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So if, for example, let's say you did all of your calculations, and it turned out that lambda star was equal to 2.3. Previously, that just seemed like some dummy number that you ignore, and you just look at whatever the associated values here are. But if you plug this in the computer and you see lambda star equals 2.3, what that means is for a tiny change in budget, like let's say your budget increases from 10,000 to 10,001, it goes up to $10,001. So you increase your budget by just a little bit, a little dB. Then the ratio of the change in the maximizing revenue to that dB is about 2.3. So what that would mean is increasing your budget by $1 is gonna increase m star over here. It would mean that m star increases by about $2.30 for every dollar that you increase your budget.
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Meaning of Lagrange multiplier.mp3
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So you increase your budget by just a little bit, a little dB. Then the ratio of the change in the maximizing revenue to that dB is about 2.3. So what that would mean is increasing your budget by $1 is gonna increase m star over here. It would mean that m star increases by about $2.30 for every dollar that you increase your budget. And that's information you'd wanna know, right? If you see that this lambda star is a number bigger than one, you'd say, hey, maybe we should increase our budget. We increase it from $10,000 to 10,001, and we're making more money.
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Meaning of Lagrange multiplier.mp3
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It would mean that m star increases by about $2.30 for every dollar that you increase your budget. And that's information you'd wanna know, right? If you see that this lambda star is a number bigger than one, you'd say, hey, maybe we should increase our budget. We increase it from $10,000 to 10,001, and we're making more money. So maybe as long as lambda star is greater than one, you should keep doing whatever it takes to increase that budget. So this fact is quite surprising, I think, and it seems like it totally comes out of nowhere. So what I'm gonna do in the next video is prove this to you, is prove why this is true, why this lambda star value happens to be the rate of change for the maximum value of the thing we're trying to maximize with respect to this constant, with respect to whatever constant you set your constraint function equal to.
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Meaning of Lagrange multiplier.mp3
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We increase it from $10,000 to 10,001, and we're making more money. So maybe as long as lambda star is greater than one, you should keep doing whatever it takes to increase that budget. So this fact is quite surprising, I think, and it seems like it totally comes out of nowhere. So what I'm gonna do in the next video is prove this to you, is prove why this is true, why this lambda star value happens to be the rate of change for the maximum value of the thing we're trying to maximize with respect to this constant, with respect to whatever constant you set your constraint function equal to. For right now, though, I just want you to kind of try to sit back and digest what this means in the context of this specific economic example. And even if you never looked into the proof or never understood it there, I think this is an interesting and even useful tidbit of knowledge to have about Lagrange multipliers. So with that, I'll see you next video.
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Meaning of Lagrange multiplier.mp3
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Some type of substance. It gives us the mass density at any point in three dimensions. And let's say we have another function. This is a scalar function. It just gives us a number for any point in 3D. And then let's say we have another function, v, which is a vector function. It gives us a vector for any point in three dimensions, and this right over here tells us the velocity of that same fluid or gas or whatever we're talking about.
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Conceptual understanding of flux in three dimensions Multivariable Calculus Khan Academy.mp3
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This is a scalar function. It just gives us a number for any point in 3D. And then let's say we have another function, v, which is a vector function. It gives us a vector for any point in three dimensions, and this right over here tells us the velocity of that same fluid or gas or whatever we're talking about. Now let's imagine another function. This might all look a little bit familiar, because we went through a very similar exercise in two dimensions when we talked about line integrals. Now we're just extending it to three dimensions.
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Conceptual understanding of flux in three dimensions Multivariable Calculus Khan Academy.mp3
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It gives us a vector for any point in three dimensions, and this right over here tells us the velocity of that same fluid or gas or whatever we're talking about. Now let's imagine another function. This might all look a little bit familiar, because we went through a very similar exercise in two dimensions when we talked about line integrals. Now we're just extending it to three dimensions. Let's say we have a function f, and it is equal to the product of rho and v. So for any point in x, y, z, this will give us a vector, and then we'll multiply it times this scalar right over here for that same point in three dimensions. So it's equal to rho times v. Let me use that same color that I used for v before. Rho times v. And there's a couple of ways you could conceptualize this.
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Conceptual understanding of flux in three dimensions Multivariable Calculus Khan Academy.mp3
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Now we're just extending it to three dimensions. Let's say we have a function f, and it is equal to the product of rho and v. So for any point in x, y, z, this will give us a vector, and then we'll multiply it times this scalar right over here for that same point in three dimensions. So it's equal to rho times v. Let me use that same color that I used for v before. Rho times v. And there's a couple of ways you could conceptualize this. You could view this as it obviously maintains the direction of the velocity, but now its magnitude, one way to think about it, is kind of the momentum density. If that doesn't make too much sense, you don't have to worry too much about it. Hopefully, as we use these two functions and we think a little bit more about them relative to a surface, it'll make a little bit more conceptual sense.
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Conceptual understanding of flux in three dimensions Multivariable Calculus Khan Academy.mp3
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Rho times v. And there's a couple of ways you could conceptualize this. You could view this as it obviously maintains the direction of the velocity, but now its magnitude, one way to think about it, is kind of the momentum density. If that doesn't make too much sense, you don't have to worry too much about it. Hopefully, as we use these two functions and we think a little bit more about them relative to a surface, it'll make a little bit more conceptual sense. Now, what I want to do is think about what it means given this function f to evaluate the surface integral over some surface. So we're going to evaluate over some surface. We're going to evaluate f dot n, where n is the unit normal vector at each point on that surface, dS, d capital S, d surface.
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Conceptual understanding of flux in three dimensions Multivariable Calculus Khan Academy.mp3
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Hopefully, as we use these two functions and we think a little bit more about them relative to a surface, it'll make a little bit more conceptual sense. Now, what I want to do is think about what it means given this function f to evaluate the surface integral over some surface. So we're going to evaluate over some surface. We're going to evaluate f dot n, where n is the unit normal vector at each point on that surface, dS, d capital S, d surface. So let's think about what this is saying. So first let me draw my axes. So I have my z-axis.
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Conceptual understanding of flux in three dimensions Multivariable Calculus Khan Academy.mp3
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We're going to evaluate f dot n, where n is the unit normal vector at each point on that surface, dS, d capital S, d surface. So let's think about what this is saying. So first let me draw my axes. So I have my z-axis. This could be my x-axis. And let's say that this right over here is my y-axis. And let's say my surface, I'll use that same color, my surface looks something like that.
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Conceptual understanding of flux in three dimensions Multivariable Calculus Khan Academy.mp3
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So I have my z-axis. This could be my x-axis. And let's say that this right over here is my y-axis. And let's say my surface, I'll use that same color, my surface looks something like that. So that is my surface. That is the surface in question. That is S. Now let's think about the units, and hopefully that'll give us a conceptual understanding of what this thing right over here is measuring.
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Conceptual understanding of flux in three dimensions Multivariable Calculus Khan Academy.mp3
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And let's say my surface, I'll use that same color, my surface looks something like that. So that is my surface. That is the surface in question. That is S. Now let's think about the units, and hopefully that'll give us a conceptual understanding of what this thing right over here is measuring. It's completely analogous to what we did in the two-dimensional case with line integrals. So we have a dS. A dS is a little chunk of area of that surface.
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Conceptual understanding of flux in three dimensions Multivariable Calculus Khan Academy.mp3
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That is S. Now let's think about the units, and hopefully that'll give us a conceptual understanding of what this thing right over here is measuring. It's completely analogous to what we did in the two-dimensional case with line integrals. So we have a dS. A dS is a little chunk of area of that surface. So that is dS. So this is going to be area. And if we want to pick particular units, this could be square meters.
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Conceptual understanding of flux in three dimensions Multivariable Calculus Khan Academy.mp3
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A dS is a little chunk of area of that surface. So that is dS. So this is going to be area. And if we want to pick particular units, this could be square meters. And I think when you do particular units, it starts to make a little bit more concrete sense. Now, the normal vector at that dS, the normal vector is going to point right out of it. It's literally normal to that plane.
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Conceptual understanding of flux in three dimensions Multivariable Calculus Khan Academy.mp3
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And if we want to pick particular units, this could be square meters. And I think when you do particular units, it starts to make a little bit more concrete sense. Now, the normal vector at that dS, the normal vector is going to point right out of it. It's literally normal to that plane. It's literally normal to that plane. It has a magnitude 1. So that is our unit normal vector.
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Conceptual understanding of flux in three dimensions Multivariable Calculus Khan Academy.mp3
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It's literally normal to that plane. It's literally normal to that plane. It has a magnitude 1. So that is our unit normal vector. And F is defined throughout this three-dimensional space. You give me any x, y, z, I'll know its mass density, I'll know its velocity, and I'll get some F. I'll get some F at any point in three-dimensional space, including on the surface, including right over here. So right over here, F might look something like this.
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Conceptual understanding of flux in three dimensions Multivariable Calculus Khan Academy.mp3
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So that is our unit normal vector. And F is defined throughout this three-dimensional space. You give me any x, y, z, I'll know its mass density, I'll know its velocity, and I'll get some F. I'll get some F at any point in three-dimensional space, including on the surface, including right over here. So right over here, F might look something like this. So that is F right at that point. So what does all this mean? Well, when you take the dot product of two vectors, it's essentially saying how much do they go together.
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Conceptual understanding of flux in three dimensions Multivariable Calculus Khan Academy.mp3
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So right over here, F might look something like this. So that is F right at that point. So what does all this mean? Well, when you take the dot product of two vectors, it's essentially saying how much do they go together. And since N is a unit vector, since it has a magnitude 1, this is essentially saying what is the magnitude of the component of F that's going in the direction of N? Or what is the magnitude of the component of F that is normal to the surface? How much of F is normal to the surface?
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Conceptual understanding of flux in three dimensions Multivariable Calculus Khan Academy.mp3
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Well, when you take the dot product of two vectors, it's essentially saying how much do they go together. And since N is a unit vector, since it has a magnitude 1, this is essentially saying what is the magnitude of the component of F that's going in the direction of N? Or what is the magnitude of the component of F that is normal to the surface? How much of F is normal to the surface? So the component of F that is normal to the surface might look something like that. And this right over here will essentially just give the magnitude of that. And it's just going to keep the units of F. And right over here just specifies the direction.
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Conceptual understanding of flux in three dimensions Multivariable Calculus Khan Academy.mp3
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How much of F is normal to the surface? So the component of F that is normal to the surface might look something like that. And this right over here will essentially just give the magnitude of that. And it's just going to keep the units of F. And right over here just specifies the direction. It has no units associated with it. It's dimensionless. F's units are going to be units of mass density.
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And it's just going to keep the units of F. And right over here just specifies the direction. It has no units associated with it. It's dimensionless. F's units are going to be units of mass density. So it could be, let's say, it could be kilogram per meter cubed. Well, that's actually just the rho part. So it's mass density times velocity times meters per second.
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F's units are going to be units of mass density. So it could be, let's say, it could be kilogram per meter cubed. Well, that's actually just the rho part. So it's mass density times velocity times meters per second. Let me write in those colors just so we have clear what's happening here. So the units of F are going to be the units of rho, which are going to be kilogram per cubic meter. That's mass density.
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Conceptual understanding of flux in three dimensions Multivariable Calculus Khan Academy.mp3
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So it's mass density times velocity times meters per second. Let me write in those colors just so we have clear what's happening here. So the units of F are going to be the units of rho, which are going to be kilogram per cubic meter. That's mass density. Times the units of V, which is meters per second. And we're going to multiply that times meters squared. So what you have is you have a meter and then a meter squared in the numerator.
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That's mass density. Times the units of V, which is meters per second. And we're going to multiply that times meters squared. So what you have is you have a meter and then a meter squared in the numerator. That's meters cubed in the numerator. And meters cubed in the denominator. That cancels out.
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So what you have is you have a meter and then a meter squared in the numerator. That's meters cubed in the numerator. And meters cubed in the denominator. That cancels out. And so the units that we get for this are kilogram per second. And so the way to conceptualize it, given how we've defined F, what we say F represents, the way to conceptualize this, this is saying how much mass, how much mass given this mass density, this velocity, is going directly out of this little ds, this little infinitesimally chunk of surface in a given amount of time. And then if we were to add up all the ds's, and this is what essentially that surface integral is, we're essentially saying how much mass in kilograms per second, that's what we picked, how much mass is traveling across this surface at any given moment in time.
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Conceptual understanding of flux in three dimensions Multivariable Calculus Khan Academy.mp3
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That cancels out. And so the units that we get for this are kilogram per second. And so the way to conceptualize it, given how we've defined F, what we say F represents, the way to conceptualize this, this is saying how much mass, how much mass given this mass density, this velocity, is going directly out of this little ds, this little infinitesimally chunk of surface in a given amount of time. And then if we were to add up all the ds's, and this is what essentially that surface integral is, we're essentially saying how much mass in kilograms per second, that's what we picked, how much mass is traveling across this surface at any given moment in time. And this is really the same idea we do with the line integrals. This is essentially the flux through a two-dimensional surface. So this is the flux through a 2D surface.
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Conceptual understanding of flux in three dimensions Multivariable Calculus Khan Academy.mp3
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And then if we were to add up all the ds's, and this is what essentially that surface integral is, we're essentially saying how much mass in kilograms per second, that's what we picked, how much mass is traveling across this surface at any given moment in time. And this is really the same idea we do with the line integrals. This is essentially the flux through a two-dimensional surface. So this is the flux through a 2D surface. And this isn't like some crazy abstract thing. I mean, you could imagine something like water vapor in your bathroom. Water vapor in your bathroom, and I like to imagine that because that's actually visible, especially when sunlight is shining through it.
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Conceptual understanding of flux in three dimensions Multivariable Calculus Khan Academy.mp3
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So this is the flux through a 2D surface. And this isn't like some crazy abstract thing. I mean, you could imagine something like water vapor in your bathroom. Water vapor in your bathroom, and I like to imagine that because that's actually visible, especially when sunlight is shining through it. And we've all seen water vapor in our bathroom when you have a ray of sunlight, and you can see how the particles are traveling, and you see they have a certain density at different points. And so you can imagine you care about the surface of your, maybe you have a window in the bathroom. Maybe you have a window, and so if the surface was the window, and the window, let's say the window is open, so it's kind of a, there's nothing physical there, it's just kind of a rectangular surface that things can pass freely through.
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Conceptual understanding of flux in three dimensions Multivariable Calculus Khan Academy.mp3
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Water vapor in your bathroom, and I like to imagine that because that's actually visible, especially when sunlight is shining through it. And we've all seen water vapor in our bathroom when you have a ray of sunlight, and you can see how the particles are traveling, and you see they have a certain density at different points. And so you can imagine you care about the surface of your, maybe you have a window in the bathroom. Maybe you have a window, and so if the surface was the window, and the window, let's say the window is open, so it's kind of a, there's nothing physical there, it's just kind of a rectangular surface that things can pass freely through. If an F was essentially the mass density of the water vapor times the velocity of the water vapor, then this thing right over here will essentially tell you the mass of water vapor that is traveling through that window at any given moment of time. Another way to think about it is imagine a river, and I'm going to conceptualize this river as kind of just a section of the river. I'm conceptualizing it, this is kind of a river, obviously this would be the surface that we normally see, but obviously it has some depth, it's three-dimensional in nature.
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Conceptual understanding of flux in three dimensions Multivariable Calculus Khan Academy.mp3
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Maybe you have a window, and so if the surface was the window, and the window, let's say the window is open, so it's kind of a, there's nothing physical there, it's just kind of a rectangular surface that things can pass freely through. If an F was essentially the mass density of the water vapor times the velocity of the water vapor, then this thing right over here will essentially tell you the mass of water vapor that is traveling through that window at any given moment of time. Another way to think about it is imagine a river, and I'm going to conceptualize this river as kind of just a section of the river. I'm conceptualizing it, this is kind of a river, obviously this would be the surface that we normally see, but obviously it has some depth, it's three-dimensional in nature. And so we would know the density, maybe it's constant, you know the density and you know the velocity at any point, that's what F gives us. So that tells us, as we said, we could view that as the momentum density at any given point in time. And maybe our surface is some type of a net.
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Conceptual understanding of flux in three dimensions Multivariable Calculus Khan Academy.mp3
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I'm conceptualizing it, this is kind of a river, obviously this would be the surface that we normally see, but obviously it has some depth, it's three-dimensional in nature. And so we would know the density, maybe it's constant, you know the density and you know the velocity at any point, that's what F gives us. So that tells us, as we said, we could view that as the momentum density at any given point in time. And maybe our surface is some type of a net. And the net doesn't even have to be rectangular, it could be some weird-shaped net, but I'll do it rectangular just because it's easy to draw. It's some type of net that in no way impedes the flow of the fluid, then once again, when you evaluate this integral, it will tell you the mass of fluid that is flowing through that net at any given moment of time. So hopefully this makes a little bit of conceptual sense now.
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Conceptual understanding of flux in three dimensions Multivariable Calculus Khan Academy.mp3
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We have our usual setup here for this constrained optimization situation. We have a function we want to maximize, which I'm thinking of as revenues for some company, a constraint, which I'm thinking of as some kind of budget for that company, and as you know if you've gotten to this video, one way to solve this constrained optimization problem is to define this function here, the Lagrangian, which involves taking this function that you're trying to maximize, in this case the revenue, and subtracting a new variable, lambda, what's called the Lagrange multiplier, times this quantity, which is the budget function, you know, however much you spend as a function of your input parameters, minus the budget itself, which you might think of as $10,000 in our example. So that's all the usual setup, and the crazy fact, which I just declared, is that when you set this gradient equal to zero, and you find some solution, and there will be three variables in this solution, h star, s star, and lambda star, that this lambda star is not meaningless, it's not just a proportionality constant between these gradient vectors, but it will actually tell you how much the maximum possible revenue changes as a function of your budget. And the way to start writing all of that in formulas would be to make explicit the fact that if you consider this value, you know, the $10,000 that is your budget, which I'm calling b, a variable, and not a constant, then you have to acknowledge that h star and s star are dependent on b, right? It's a very implicit relationship, something that's kind of hard to think about at first, because as you change b, it changes what the Lagrangian is, which is going to change where its gradient equals zero, which changes what h star, s star, and lambda star are, but in principle, they are some function of that budget, of b. And the maximum possible revenue is whatever you get when you just plug in that solution to your function r, and the claim I made that I just pulled out of the hat is that lambda star, the lambda value that comes packaged in with these two when you set the gradient of the Lagrangian equal to zero, equals the derivative of this maximum value, thought of as a function of b, maybe I should emphasize that, you know, we're thinking of this maximum value as a function of b, with respect to b. So that's kind of a mouthful, it takes a lot just to even phrase what's going on, but in the context of an economic example, it has a very clear, precise meaning, which is if you increase your budget by a dollar, right, if you increase it from $10,000 to $10,001, you're wondering for that tiny change in budget, that tiny db, what is the ratio of the resulting change in revenue?
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Proof for the meaning of Lagrange multipliers Multivariable Calculus Khan Academy.mp3
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And the way to start writing all of that in formulas would be to make explicit the fact that if you consider this value, you know, the $10,000 that is your budget, which I'm calling b, a variable, and not a constant, then you have to acknowledge that h star and s star are dependent on b, right? It's a very implicit relationship, something that's kind of hard to think about at first, because as you change b, it changes what the Lagrangian is, which is going to change where its gradient equals zero, which changes what h star, s star, and lambda star are, but in principle, they are some function of that budget, of b. And the maximum possible revenue is whatever you get when you just plug in that solution to your function r, and the claim I made that I just pulled out of the hat is that lambda star, the lambda value that comes packaged in with these two when you set the gradient of the Lagrangian equal to zero, equals the derivative of this maximum value, thought of as a function of b, maybe I should emphasize that, you know, we're thinking of this maximum value as a function of b, with respect to b. So that's kind of a mouthful, it takes a lot just to even phrase what's going on, but in the context of an economic example, it has a very clear, precise meaning, which is if you increase your budget by a dollar, right, if you increase it from $10,000 to $10,001, you're wondering for that tiny change in budget, that tiny db, what is the ratio of the resulting change in revenue? So in a sense, this lambda star tells you for every dollar that you increase the budget, how much can your revenue increase if you're always maximizing it? So, why on earth is this true, right? This just seems like it comes out of nowhere.
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Proof for the meaning of Lagrange multipliers Multivariable Calculus Khan Academy.mp3
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So that's kind of a mouthful, it takes a lot just to even phrase what's going on, but in the context of an economic example, it has a very clear, precise meaning, which is if you increase your budget by a dollar, right, if you increase it from $10,000 to $10,001, you're wondering for that tiny change in budget, that tiny db, what is the ratio of the resulting change in revenue? So in a sense, this lambda star tells you for every dollar that you increase the budget, how much can your revenue increase if you're always maximizing it? So, why on earth is this true, right? This just seems like it comes out of nowhere. Well, there are a couple clever observations that go into proving this. The first, the first is to notice what happens if we evaluate this Lagrangian function itself at this critical point, when you input h star, s star, and lambda star. And remember, the way that these guys are defined is that you look at all of the values where the gradient of the Lagrangian equals the zero vector, and then if you get multiple options, you know, sometimes when you set the gradient equal to zero, you get multiple solutions, and whichever one maximizes r, that is h star, s star, lambda star.
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Proof for the meaning of Lagrange multipliers Multivariable Calculus Khan Academy.mp3
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This just seems like it comes out of nowhere. Well, there are a couple clever observations that go into proving this. The first, the first is to notice what happens if we evaluate this Lagrangian function itself at this critical point, when you input h star, s star, and lambda star. And remember, the way that these guys are defined is that you look at all of the values where the gradient of the Lagrangian equals the zero vector, and then if you get multiple options, you know, sometimes when you set the gradient equal to zero, you get multiple solutions, and whichever one maximizes r, that is h star, s star, lambda star. So now, I'm just asking if you plug that not into the gradient of the Lagrangian, but to the Lagrangian itself, what do you get? Well, you're going to get, we just look at its definition up here, r evaluated at h star and s star, right? And we subtract off lambda star times b of h star, s star, minus the constant that is your budget, you know, something you might think of as $10,000, whatever you set little b equal to.
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Proof for the meaning of Lagrange multipliers Multivariable Calculus Khan Academy.mp3
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And remember, the way that these guys are defined is that you look at all of the values where the gradient of the Lagrangian equals the zero vector, and then if you get multiple options, you know, sometimes when you set the gradient equal to zero, you get multiple solutions, and whichever one maximizes r, that is h star, s star, lambda star. So now, I'm just asking if you plug that not into the gradient of the Lagrangian, but to the Lagrangian itself, what do you get? Well, you're going to get, we just look at its definition up here, r evaluated at h star and s star, right? And we subtract off lambda star times b of h star, s star, minus the constant that is your budget, you know, something you might think of as $10,000, whatever you set little b equal to. Okay, Grant, you might say, why does this tell us anything? You're just plugging in stars instead of the usual variables. But the key is that if you plug in h star and s star, this value has to equal zero because h star and s star have to satisfy the constraint.
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Proof for the meaning of Lagrange multipliers Multivariable Calculus Khan Academy.mp3
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And we subtract off lambda star times b of h star, s star, minus the constant that is your budget, you know, something you might think of as $10,000, whatever you set little b equal to. Okay, Grant, you might say, why does this tell us anything? You're just plugging in stars instead of the usual variables. But the key is that if you plug in h star and s star, this value has to equal zero because h star and s star have to satisfy the constraint. Remember, one of the cool parts about this Lagrangian function as a whole is that when you take its partial derivative with respect to lambda, all that's left is this constraint function minus the constraint portion. When you set the gradient of the Lagrangian equal to the zero vector, one component of that is to set the partial derivative with respect to lambda equal to zero. And if you remember from the Lagrangian video, all that really boils down to is the fact that the constraint holds, right?
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Proof for the meaning of Lagrange multipliers Multivariable Calculus Khan Academy.mp3
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But the key is that if you plug in h star and s star, this value has to equal zero because h star and s star have to satisfy the constraint. Remember, one of the cool parts about this Lagrangian function as a whole is that when you take its partial derivative with respect to lambda, all that's left is this constraint function minus the constraint portion. When you set the gradient of the Lagrangian equal to the zero vector, one component of that is to set the partial derivative with respect to lambda equal to zero. And if you remember from the Lagrangian video, all that really boils down to is the fact that the constraint holds, right? Which would be your budget achieves $10,000. When you plug in the appropriate h star and s star to this value, you're hitting this constrained amount of money that you can spend. So by virtue of how h star and s star are defined, the fact that they are solutions to the constrained optimization problem means this whole portion goes to zero.
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Proof for the meaning of Lagrange multipliers Multivariable Calculus Khan Academy.mp3
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And if you remember from the Lagrangian video, all that really boils down to is the fact that the constraint holds, right? Which would be your budget achieves $10,000. When you plug in the appropriate h star and s star to this value, you're hitting this constrained amount of money that you can spend. So by virtue of how h star and s star are defined, the fact that they are solutions to the constrained optimization problem means this whole portion goes to zero. So we can just kind of cancel all that out. And what's left, what's left here is the maximum possible revenue, right? So evidently, when you evaluate the Lagrangian at this critical point, at h star, s star, and lambda star, it equals m star.
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Proof for the meaning of Lagrange multipliers Multivariable Calculus Khan Academy.mp3
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So by virtue of how h star and s star are defined, the fact that they are solutions to the constrained optimization problem means this whole portion goes to zero. So we can just kind of cancel all that out. And what's left, what's left here is the maximum possible revenue, right? So evidently, when you evaluate the Lagrangian at this critical point, at h star, s star, and lambda star, it equals m star. It equals the maximum possible value for the function you're trying to maximize. So ultimately what we want is to understand how that maximum value changes when you consider it a function of the budget. So evidently, what we can look for is to just ask how the Lagrangian changes as you consider it a function of the budget.
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Proof for the meaning of Lagrange multipliers Multivariable Calculus Khan Academy.mp3
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So evidently, when you evaluate the Lagrangian at this critical point, at h star, s star, and lambda star, it equals m star. It equals the maximum possible value for the function you're trying to maximize. So ultimately what we want is to understand how that maximum value changes when you consider it a function of the budget. So evidently, what we can look for is to just ask how the Lagrangian changes as you consider it a function of the budget. Now this is an interesting thing to observe, because if we just look up at the definition of the Lagrangian, if you just look at this formula, if I told you to take the derivative of this with respect to little b, right? You know, how much does this change with respect to little b? You would notice that this goes to zero, it doesn't have a little b.
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Proof for the meaning of Lagrange multipliers Multivariable Calculus Khan Academy.mp3
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So evidently, what we can look for is to just ask how the Lagrangian changes as you consider it a function of the budget. Now this is an interesting thing to observe, because if we just look up at the definition of the Lagrangian, if you just look at this formula, if I told you to take the derivative of this with respect to little b, right? You know, how much does this change with respect to little b? You would notice that this goes to zero, it doesn't have a little b. This would also go to zero. And all you'd be left with would be negative lambda times negative b, and the derivative of that with respect to b would be lambda. So you might say, oh yeah, of course, of course, this, the derivative of that Lagrangian with respect to b, once we work it all out, the only term that was left there was the lambda.
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Proof for the meaning of Lagrange multipliers Multivariable Calculus Khan Academy.mp3
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You would notice that this goes to zero, it doesn't have a little b. This would also go to zero. And all you'd be left with would be negative lambda times negative b, and the derivative of that with respect to b would be lambda. So you might say, oh yeah, of course, of course, this, the derivative of that Lagrangian with respect to b, once we work it all out, the only term that was left there was the lambda. And that's compelling, but ultimately, it's not entirely right. That overlooks the fact that L is not actually defined as a function of b. When we defined the Lagrangian, we were considering b to be a constant.
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Proof for the meaning of Lagrange multipliers Multivariable Calculus Khan Academy.mp3
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So you might say, oh yeah, of course, of course, this, the derivative of that Lagrangian with respect to b, once we work it all out, the only term that was left there was the lambda. And that's compelling, but ultimately, it's not entirely right. That overlooks the fact that L is not actually defined as a function of b. When we defined the Lagrangian, we were considering b to be a constant. So if you really want to consider this to be a function that involves b, the way we should write it, and I'll go ahead and erase this guy, the way we should write this Lagrangian is to say, you're a function of h star, which itself is dependent on b, and s star, which is also a function of b, as soon as we start considering b a variable and not a constant, we have to acknowledge that this critical point, h star, s star, and lambda star, depends on the value of b. So likewise, that lambda star, lambda star, is also going to be a function of b. And then we can consider it as a fourth variable, so we're adding on yet another variable to this function, the value of b itself.
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Proof for the meaning of Lagrange multipliers Multivariable Calculus Khan Academy.mp3
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When we defined the Lagrangian, we were considering b to be a constant. So if you really want to consider this to be a function that involves b, the way we should write it, and I'll go ahead and erase this guy, the way we should write this Lagrangian is to say, you're a function of h star, which itself is dependent on b, and s star, which is also a function of b, as soon as we start considering b a variable and not a constant, we have to acknowledge that this critical point, h star, s star, and lambda star, depends on the value of b. So likewise, that lambda star, lambda star, is also going to be a function of b. And then we can consider it as a fourth variable, so we're adding on yet another variable to this function, the value of b itself. The value of b itself here. So now, when we want to know what is the value of the Lagrangian at the critical point, h star, s star, lambda star, as a function of b, so that can be kind of confusing, what you basically have is this function that only really depends on one value, right? It only depends on b, but it kind of goes through a four variable function.
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Proof for the meaning of Lagrange multipliers Multivariable Calculus Khan Academy.mp3
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And then we can consider it as a fourth variable, so we're adding on yet another variable to this function, the value of b itself. The value of b itself here. So now, when we want to know what is the value of the Lagrangian at the critical point, h star, s star, lambda star, as a function of b, so that can be kind of confusing, what you basically have is this function that only really depends on one value, right? It only depends on b, but it kind of goes through a four variable function. And so just to make it explicit, this would equal the value of r as a function of h star and s star, and each one of those is a function of little b, right? So this term is saying what's your revenue evaluated on the maximizing h and s for the given budget, and then you subtract off lambda star, oh here, I should probably, I'm not going to have room here, am I? So what you subtract off minus lambda star at b of h star and s star, but each of these guys is also a function of little b, little b, minus little b, right?
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Proof for the meaning of Lagrange multipliers Multivariable Calculus Khan Academy.mp3
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It only depends on b, but it kind of goes through a four variable function. And so just to make it explicit, this would equal the value of r as a function of h star and s star, and each one of those is a function of little b, right? So this term is saying what's your revenue evaluated on the maximizing h and s for the given budget, and then you subtract off lambda star, oh here, I should probably, I'm not going to have room here, am I? So what you subtract off minus lambda star at b of h star and s star, but each of these guys is also a function of little b, little b, minus little b, right? So you have this large, kind of complicated multivariable function, it's defined in terms of h stars and s stars, which are themselves very implicit, right? We just say, by definition, these are whatever values make the gradient of L equal zero, so very hard to think about what that means concretely, but all of this is really just dependent on the single value little b. And from here, if we want to evaluate the derivative of L, we want to evaluate the derivative of this Lagrangian with respect to little b, which is really the only thing it depends on, it's just via all of these other variables, we use the multivariable chain rule.
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Proof for the meaning of Lagrange multipliers Multivariable Calculus Khan Academy.mp3
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So what you subtract off minus lambda star at b of h star and s star, but each of these guys is also a function of little b, little b, minus little b, right? So you have this large, kind of complicated multivariable function, it's defined in terms of h stars and s stars, which are themselves very implicit, right? We just say, by definition, these are whatever values make the gradient of L equal zero, so very hard to think about what that means concretely, but all of this is really just dependent on the single value little b. And from here, if we want to evaluate the derivative of L, we want to evaluate the derivative of this Lagrangian with respect to little b, which is really the only thing it depends on, it's just via all of these other variables, we use the multivariable chain rule. And at this point, if you don't know the multivariable chain rule, I have a video on that, definitely pause, go take a look, make sure that it all makes sense, but right here, I'm just going to be assuming that you know what the multivariable chain rule is. So what it is, is we take the partial, we're going to look at the partial derivatives with respect to all four of these inputs. So we'll start with the partial derivative of L with respect to h star, with respect to h star, and we're going to multiply that by the derivative of h star with respect to b.
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Proof for the meaning of Lagrange multipliers Multivariable Calculus Khan Academy.mp3
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And from here, if we want to evaluate the derivative of L, we want to evaluate the derivative of this Lagrangian with respect to little b, which is really the only thing it depends on, it's just via all of these other variables, we use the multivariable chain rule. And at this point, if you don't know the multivariable chain rule, I have a video on that, definitely pause, go take a look, make sure that it all makes sense, but right here, I'm just going to be assuming that you know what the multivariable chain rule is. So what it is, is we take the partial, we're going to look at the partial derivatives with respect to all four of these inputs. So we'll start with the partial derivative of L with respect to h star, with respect to h star, and we're going to multiply that by the derivative of h star with respect to b. And this might seem like a very hard thing to think about, like how do we know how h star changes as b changes? But don't worry about it, you'll see something magic happen in just a moment. And then we add in partial derivative of L with respect to that second variable, s star, with respect to whatever the second variable is, multiplied by the derivative of s star with respect to b.
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Proof for the meaning of Lagrange multipliers Multivariable Calculus Khan Academy.mp3
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So we'll start with the partial derivative of L with respect to h star, with respect to h star, and we're going to multiply that by the derivative of h star with respect to b. And this might seem like a very hard thing to think about, like how do we know how h star changes as b changes? But don't worry about it, you'll see something magic happen in just a moment. And then we add in partial derivative of L with respect to that second variable, s star, with respect to whatever the second variable is, multiplied by the derivative of s star with respect to b. You can see how you really need to know what the multivariable chain rule is, right? This would all seem kind of out of the blue. So what we now add in is the partial derivative of L with respect to that lambda star, with respect to lambda star, multiplied by the derivative of lambda star with respect to little b.
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Proof for the meaning of Lagrange multipliers Multivariable Calculus Khan Academy.mp3
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And then we add in partial derivative of L with respect to that second variable, s star, with respect to whatever the second variable is, multiplied by the derivative of s star with respect to b. You can see how you really need to know what the multivariable chain rule is, right? This would all seem kind of out of the blue. So what we now add in is the partial derivative of L with respect to that lambda star, with respect to lambda star, multiplied by the derivative of lambda star with respect to little b. And then finally, finally, we take the partial derivative of this Lagrangian with respect to that little b, which we're now considering a variable in there, right? We're no longer considering that b a constant, multiplied by, well something kind of silly, the derivative of b with respect to itself. So now, if you're thinking that this is going to be horrifying to compute, I can understand where you're coming from.
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Proof for the meaning of Lagrange multipliers Multivariable Calculus Khan Academy.mp3
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So what we now add in is the partial derivative of L with respect to that lambda star, with respect to lambda star, multiplied by the derivative of lambda star with respect to little b. And then finally, finally, we take the partial derivative of this Lagrangian with respect to that little b, which we're now considering a variable in there, right? We're no longer considering that b a constant, multiplied by, well something kind of silly, the derivative of b with respect to itself. So now, if you're thinking that this is going to be horrifying to compute, I can understand where you're coming from. You have to know the derivative of lambda star with respect to b, you have to somehow intimately be familiar with how this lambda star changes as you change b, and like I said, that's such an implicit relationship, right? We just said that lambda star is by definition whatever the solution to this gradient equation is. So somehow you're supposed to know how that changes when you slightly alter b over here.
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Proof for the meaning of Lagrange multipliers Multivariable Calculus Khan Academy.mp3
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So now, if you're thinking that this is going to be horrifying to compute, I can understand where you're coming from. You have to know the derivative of lambda star with respect to b, you have to somehow intimately be familiar with how this lambda star changes as you change b, and like I said, that's such an implicit relationship, right? We just said that lambda star is by definition whatever the solution to this gradient equation is. So somehow you're supposed to know how that changes when you slightly alter b over here. Well, you don't really have to worry about things, because by definition h star, s star, and lambda star are whatever values make the gradient of L equal to 0. But if you think about that, what does it mean for the gradient of L to equal the 0 vector? Well, what it means is that when you take its derivative with respect to that first variable, h star, it equals 0.
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Proof for the meaning of Lagrange multipliers Multivariable Calculus Khan Academy.mp3
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So somehow you're supposed to know how that changes when you slightly alter b over here. Well, you don't really have to worry about things, because by definition h star, s star, and lambda star are whatever values make the gradient of L equal to 0. But if you think about that, what does it mean for the gradient of L to equal the 0 vector? Well, what it means is that when you take its derivative with respect to that first variable, h star, it equals 0. When you take its derivative with respect to the second variable, that equals 0 as well. And with respect to this third variable, that's going to equal 0. By definition, h star, s star, and lambda star are whatever values make it the case so that when you plug them in, the partial derivative of the Lagrangian with respect to any one of those variables equals 0.
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Proof for the meaning of Lagrange multipliers Multivariable Calculus Khan Academy.mp3
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Well, what it means is that when you take its derivative with respect to that first variable, h star, it equals 0. When you take its derivative with respect to the second variable, that equals 0 as well. And with respect to this third variable, that's going to equal 0. By definition, h star, s star, and lambda star are whatever values make it the case so that when you plug them in, the partial derivative of the Lagrangian with respect to any one of those variables equals 0. So we don't even have to worry about most of this equation. The only part that matters here is the partial derivative of L with respect to b that we're now considering a variable multiplied by, well, what's db db? What is the rate of change of a variable with respect to itself?
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Proof for the meaning of Lagrange multipliers Multivariable Calculus Khan Academy.mp3
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By definition, h star, s star, and lambda star are whatever values make it the case so that when you plug them in, the partial derivative of the Lagrangian with respect to any one of those variables equals 0. So we don't even have to worry about most of this equation. The only part that matters here is the partial derivative of L with respect to b that we're now considering a variable multiplied by, well, what's db db? What is the rate of change of a variable with respect to itself? It's 1. It is 1. So all of this stuff, this entire multivariable chain rule boils down to a single innocent-looking factor which is the partial derivative of L partial derivative of L with respect to little b.
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Proof for the meaning of Lagrange multipliers Multivariable Calculus Khan Academy.mp3
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What is the rate of change of a variable with respect to itself? It's 1. It is 1. So all of this stuff, this entire multivariable chain rule boils down to a single innocent-looking factor which is the partial derivative of L partial derivative of L with respect to little b. And now, there's something very subtle here, right? Because this might seem obvious. I'm saying the derivative of L with respect to b equals the derivative of L with respect to b.
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Proof for the meaning of Lagrange multipliers Multivariable Calculus Khan Academy.mp3
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So all of this stuff, this entire multivariable chain rule boils down to a single innocent-looking factor which is the partial derivative of L partial derivative of L with respect to little b. And now, there's something very subtle here, right? Because this might seem obvious. I'm saying the derivative of L with respect to b equals the derivative of L with respect to b. But maybe I should give a different notation here, right? Because here when I'm taking the derivative, really I'm considering L as a single variable function, right? I'm considering not what happens as you can freely change all four of these variables.
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Proof for the meaning of Lagrange multipliers Multivariable Calculus Khan Academy.mp3
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I'm saying the derivative of L with respect to b equals the derivative of L with respect to b. But maybe I should give a different notation here, right? Because here when I'm taking the derivative, really I'm considering L as a single variable function, right? I'm considering not what happens as you can freely change all four of these variables. Three of them are locked into place by b. So maybe I should really give that a different name. I should call that L star.
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Proof for the meaning of Lagrange multipliers Multivariable Calculus Khan Academy.mp3
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I'm considering not what happens as you can freely change all four of these variables. Three of them are locked into place by b. So maybe I should really give that a different name. I should call that L star. L star is a single variable function. Whereas this L is a multivariable function. This is the function where you can freely change the values of h and s and lambda and b as you put them in.
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Proof for the meaning of Lagrange multipliers Multivariable Calculus Khan Academy.mp3
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I should call that L star. L star is a single variable function. Whereas this L is a multivariable function. This is the function where you can freely change the values of h and s and lambda and b as you put them in. So if we kind of scroll up to look at its definition, which I've written all over I guess here, let me actually rewrite its definition, right? I think that'll be useful. I'm going to rewrite that L, if I consider it as a four variable function of h s lambda and b that what that equals is r evaluated at h and s minus lambda multiplied by this constraint function b evaluated at h and s minus little b.
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Proof for the meaning of Lagrange multipliers Multivariable Calculus Khan Academy.mp3
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This is the function where you can freely change the values of h and s and lambda and b as you put them in. So if we kind of scroll up to look at its definition, which I've written all over I guess here, let me actually rewrite its definition, right? I think that'll be useful. I'm going to rewrite that L, if I consider it as a four variable function of h s lambda and b that what that equals is r evaluated at h and s minus lambda multiplied by this constraint function b evaluated at h and s minus little b. And this is now when I'm considering little b to be a variable. So this is the Lagrangian when you consider all four of these to be freely changing as you want. Whereas the thing up here that I'm considering a single variable function has three of its inputs locked into place.
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Proof for the meaning of Lagrange multipliers Multivariable Calculus Khan Academy.mp3
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I'm going to rewrite that L, if I consider it as a four variable function of h s lambda and b that what that equals is r evaluated at h and s minus lambda multiplied by this constraint function b evaluated at h and s minus little b. And this is now when I'm considering little b to be a variable. So this is the Lagrangian when you consider all four of these to be freely changing as you want. Whereas the thing up here that I'm considering a single variable function has three of its inputs locked into place. So effectively it's just a single variable function with respect to b. So it's actually quite miraculous that the single variable derivative of that L here, I should L star with respect to b ends up being the same as the partial derivative of L. This L where you're free to change all the variables that these should be the same. Usually in any usual circumstance all of these other terms would have come into play somehow.
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Proof for the meaning of Lagrange multipliers Multivariable Calculus Khan Academy.mp3
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Whereas the thing up here that I'm considering a single variable function has three of its inputs locked into place. So effectively it's just a single variable function with respect to b. So it's actually quite miraculous that the single variable derivative of that L here, I should L star with respect to b ends up being the same as the partial derivative of L. This L where you're free to change all the variables that these should be the same. Usually in any usual circumstance all of these other terms would have come into play somehow. But what's special here is that by the definition of this L star the specific way in which these h star, s star, and lambda stars are locked into place happens to be one in which all of these partial derivatives go to zero. So that's pretty subtle and I think it's quite clever and what it leaves us with is that we just have to evaluate this partial derivative which is quite simple because we look down here and you say what's the partial derivative of L with respect to b? Well this R has no b's in it so don't need to care about that.
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Proof for the meaning of Lagrange multipliers Multivariable Calculus Khan Academy.mp3
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Usually in any usual circumstance all of these other terms would have come into play somehow. But what's special here is that by the definition of this L star the specific way in which these h star, s star, and lambda stars are locked into place happens to be one in which all of these partial derivatives go to zero. So that's pretty subtle and I think it's quite clever and what it leaves us with is that we just have to evaluate this partial derivative which is quite simple because we look down here and you say what's the partial derivative of L with respect to b? Well this R has no b's in it so don't need to care about that. This term over here it's partial derivative is negative one, right, just because there's a b here and that's multiplied by the constant lambda so that all just equals lambda. But if we're in the situation where lambda is locked into place as a function of little b then we'd write lambda star as a function of little b. Right?
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Proof for the meaning of Lagrange multipliers Multivariable Calculus Khan Academy.mp3
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Well this R has no b's in it so don't need to care about that. This term over here it's partial derivative is negative one, right, just because there's a b here and that's multiplied by the constant lambda so that all just equals lambda. But if we're in the situation where lambda is locked into place as a function of little b then we'd write lambda star as a function of little b. Right? So if that feels a little notationally confusing I'm right there with you, but the important part here the important thing to remember is that we just started considering b as a variable right, and we were looking at the h star, s star, and lambda star as they depended on that variable. We made the observation that the Lagrangian evaluated at that critical point equals the revenue evaluated at that critical point. The rest of the stuff just cancels out.
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Proof for the meaning of Lagrange multipliers Multivariable Calculus Khan Academy.mp3
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Hey everyone. So in the last video I was talking about divergence and kind of laying down the intuition that we need for it. Where you're imagining a vector field as representing some kind of fluid flow where particles move according to the vector that they're attached to in that point in time. And as they move to a different point, the vector they're attached to is different, so their velocity changes in some way. And the key question that we want to think about is if you have a given point somewhere in space, does fluid tend to flow towards that point or does it tend to flow more away from it? Does it diverge away from that point? And what I want to do here is start kind of closing our grasp on that intuition a little bit more tightly as if we are trying to discover the formula for divergence ourselves.
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Divergence intuition, part 2.mp3
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And as they move to a different point, the vector they're attached to is different, so their velocity changes in some way. And the key question that we want to think about is if you have a given point somewhere in space, does fluid tend to flow towards that point or does it tend to flow more away from it? Does it diverge away from that point? And what I want to do here is start kind of closing our grasp on that intuition a little bit more tightly as if we are trying to discover the formula for divergence ourselves. Because ultimately that's what I'm gonna get to, a formula for divergence. But I want it to be something that's not just plopped down in front of you, but something that you actually feel deep in your bones. So a vector field like the one I have pictured above is given as a function, a multivariable function, with a two-dimensional input, since it's a two-dimensional vector field, and then some kind of two-dimensional output.
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Divergence intuition, part 2.mp3
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And what I want to do here is start kind of closing our grasp on that intuition a little bit more tightly as if we are trying to discover the formula for divergence ourselves. Because ultimately that's what I'm gonna get to, a formula for divergence. But I want it to be something that's not just plopped down in front of you, but something that you actually feel deep in your bones. So a vector field like the one I have pictured above is given as a function, a multivariable function, with a two-dimensional input, since it's a two-dimensional vector field, and then some kind of two-dimensional output. And it's common to, whoop, it's common to write p and q as the functions for the components of the output. So p and q are each just scalar-valued functions, and you think of them as the components of your vector-valued output. And the divergence is kind of like a derivative, where you might denote it by just div, and in the same way that your derivative, you have this operator, and what it does is it takes in a function, and what you get is a whole new function.
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Divergence intuition, part 2.mp3
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So a vector field like the one I have pictured above is given as a function, a multivariable function, with a two-dimensional input, since it's a two-dimensional vector field, and then some kind of two-dimensional output. And it's common to, whoop, it's common to write p and q as the functions for the components of the output. So p and q are each just scalar-valued functions, and you think of them as the components of your vector-valued output. And the divergence is kind of like a derivative, where you might denote it by just div, and in the same way that your derivative, you have this operator, and what it does is it takes in a function, and what you get is a whole new function. This div operator, you think of as taking in a vector field of some kind, and you get a new function. And the new function you get will be scalar-valued. It'll be something that just takes in points in space and outputs a number.
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Divergence intuition, part 2.mp3
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And the divergence is kind of like a derivative, where you might denote it by just div, and in the same way that your derivative, you have this operator, and what it does is it takes in a function, and what you get is a whole new function. This div operator, you think of as taking in a vector field of some kind, and you get a new function. And the new function you get will be scalar-valued. It'll be something that just takes in points in space and outputs a number. Because what you're thinking, the thing that it's trying to do is take in some specific point with x, y coordinates and just give you a single number to tell you, hey, does fluid tend to diverge away from it? How much? Or does it tend to flow towards it, and how much?
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Divergence intuition, part 2.mp3
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It'll be something that just takes in points in space and outputs a number. Because what you're thinking, the thing that it's trying to do is take in some specific point with x, y coordinates and just give you a single number to tell you, hey, does fluid tend to diverge away from it? How much? Or does it tend to flow towards it, and how much? So this is the kind of the form of the thing that we're going for. So here what we're gonna do is just start thinking about cases where this divergence is positive or negative or zero and what that should look like. So for example, let's say we want cases where the divergence of our vector field at a specific point, at a specific point x, y, is positive.
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Divergence intuition, part 2.mp3
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Or does it tend to flow towards it, and how much? So this is the kind of the form of the thing that we're going for. So here what we're gonna do is just start thinking about cases where this divergence is positive or negative or zero and what that should look like. So for example, let's say we want cases where the divergence of our vector field at a specific point, at a specific point x, y, is positive. What might that look like? So one case would be where your point, nothing is happening at that point, the vector attached to it is zero, and everyone around it is going away. This is kind of the extreme example of positive divergence.
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Divergence intuition, part 2.mp3
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So for example, let's say we want cases where the divergence of our vector field at a specific point, at a specific point x, y, is positive. What might that look like? So one case would be where your point, nothing is happening at that point, the vector attached to it is zero, and everyone around it is going away. This is kind of the extreme example of positive divergence. And I animated this in the last video where we have all of the vectors pointing away from the origin, and if you look at a region around that origin, all the fluid particles kind of go out of that region. And that's the quintessential positive divergence example. But it doesn't have to look like that.
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Divergence intuition, part 2.mp3
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This is kind of the extreme example of positive divergence. And I animated this in the last video where we have all of the vectors pointing away from the origin, and if you look at a region around that origin, all the fluid particles kind of go out of that region. And that's the quintessential positive divergence example. But it doesn't have to look like that. It actually, I mean, you could have something where there is a little bit of movement at your point, and then maybe there's movement towards it as well from one side, and vectors are kind of going towards it, but they're going away from it even more rapidly on the other side. So if you think of any kind of actual region around your point, you're saying, sure, fluid is going into that region a little bit, but it's much more counterbalanced by how quickly it's going out. So these are the kind of situations you might see for positive divergence.
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Divergence intuition, part 2.mp3
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But it doesn't have to look like that. It actually, I mean, you could have something where there is a little bit of movement at your point, and then maybe there's movement towards it as well from one side, and vectors are kind of going towards it, but they're going away from it even more rapidly on the other side. So if you think of any kind of actual region around your point, you're saying, sure, fluid is going into that region a little bit, but it's much more counterbalanced by how quickly it's going out. So these are the kind of situations you might see for positive divergence. Now negative divergence, negative divergence, let's think about what examples of that might look like. Divergence of V at a given point, and really it's something that takes in all points of the plane, but we're just looking at specific points. So if the divergence is negative, well, the quintessential example here is that nothing happens at your point, but all of the vectors around it are kind of flowing in towards it.
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Divergence intuition, part 2.mp3
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So these are the kind of situations you might see for positive divergence. Now negative divergence, negative divergence, let's think about what examples of that might look like. Divergence of V at a given point, and really it's something that takes in all points of the plane, but we're just looking at specific points. So if the divergence is negative, well, the quintessential example here is that nothing happens at your point, but all of the vectors around it are kind of flowing in towards it. And this is the thing where I animated where we took this and we flipped all of the vectors and said, ah, there, if you start playing the fluid flow, then the density in any region around the origin increases a lot. All of the fluid particles tend to converge towards that center. But again, this isn't the only example that you might have.
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Divergence intuition, part 2.mp3
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So if the divergence is negative, well, the quintessential example here is that nothing happens at your point, but all of the vectors around it are kind of flowing in towards it. And this is the thing where I animated where we took this and we flipped all of the vectors and said, ah, there, if you start playing the fluid flow, then the density in any region around the origin increases a lot. All of the fluid particles tend to converge towards that center. But again, this isn't the only example that you might have. You could have a little bit of activity at your point itself, and maybe it is the case that things do flow away from it a little bit as you're going away. And some of the fluid particles are going away, and it's just the case that the fluid particles flowing in towards it from another direction heavily counterbalance that. Because then if you're looking at any kind of region around your point, you say fluid particles are coming in quite rapidly, a lot of particles per time, but they're not leaving too rapidly around the other end.
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Divergence intuition, part 2.mp3
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But again, this isn't the only example that you might have. You could have a little bit of activity at your point itself, and maybe it is the case that things do flow away from it a little bit as you're going away. And some of the fluid particles are going away, and it's just the case that the fluid particles flowing in towards it from another direction heavily counterbalance that. Because then if you're looking at any kind of region around your point, you say fluid particles are coming in quite rapidly, a lot of particles per time, but they're not leaving too rapidly around the other end. So kind of loosely, intuitively, this is what a negative divergence case might look like. And finally, another case that we wanna start thinking about as we're tightening our grasp on this intuition is what happens, or what does it look like if the divergence of your function at a specific point is zero? Right, if it's just absolutely zero.
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Divergence intuition, part 2.mp3
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Because then if you're looking at any kind of region around your point, you say fluid particles are coming in quite rapidly, a lot of particles per time, but they're not leaving too rapidly around the other end. So kind of loosely, intuitively, this is what a negative divergence case might look like. And finally, another case that we wanna start thinking about as we're tightening our grasp on this intuition is what happens, or what does it look like if the divergence of your function at a specific point is zero? Right, if it's just absolutely zero. And one thing this could look like is, you know, you have something going on, but nothing really changes, and all of the fluid just kind of flows in and it flows out, and on the whole it balances. You know, if you take any kind of region, the amount flowing in is balanced with the amount flowing out. But it could also look like you have fluid flowing in, kind of from one dimension, but it's canceled out by flowing away from the point in a manner that sort of perfectly balances it in another direction.
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Divergence intuition, part 2.mp3
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Right, if it's just absolutely zero. And one thing this could look like is, you know, you have something going on, but nothing really changes, and all of the fluid just kind of flows in and it flows out, and on the whole it balances. You know, if you take any kind of region, the amount flowing in is balanced with the amount flowing out. But it could also look like you have fluid flowing in, kind of from one dimension, but it's canceled out by flowing away from the point in a manner that sort of perfectly balances it in another direction. So these are the loose pictures that I want you to have in the back of your mind as we start looking for the actual formula for divergence. And what I'll do in the next video or two is start looking at these functions p and q and thinking about the partial derivative properties that they have that will correspond with, you know, these positive divergence images that you should have in your head, or the negative divergence images that you should have in your head. So with that, I'll see you next video.
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Divergence intuition, part 2.mp3
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So somehow our whole function, our function takes things from this two-dimensional space and plugs it onto this output. T, you're thinking of as just another number line up here, so t, and then you've got two separate functions here, you know, x of t and y of t, x of t and y of t, each of which take that same value for a specific input. You know, it's not that they're acting on different inputs, x of some other input t and y of some other input, it's the same one, and then they move that somewhere to this output space, which itself gets moved over. And in this way, you're thinking of it as just a single variable function that goes from t and ultimately outputs f, it's just that there's multidimensional stuff happening in between. And now if we start thinking about the derivative of it, what does that mean? What does that mean for the conception of the picture that we have going on here? Well, that bottom part, that dt, you're thinking of as a tiny change to t, right?
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Multivariable chain rule intuition.mp3
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And in this way, you're thinking of it as just a single variable function that goes from t and ultimately outputs f, it's just that there's multidimensional stuff happening in between. And now if we start thinking about the derivative of it, what does that mean? What does that mean for the conception of the picture that we have going on here? Well, that bottom part, that dt, you're thinking of as a tiny change to t, right? So you're thinking of it as kind of a nudge. I'll draw it as a sizable line here for like moving from some original input over, but you might in principle think of it as a very, very tiny nudge in t. And over here, you'd say, well, that's gonna move your intermediary output in the xy-plane to, you know, maybe it'll move it in some amount. Again, imagine this is a very small nudge, I'm gonna give it some size here just so I can write into it.
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Multivariable chain rule intuition.mp3
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Well, that bottom part, that dt, you're thinking of as a tiny change to t, right? So you're thinking of it as kind of a nudge. I'll draw it as a sizable line here for like moving from some original input over, but you might in principle think of it as a very, very tiny nudge in t. And over here, you'd say, well, that's gonna move your intermediary output in the xy-plane to, you know, maybe it'll move it in some amount. Again, imagine this is a very small nudge, I'm gonna give it some size here just so I can write into it. And then whatever that nudge in the output space, and it's a nudge in some direction, that's gonna correspond to some change in f, some change based on, you know, based on the differential properties of the multivariable function itself. And if we think about this, this change, you might break it into components and say this shift here has some kind of dx, some kind of shift in the x direction and some kind of dy, some shift in the y direction. But you can actually reason about what these should be because it's not just an arbitrary change in x or an arbitrary change in y, it's the one that was caused by dt.
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Multivariable chain rule intuition.mp3
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Again, imagine this is a very small nudge, I'm gonna give it some size here just so I can write into it. And then whatever that nudge in the output space, and it's a nudge in some direction, that's gonna correspond to some change in f, some change based on, you know, based on the differential properties of the multivariable function itself. And if we think about this, this change, you might break it into components and say this shift here has some kind of dx, some kind of shift in the x direction and some kind of dy, some shift in the y direction. But you can actually reason about what these should be because it's not just an arbitrary change in x or an arbitrary change in y, it's the one that was caused by dt. So if I go over here, I might say that dx is caused by that dt, and the whole meaning of the derivative, the whole meaning of the single variable derivative would be that when we take dx dt, this is the factor that tells us, you know, a tiny nudge in t, how much does that change the x component? And if you want, you could think of this as kind of canceling out the dts and you're just left with x. But really, you're saying there's a tiny nudge in t and that results in a change in x, and this derivative is what tells you the ratio between those sizes.
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Multivariable chain rule intuition.mp3
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But you can actually reason about what these should be because it's not just an arbitrary change in x or an arbitrary change in y, it's the one that was caused by dt. So if I go over here, I might say that dx is caused by that dt, and the whole meaning of the derivative, the whole meaning of the single variable derivative would be that when we take dx dt, this is the factor that tells us, you know, a tiny nudge in t, how much does that change the x component? And if you want, you could think of this as kind of canceling out the dts and you're just left with x. But really, you're saying there's a tiny nudge in t and that results in a change in x, and this derivative is what tells you the ratio between those sizes. And similarly, similarly, that change in y here, that change in y is gonna be somehow proportional to the change in t, and that proportion is given by the derivative of y with respect to t, that's the whole point of the derivative, oh, no, no, no, with respect to t. And again, you can kind of think of it as if you're canceling out the t's, and this is why the fractional writing, this Leibniz notation is actually pretty helpful. You know, people will say, oh, mathematicians would like shake their heads at the idea of treating these like fractions, but not only is it a useful thing to do because it is a helpful mnemonic, it's reflective of what you're gonna do when you make a very formal argument, and I think I'll do that in one of the following videos. I'll describe this in a much more formal way that's a little bit more airtight than the kind of hand-waving, nudging around, but the intuition you get from just writing this as a fraction is basically the scaffolding for that formal argument, so it's a fine thing to do.
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Multivariable chain rule intuition.mp3
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But really, you're saying there's a tiny nudge in t and that results in a change in x, and this derivative is what tells you the ratio between those sizes. And similarly, similarly, that change in y here, that change in y is gonna be somehow proportional to the change in t, and that proportion is given by the derivative of y with respect to t, that's the whole point of the derivative, oh, no, no, no, with respect to t. And again, you can kind of think of it as if you're canceling out the t's, and this is why the fractional writing, this Leibniz notation is actually pretty helpful. You know, people will say, oh, mathematicians would like shake their heads at the idea of treating these like fractions, but not only is it a useful thing to do because it is a helpful mnemonic, it's reflective of what you're gonna do when you make a very formal argument, and I think I'll do that in one of the following videos. I'll describe this in a much more formal way that's a little bit more airtight than the kind of hand-waving, nudging around, but the intuition you get from just writing this as a fraction is basically the scaffolding for that formal argument, so it's a fine thing to do. I don't think mathematicians are shaking their head every time that a student or a teacher does this, but anyway, so this is kind of what gives you what that dx is, what that dy is, and then over here, if you're saying, how much does that change the ultimate output of the f, you could say, well, your nudge of size dx over here, you're wondering how much that changes the output of f, that's the meaning of the partial derivative, right? If we say we have the partial derivative with respect to x, what that means is that if you take a tiny nudge of size x, this is giving you the ratio between that and the ultimate change to the output that you want. You could kind of think of it like this partial x is canceling out with that dx if you wanted, or you could just say, this is a tiny nudge in x, this is gonna result in some change in f, I'm not sure what, but the meaning of the derivative is the ratio between those two, and that's what lets you figure it out.
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Multivariable chain rule intuition.mp3
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I'll describe this in a much more formal way that's a little bit more airtight than the kind of hand-waving, nudging around, but the intuition you get from just writing this as a fraction is basically the scaffolding for that formal argument, so it's a fine thing to do. I don't think mathematicians are shaking their head every time that a student or a teacher does this, but anyway, so this is kind of what gives you what that dx is, what that dy is, and then over here, if you're saying, how much does that change the ultimate output of the f, you could say, well, your nudge of size dx over here, you're wondering how much that changes the output of f, that's the meaning of the partial derivative, right? If we say we have the partial derivative with respect to x, what that means is that if you take a tiny nudge of size x, this is giving you the ratio between that and the ultimate change to the output that you want. You could kind of think of it like this partial x is canceling out with that dx if you wanted, or you could just say, this is a tiny nudge in x, this is gonna result in some change in f, I'm not sure what, but the meaning of the derivative is the ratio between those two, and that's what lets you figure it out. And similarly, that's just, you might call this the change in f caused by x, like due to x, due to, I should say, dx. But that's not the only thing changing the value of f, right, that's not the only change happening in the input space. You also have another change in f, and this one I might say is due to dy, due to that tiny shift in y, and what that's gonna be, we know it's gonna be proportional to that tiny shift in y, and the proportionality constant, this is the meaning of the partial derivative, that when you nudge y in some way, it results in some kind of nudge in f, and the ratio between those two is what the derivative gives.
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Multivariable chain rule intuition.mp3
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You could kind of think of it like this partial x is canceling out with that dx if you wanted, or you could just say, this is a tiny nudge in x, this is gonna result in some change in f, I'm not sure what, but the meaning of the derivative is the ratio between those two, and that's what lets you figure it out. And similarly, that's just, you might call this the change in f caused by x, like due to x, due to, I should say, dx. But that's not the only thing changing the value of f, right, that's not the only change happening in the input space. You also have another change in f, and this one I might say is due to dy, due to that tiny shift in y, and what that's gonna be, we know it's gonna be proportional to that tiny shift in y, and the proportionality constant, this is the meaning of the partial derivative, that when you nudge y in some way, it results in some kind of nudge in f, and the ratio between those two is what the derivative gives. So ultimately, if you put this all together, what you'd say is there's two different things causing an ultimate change to f. So if you put these together, and you wanna know what the total change in f is, so I might go over here and say the total change in f, one of them is caused by partial f, partial x, and I can multiply it by dx here, but really, we know that dx, the change there, was in turn caused by dt, so that in turn is caused by the change in the x component that was due to dt, that was of course of size dt, and then for similar reasons, the other way that this changes in the y direction is a partial of f with respect to y, but what caused that initial shift in y? You'd say, well, that was a shift in y that was due to t, and that size of dy dt times dt, you could think of it. So slight nudge in t causes a change in y, that change in y causes a change in f, and when you add those two together, that's everything that's going on, that's everything that influences the ultimate change in f. So then if you take this whole expression and you divide everything out by dt, so I kind of erase it from this side and put it over here, dt, this is your multivariable chain rule, and of course, I've just written the same thing again, but hopefully this gives a little bit of an intuition for how you're composing different nudges and why you wanna think about it that way.
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Multivariable chain rule intuition.mp3
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So let's say you have yourself some kind of multivariable function, and this time let's say it's got some very high dimensional input. So x1, x2, on and on and on and on, up to x sub n for some large number n. In the last couple videos I told you about the Laplacian operator, which is a way of taking in your scalar valued function f, and it gives you a new scalar valued function. It's kind of like a second derivative thing, because it takes the divergence of the gradient of your function f. So the gradient of f gives you a vector field, and the divergence of that gives you a scalar field. And what I want to show you here is another formula that you might commonly see for this Laplacian. So first let's kind of abstractly write out what the gradient of f will look like. So we start by taking this del operator, which is gonna be a vector full of partial differential operators, partial with respect to x1, partial with respect to x2, and kind of on and on and on, up to partial with respect to whatever that last input variable is. You take that whole thing, and then you just kind of imagine multiplying it by your function.
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Explicit Laplacian formula.mp3
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And what I want to show you here is another formula that you might commonly see for this Laplacian. So first let's kind of abstractly write out what the gradient of f will look like. So we start by taking this del operator, which is gonna be a vector full of partial differential operators, partial with respect to x1, partial with respect to x2, and kind of on and on and on, up to partial with respect to whatever that last input variable is. You take that whole thing, and then you just kind of imagine multiplying it by your function. So what you end up getting is all the different partial derivatives of f, right? It's partial of f with respect to the first variable, and then kind of on and on and on, up until you get the partial derivative of f with respect to that last variable, x sub n. And the divergence of that, and just to save myself some writing, I'm gonna say you take that nabla operator, and then you imagine taking the dot product between that whole operator and this gradient vector that you have here. What you end up getting is, well you start by multiplying the first components, which involves taking the partial derivative with respect to x1, that first variable, of the partial derivative of f with respect to that same variable.
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Explicit Laplacian formula.mp3
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You take that whole thing, and then you just kind of imagine multiplying it by your function. So what you end up getting is all the different partial derivatives of f, right? It's partial of f with respect to the first variable, and then kind of on and on and on, up until you get the partial derivative of f with respect to that last variable, x sub n. And the divergence of that, and just to save myself some writing, I'm gonna say you take that nabla operator, and then you imagine taking the dot product between that whole operator and this gradient vector that you have here. What you end up getting is, well you start by multiplying the first components, which involves taking the partial derivative with respect to x1, that first variable, of the partial derivative of f with respect to that same variable. So it looks like the second partial derivative of f with respect to that first variable. So second partial derivative of f with respect to x1, that first variable. And then you imagine kind of adding what the product of these next two items will be, and for very similar reasons, that's gonna look like the second partial derivative of f with respect to that second variable, partial x2 squared.
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Explicit Laplacian formula.mp3
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What you end up getting is, well you start by multiplying the first components, which involves taking the partial derivative with respect to x1, that first variable, of the partial derivative of f with respect to that same variable. So it looks like the second partial derivative of f with respect to that first variable. So second partial derivative of f with respect to x1, that first variable. And then you imagine kind of adding what the product of these next two items will be, and for very similar reasons, that's gonna look like the second partial derivative of f with respect to that second variable, partial x2 squared. And you do this to all of them, and you're adding them all up until you find yourself doing it to the last one. So you've got plus and then a whole bunch of things, and you'll be taking the second partial derivative of f with respect to that last variable, partial of x sub n. This is another format in which you might see the Laplacian, and oftentimes it's written kind of compactly. So people will say the Laplacian of your function f is equal to, and then using sigma notation, you'd say the sum from i is equal to one up to, you know, one, two, three, up to n. So the sum from that up to n of your second partial derivatives, partial squared of f with respect to that ith variable.
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Explicit Laplacian formula.mp3
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And then you imagine kind of adding what the product of these next two items will be, and for very similar reasons, that's gonna look like the second partial derivative of f with respect to that second variable, partial x2 squared. And you do this to all of them, and you're adding them all up until you find yourself doing it to the last one. So you've got plus and then a whole bunch of things, and you'll be taking the second partial derivative of f with respect to that last variable, partial of x sub n. This is another format in which you might see the Laplacian, and oftentimes it's written kind of compactly. So people will say the Laplacian of your function f is equal to, and then using sigma notation, you'd say the sum from i is equal to one up to, you know, one, two, three, up to n. So the sum from that up to n of your second partial derivatives, partial squared of f with respect to that ith variable. So, you know, if you were thinking in terms of three variables, often x1, x2, x3, we instead write x, y, z, but it's common to more generally just say x sub i. So this here is kind of the alternate formula that you might see for the Laplacian. Personally, I always like to think about it as taking the divergence of the gradient of f, because you're thinking about the gradient field and the divergence of that kind of corresponds to maxima and minima of your original function, which is what I talked about in the initial intuition of Laplacian video.
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Explicit Laplacian formula.mp3
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So people will say the Laplacian of your function f is equal to, and then using sigma notation, you'd say the sum from i is equal to one up to, you know, one, two, three, up to n. So the sum from that up to n of your second partial derivatives, partial squared of f with respect to that ith variable. So, you know, if you were thinking in terms of three variables, often x1, x2, x3, we instead write x, y, z, but it's common to more generally just say x sub i. So this here is kind of the alternate formula that you might see for the Laplacian. Personally, I always like to think about it as taking the divergence of the gradient of f, because you're thinking about the gradient field and the divergence of that kind of corresponds to maxima and minima of your original function, which is what I talked about in the initial intuition of Laplacian video. But this formula is probably a little more straightforward when it comes to actual computations, and, oh wait, sorry, I forgot a squared there, didn't I? So, partial x squared. So this is second derivative.
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Explicit Laplacian formula.mp3
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Personally, I always like to think about it as taking the divergence of the gradient of f, because you're thinking about the gradient field and the divergence of that kind of corresponds to maxima and minima of your original function, which is what I talked about in the initial intuition of Laplacian video. But this formula is probably a little more straightforward when it comes to actual computations, and, oh wait, sorry, I forgot a squared there, didn't I? So, partial x squared. So this is second derivative. Yeah, so summing these second partial derivatives. And you can probably see this is kind of a more straightforward way to compute a given example that you might come across, and it also makes clear how the Laplacian is kind of an extension of the idea of a second derivative. See you next video.
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Explicit Laplacian formula.mp3
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We can now express this as a double integral over the domain of the parameters that we care about. And we're going to do that in this video. And then in the next series of videos, we'll do the same thing for this expression. We're actually going to do that using Green's Theorem. What we're going to do is we're going to see we get the same expressions, which will show us that Stokes' Theorem is true, at least for this special class of surfaces that we are studying right here. But they're pretty general. Now let's now try to do that.
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Stokes' theorem proof part 3 Multivariable Calculus Khan Academy.mp3
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We're actually going to do that using Green's Theorem. What we're going to do is we're going to see we get the same expressions, which will show us that Stokes' Theorem is true, at least for this special class of surfaces that we are studying right here. But they're pretty general. Now let's now try to do that. So our surface integral, I'm just going to rewrite it down here. It's the surface integral over our surface of the curl of f. Let me go a little bit lower. So we have our surface integral of the curl of f dot ds.
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Stokes' theorem proof part 3 Multivariable Calculus Khan Academy.mp3
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Now let's now try to do that. So our surface integral, I'm just going to rewrite it down here. It's the surface integral over our surface of the curl of f. Let me go a little bit lower. 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.
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Stokes' theorem proof part 3 Multivariable Calculus Khan Academy.mp3
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