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I'll do it over here. If we were to draw our x and y axes, this associates a vector, a force vector. Let's say this is actually a force vector with every point in our xy-plane. So this is x, this is y. So if we're at the point, for example, 1, 0, what will the vector look like that's associated with that point? Well, a... | Using a line integral to find the work done by a vector field example Khan Academy.mp3 |
So this is x, this is y. So if we're at the point, for example, 1, 0, what will the vector look like that's associated with that point? Well, at 1, 0, y is 0. So this will be 0i minus 1j. Minus 1j looks like this. So minus 1j will look like that. At x is equal to 2, I'm just picking points at random ones. | Using a line integral to find the work done by a vector field example Khan Academy.mp3 |
So this will be 0i minus 1j. Minus 1j looks like this. So minus 1j will look like that. At x is equal to 2, I'm just picking points at random ones. It'll be easy. y is still 0. And now the force vector here would be minus 2j. | Using a line integral to find the work done by a vector field example Khan Academy.mp3 |
At x is equal to 2, I'm just picking points at random ones. It'll be easy. y is still 0. And now the force vector here would be minus 2j. So it would look something like this. Minus 2j, something like that. Likewise, if we were to go here where y is equal to 1 and x is equal to 0. | Using a line integral to find the work done by a vector field example Khan Academy.mp3 |
And now the force vector here would be minus 2j. So it would look something like this. Minus 2j, something like that. Likewise, if we were to go here where y is equal to 1 and x is equal to 0. When y is equal to 1, we have 1i minus 0j. So then our vector is going to look like that at this point. Our vector will look li... | Using a line integral to find the work done by a vector field example Khan Academy.mp3 |
Likewise, if we were to go here where y is equal to 1 and x is equal to 0. When y is equal to 1, we have 1i minus 0j. So then our vector is going to look like that at this point. Our vector will look like that. If we were to go to 2, you could get the picture. You can keep plotting these points. You just want to get a ... | Using a line integral to find the work done by a vector field example Khan Academy.mp3 |
Our vector will look like that. If we were to go to 2, you could get the picture. You can keep plotting these points. You just want to get a sense of what it looks like. If you go here, the vector is going to look like that. If you go maybe this point right here, the vector is going to look like that. I think you get t... | Using a line integral to find the work done by a vector field example Khan Academy.mp3 |
You just want to get a sense of what it looks like. If you go here, the vector is going to look like that. If you go maybe this point right here, the vector is going to look like that. I think you get the general idea. I could keep filling in the space for this entire field all over just to make it symmetric. If I was ... | Using a line integral to find the work done by a vector field example Khan Academy.mp3 |
I think you get the general idea. I could keep filling in the space for this entire field all over just to make it symmetric. If I was right over here, the vector is going to look like that. You get the idea. I could just fill in all of the points if I had to. Now, in that field, I have some particle moving. And let's ... | Using a line integral to find the work done by a vector field example Khan Academy.mp3 |
You get the idea. I could just fill in all of the points if I had to. Now, in that field, I have some particle moving. And let's say its path is described by the curve C. And the parametrization of it is x of t is equal to cosine of t. And y of t is equal to sine of t. And the path will occur from t. Let's say 0 is les... | Using a line integral to find the work done by a vector field example Khan Academy.mp3 |
And let's say its path is described by the curve C. And the parametrization of it is x of t is equal to cosine of t. And y of t is equal to sine of t. And the path will occur from t. Let's say 0 is less than or equal to t. It's less than or equal to 2 pi. You might already recognize what this would be. This parametriza... | Using a line integral to find the work done by a vector field example Khan Academy.mp3 |
So the path that this guy is going to go is going to start here. t, in this case, you can imagine it's the angle of the circle. But you could also imagine t as time. So time equals 0. We're going to be over here. Then at time of pi over 2, we're going to have traveled a quarter of the circle to there. So we're moving i... | Using a line integral to find the work done by a vector field example Khan Academy.mp3 |
So time equals 0. We're going to be over here. Then at time of pi over 2, we're going to have traveled a quarter of the circle to there. So we're moving in that direction. And then at time after pi seconds, we would have gotten right there. And then all the way after 2 pi seconds, we would have gotten all the way aroun... | Using a line integral to find the work done by a vector field example Khan Academy.mp3 |
So we're moving in that direction. And then at time after pi seconds, we would have gotten right there. And then all the way after 2 pi seconds, we would have gotten all the way around the circle. So our path, our curve, is one counterclockwise rotation around the circle, so to speak. So what is the work done by this f... | Using a line integral to find the work done by a vector field example Khan Academy.mp3 |
So our path, our curve, is one counterclockwise rotation around the circle, so to speak. So what is the work done by this field on this curve? So the work we learned in the previous video is equal to the line integral over this contour of our vector field dotted with the differential of our movement. Well, I haven't ev... | Using a line integral to find the work done by a vector field example Khan Academy.mp3 |
Well, I haven't even defined r yet. I mean, I kind of have just the parametrization here. So we need to have a vector function. We need to have some r that defines this path. This is just a standard parametrization. But if I wanted to write it as a vector function of t, we would write that r of t is equal to x of t, wh... | Using a line integral to find the work done by a vector field example Khan Academy.mp3 |
We need to have some r that defines this path. This is just a standard parametrization. But if I wanted to write it as a vector function of t, we would write that r of t is equal to x of t, which is cosine of t, times i plus y of t times j, which is just sine of t times j. And likewise, this is for 0 is less than or eq... | Using a line integral to find the work done by a vector field example Khan Academy.mp3 |
And likewise, this is for 0 is less than or equal to t, which is less than or equal to 2 pi. This and this are equivalent. The reason why I took the pain of doing this is so now I can take its vector function derivative and I can figure out its differential. And then I can actually take the dot product with this thing ... | Using a line integral to find the work done by a vector field example Khan Academy.mp3 |
And then I can actually take the dot product with this thing over here. So let's do all of that and actually calculate this line integral and figure out the work done by this field. And one thing might already pop in your mind. We're going in a counterclockwise direction. But at every point where we're passing through,... | Using a line integral to find the work done by a vector field example Khan Academy.mp3 |
We're going in a counterclockwise direction. But at every point where we're passing through, it looks like the field is going exactly opposite the direction of our motion. For example, here we're moving upwards. The field is pulling us backwards. Here we're moving to the top left. The field is moving us to the bottom r... | Using a line integral to find the work done by a vector field example Khan Academy.mp3 |
The field is pulling us backwards. Here we're moving to the top left. The field is moving us to the bottom right. Here we're moving exactly to the left. The field is pulling us to the right. So it looks like the field is always doing the exact opposite of what we're trying to do. It's hindering our ability to move. | Using a line integral to find the work done by a vector field example Khan Academy.mp3 |
Here we're moving exactly to the left. The field is pulling us to the right. So it looks like the field is always doing the exact opposite of what we're trying to do. It's hindering our ability to move. So I'll give you a little intuition. This will probably deal with negative work. For example, if I lift something off... | Using a line integral to find the work done by a vector field example Khan Academy.mp3 |
It's hindering our ability to move. So I'll give you a little intuition. This will probably deal with negative work. For example, if I lift something off the ground, I have to apply force to fight gravity. I'm doing positive work, but gravity's doing negative work on that. We're just going to do the math here just to m... | Using a line integral to find the work done by a vector field example Khan Academy.mp3 |
For example, if I lift something off the ground, I have to apply force to fight gravity. I'm doing positive work, but gravity's doing negative work on that. We're just going to do the math here just to make you comfortable with this idea. But it's interesting to think about what's exactly going on even here. The field ... | Using a line integral to find the work done by a vector field example Khan Academy.mp3 |
But it's interesting to think about what's exactly going on even here. The field is, let me, the field, I'm doing that pink color. So let me stick to that. The field is pushing in that direction. So it's always going opposite the motion. But let's just do the math to make everything in the last video a little bit more ... | Using a line integral to find the work done by a vector field example Khan Academy.mp3 |
The field is pushing in that direction. So it's always going opposite the motion. But let's just do the math to make everything in the last video a little bit more concrete. So a good place to start is the derivative of our position vector function with respect to t. So let me switch. So we have dr dt, which we could a... | Using a line integral to find the work done by a vector field example Khan Academy.mp3 |
So a good place to start is the derivative of our position vector function with respect to t. So let me switch. So we have dr dt, which we could also write as r prime of t. This is equal to the derivative of x of t with respect to t, which is minus sine of t times i, plus the derivative of y of t with respect to t. Tha... | Using a line integral to find the work done by a vector field example Khan Academy.mp3 |
So we get dr is equal to, we could write it this way. We could actually even just put the d, well, let me just do it. So it's minus sine of t dt. I'm just multiplying each of these terms by dt, distributive property, times the unit vector i, plus cosine of t dt times the unit vector j. So we have this piece now. And th... | Using a line integral to find the work done by a vector field example Khan Academy.mp3 |
I'm just multiplying each of these terms by dt, distributive property, times the unit vector i, plus cosine of t dt times the unit vector j. So we have this piece now. And then we want to take the dot product with this over here. But let me rewrite our vector field in terms of t, so to speak. So what's our field going ... | Using a line integral to find the work done by a vector field example Khan Academy.mp3 |
But let me rewrite our vector field in terms of t, so to speak. So what's our field going to be doing at any point t? We don't have to worry about every point. We don't have to worry, for example, that over here, the vector field is going to be doing something like that. Because that's not on our path. That force never... | Using a line integral to find the work done by a vector field example Khan Academy.mp3 |
We don't have to worry, for example, that over here, the vector field is going to be doing something like that. Because that's not on our path. That force never had an impact on the particle. We only care about what happens along our path. So we can find a function that we can essentially substitute y and x for their r... | Using a line integral to find the work done by a vector field example Khan Academy.mp3 |
We only care about what happens along our path. So we can find a function that we can essentially substitute y and x for their relative functions with respect to t. And then we'll have the force from the field at any point or any time t. So let's do that. So this guy right here, if I were to write it as a function of t... | Using a line integral to find the work done by a vector field example Khan Academy.mp3 |
Sine of t times i, plus, or actually minus, x of t. x is a function of t. So minus cosine of t times j. And now all of it seems a little bit more straightforward. If we want to find this line integral, this line integral is going to be the same thing as the integral. Let me pick a nice, soothing color. Maybe this is a ... | Using a line integral to find the work done by a vector field example Khan Academy.mp3 |
Let me pick a nice, soothing color. Maybe this is a nice one. The integral from t is equal to 0 to t is equal to 2 pi of f dot dr. Now, when you take the dot product, you just multiply the corresponding components and add it up. So you take the product of the sine of t with the minus sine of t dt, I get minus sine squa... | Using a line integral to find the work done by a vector field example Khan Academy.mp3 |
So you take the product of the sine of t with the minus sine of t dt, I get minus sine squared t dt. And then you're going to add that to. So you're going to have that plus. Let me write that dt a little bit. That was a wacky looking dt. And then you're going to have that plus. These two guys multiplied by each other. | Using a line integral to find the work done by a vector field example Khan Academy.mp3 |
Let me write that dt a little bit. That was a wacky looking dt. And then you're going to have that plus. These two guys multiplied by each other. So that's, well, there's a minus sign here. So plus, let me just change this to minus, minus cosine squared dt. And if we factor out a minus sign and a dt, what is this going... | Using a line integral to find the work done by a vector field example Khan Academy.mp3 |
These two guys multiplied by each other. So that's, well, there's a minus sign here. So plus, let me just change this to minus, minus cosine squared dt. And if we factor out a minus sign and a dt, what is this going to be equal to? This is going to be equal to the integral from 0 to 2 pi of, we could say, sine squared ... | Using a line integral to find the work done by a vector field example Khan Academy.mp3 |
And if we factor out a minus sign and a dt, what is this going to be equal to? This is going to be equal to the integral from 0 to 2 pi of, we could say, sine squared plus, I want to put the t there, sine squared of t plus cosine squared of t. Actually, let me take the minus sign out to the front. So if we just factor ... | Using a line integral to find the work done by a vector field example Khan Academy.mp3 |
So the minus sign out there. And we factor dt out. I did a couple of steps in there, but I think you got it. And this is just algebra at this point. Factoring out a minus sign, so this becomes positive. And then you have a dt and a dt, factor that out, and you get this. You could multiply this out and you'll get what w... | Using a line integral to find the work done by a vector field example Khan Academy.mp3 |
And this is just algebra at this point. Factoring out a minus sign, so this becomes positive. And then you have a dt and a dt, factor that out, and you get this. You could multiply this out and you'll get what we originally have, if that confuses you at all. And the reason why I did that, we know what sine squared of a... | Using a line integral to find the work done by a vector field example Khan Academy.mp3 |
You could multiply this out and you'll get what we originally have, if that confuses you at all. And the reason why I did that, we know what sine squared of anything plus cosine squared of that same anything is. That falls right out of the unit circle definition of our trig function. So this is just 1. So our whole int... | Using a line integral to find the work done by a vector field example Khan Academy.mp3 |
So this is just 1. So our whole integral has been reduced to the minus integral from 0 to 2 pi of dt. And we've seen this before, we can say this is of 1, if you want to put something there. Then the antiderivative of 1 is just going to be equal to minus. And that minus sign is just the same minus sign that we're carry... | Using a line integral to find the work done by a vector field example Khan Academy.mp3 |
Then the antiderivative of 1 is just going to be equal to minus. And that minus sign is just the same minus sign that we're carrying forward. The antiderivative of 1 is just t. And we're going to evaluate it from 2 pi to 0, or from 0 to 2 pi. So this is equal to minus that minus sign right there. 2 pi minus t at 0. So ... | Using a line integral to find the work done by a vector field example Khan Academy.mp3 |
So this is equal to minus that minus sign right there. 2 pi minus t at 0. So minus 0. So this is just equal to minus 2 pi. And there you have it. We figured out the work that this field did on the particle, or whatever, as whatever thing, was moving around in this counterclockwise fashion. And our intuition held up. | Using a line integral to find the work done by a vector field example Khan Academy.mp3 |
So this is just equal to minus 2 pi. And there you have it. We figured out the work that this field did on the particle, or whatever, as whatever thing, was moving around in this counterclockwise fashion. And our intuition held up. We actually got a negative number for the work done. And that's because at all times, th... | Using a line integral to find the work done by a vector field example Khan Academy.mp3 |
And essentially what this is, it's just a way to package all the information of the second derivatives of a function. So let's say you have some kind of multivariable function, like, I don't know, like the example we had in the last video, e to the x halves multiplied by sine of y. So some kind of multivariable functio... | The Hessian matrix Multivariable calculus Khan Academy.mp3 |
What the Hessian matrix is, and it's often denoted with an H, or kind of a bold-faced H, is it's a matrix, incidentally enough, that contains all the second partial derivatives of f. So the first component is gonna be the partial derivative of f with respect to x, kind of twice in a row. And everything in this first co... | The Hessian matrix Multivariable calculus Khan Academy.mp3 |
So that's kind of the first column of the matrix. And then up here, it's the partial derivative where first you do it with respect to y, and then you do it with respect to x. And then over here it's where you do it with respect to y, both times in a row. So partial with respect to y, both times in a row. So let's go ah... | The Hessian matrix Multivariable calculus Khan Academy.mp3 |
So partial with respect to y, both times in a row. So let's go ahead and actually compute this and think about what this would look like in the case of our specific function here. So in order to get all the second partial derivatives, we first should just kind of keep a record of the first partial derivatives. So the p... | The Hessian matrix Multivariable calculus Khan Academy.mp3 |
So the partial derivative of f with respect to x, the only place x shows up is in this e to the x halves. Kind of bring down that 1 1 2, e to the x halves, and sine of y just looks like a constant as far as x is concerned, sine of y. And then the partial derivative with respect to y, partial derivative of f with respec... | The Hessian matrix Multivariable calculus Khan Academy.mp3 |
Now e to the x halves looks like a constant and it's being multiplied by something that has a y in it, e to the x halves. And the derivative of sine of y, since we're doing it with respect to y, is cosine of y, cosine of y. So these terms won't be included in the Hessian itself, but we're kind of just kind of keeping a... | The Hessian matrix Multivariable calculus Khan Academy.mp3 |
So up here is when we did it with respect to x. If we did it with respect to x again, we kind of bring down another half, so that becomes 1 4th by e to the x halves. And that sine of y just still looks like a constant. Sine of y. And then this mixed partial derivative where we do it with respect to x, then y. So we did... | The Hessian matrix Multivariable calculus Khan Academy.mp3 |
Sine of y. And then this mixed partial derivative where we do it with respect to x, then y. So we did it with respect to x here. When we differentiate this with respect to y, the 1 1 2 e to the x halves just looks like a constant, but then derivative of sine of y ends up as cosine of y. And then up here, it's gonna be ... | The Hessian matrix Multivariable calculus Khan Academy.mp3 |
When we differentiate this with respect to y, the 1 1 2 e to the x halves just looks like a constant, but then derivative of sine of y ends up as cosine of y. And then up here, it's gonna be the same thing, but let's kind of see how, when you do it in the other direction, when you do it first with respect to y, then x.... | The Hessian matrix Multivariable calculus Khan Academy.mp3 |
If we took this derivative with respect to x, you'd have the half would come down, so that would be 1 1 2 e to the x halves multiplied by cosine of y because that just looks like a constant since we're doing it with respect to x the second time. So that would be cosine of y. And it shouldn't feel like a surprise that b... | The Hessian matrix Multivariable calculus Khan Academy.mp3 |
With most functions, that's the case. Technically not all functions. You can come up with some crazy things where this won't be symmetric, where you'll have different terms than the diagonal. But for the most part, those you can kind of expect to be the same. And then this last term here where we do it with respect to ... | The Hessian matrix Multivariable calculus Khan Academy.mp3 |
But for the most part, those you can kind of expect to be the same. And then this last term here where we do it with respect to y twice, we now think of taking the derivative of this whole term with respect to y, that e to the x halves looks like a constant, and derivative of cosine is negative sine, negative sine of y... | The Hessian matrix Multivariable calculus Khan Academy.mp3 |
This is the Hessian of f. And sometimes people will write it as Hessian of f, kind of specifying what function it's of. And you could think of this, I mean, you could think of it as a matrix-valued function, which feels kind of weird, but you plug in two different values, x and y, and you'll get a matrix. So it's this ... | The Hessian matrix Multivariable calculus Khan Academy.mp3 |
And the nice thing about writing it like this is that you can actually extend it so that rather than just for functions that have two variables, let's say you had a function, you know, kind of like clear this up, let's say you had a function that had three variables or four variables or kind of any number. So let's say... | The Hessian matrix Multivariable calculus Khan Academy.mp3 |
And then over here it would be the second partial derivative of f, where first you did it with respect to, first you did it with respect to y, and then you do it with respect to z. I'll clear up even more room here. Because you'd have another column where you'd have the second partial derivative where this time in ever... | The Hessian matrix Multivariable calculus Khan Academy.mp3 |
And then as the very last component, you'd have the second partial derivative where first you do it with respect to, well I guess you do it with respect to z twice. So this whole thing, this three by three matrix would be the Hessian of a three variable function. And you can see how you could extend this pattern where ... | The Hessian matrix Multivariable calculus Khan Academy.mp3 |
So here I want to talk about the gradient in the context of a contour map. So let's say we have a multivariable function, a two variable function, f of x, y, and this one is just going to equal x times y. So we can visualize this with a contour map just on the x, y plane. So what I'm going to do is I'm going to go over... | Gradient and contour maps.mp3 |
So what I'm going to do is I'm going to go over here, I'm going to draw my y axis and my x axis. Alright, so this right here represents x values, this represents y values, and this is entirely the input space. And I have a video on contour maps if you are unfamiliar with them or are feeling uncomfortable. And the conto... | Gradient and contour maps.mp3 |
And the contour map for x times y looks something like this. And each one of these lines represents a constant value. So you might be thinking that you have, you know, let's say you want to, the constant value for f of x times y is equal to two would be one of these lines, that would be what one of these lines represen... | Gradient and contour maps.mp3 |
And a way you could think about that for this specific function is you're saying, hey, when is x times y equal to two? And that's kind of like the graph y equals two over x. And that's where you would see something like this. So all of these lines, they're representing constant values for the function. And now I want t... | Gradient and contour maps.mp3 |
So all of these lines, they're representing constant values for the function. And now I want to take a look at the gradient field. And the gradient, if you'll remember, is just a vector full of the partial derivatives of f. And let's just actually write it out. The gradient of f with our little del symbol is a function... | Gradient and contour maps.mp3 |
The gradient of f with our little del symbol is a function of x and y. And it's a vector-valued function whose first coordinate is the partial derivative of f with respect to x. And the second component is the partial derivative with respect to y. So when we actually do this for our function, we take the partial deriva... | Gradient and contour maps.mp3 |
So when we actually do this for our function, we take the partial derivative with respect to x, it takes a look, x looks like a variable, y looks like a constant, the derivative of this whole thing is just equal to that constant, y. And then kind of the reverse for when you take the partial derivative with respect to y... | Gradient and contour maps.mp3 |
You know, at every given point, xy, so you kind of go like x equals two, y equals one, let's say, so that would be x equals two, y equals one. You would plug in the vector and see what should be output. And at this point, the point is two, one, the desired output kind of swaps those. So we're looking somehow to draw th... | Gradient and contour maps.mp3 |
So we're looking somehow to draw the vector one, two. So you would expect to see the vector that has an x component of one and a y component of two. Something like that. But it's probably gonna be scaled down because of the way we usually draw vector fields. And the entire field looks like this. So I'll go ahead and er... | Gradient and contour maps.mp3 |
But it's probably gonna be scaled down because of the way we usually draw vector fields. And the entire field looks like this. So I'll go ahead and erase what I had going on since this is a little bit clearer. And remember, we scaled down all the vectors, but color represents length. So red here is super long, blue is ... | Gradient and contour maps.mp3 |
And remember, we scaled down all the vectors, but color represents length. So red here is super long, blue is gonna be kind of short. And one thing worth noticing, if you take a look at all of the given points around, if the vector is crossing a contour line, it's perpendicular to that contour line, wherever you go. Yo... | Gradient and contour maps.mp3 |
You know, you go down here, this vector's perpendicular to the contour line. Over here, perpendicular to the contour line. And this happens everywhere, and it's for a very good reason. And it's also super useful. So let's just think about what that reason should be. Let's zoom in on a point. So I'm gonna clear up our f... | Gradient and contour maps.mp3 |
And it's also super useful. So let's just think about what that reason should be. Let's zoom in on a point. So I'm gonna clear up our function here, clear up all of the information about it, and just zoom in on one of those points. So let's say like right here. I'll take that guy and kind of imagine zooming in and sayi... | Gradient and contour maps.mp3 |
So I'm gonna clear up our function here, clear up all of the information about it, and just zoom in on one of those points. So let's say like right here. I'll take that guy and kind of imagine zooming in and saying what's going on in that region. So you've got some kind of contour line, and it's swooping down like this... | Gradient and contour maps.mp3 |
So you've got some kind of contour line, and it's swooping down like this. And that represents some kind of value. Let's say that represents the value f equals two. And you know, it might not be a perfect straight line, but the more you zoom in, the more it looks like a straight line. And when you want to interpret the... | Gradient and contour maps.mp3 |
And you know, it might not be a perfect straight line, but the more you zoom in, the more it looks like a straight line. And when you want to interpret the gradient vector, if you remember in the video about how to interpret the gradient in the context of a graph, I said it points in the direction of steepest ascent. S... | Gradient and contour maps.mp3 |
And there's two ways of thinking about that. One is to look at all of these different directions and say which one increases x the most. But another way of doing it would be to get rid of them all and just take a look at another contour line that represents a slight increase. So let's say you're taking a look at a cont... | Gradient and contour maps.mp3 |
So let's say you're taking a look at a contour line, another contour line, something like this. And maybe that represents something that's right next to it, like 2.1. That represents another value that's very close. And if I were a better artist and this was more representative, this would be, it would look like a line... | Gradient and contour maps.mp3 |
And if I were a better artist and this was more representative, this would be, it would look like a line that's parallel to the original one. Because if you change the output by just a little bit, the set of endpoints that look like it is pretty much the same but just shifted over a bit. So another way we can think abo... | Gradient and contour maps.mp3 |
And this time, instead of thinking of the fastest as constant length vectors, what increases it the most, we'll be thinking constant increase in the output, which one does it with the shortest distance? And if you think of them as being roughly parallel lines, it shouldn't be hard to convince yourself that the shortest... | Gradient and contour maps.mp3 |
Because if you think about these as lines, and the more you zoom in, the more they pretty much look like parallel lines, the path that connects one to the other is gonna be perpendicular to both of them. So because of this interpretation of the gradient as the direction of steepest descent, it's a natural consequence t... | Gradient and contour maps.mp3 |
And we landed at the idea that the divergence of v, you know, when you take the divergence of this vector-valued function, it should definitely have something to do with the partial derivative of p, that x component of the output with respect to x. And here I want to do the opposite and say, okay, what if we look at fu... | Divergence formula, part 2.mp3 |
So kind of doing the same thing we did last time, if we start thinking about cases where the divergence of our function at a given point should be positive, and an example of that, you might be saying, nothing is happening at the point itself, so q itself would be zero, but then below it, things are kind of going away,... | Divergence formula, part 2.mp3 |
So now you're starting to get this idea of partial q with respect to y, you know, as we change that y and move up in space, the value of q should be positive. So positive divergence seems to correspond to a positive value here. And the thinking is actually going to be almost identical to what we did in the last video w... | Divergence formula, part 2.mp3 |
And again, you have this idea of q starts off small, so here it's kind of q starts off small, maybe it's kind of like near zero, and then here q is something positive, and then here it's even more positive, and I'm sort of making up notation here, but I want the idea of kind of small, and then medium-sized, and then bi... | Divergence formula, part 2.mp3 |
What would positive and negative and zero divergence all look like? But the upshot of it all, pretty much for the same reasons I went through in the last video, is this partial derivative with respect to y corresponds to the divergence. And when we combine this with our conclusions about the x component, that actually ... | Divergence formula, part 2.mp3 |
So just to write it all out, if we have a vector-valued function of x and y, and it's got both of its components, you've got p as the x component of the output, that first component of the output, and q, and we're looking at both of these at once, the way that we compute divergence, the definition of divergence of this... | Divergence formula, part 2.mp3 |
And hopefully by now, this isn't just kind of a formula that I'm plopping down for you, but it's something that makes intuitive sense. When you see this term, this partial p with respect to x, you're thinking about, oh, yes, yes, because if you have flow that's kind of increasing as you move in the x direction, that's ... | Divergence formula, part 2.mp3 |
So both of these correspond to that idea of divergence that we're going for. And if you just add them up, this gives you everything you need to know. And one thing that's pretty neat and maybe kind of surprising is that the way we just came across this formula and started to think about it was in the simplified case wh... | Divergence formula, part 2.mp3 |
But in reality, as we know, vector fields can look much more complicated. And maybe you have something where, you know, it's not just in the x direction. There's lots of things going on and you need to account for all of those. And evidently, just looking at the change in the x component with respect to x or the change... | Divergence formula, part 2.mp3 |
And evidently, just looking at the change in the x component with respect to x or the change in the y component of the output with respect to the y component of the input gives you all the information you need to know. And basically, what's going on here is that any fluid flow can just be broken down into the x and y c... | Divergence formula, part 2.mp3 |
So what I want to talk about here is how to interpret the directional derivative in terms of graphs. So I have here the graph of a function, a multivariable function. It's f of x, y is equal to x squared times y. In the last couple videos, I talked about what the directional derivative is, how you can formally define i... | Directional derivatives and slope.mp3 |
In the last couple videos, I talked about what the directional derivative is, how you can formally define it, how you compute it using the gradient. And generally the way that you, the setup that you might have is you'll have some kind of vector, and this is a vector in the input space. So in this case, it's gonna be i... | Directional derivatives and slope.mp3 |
And in this case, I'll just take the vector, I'll take the vector one, one. Okay. And the directional derivative, which we denote by kind of taking the gradient symbol, except you stick the name of that vector down in the lower part there, the directional derivative of your function, it'll still take the same inputs. T... | Directional derivatives and slope.mp3 |
This is kind of a measure of how the function changes when the input moves in that direction. So I'll show you what I mean. I mean, you can imagine slicing this graph by some kind of plane, but that plane doesn't necessarily have to be parallel to the x or y axes. You know, that's what we did for the partial derivative... | Directional derivatives and slope.mp3 |
You know, that's what we did for the partial derivative. We'd take a plane that represented a constant x value or a constant y value, but this is gonna be a plane that kind of tells you what movement in the direction of your vector looks like. And like I have a number of other times, I'm gonna go ahead and slice the gr... | Directional derivatives and slope.mp3 |
And just to make it clear, I'm gonna color in where the graph intersects that slice. And this vector here, this little v, you'd be thinking of as living on the xy plane, and it's determining the direction of this plane that we're slicing things with. So on the xy plane, you've got this vector, it's one, one. It kind of... | Directional derivatives and slope.mp3 |
It kind of points in that diagonal direction. And you take the whole plane and you slice your graph. And if we want to interpret the directional derivative here, I'm gonna go ahead and fill in an actual value. So let's say we wanted to do it at like negative one, one. Negative one, negative one, because I guess I chose... | Directional derivatives and slope.mp3 |
So let's say we wanted to do it at like negative one, one. Negative one, negative one, because I guess I chose a plane that passes through the origin, so I've got to make sure that the point I'm evaluating it actually goes along this plane. But you could imagine one that points in the same direction, but you kind of sl... | Directional derivatives and slope.mp3 |
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