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And so this hopefully makes it clear what's going on here. This is telling us how much mass is crossing that little ds, that little section of the curve per second. And if you were to add up, and that's what integrals are all about, adding up an infinite number of these infinitesimally small ds's, if you were to add all of that up, you're going to get the value of this entire integral is going to be in kilograms per second. And it's essentially going to say how much mass is exiting this curve at any given point or at any given time. So this is mass, so this whole integral, so the whole integral, so the whole integral, let me rewrite it, of F dot n ds tells us the mass exiting the curve per second. And this should also be consistent. In the last video, we saw that this is equivalent to, and this is where we kind of view it as a two-dimensional divergence theorem.
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Conceptual clarification for 2D divergence theorem Multivariable Calculus Khan Academy.mp3
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And it's essentially going to say how much mass is exiting this curve at any given point or at any given time. So this is mass, so this whole integral, so the whole integral, so the whole integral, let me rewrite it, of F dot n ds tells us the mass exiting the curve per second. And this should also be consistent. In the last video, we saw that this is equivalent to, and this is where we kind of view it as a two-dimensional divergence theorem. In the last video, we saw that this is equivalent to the double integral over the area of the divergence of F, which is essentially just, well, I could write it two ways, the divergence of F, and this right over here, that's just the partial of, let me write it down here so I have some space, the partial of the i component with respect to x, let me write it over here, I don't want to do this too fast and loose. So this right over here is going to be the partial of rho m, let me write it like this, rho m with respect to x plus the partial of the y component, rho n, with respect to y, times each little chunk of area, times each little chunk of area. Well, what are the units of this going to be right over here?
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Conceptual clarification for 2D divergence theorem Multivariable Calculus Khan Academy.mp3
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In the last video, we saw that this is equivalent to, and this is where we kind of view it as a two-dimensional divergence theorem. In the last video, we saw that this is equivalent to the double integral over the area of the divergence of F, which is essentially just, well, I could write it two ways, the divergence of F, and this right over here, that's just the partial of, let me write it down here so I have some space, the partial of the i component with respect to x, let me write it over here, I don't want to do this too fast and loose. So this right over here is going to be the partial of rho m, let me write it like this, rho m with respect to x plus the partial of the y component, rho n, with respect to y, times each little chunk of area, times each little chunk of area. Well, what are the units of this going to be right over here? We know what rho m is, rho m gives us kilogram per meter second. But if we were to take, essentially, the derivative with respect to meters again, the units here are going to give, the units for either of these, the units for either of these characters are going to be kilograms per meter second per second, because we're taking the derivative with respect, sorry, per meter, we're taking the derivative with respect to another unit of distance. So you're going to take per meter, so you're going to have another meter right here in the denominator, that's going to be the units here, and then you're multiplying it times an area.
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Conceptual clarification for 2D divergence theorem Multivariable Calculus Khan Academy.mp3
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Well, what are the units of this going to be right over here? We know what rho m is, rho m gives us kilogram per meter second. But if we were to take, essentially, the derivative with respect to meters again, the units here are going to give, the units for either of these, the units for either of these characters are going to be kilograms per meter second per second, because we're taking the derivative with respect, sorry, per meter, we're taking the derivative with respect to another unit of distance. So you're going to take per meter, so you're going to have another meter right here in the denominator, that's going to be the units here, and then you're multiplying it times an area. You're multiplying it times an area, so that would be meter squared, that's right, this right over here is square meters, they cancel it out, and once again, this whole part right over here that you're summing up gives us kilograms per second. So you're having a bunch of kilograms per second and you're just adding them up throughout the entire, throughout this entire area right over here. So hopefully this makes a little bit more sense about kind of how to conceptualize this vector function f. If it confuses you, try your best to ignore it, I guess, but for me at least, this helped me out conceptual, having a stronger conception of what vector f could kind of represent.
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Conceptual clarification for 2D divergence theorem Multivariable Calculus Khan Academy.mp3
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Then the way to think about maximizing this function is to try to increase that value of c as much as you can without it falling off the circle. And the key observation is that happens when they're tangent. So you might kind of draw this out in a little sketch and say there's some curve representing your constraint, which in this case would be kind of where our circle is, and then the curve representing the contour would just kiss that curve, just barely touch it in some way. Now, that's pretty, but in terms of solving the problem, we still have some work to do. And the main tool we're gonna use here is the gradient. So let me go ahead and draw a lot more contour lines than there already are for x squared times y, so this is many of the contour lines, and I'll draw the gradient field, the gradient field of f. So I've made a video about the relationship between the gradient and contour lines, and the upshot of it is that these gradient vectors, every time they pass through a contour line, they're perpendicular to it. And the basic reason for that is if you walk along the contour line, the function isn't changing value, so if you want it to change most rapidly, you know, it kind of makes sense you should walk in the perpendicular direction, so that no component of the walk that you're taking is, you know, useless, is along the line where the function doesn't change.
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Lagrange multipliers, using tangency to solve constrained optimization.mp3
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Now, that's pretty, but in terms of solving the problem, we still have some work to do. And the main tool we're gonna use here is the gradient. So let me go ahead and draw a lot more contour lines than there already are for x squared times y, so this is many of the contour lines, and I'll draw the gradient field, the gradient field of f. So I've made a video about the relationship between the gradient and contour lines, and the upshot of it is that these gradient vectors, every time they pass through a contour line, they're perpendicular to it. And the basic reason for that is if you walk along the contour line, the function isn't changing value, so if you want it to change most rapidly, you know, it kind of makes sense you should walk in the perpendicular direction, so that no component of the walk that you're taking is, you know, useless, is along the line where the function doesn't change. But again, there's a whole video on that that's worth checking out if this feels unfamiliar. For our purposes, what it means is that when we're considering this point of tangency, the gradient of f at that point is gonna be some vector perpendicular to both of the curves at that point, so that little vector represents the gradient of f at this point on the plane. And we can do something very similar to understand the other curve.
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Lagrange multipliers, using tangency to solve constrained optimization.mp3
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And the basic reason for that is if you walk along the contour line, the function isn't changing value, so if you want it to change most rapidly, you know, it kind of makes sense you should walk in the perpendicular direction, so that no component of the walk that you're taking is, you know, useless, is along the line where the function doesn't change. But again, there's a whole video on that that's worth checking out if this feels unfamiliar. For our purposes, what it means is that when we're considering this point of tangency, the gradient of f at that point is gonna be some vector perpendicular to both of the curves at that point, so that little vector represents the gradient of f at this point on the plane. And we can do something very similar to understand the other curve. Right now, I've just written it as a constraint x squared plus y squared equals one, but, you know, to give that function a name, let's say that we've defined g of xy to be x squared plus y squared, x squared plus y squared. In that case, this constraint is pretty much just one of the contour lines for the function g. And we can take a look at that. If we go over here and we look at all of the other contour lines for this function g, and it should make sense that they're circles because this function is x squared plus y squared.
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Lagrange multipliers, using tangency to solve constrained optimization.mp3
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And we can do something very similar to understand the other curve. Right now, I've just written it as a constraint x squared plus y squared equals one, but, you know, to give that function a name, let's say that we've defined g of xy to be x squared plus y squared, x squared plus y squared. In that case, this constraint is pretty much just one of the contour lines for the function g. And we can take a look at that. If we go over here and we look at all of the other contour lines for this function g, and it should make sense that they're circles because this function is x squared plus y squared. And if we took a look at the gradient of g, and we go over and ask about the gradient of g, it has that same property, that every gradient vector, if it passes through a contour line, is perpendicular to it. So over on our drawing here, the gradient vector of g would also be perpendicular to both these curves. And, you know, maybe in this case, it's not as long as the gradient of f, or maybe it's longer.
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Lagrange multipliers, using tangency to solve constrained optimization.mp3
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If we go over here and we look at all of the other contour lines for this function g, and it should make sense that they're circles because this function is x squared plus y squared. And if we took a look at the gradient of g, and we go over and ask about the gradient of g, it has that same property, that every gradient vector, if it passes through a contour line, is perpendicular to it. So over on our drawing here, the gradient vector of g would also be perpendicular to both these curves. And, you know, maybe in this case, it's not as long as the gradient of f, or maybe it's longer. There's no reason that it would be the same length, but the important fact is that it's proportional. And the way that we're gonna write this in formulas is to say that the gradient of f evaluated, let's see, evaluated at whatever the maximizing value of x and y are. So we should give that a name probably.
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Lagrange multipliers, using tangency to solve constrained optimization.mp3
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And, you know, maybe in this case, it's not as long as the gradient of f, or maybe it's longer. There's no reason that it would be the same length, but the important fact is that it's proportional. And the way that we're gonna write this in formulas is to say that the gradient of f evaluated, let's see, evaluated at whatever the maximizing value of x and y are. So we should give that a name probably. Maybe, you know, x sub m, y sub m. The specific values of x and y that are gonna be at this point that maximizes the function subject to our constraint. So that's gonna be related to the gradient of g. It's not gonna be quite equal, so I'll leave some room here. Related to the gradient of g evaluated at that same point, xm, ym.
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So we should give that a name probably. Maybe, you know, x sub m, y sub m. The specific values of x and y that are gonna be at this point that maximizes the function subject to our constraint. So that's gonna be related to the gradient of g. It's not gonna be quite equal, so I'll leave some room here. Related to the gradient of g evaluated at that same point, xm, ym. And like I said, they're not equal. They're proportional. So we need to have some kind of proportionality constant in there.
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Lagrange multipliers, using tangency to solve constrained optimization.mp3
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Related to the gradient of g evaluated at that same point, xm, ym. And like I said, they're not equal. They're proportional. So we need to have some kind of proportionality constant in there. And you almost always use the variable lambda. And this guy has a fancy name. It's called a Lagrange multiplier.
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Lagrange multipliers, using tangency to solve constrained optimization.mp3
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So we need to have some kind of proportionality constant in there. And you almost always use the variable lambda. And this guy has a fancy name. It's called a Lagrange multiplier. Lagrange. Lagrange was one of those famous French mathematicians. I always get him confused with some of the other French mathematicians at the time, like Legendre or Laplace.
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Lagrange multipliers, using tangency to solve constrained optimization.mp3
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It's called a Lagrange multiplier. Lagrange. Lagrange was one of those famous French mathematicians. I always get him confused with some of the other French mathematicians at the time, like Legendre or Laplace. There's a whole bunch of things. Let's see, multiplier. Distracting myself talking here.
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Lagrange multipliers, using tangency to solve constrained optimization.mp3
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I always get him confused with some of the other French mathematicians at the time, like Legendre or Laplace. There's a whole bunch of things. Let's see, multiplier. Distracting myself talking here. So Lagrange multiplier. So there's a number of things in multivariable calculus named after Lagrange. And this is one of the big ones.
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Lagrange multipliers, using tangency to solve constrained optimization.mp3
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Distracting myself talking here. So Lagrange multiplier. So there's a number of things in multivariable calculus named after Lagrange. And this is one of the big ones. This is a technique that he kind of developed or at the very least popularized. And the core idea is to just set these gradients equal to each other, because that represents when the contour line for one function is tangent to the contour line of another. So this, this is something that we can actually work with.
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Lagrange multipliers, using tangency to solve constrained optimization.mp3
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And this is one of the big ones. This is a technique that he kind of developed or at the very least popularized. And the core idea is to just set these gradients equal to each other, because that represents when the contour line for one function is tangent to the contour line of another. So this, this is something that we can actually work with. And let's start working it out, right? Let's see what this translates to in formulas. So I already have g written here, so let's go ahead and just evaluate what the gradient of g should be.
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Lagrange multipliers, using tangency to solve constrained optimization.mp3
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So this, this is something that we can actually work with. And let's start working it out, right? Let's see what this translates to in formulas. So I already have g written here, so let's go ahead and just evaluate what the gradient of g should be. And that's the gradient of x squared plus y squared. And the way that we take our gradient is it's gonna be a vector whose components are all the partial derivatives. So the first component is the partial derivative with respect to x.
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Lagrange multipliers, using tangency to solve constrained optimization.mp3
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So I already have g written here, so let's go ahead and just evaluate what the gradient of g should be. And that's the gradient of x squared plus y squared. And the way that we take our gradient is it's gonna be a vector whose components are all the partial derivatives. So the first component is the partial derivative with respect to x. So we treat x as a variable. Y looks like a constant. The derivative is two x.
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Lagrange multipliers, using tangency to solve constrained optimization.mp3
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So the first component is the partial derivative with respect to x. So we treat x as a variable. Y looks like a constant. The derivative is two x. The second component, the partial derivative with respect to y. So now we're treating y as the variable. X is the constant.
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Lagrange multipliers, using tangency to solve constrained optimization.mp3
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The derivative is two x. The second component, the partial derivative with respect to y. So now we're treating y as the variable. X is the constant. So the derivative looks like two y. Okay, so that's the gradient of g. And then the gradient of f, gradient of f, is gonna look like gradient of, let's see, what is x? What is f?
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Lagrange multipliers, using tangency to solve constrained optimization.mp3
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X is the constant. So the derivative looks like two y. Okay, so that's the gradient of g. And then the gradient of f, gradient of f, is gonna look like gradient of, let's see, what is x? What is f? It's x squared times y. So x squared times y. We do the same thing.
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Lagrange multipliers, using tangency to solve constrained optimization.mp3
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What is f? It's x squared times y. So x squared times y. We do the same thing. First component, partial derivative with respect to x. X looks like a variable, so its derivative is two times x. And then that y looks like a constant when we're up here. But then partial derivative with respect to y.
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Lagrange multipliers, using tangency to solve constrained optimization.mp3
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We do the same thing. First component, partial derivative with respect to x. X looks like a variable, so its derivative is two times x. And then that y looks like a constant when we're up here. But then partial derivative with respect to y. That y looks like a variable. That x squared just looks like a constant sitting in front of it. So that's what we get.
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Lagrange multipliers, using tangency to solve constrained optimization.mp3
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But then partial derivative with respect to y. That y looks like a variable. That x squared just looks like a constant sitting in front of it. So that's what we get. And now if we kind of work out this Lagrange multiplier expression using these two vectors, what we have written, what we're gonna have written is that this vector, two xy x squared, is proportional with a proportionality constant lambda to the gradient vector for g. So two x, two y. And if you want, you can think about this as two separate equations. I mean, right now it's one equation with vectors, but really what this is saying is you've got two separate equations.
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Lagrange multipliers, using tangency to solve constrained optimization.mp3
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So that's what we get. And now if we kind of work out this Lagrange multiplier expression using these two vectors, what we have written, what we're gonna have written is that this vector, two xy x squared, is proportional with a proportionality constant lambda to the gradient vector for g. So two x, two y. And if you want, you can think about this as two separate equations. I mean, right now it's one equation with vectors, but really what this is saying is you've got two separate equations. Two times xy is equal to lambda. Ah, gotta change colors a lot here. Lambda times two x. Gonna be a stickler for color.
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Lagrange multipliers, using tangency to solve constrained optimization.mp3
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I mean, right now it's one equation with vectors, but really what this is saying is you've got two separate equations. Two times xy is equal to lambda. Ah, gotta change colors a lot here. Lambda times two x. Gonna be a stickler for color. Keep red all of the things associated with g. And then the second equation is that x squared is equal to lambda times two y. And this might seem like a problem because we have three unknowns, x, y, and this new lambda that we introduced. Kind of shot ourselves in the foot by giving ourselves a new variable to deal with.
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Lagrange multipliers, using tangency to solve constrained optimization.mp3
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Lambda times two x. Gonna be a stickler for color. Keep red all of the things associated with g. And then the second equation is that x squared is equal to lambda times two y. And this might seem like a problem because we have three unknowns, x, y, and this new lambda that we introduced. Kind of shot ourselves in the foot by giving ourselves a new variable to deal with. But we only have two equations. So in order to solve this, we're gonna need three equations. And the third equation is something that we've known the whole time.
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Lagrange multipliers, using tangency to solve constrained optimization.mp3
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Kind of shot ourselves in the foot by giving ourselves a new variable to deal with. But we only have two equations. So in order to solve this, we're gonna need three equations. And the third equation is something that we've known the whole time. It's been part of the original problem. It's the constraint itself, x squared plus y. X squared plus y squared, excuse me, equals one. So that, that third equation, x squared plus y squared is equal to one.
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Lagrange multipliers, using tangency to solve constrained optimization.mp3
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And the third equation is something that we've known the whole time. It's been part of the original problem. It's the constraint itself, x squared plus y. X squared plus y squared, excuse me, equals one. So that, that third equation, x squared plus y squared is equal to one. So these are the three equations that characterize our constrained optimization problem. The bottom one just tells you that we have to be on this unit circle here. So let me just highlight it.
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Lagrange multipliers, using tangency to solve constrained optimization.mp3
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So that, that third equation, x squared plus y squared is equal to one. So these are the three equations that characterize our constrained optimization problem. The bottom one just tells you that we have to be on this unit circle here. So let me just highlight it. We have to be on this unit circle. And then these top two tell us what's necessary in order for our contour lines, the contour of f and the contour of g, to be perfectly tangent with each other. So in the next video, I'll go ahead and solve this.
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Lagrange multipliers, using tangency to solve constrained optimization.mp3
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So let me just highlight it. We have to be on this unit circle. And then these top two tell us what's necessary in order for our contour lines, the contour of f and the contour of g, to be perfectly tangent with each other. So in the next video, I'll go ahead and solve this. At this point, it's pretty much just algebra to deal with, but it's worth going through. And then the next couple ones, I'll talk about a way that you can encapsulate all three of these equations into one expression, and also a little bit about the interpretation of this lambda that we introduced. Because it's not actually just a dummy variable.
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Lagrange multipliers, using tangency to solve constrained optimization.mp3
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Now that we've explored Stokes' theorem a little bit, I want to talk about the situations in which we can use it. And we'll see it's a pretty general theorem, but we do have to think about what type of surfaces and what type of boundaries of those surfaces we are actually dealing with. And in the case of Stokes, we need surfaces that are piecewise smooth. Smooth, piecewise smooth surfaces. So this surface right over here, it is actually smooth, not just even piecewise smooth. Sounds like a very fancy term, but all the smooth part means, all the smooth part means is that you have continuous derivatives. And since we're talking about a surface, we're gonna think about continuous partial derivatives, regardless of which direction you pick.
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Conditions for stokes theorem Multivariable Calculus Khan Academy.mp3
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Smooth, piecewise smooth surfaces. So this surface right over here, it is actually smooth, not just even piecewise smooth. Sounds like a very fancy term, but all the smooth part means, all the smooth part means is that you have continuous derivatives. And since we're talking about a surface, we're gonna think about continuous partial derivatives, regardless of which direction you pick. So this is continuous derivatives. And another way to think about that conceptually is that if you pick a direction on the surface, if you say that we go in that direction, the slope in that direction changes gradually. It doesn't jump around.
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Conditions for stokes theorem Multivariable Calculus Khan Academy.mp3
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And since we're talking about a surface, we're gonna think about continuous partial derivatives, regardless of which direction you pick. So this is continuous derivatives. And another way to think about that conceptually is that if you pick a direction on the surface, if you say that we go in that direction, the slope in that direction changes gradually. It doesn't jump around. If you pick this direction right over here, the slope is changing gradually. So we have a continuous derivative. Now you're like, well, what does the piecewise mean?
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Conditions for stokes theorem Multivariable Calculus Khan Academy.mp3
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It doesn't jump around. If you pick this direction right over here, the slope is changing gradually. So we have a continuous derivative. Now you're like, well, what does the piecewise mean? Well, the piecewise actually allows us to use Stokes' theorem with more surfaces. Because if we have a surface that looks like, let's say a surface that looks like this, let's say it looks like a cup. So this is the opening of the top of the cup.
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Conditions for stokes theorem Multivariable Calculus Khan Academy.mp3
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Now you're like, well, what does the piecewise mean? Well, the piecewise actually allows us to use Stokes' theorem with more surfaces. Because if we have a surface that looks like, let's say a surface that looks like this, let's say it looks like a cup. So this is the opening of the top of the cup. Let's say it has no opening on top. So we can see the backside of the cup. And this is the side of the cup.
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Conditions for stokes theorem Multivariable Calculus Khan Academy.mp3
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So this is the opening of the top of the cup. Let's say it has no opening on top. So we can see the backside of the cup. And this is the side of the cup. And then this right over here is the bottom of the cup. And if it was transparent, we could actually see through it. So a surface like this is not entirely smooth because it has edges.
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Conditions for stokes theorem Multivariable Calculus Khan Academy.mp3
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And this is the side of the cup. And then this right over here is the bottom of the cup. And if it was transparent, we could actually see through it. So a surface like this is not entirely smooth because it has edges. There are points right over here. So this edge right over here, if we pick this, let's say we pick this direction to go. And if we go in this direction along the bottom, then right when we get to the edge, then all of a sudden the slope changes dramatically.
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Conditions for stokes theorem Multivariable Calculus Khan Academy.mp3
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So a surface like this is not entirely smooth because it has edges. There are points right over here. So this edge right over here, if we pick this, let's say we pick this direction to go. And if we go in this direction along the bottom, then right when we get to the edge, then all of a sudden the slope changes dramatically. It jumps. So the slope is not continuous at that edge. The slope jumps and we start going straight up.
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Conditions for stokes theorem Multivariable Calculus Khan Academy.mp3
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And if we go in this direction along the bottom, then right when we get to the edge, then all of a sudden the slope changes dramatically. It jumps. So the slope is not continuous at that edge. The slope jumps and we start going straight up. And so this entire surface is not smooth, but the piecewise actually gives us an out. This tells us that it's okay as long as we can break the surface up into pieces that are smooth. And so this cup, we can break it up.
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The slope jumps and we start going straight up. And so this entire surface is not smooth, but the piecewise actually gives us an out. This tells us that it's okay as long as we can break the surface up into pieces that are smooth. And so this cup, we can break it up. And we've been doing this when we've been tackling surface integrals. We can break it up into the bottom part, which is a smooth surface. It has a continuous derivative.
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Conditions for stokes theorem Multivariable Calculus Khan Academy.mp3
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And so this cup, we can break it up. And we've been doing this when we've been tackling surface integrals. We can break it up into the bottom part, which is a smooth surface. It has a continuous derivative. And the sides, the sides, the side, which kind of wraps around, is also a smooth surface. So most things that you will encounter in a traditional calculus course actually do, especially surfaces, do fit piecewise smooth. And the things that don't actually are fairly hard to visualize.
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Conditions for stokes theorem Multivariable Calculus Khan Academy.mp3
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It has a continuous derivative. And the sides, the sides, the side, which kind of wraps around, is also a smooth surface. So most things that you will encounter in a traditional calculus course actually do, especially surfaces, do fit piecewise smooth. And the things that don't actually are fairly hard to visualize. And I imagine these ultra pointy, fractally looking things where it's hard to break it up into pieces that are actually smooth. That's for the surface part, but we also have to care about the boundary in order to apply Stokes' theorem. And that's that right over there.
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Conditions for stokes theorem Multivariable Calculus Khan Academy.mp3
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And the things that don't actually are fairly hard to visualize. And I imagine these ultra pointy, fractally looking things where it's hard to break it up into pieces that are actually smooth. That's for the surface part, but we also have to care about the boundary in order to apply Stokes' theorem. And that's that right over there. The boundary needs to be a simple, which means it doesn't cross itself, a simple, closed, piecewise smooth boundary. So once again, simple and closed, that just means, so this is not a simple boundary if it's really crossing itself, if it intersects itself, although you could break it up into two simple boundaries. But something like this is a simple boundary.
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Conditions for stokes theorem Multivariable Calculus Khan Academy.mp3
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And that's that right over there. The boundary needs to be a simple, which means it doesn't cross itself, a simple, closed, piecewise smooth boundary. So once again, simple and closed, that just means, so this is not a simple boundary if it's really crossing itself, if it intersects itself, although you could break it up into two simple boundaries. But something like this is a simple boundary. So that's a simple boundary right over there. It also has to be closed, which really means it just loops in on itself. You just won't have something like that.
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Conditions for stokes theorem Multivariable Calculus Khan Academy.mp3
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But something like this is a simple boundary. So that's a simple boundary right over there. It also has to be closed, which really means it just loops in on itself. You just won't have something like that. It actually has to close. It actually has to loop in on itself in order to use Stokes' theorem. And once again, it has to be piecewise smooth, but now we're talking about a path or a line or a curve like this.
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Conditions for stokes theorem Multivariable Calculus Khan Academy.mp3
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You just won't have something like that. It actually has to close. It actually has to loop in on itself in order to use Stokes' theorem. And once again, it has to be piecewise smooth, but now we're talking about a path or a line or a curve like this. And piecewise smooth just means that you can break it up into sections where the derivative is continuous. The way I've drawn this one, this one, and this one, the slope is changing gradually. So over there, the slope is like that.
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Conditions for stokes theorem Multivariable Calculus Khan Academy.mp3
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And once again, it has to be piecewise smooth, but now we're talking about a path or a line or a curve like this. And piecewise smooth just means that you can break it up into sections where the derivative is continuous. The way I've drawn this one, this one, and this one, the slope is changing gradually. So over there, the slope is like that. And it's changing gradually as we go around this path. Something that is not smooth, a path that is not smooth might look something like this, might look something like that. And the places that aren't smooth are at the edges, not smooth there, not smooth there, not smooth there, not smooth there.
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Conditions for stokes theorem Multivariable Calculus Khan Academy.mp3
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So over there, the slope is like that. And it's changing gradually as we go around this path. Something that is not smooth, a path that is not smooth might look something like this, might look something like that. And the places that aren't smooth are at the edges, not smooth there, not smooth there, not smooth there, not smooth there. But we just have to be simple, close. So this is simple and closed, and it's not smooth, but it is piecewise smooth. We can break it up into this section of the path, which is that line right over there is smooth.
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Conditions for stokes theorem Multivariable Calculus Khan Academy.mp3
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And the places that aren't smooth are at the edges, not smooth there, not smooth there, not smooth there, not smooth there. But we just have to be simple, close. So this is simple and closed, and it's not smooth, but it is piecewise smooth. We can break it up into this section of the path, which is that line right over there is smooth. That line right over there is smooth. That line is smooth, and that line is smooth. And we've done that when we've evaluated line integrals.
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Conditions for stokes theorem Multivariable Calculus Khan Academy.mp3
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In this video, I will attempt to prove, or actually this in the next several videos, attempt to prove a special case version of Stokes' Theorem, or essentially Stokes' Theorem for a special case. And I'm doing this because the proof will be a little bit simpler, but at the same time, it's pretty convincing. And the special case we're going to assume is that the surface we're dealing with is a function of x and y. So if you give me any particular x and y, it only determines one point on that surface. So a surface like this would be the case. So it's kind of a mapping of this region in the xy plane into three dimensions. So for any xy, we can figure out the height.
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Stokes' theorem proof part 1 Multivariable Calculus Khan Academy.mp3
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So if you give me any particular x and y, it only determines one point on that surface. So a surface like this would be the case. So it's kind of a mapping of this region in the xy plane into three dimensions. So for any xy, we can figure out the height. So essentially z is going to be a function of x and y, and we can get a point on the surface. So this proof would not apply to a surface that's like a sphere or something like that, where any point on the xy plane could actually determine two points on our surface. But this is a pretty good start.
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Stokes' theorem proof part 1 Multivariable Calculus Khan Academy.mp3
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So for any xy, we can figure out the height. So essentially z is going to be a function of x and y, and we can get a point on the surface. So this proof would not apply to a surface that's like a sphere or something like that, where any point on the xy plane could actually determine two points on our surface. But this is a pretty good start. The other thing that we are going to assume, we are going to assume that z, which is essentially a function of x and y, that this function of x and y has continuous second-order derivatives. So continuous second derivatives. And the reason why I'm going to make that assumption is it's going to help us in our proof later on.
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Stokes' theorem proof part 1 Multivariable Calculus Khan Academy.mp3
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But this is a pretty good start. The other thing that we are going to assume, we are going to assume that z, which is essentially a function of x and y, that this function of x and y has continuous second-order derivatives. So continuous second derivatives. And the reason why I'm going to make that assumption is it's going to help us in our proof later on. It's going to allow us to say that the partial of our surface, or the partial of z with respect to x, and then taking the derivative of that with respect to y, is going to be the same as the partial of z with respect to y, and then taking the derivative of that with respect to x. And in order to be able to make this statement, we have to assume that z, or this z right over here, z is a function of x and y, has continuous second-order derivatives. And over here, we've just written our vector field F that we're going to deal with when we're trying to play with Stokes' Theorem, and we'll assume that it has continuous first-order derivatives.
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Stokes' theorem proof part 1 Multivariable Calculus Khan Academy.mp3
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And the reason why I'm going to make that assumption is it's going to help us in our proof later on. It's going to allow us to say that the partial of our surface, or the partial of z with respect to x, and then taking the derivative of that with respect to y, is going to be the same as the partial of z with respect to y, and then taking the derivative of that with respect to x. And in order to be able to make this statement, we have to assume that z, or this z right over here, z is a function of x and y, has continuous second-order derivatives. And over here, we've just written our vector field F that we're going to deal with when we're trying to play with Stokes' Theorem, and we'll assume that it has continuous first-order derivatives. Now with that out of the way, let's think about what Stokes' Theorem tells us, and then we'll think about, for this particular case, how we can write it out, and then hopefully we will see that the two things are equal. So let me write it out. So Stokes' Theorem tells us that F dot dr over some path, and the path that we care about is essentially this path right over here.
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Stokes' theorem proof part 1 Multivariable Calculus Khan Academy.mp3
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And over here, we've just written our vector field F that we're going to deal with when we're trying to play with Stokes' Theorem, and we'll assume that it has continuous first-order derivatives. Now with that out of the way, let's think about what Stokes' Theorem tells us, and then we'll think about, for this particular case, how we can write it out, and then hopefully we will see that the two things are equal. So let me write it out. So Stokes' Theorem tells us that F dot dr over some path, and the path that we care about is essentially this path right over here. I'll do it in blue. It's this path right over here. This is the boundary of our surface.
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Stokes' theorem proof part 1 Multivariable Calculus Khan Academy.mp3
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So Stokes' Theorem tells us that F dot dr over some path, and the path that we care about is essentially this path right over here. I'll do it in blue. It's this path right over here. This is the boundary of our surface. So this is C right over here. Stokes' Theorem tells us that this should be the same thing. This should be equivalent to the surface integral over our surface of curl of F dot ds, dot dotted with the surface itself.
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Stokes' theorem proof part 1 Multivariable Calculus Khan Academy.mp3
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This is the boundary of our surface. So this is C right over here. Stokes' Theorem tells us that this should be the same thing. This should be equivalent to the surface integral over our surface of curl of F dot ds, dot dotted with the surface itself. And so in this video, I want to focus, or probably this and the next video, I want to focus on the second half. I want to focus this. I want to see how we can express this given the assumptions we've made.
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Stokes' theorem proof part 1 Multivariable Calculus Khan Academy.mp3
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This should be equivalent to the surface integral over our surface of curl of F dot ds, dot dotted with the surface itself. And so in this video, I want to focus, or probably this and the next video, I want to focus on the second half. I want to focus this. I want to see how we can express this given the assumptions we've made. And then after that, we're gonna see how we can express this given the same assumptions, and then hopefully we'll find that we get them to be equal to each other. So let's just start figuring out what the curl of F is equal to. So the curl of F is equal to, you could view it as the del operator crossed with our vector field F, which is equal to, we can write our components.
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Stokes' theorem proof part 1 Multivariable Calculus Khan Academy.mp3
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I want to see how we can express this given the assumptions we've made. And then after that, we're gonna see how we can express this given the same assumptions, and then hopefully we'll find that we get them to be equal to each other. So let's just start figuring out what the curl of F is equal to. So the curl of F is equal to, you could view it as the del operator crossed with our vector field F, which is equal to, we can write our components. So I, I'll do them in different colors, I, J, and K components. I, J, and K components. And then I need to write my del operator, or my partial operators, I guess I should call them.
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Stokes' theorem proof part 1 Multivariable Calculus Khan Academy.mp3
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So the curl of F is equal to, you could view it as the del operator crossed with our vector field F, which is equal to, we can write our components. So I, I'll do them in different colors, I, J, and K components. I, J, and K components. And then I need to write my del operator, or my partial operators, I guess I should call them. So the partial with respect to X, the partial with respect to Y, partial with respect to Z, and then I have to write the I, J, and K components of my vector field F. And I will do that in green, well, I'll do it in blue. And so we have P, which is a function of X, Y, Z, Q, which is a function of X, Y, and Z, and then R, which is a function of X, Y, and Z. And so this is going to evaluate as, it's going to be I, I times, so blank out that column, that row, it's gonna be the partial of R with respect to Y, partial of R with respect to Y, minus the partial of Q with respect to Z, minus the partial of Q with respect to Z.
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Stokes' theorem proof part 1 Multivariable Calculus Khan Academy.mp3
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And then I need to write my del operator, or my partial operators, I guess I should call them. So the partial with respect to X, the partial with respect to Y, partial with respect to Z, and then I have to write the I, J, and K components of my vector field F. And I will do that in green, well, I'll do it in blue. And so we have P, which is a function of X, Y, Z, Q, which is a function of X, Y, and Z, and then R, which is a function of X, Y, and Z. And so this is going to evaluate as, it's going to be I, I times, so blank out that column, that row, it's gonna be the partial of R with respect to Y, partial of R with respect to Y, minus the partial of Q with respect to Z, minus the partial of Q with respect to Z. And then checkerboard pattern, minus J, let me make that hat a little bit better, minus J, and then times the partial of R with respect to X, partial of R with respect to X, minus the partial of P with respect to Z, minus the partial of P with respect to Z. And then finally, plus K, plus K, and that's going to be times the partial of X, sorry, the partial of Q with respect to X, partial of Q with respect to X, minus the partial of P with respect to Y, minus the partial of P with respect to Y. So we figured out the curl of F, and I'll leave you there, I'm trying to make shorter videos now.
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Stokes' theorem proof part 1 Multivariable Calculus Khan Academy.mp3
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And here, I'd like to do the same, but for three dimensions. So a three-dimensional vector field is given by a function, a certain multivariable function, that has a three-dimensional input, given with coordinates x, y, and z, and then a three-dimensional vector output that has expressions that are somehow dependent on x, y, and z. I'll just put dots in here for now, but we'll fill this in with an example in just a moment. And the way that this works, just like with the two-dimensional vector field, you're gonna choose a sample of various points in three-dimensional space. And for each one of those points, you consider what the output of the function is, and that's gonna be some three-dimensional vector, and you draw that vector off of the point itself. So to start off, let's take a very simple example, one where the vector that it outputs is actually just a constant. So in this case, I'll make that constant the vector one, zero, zero. So what this vector is, it's just got a unit length in the x direction, so this is the x-axis.
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3d vector fields, introduction Multivariable calculus Khan Academy.mp3
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And for each one of those points, you consider what the output of the function is, and that's gonna be some three-dimensional vector, and you draw that vector off of the point itself. So to start off, let's take a very simple example, one where the vector that it outputs is actually just a constant. So in this case, I'll make that constant the vector one, zero, zero. So what this vector is, it's just got a unit length in the x direction, so this is the x-axis. So all of the vectors are gonna end up looking something like this, where it's a vector that has length one in the x direction. And when we do this at every possible point, well, not every possible point, but a sample of a whole bunch of points, whoops, we get a vector field that looks like this. At any given point in space, we get one of these little blue vectors, and all of them are the same.
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3d vector fields, introduction Multivariable calculus Khan Academy.mp3
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So what this vector is, it's just got a unit length in the x direction, so this is the x-axis. So all of the vectors are gonna end up looking something like this, where it's a vector that has length one in the x direction. And when we do this at every possible point, well, not every possible point, but a sample of a whole bunch of points, whoops, we get a vector field that looks like this. At any given point in space, we get one of these little blue vectors, and all of them are the same. They're just copies of each other, each pointing with unit length in the x direction. So as vector fields go, this is relatively boring, but we can make it a little bit more exciting if we make the input start to depend somehow on the actual input. So what I'll do to start, I'll just make the input y, zero, zero.
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3d vector fields, introduction Multivariable calculus Khan Academy.mp3
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At any given point in space, we get one of these little blue vectors, and all of them are the same. They're just copies of each other, each pointing with unit length in the x direction. So as vector fields go, this is relatively boring, but we can make it a little bit more exciting if we make the input start to depend somehow on the actual input. So what I'll do to start, I'll just make the input y, zero, zero. So they're still just gonna point in the x direction, but now it's gonna depend on the y value. So let's think for a second before I change the image what that's gonna mean. The y-axis is this one here, so now the z-axis is pointing straight in our face, that's the y.
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3d vector fields, introduction Multivariable calculus Khan Academy.mp3
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So what I'll do to start, I'll just make the input y, zero, zero. So they're still just gonna point in the x direction, but now it's gonna depend on the y value. So let's think for a second before I change the image what that's gonna mean. The y-axis is this one here, so now the z-axis is pointing straight in our face, that's the y. So as y increases value to one, two, three, the length of these vectors are gonna increase. It's gonna be a stronger vector in the x direction, a very strong vector in the x direction. And if y is negative, these vectors are gonna point in the opposite direction.
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3d vector fields, introduction Multivariable calculus Khan Academy.mp3
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The y-axis is this one here, so now the z-axis is pointing straight in our face, that's the y. So as y increases value to one, two, three, the length of these vectors are gonna increase. It's gonna be a stronger vector in the x direction, a very strong vector in the x direction. And if y is negative, these vectors are gonna point in the opposite direction. So let's see what that looks like. There we go. So in this vector field, color and length are used to indicate how the magnitude of the vector.
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3d vector fields, introduction Multivariable calculus Khan Academy.mp3
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And if y is negative, these vectors are gonna point in the opposite direction. So let's see what that looks like. There we go. So in this vector field, color and length are used to indicate how the magnitude of the vector. So red vectors are very long, blue vectors are pretty short, and at zero we don't even see any because those are vectors with zero length. And just like with two-dimensional vector fields, when you draw them, you lie a little bit. This one should have a length of one, right?
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3d vector fields, introduction Multivariable calculus Khan Academy.mp3
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So in this vector field, color and length are used to indicate how the magnitude of the vector. So red vectors are very long, blue vectors are pretty short, and at zero we don't even see any because those are vectors with zero length. And just like with two-dimensional vector fields, when you draw them, you lie a little bit. This one should have a length of one, right? Because when y is equal to one, this should have a unit length, but it's made really, really small. And this one up here, where y is five or six, should be a really long vector, but we're lying a little bit because if we actually drew them to scale, it would really clutter up the image. So a couple things to notice about this one.
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3d vector fields, introduction Multivariable calculus Khan Academy.mp3
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This one should have a length of one, right? Because when y is equal to one, this should have a unit length, but it's made really, really small. And this one up here, where y is five or six, should be a really long vector, but we're lying a little bit because if we actually drew them to scale, it would really clutter up the image. So a couple things to notice about this one. Since the output doesn't depend on x or z, if you move in the x direction, which is back and forth here, the vectors don't change. And if you move in the z direction, which is up and down, the vectors also don't change. They only change as you move in the y direction.
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3d vector fields, introduction Multivariable calculus Khan Academy.mp3
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So a couple things to notice about this one. Since the output doesn't depend on x or z, if you move in the x direction, which is back and forth here, the vectors don't change. And if you move in the z direction, which is up and down, the vectors also don't change. They only change as you move in the y direction. Okay, so this is, we're starting to get a feel for how the output can depend on the input. Now let's do something a little bit different. Let's say that all three of the components of the input depend on x, y, and z, but I'm just gonna make it kind of an identity function.
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3d vector fields, introduction Multivariable calculus Khan Academy.mp3
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They only change as you move in the y direction. Okay, so this is, we're starting to get a feel for how the output can depend on the input. Now let's do something a little bit different. Let's say that all three of the components of the input depend on x, y, and z, but I'm just gonna make it kind of an identity function. At a given point x, y, z, you output the vector itself, x, y, z. So let's think about what this would actually mean. And let's say you've got a given point, some point floating off in space.
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3d vector fields, introduction Multivariable calculus Khan Academy.mp3
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Let's say that all three of the components of the input depend on x, y, and z, but I'm just gonna make it kind of an identity function. At a given point x, y, z, you output the vector itself, x, y, z. So let's think about what this would actually mean. And let's say you've got a given point, some point floating off in space. What is the output vector for that? Well, the point has a certain x component, a certain y component, and a z component. And the vector that corresponds to x, y, z is gonna be the one from the origin to that point itself.
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3d vector fields, introduction Multivariable calculus Khan Academy.mp3
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And let's say you've got a given point, some point floating off in space. What is the output vector for that? Well, the point has a certain x component, a certain y component, and a z component. And the vector that corresponds to x, y, z is gonna be the one from the origin to that point itself. Let me just draw that here. From the origin to the point itself. And because of how we do vector fields, you move this so that instead of stemming from the origin, it actually stems from the point.
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3d vector fields, introduction Multivariable calculus Khan Academy.mp3
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And the vector that corresponds to x, y, z is gonna be the one from the origin to that point itself. Let me just draw that here. From the origin to the point itself. And because of how we do vector fields, you move this so that instead of stemming from the origin, it actually stems from the point. But the main thing to take away here is it's gonna point directly away from the origin. And the farther away the point is, the longer this vector will be. So with that, let's take a look at the vector field itself.
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3d vector fields, introduction Multivariable calculus Khan Academy.mp3
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And because of how we do vector fields, you move this so that instead of stemming from the origin, it actually stems from the point. But the main thing to take away here is it's gonna point directly away from the origin. And the farther away the point is, the longer this vector will be. So with that, let's take a look at the vector field itself. Here we go. So again, you kind of lie when you draw these. Like the vectors, these red guys that are out at the end, they should be really long.
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3d vector fields, introduction Multivariable calculus Khan Academy.mp3
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So with that, let's take a look at the vector field itself. Here we go. So again, you kind of lie when you draw these. Like the vectors, these red guys that are out at the end, they should be really long. Because this vector should be as long as that point is away from the origin. But to give a cleaner vector field, you scale things down. And notice the blue ones that are close to the center here are actually really, really short guys.
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3d vector fields, introduction Multivariable calculus Khan Academy.mp3
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Like the vectors, these red guys that are out at the end, they should be really long. Because this vector should be as long as that point is away from the origin. But to give a cleaner vector field, you scale things down. And notice the blue ones that are close to the center here are actually really, really short guys. And all of these are pointing directly away from the origin. And this is one of those vector fields that is actually pretty, a good one to have a strong intuition of. Because it comes up now and then, thinking about what the identity function looks like as a vector field itself.
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3d vector fields, introduction Multivariable calculus Khan Academy.mp3
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And the way we did it is we integrated with respect to x first. We said, well, let's pick a y and let's just figure out the area under the curve. And so we integrated with respect to x first, and then we integrated with respect to y. But we could have done it the other way around. So let's do that and just make sure we got the right answer. So let me erase a lot of this. So remember, our answer was 2 thirds when we integrated with respect to x first and then with respect to y.
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Double integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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But we could have done it the other way around. So let's do that and just make sure we got the right answer. So let me erase a lot of this. So remember, our answer was 2 thirds when we integrated with respect to x first and then with respect to y. But I will show you that we can integrate the other way around. That's good. We can get the same answer in two different ways.
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Double integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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So remember, our answer was 2 thirds when we integrated with respect to x first and then with respect to y. But I will show you that we can integrate the other way around. That's good. We can get the same answer in two different ways. So let me redraw that graph because I want to give you the intuition again. So that's my x-axis, y-axis, z-axis. And then this is my xy-plane down here.
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Double integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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We can get the same answer in two different ways. So let me redraw that graph because I want to give you the intuition again. So that's my x-axis, y-axis, z-axis. And then this is my xy-plane down here. y goes from 0 to 1, x goes from 0 to 2. And this, you could do this x equals 1, this is x equals 2, this is y equals 1. And then the graph, I will do my best to draw it.
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Double integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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And then this is my xy-plane down here. y goes from 0 to 1, x goes from 0 to 2. And this, you could do this x equals 1, this is x equals 2, this is y equals 1. And then the graph, I will do my best to draw it. It looks something, let me do it in a more, get some contrast going here. So the graph looks something like this. Let me see if I can draw it.
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Double integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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And then the graph, I will do my best to draw it. It looks something, let me do it in a more, get some contrast going here. So the graph looks something like this. Let me see if I can draw it. This side, it looks something like that. And then it comes down like that straight. And then the volume we care about is actually this volume underneath the graph.
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Double integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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Let me see if I can draw it. This side, it looks something like that. And then it comes down like that straight. And then the volume we care about is actually this volume underneath the graph. This is the top of the surface on that side. And we care about this volume underneath the surface. And then when we draw the bottom of the surface, let me do it in a darker color.
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Double integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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And then the volume we care about is actually this volume underneath the graph. This is the top of the surface on that side. And we care about this volume underneath the surface. And then when we draw the bottom of the surface, let me do it in a darker color. Looks something like this. It looks something like this, this is the bottom underneath the surface. I could even shade it a little bit just to show you that it's underneath the surface.
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Double integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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And then when we draw the bottom of the surface, let me do it in a darker color. Looks something like this. It looks something like this, this is the bottom underneath the surface. I could even shade it a little bit just to show you that it's underneath the surface. Hopefully that's a decent rendering of it. Let's look back at what we had before. It's like a page that I just flipped up at this point.
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Double integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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I could even shade it a little bit just to show you that it's underneath the surface. Hopefully that's a decent rendering of it. Let's look back at what we had before. It's like a page that I just flipped up at this point. And we care about this volume, kind of the colored area under there. So let's figure out how to do it. Last time we integrated with respect to x first, let's integrate with respect to y first.
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Double integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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It's like a page that I just flipped up at this point. And we care about this volume, kind of the colored area under there. So let's figure out how to do it. Last time we integrated with respect to x first, let's integrate with respect to y first. So let's hold x constant. So if we hold x constant, what we could do is for a given x, let's pick an x. So if we pick a given x, let's pick the x here.
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Double integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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Last time we integrated with respect to x first, let's integrate with respect to y first. So let's hold x constant. So if we hold x constant, what we could do is for a given x, let's pick an x. So if we pick a given x, let's pick the x here. Then what we can do, for a given x, you can view that function of x and y. If x is a constant, let's say if x is 1, then z is just equal to y squared. That's easy to figure out the area under.
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Double integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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So if we pick a given x, let's pick the x here. Then what we can do, for a given x, you can view that function of x and y. If x is a constant, let's say if x is 1, then z is just equal to y squared. That's easy to figure out the area under. As we can see, that x isn't a constant, but we can treat it as a constant. So for example, for any given x, we would have a curve like this. And what we could do is we could try to figure out the area of this curve first.
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Double integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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That's easy to figure out the area under. As we can see, that x isn't a constant, but we can treat it as a constant. So for example, for any given x, we would have a curve like this. And what we could do is we could try to figure out the area of this curve first. So how do we do that? Well, we just said we could kind of view this function up here as z is equal to xy squared, because that's exactly what it is. But we're holding x constant.
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Double integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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And what we could do is we could try to figure out the area of this curve first. So how do we do that? Well, we just said we could kind of view this function up here as z is equal to xy squared, because that's exactly what it is. But we're holding x constant. We're treating it like a constant. To figure out that area, we could take a dy, a change in y, multiply it by the height, which is xy squared. So we take xy squared, multiply it by dy.
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Double integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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But we're holding x constant. We're treating it like a constant. To figure out that area, we could take a dy, a change in y, multiply it by the height, which is xy squared. So we take xy squared, multiply it by dy. And if we want this entire area, we integrate it from y is equal to 0 to y is equal to 1. Fair enough. Now once we have that area, if we want the volume underneath this entire surface, what we could do is we could multiply this area times dx and get some depth going.
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Double integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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So we take xy squared, multiply it by dy. And if we want this entire area, we integrate it from y is equal to 0 to y is equal to 1. Fair enough. Now once we have that area, if we want the volume underneath this entire surface, what we could do is we could multiply this area times dx and get some depth going. Let me pick a nice color. This green. So let's say that's our dx.
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Double integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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Now once we have that area, if we want the volume underneath this entire surface, what we could do is we could multiply this area times dx and get some depth going. Let me pick a nice color. This green. So let's say that's our dx. So if we multiply that times dx, we would get some depth. Let me do a darker color, get some contrast. Sometimes I feel like that guy who paints on PBS.
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Double integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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So let's say that's our dx. So if we multiply that times dx, we would get some depth. Let me do a darker color, get some contrast. Sometimes I feel like that guy who paints on PBS. dx. So now we have the volume of this. You could kind of view it the area under the curve times a dx, so we have some depth here.
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Double integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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Sometimes I feel like that guy who paints on PBS. dx. So now we have the volume of this. You could kind of view it the area under the curve times a dx, so we have some depth here. So it's times dx. And if we want to figure out the entire volume under this surface, between the surface and the xy plane, given this constraint to our domain, we just integrate from x is equal to 0 to 2. All right.
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Double integrals 3 Double and triple integrals Multivariable Calculus Khan Academy.mp3
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