parameterize a traditional unit circle. So we could set x is equal to, let's set it, and our radius is 1. So x is equal to cosine of, I'll use the parameter u, is equal to x is equal to cosine of u. And then let's say y is equal to sine of u.
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Surface integral ex3 part 1 Parameterizing the outside surface Khan Academy.mp3
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So we can really parameterize it the way that we would parameterize a traditional unit circle. So we could set x is equal to, let's set it, and our radius is 1. So x is equal to cosine of, I'll use the parameter u, is equal to x is equal to cosine of u. And then let's say y is equal to sine of u. And then u is the angle between the positive x-axis and wherever we are essentially on that unit circle. So that right over there, that angle right over there is u. And so this would give us for u, so as long as u is between 0 and 2 pi, that we're essentially going around this unit circle.
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Surface integral ex3 part 1 Parameterizing the outside surface Khan Academy.mp3
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And then let's say y is equal to sine of u. And then u is the angle between the positive x-axis and wherever we are essentially on that unit circle. So that right over there, that angle right over there is u. And so this would give us for u, so as long as u is between 0 and 2 pi, that we're essentially going around this unit circle. So those are all the possible x and y values that we can take on. And then the z value is what takes us up above that boundary and gets it someplace along the surface. But this is interesting.
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Surface integral ex3 part 1 Parameterizing the outside surface Khan Academy.mp3
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And so this would give us for u, so as long as u is between 0 and 2 pi, that we're essentially going around this unit circle. So those are all the possible x and y values that we can take on. And then the z value is what takes us up above that boundary and gets it someplace along the surface. But this is interesting. Because the z value, it can take on obviously a bunch of different values. But it always has to be below this plane right over here. So for this z value, I'm going to introduce a new parameter.
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Surface integral ex3 part 1 Parameterizing the outside surface Khan Academy.mp3
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But this is interesting. Because the z value, it can take on obviously a bunch of different values. But it always has to be below this plane right over here. So for this z value, I'm going to introduce a new parameter. Let's call z v. That's the second parameter. And v is going to be, it's definitely going to be greater than 0. z and v, they're the same thing. They're always going to be positive.
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Surface integral ex3 part 1 Parameterizing the outside surface Khan Academy.mp3
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So for this z value, I'm going to introduce a new parameter. Let's call z v. That's the second parameter. And v is going to be, it's definitely going to be greater than 0. z and v, they're the same thing. They're always going to be positive. So they're definitely going to be greater than or equal to 0. But it's not less than or equal to some constant thing. This has kind of a variable roof on it.
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Surface integral ex3 part 1 Parameterizing the outside surface Khan Academy.mp3
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They're always going to be positive. So they're definitely going to be greater than or equal to 0. But it's not less than or equal to some constant thing. This has kind of a variable roof on it. And so it's always going to be less than or equal to this plane right over here. And so we could say, is less than or equal to 1 minus x. I mean, we could have said, so we know that z is less than or equal to 1 minus x. But if we use the new parameters, v is z.
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Surface integral ex3 part 1 Parameterizing the outside surface Khan Academy.mp3
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This has kind of a variable roof on it. And so it's always going to be less than or equal to this plane right over here. And so we could say, is less than or equal to 1 minus x. I mean, we could have said, so we know that z is less than or equal to 1 minus x. But if we use the new parameters, v is z. And 1 minus x is the same thing as 1 minus cosine u. And so now we have our parameterization. We are ready to actually evaluate the surface integral.
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Surface integral ex3 part 1 Parameterizing the outside surface Khan Academy.mp3
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But if we use the new parameters, v is z. And 1 minus x is the same thing as 1 minus cosine u. And so now we have our parameterization. We are ready to actually evaluate the surface integral. And to do that, first let's do the cross product. We want to figure out what ds is. And we have to take the magnitude of the cross product of the partial of our parameterization with respect to u, cross with the partial with respect to v. Actually, let me just write all this stuff down instead of trying to do shortcuts.
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Surface integral ex3 part 1 Parameterizing the outside surface Khan Academy.mp3
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We are ready to actually evaluate the surface integral. And to do that, first let's do the cross product. We want to figure out what ds is. And we have to take the magnitude of the cross product of the partial of our parameterization with respect to u, cross with the partial with respect to v. Actually, let me just write all this stuff down instead of trying to do shortcuts. So ds, and this is all, let me do it in blue still, because we're talking about surface 2. ds is going to be equal to the magnitude of the cross product of our parameterization with the partial of our parameterization with respect to u, crossed with our partial of our parameterization with respect to v, du, dv. So let's write the partial of our parameterization with respect to u. And you might say, wait, where's our parameterization?
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Surface integral ex3 part 1 Parameterizing the outside surface Khan Academy.mp3
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And we have to take the magnitude of the cross product of the partial of our parameterization with respect to u, cross with the partial with respect to v. Actually, let me just write all this stuff down instead of trying to do shortcuts. So ds, and this is all, let me do it in blue still, because we're talking about surface 2. ds is going to be equal to the magnitude of the cross product of our parameterization with the partial of our parameterization with respect to u, crossed with our partial of our parameterization with respect to v, du, dv. So let's write the partial of our parameterization with respect to u. And you might say, wait, where's our parameterization? Well, it's right over here. I just didn't write it in the traditional ijk form. But I can.
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Surface integral ex3 part 1 Parameterizing the outside surface Khan Academy.mp3
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And you might say, wait, where's our parameterization? Well, it's right over here. I just didn't write it in the traditional ijk form. But I can. I could write r is equal to, maybe I call it r2, because we're talking about surface 2. I shouldn't use that orange color, because I used that for surface 1. I'll use that in a still bluish color.
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Surface integral ex3 part 1 Parameterizing the outside surface Khan Academy.mp3
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But I can. I could write r is equal to, maybe I call it r2, because we're talking about surface 2. I shouldn't use that orange color, because I used that for surface 1. I'll use that in a still bluish color. So r for surface 2 is equal to cosine of ui plus sine of uj plus vk. And this is the range that the u's and the v's can take on. If I want to take the partial of r with respect to u, I would get, let's see, the derivative of cosine u with respect to u is negative sine ui.
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Surface integral ex3 part 1 Parameterizing the outside surface Khan Academy.mp3
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I'll use that in a still bluish color. So r for surface 2 is equal to cosine of ui plus sine of uj plus vk. And this is the range that the u's and the v's can take on. If I want to take the partial of r with respect to u, I would get, let's see, the derivative of cosine u with respect to u is negative sine ui. Derivative of sine of u with respect to u is plus cosine uj. And the derivative of v with respect to u is just 0. So that's our partial with respect to u.
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Surface integral ex3 part 1 Parameterizing the outside surface Khan Academy.mp3
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If I want to take the partial of r with respect to u, I would get, let's see, the derivative of cosine u with respect to u is negative sine ui. Derivative of sine of u with respect to u is plus cosine uj. And the derivative of v with respect to u is just 0. So that's our partial with respect to u. Our partial, we'll do this in a different color, just for the sake of it. Our partial with respect to v is equal to, well, this is going to be 0, this is going to be 0, and we're just going to have 1k. So it's just going to be equal to k. And so now we're ready to at least evaluate this cross product.
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Surface integral ex3 part 1 Parameterizing the outside surface Khan Academy.mp3
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So that's our partial with respect to u. Our partial, we'll do this in a different color, just for the sake of it. Our partial with respect to v is equal to, well, this is going to be 0, this is going to be 0, and we're just going to have 1k. So it's just going to be equal to k. And so now we're ready to at least evaluate this cross product. So we'll evaluate the cross product, and then we'll take the magnitude of that. So just this part, so just the cross product, the inside of that, before we take the magnitude. So copy and paste.
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Surface integral ex3 part 1 Parameterizing the outside surface Khan Academy.mp3
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So it's just going to be equal to k. And so now we're ready to at least evaluate this cross product. So we'll evaluate the cross product, and then we'll take the magnitude of that. So just this part, so just the cross product, the inside of that, before we take the magnitude. So copy and paste. So just that is going to be equal to the determinant, the 3 by 3 determinant. Let me write my components i, j, k. And then for r sub u, we have this. Our i component is negative sine of u.
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Surface integral ex3 part 1 Parameterizing the outside surface Khan Academy.mp3
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So copy and paste. So just that is going to be equal to the determinant, the 3 by 3 determinant. Let me write my components i, j, k. And then for r sub u, we have this. Our i component is negative sine of u. Our j component is cosine of u. It has no k components. We'll put a 0 right there.
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Surface integral ex3 part 1 Parameterizing the outside surface Khan Academy.mp3
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Our i component is negative sine of u. Our j component is cosine of u. It has no k components. We'll put a 0 right there. And then r sub v, no i component, no j component, and it has 1 for its k coefficient. And so we can just evaluate this. And so first we think about the i component.
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Surface integral ex3 part 1 Parameterizing the outside surface Khan Academy.mp3
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We'll put a 0 right there. And then r sub v, no i component, no j component, and it has 1 for its k coefficient. And so we can just evaluate this. And so first we think about the i component. Well, we'll ignore that column, that row. It's going to be cosine of u minus 0. So it's going to be cosine of u i.
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Surface integral ex3 part 1 Parameterizing the outside surface Khan Academy.mp3
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And so first we think about the i component. Well, we'll ignore that column, that row. It's going to be cosine of u minus 0. So it's going to be cosine of u i. And then we'll have minus j times, ignore column, ignore row, negative sine of u times 1 minus 0. So it's negative j times negative sine of u. So it's just going to be plus sine of uj.
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Surface integral ex3 part 1 Parameterizing the outside surface Khan Academy.mp3
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So it's going to be cosine of u i. And then we'll have minus j times, ignore column, ignore row, negative sine of u times 1 minus 0. So it's negative j times negative sine of u. So it's just going to be plus sine of uj. And then for the k, we have this times 0 minus this times 0. So we're just going to have 0. So this is the cross product.
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Surface integral ex3 part 1 Parameterizing the outside surface Khan Academy.mp3
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So it's just going to be plus sine of uj. And then for the k, we have this times 0 minus this times 0. So we're just going to have 0. So this is the cross product. And if we were to take the magnitude of this, so now we're ready to take the magnitude of this. It's going to be equal to, the magnitude of this is going to be equal to the square root of cosine of u squared plus sine of u squared. There is no k component.
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Surface integral ex3 part 1 Parameterizing the outside surface Khan Academy.mp3
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So this is the cross product. And if we were to take the magnitude of this, so now we're ready to take the magnitude of this. It's going to be equal to, the magnitude of this is going to be equal to the square root of cosine of u squared plus sine of u squared. There is no k component. And from the unit circle, this is the most basic trig identity. This is just going to be equal to 1. And so we have this part right over here just simplifies to 1, which is nice.
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Surface integral ex3 part 1 Parameterizing the outside surface Khan Academy.mp3
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There is no k component. And from the unit circle, this is the most basic trig identity. This is just going to be equal to 1. And so we have this part right over here just simplifies to 1, which is nice. And ds simplifies to du dv for at least this surface. And so now we are ready to attempt to evaluate the integral. But I'm almost running out of time.
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Surface integral ex3 part 1 Parameterizing the outside surface Khan Academy.mp3
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So what I'd like to do here and in the following few videos is talk about how you take the partial derivative of vector-valued functions. So the kind of thing I have in mind will be a function with a multivariable input. So this specific example have a two-variable input, T and S. You could think of that as a two-dimensional space as the input or just two separate numbers. And its output will be three-dimensional. The first component, T squared minus S squared. The Y component will be S times T. And that Z component will be T times S squared minus S times T squared. Minus S times T squared.
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Computing the partial derivative of a vector-valued function.mp3
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And its output will be three-dimensional. The first component, T squared minus S squared. The Y component will be S times T. And that Z component will be T times S squared minus S times T squared. Minus S times T squared. And the way that you compute a partial derivative of a guy like this is actually relatively straightforward. If you were to just guess what it might mean, you'd probably guess right. It will look like partial of V with respect to one of its input variables, and I'll choose T with respect to T. And you just do it component-wise, which means you look at each component and you do the partial derivative to that.
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Computing the partial derivative of a vector-valued function.mp3
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Minus S times T squared. And the way that you compute a partial derivative of a guy like this is actually relatively straightforward. If you were to just guess what it might mean, you'd probably guess right. It will look like partial of V with respect to one of its input variables, and I'll choose T with respect to T. And you just do it component-wise, which means you look at each component and you do the partial derivative to that. Because each component is just a normal scalar-valued function. So you go up to the top one and you say T squared looks like a variable as far as T is concerned, and its derivative is 2T. But S squared looks like a constant, so its derivative is zero.
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Computing the partial derivative of a vector-valued function.mp3
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It will look like partial of V with respect to one of its input variables, and I'll choose T with respect to T. And you just do it component-wise, which means you look at each component and you do the partial derivative to that. Because each component is just a normal scalar-valued function. So you go up to the top one and you say T squared looks like a variable as far as T is concerned, and its derivative is 2T. But S squared looks like a constant, so its derivative is zero. S times T, when S looks like a constant and when T looks like a variable, has a derivative of S, then T times S squared, when T is the variable and S is the constant, just looks like that constant, which is S squared, minus S times T squared. So now derivative of T squared is 2T, and that constant S stays in. So that's 2 times S times T. And that's how you compute it.
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Computing the partial derivative of a vector-valued function.mp3
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But S squared looks like a constant, so its derivative is zero. S times T, when S looks like a constant and when T looks like a variable, has a derivative of S, then T times S squared, when T is the variable and S is the constant, just looks like that constant, which is S squared, minus S times T squared. So now derivative of T squared is 2T, and that constant S stays in. So that's 2 times S times T. And that's how you compute it. Probably relatively straightforward. The way you do it with respect to S is very similar. But where this gets fun and where this gets cool is how you interpret the partial derivative, right?
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Computing the partial derivative of a vector-valued function.mp3
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So that's 2 times S times T. And that's how you compute it. Probably relatively straightforward. The way you do it with respect to S is very similar. But where this gets fun and where this gets cool is how you interpret the partial derivative, right? How you interpret this value that we just found. And what that means depends a lot on how you actually visualize the function. So what I'll go ahead and do in the next video, and in the next few ones, is talk about visualizing this function.
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Computing the partial derivative of a vector-valued function.mp3
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But where this gets fun and where this gets cool is how you interpret the partial derivative, right? How you interpret this value that we just found. And what that means depends a lot on how you actually visualize the function. So what I'll go ahead and do in the next video, and in the next few ones, is talk about visualizing this function. It'll be as a parametric surface in three-dimensional space. That's why I've got my grapher program out here. And I think you'll find there's actually a very satisfying understanding of what this value means.
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Computing the partial derivative of a vector-valued function.mp3
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But we saw that if our vector field is the gradient of a scalar field, then we call it conservative. So that tells us that F is a conservative vector field. And it also tells us, and this was the big takeaway from the last video, that the line integral of F between two points, so let me draw two points here. So let me say, let me draw my coordinates just so that we know we're on the xy plane. My axes, x-axis, y-axis. Let's say I have that point and that point, and I have two different paths between those two points. So I have path one that goes something like that.
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Closed curve line integrals of conservative vector fields Multivariable Calculus Khan Academy.mp3
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So let me say, let me draw my coordinates just so that we know we're on the xy plane. My axes, x-axis, y-axis. Let's say I have that point and that point, and I have two different paths between those two points. So I have path one that goes something like that. So I'll call that c1, and it goes in that direction. And then I have, maybe in a different shade of green, c2. It goes like that.
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Closed curve line integrals of conservative vector fields Multivariable Calculus Khan Academy.mp3
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So I have path one that goes something like that. So I'll call that c1, and it goes in that direction. And then I have, maybe in a different shade of green, c2. It goes like that. They both start here and go to there. c2, we learned in the last video that the line integral is path independent between any two points. So in this case, the line integral along c1 of F dot dr is going to be equal to the line integral of c2, over the path c2 of F dot dr.
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Closed curve line integrals of conservative vector fields Multivariable Calculus Khan Academy.mp3
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It goes like that. They both start here and go to there. c2, we learned in the last video that the line integral is path independent between any two points. So in this case, the line integral along c1 of F dot dr is going to be equal to the line integral of c2, over the path c2 of F dot dr. If we have a potential in a region, and we're maybe everywhere, then the line integral between any two points is independent of the path. That's the neat thing about a conservative field. Now what I want to do in this video is do a little bit of an extension of the takeaway of the last video.
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Closed curve line integrals of conservative vector fields Multivariable Calculus Khan Academy.mp3
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