Sentence stringlengths 131 8.39k | video_title stringlengths 12 104 |
|---|---|
So you have a minus 2y just like that. And so this simplifies to 2y minus minus 2y. That's 2y plus 2y. I'm just subtracting a negative. Or this inside, and just to save space, this inside, that's just 4y. So let me just, I don't want to have to rewrite the boundaries. So that right there is the same thing as 4y. | Green's theorem example 1 Multivariable Calculus Khan Academy.mp3 |
I'm just subtracting a negative. Or this inside, and just to save space, this inside, that's just 4y. So let me just, I don't want to have to rewrite the boundaries. So that right there is the same thing as 4y. The partial of q with respect to y, 2y, minus the partial of p with respect to y, which is minus 2y. You subt... | Green's theorem example 1 Multivariable Calculus Khan Academy.mp3 |
So that right there is the same thing as 4y. The partial of q with respect to y, 2y, minus the partial of p with respect to y, which is minus 2y. You subtract a negative, you get a positive. You have 4y. So let's take the antiderivative of the inside with respect to y. And we're going to get 2y squared. Let me do it a ... | Green's theorem example 1 Multivariable Calculus Khan Academy.mp3 |
You have 4y. So let's take the antiderivative of the inside with respect to y. And we're going to get 2y squared. Let me do it a little bit lower. We're going to get 2y squared. If you take the derivative of this with respect, the partial with respect to y, you're going to get 4y. And we're going to evaluate that from ... | Green's theorem example 1 Multivariable Calculus Khan Academy.mp3 |
Let me do it a little bit lower. We're going to get 2y squared. If you take the derivative of this with respect, the partial with respect to y, you're going to get 4y. And we're going to evaluate that from y is equal to 2x squared to y is equal to 2x. And of course, we still have the outside integral still there. x goe... | Green's theorem example 1 Multivariable Calculus Khan Academy.mp3 |
And we're going to evaluate that from y is equal to 2x squared to y is equal to 2x. And of course, we still have the outside integral still there. x goes from 0 to 1 dx. This thing is going to be equal to the integral from 0 to 1. And then we evaluate it first at 2x. So you put 2x in here. 2x squared is 4x squared, rig... | Green's theorem example 1 Multivariable Calculus Khan Academy.mp3 |
This thing is going to be equal to the integral from 0 to 1. And then we evaluate it first at 2x. So you put 2x in here. 2x squared is 4x squared, right? 2 squared x squared. So 4x squared times 2 is going to be 8x squared minus, put this guy in there. 2x squared squared is 4x to the fourth times 2 is 8x to the 4th. | Green's theorem example 1 Multivariable Calculus Khan Academy.mp3 |
2x squared is 4x squared, right? 2 squared x squared. So 4x squared times 2 is going to be 8x squared minus, put this guy in there. 2x squared squared is 4x to the fourth times 2 is 8x to the 4th. Did I do that right? 2x squared, I'm going to put it there for y, substitute y with it. That squared is 4x to the fourth ti... | Green's theorem example 1 Multivariable Calculus Khan Academy.mp3 |
2x squared squared is 4x to the fourth times 2 is 8x to the 4th. Did I do that right? 2x squared, I'm going to put it there for y, substitute y with it. That squared is 4x to the fourth times 2 is 8x to the fourth. Very good. All right. Now dx, now this is just a straightforward one-dimensional integral. | Green's theorem example 1 Multivariable Calculus Khan Academy.mp3 |
That squared is 4x to the fourth times 2 is 8x to the fourth. Very good. All right. Now dx, now this is just a straightforward one-dimensional integral. This is going to be equal to the antiderivative of 8x squared is 8 thirds x to the third. And then the antiderivative of x to the fourth is minus 8 fifths x to the fif... | Green's theorem example 1 Multivariable Calculus Khan Academy.mp3 |
Now dx, now this is just a straightforward one-dimensional integral. This is going to be equal to the antiderivative of 8x squared is 8 thirds x to the third. And then the antiderivative of x to the fourth is minus 8 fifths x to the fifth. And we're going to have to evaluate that from 0 to 1. I'll give it a little line... | Green's theorem example 1 Multivariable Calculus Khan Academy.mp3 |
And we're going to have to evaluate that from 0 to 1. I'll give it a little line there sometimes. And when you put 1 in there, you get, I'll do it in a different color. We get 8 fifths times 1 to the third, which is 8 fifths, minus 8 fifths. And then we're going to have minus, when you put 0 in here, you're just going ... | Green's theorem example 1 Multivariable Calculus Khan Academy.mp3 |
We get 8 fifths times 1 to the third, which is 8 fifths, minus 8 fifths. And then we're going to have minus, when you put 0 in here, you're just going to get a bunch of 0's. Oh, sorry. I made a mistake. It's not a blunder. It's 8 thirds. 8 thirds times 1 to the third minus 8 fifths times 1 to the fifth. | Green's theorem example 1 Multivariable Calculus Khan Academy.mp3 |
I made a mistake. It's not a blunder. It's 8 thirds. 8 thirds times 1 to the third minus 8 fifths times 1 to the fifth. So that's minus 8 fifths. And then when you subtract the 0, so then minus, you evaluate 0 here, you're just going to get a bunch of 0's. You don't have to do anything else. | Green's theorem example 1 Multivariable Calculus Khan Academy.mp3 |
8 thirds times 1 to the third minus 8 fifths times 1 to the fifth. So that's minus 8 fifths. And then when you subtract the 0, so then minus, you evaluate 0 here, you're just going to get a bunch of 0's. You don't have to do anything else. So now we just have to subtract these two fractions. So let's get a common denom... | Green's theorem example 1 Multivariable Calculus Khan Academy.mp3 |
You don't have to do anything else. So now we just have to subtract these two fractions. So let's get a common denominator of 15. 8 thirds is the same thing if we multiply the numerator and denominator by 5. That is 40 fifteenths. And if we multiply this numerator and denominator by 3, that's going to be 24 over 15. So... | Green's theorem example 1 Multivariable Calculus Khan Academy.mp3 |
8 thirds is the same thing if we multiply the numerator and denominator by 5. That is 40 fifteenths. And if we multiply this numerator and denominator by 3, that's going to be 24 over 15. So minus 24 over 15. And we get it being equal to 16 over 15. So using Green's theorem, we were able to find the answer to this inte... | Green's theorem example 1 Multivariable Calculus Khan Academy.mp3 |
So far, when I've talked about the gradient of a function, and you know, let's think about this as a multivariable function with just two inputs, those are the easiest to think about. So maybe it's something like x squared plus y squared, very friendly function. When I've talked about the gradient, I've left open a mys... | Why the gradient is the direction of steepest ascent.mp3 |
We have the way of computing it, and the way that you think about computing it is you just take this vector and you just throw the partial derivatives in there, partial with respect to x and the partial with respect to y, with respect to y, and if it was a higher dimensional input, then the output would have as many va... | Why the gradient is the direction of steepest ascent.mp3 |
But then I gave you a graphical intuition. I said that it points in the direction of steepest ascent. And maybe the way you think about that is you have your input space, which in this case is the xy plane, and you think of it as somehow mapping over to the number line, to your output space. And if you have a given poi... | Why the gradient is the direction of steepest ascent.mp3 |
And if you have a given point somewhere, the question is, of all the possible directions that you can move away from this point, all those different directions you can go, which one of them, you know, like this point will land somewhere on the function, and as you move in the various directions, maybe one of them nudge... | Why the gradient is the direction of steepest ascent.mp3 |
If you want to think in terms of graphs, we could look over at the graph of f of x squared, and this is the gradient field. All of these vectors in the xy plane are the gradients, and you kind of look from below, you can maybe see why each one of these points in the direction you should move to walk uphill on that grap... | Why the gradient is the direction of steepest ascent.mp3 |
This is why you call it direction of steepest ascent. So back over here, I don't see the connection immediately, or at least when I was first learning about it, it wasn't clear why this combination of partial derivatives has anything to do with choosing the best direction. And now that we've learned about the direction... | Why the gradient is the direction of steepest ascent.mp3 |
So let's say instead of thinking about, you know, all the possible directions, and all of the possible changes to the output that they have, I'll fill in my line there. You know, let's say you just have, you've got your point where you're evaluating things, and then you just have a single vector. And let's actually mak... | Why the gradient is the direction of steepest ascent.mp3 |
Let's make it the case that this guy has a length of one. So I'll go over here and I'll just think of that guy as being v, and say that v has a length of one. So this is our vector. We know now, having learned about the directional derivative that you can tell the rate at which the function changes as you move in this ... | Why the gradient is the direction of steepest ascent.mp3 |
We know now, having learned about the directional derivative that you can tell the rate at which the function changes as you move in this direction by taking the directional derivative of your function, and let's say this point, I don't know, what's a good name for this point? Just like a, b. a, b is this point. When y... | Why the gradient is the direction of steepest ascent.mp3 |
So evaluating that at your point, a, b, together with whatever the vector is, whatever that value is. And in this case, we're thinking of v as a univector. So this, this is how you tell the rate of change, and when I originally introduced the directional derivative, I gave kind of an indication why. You know, if you im... | Why the gradient is the direction of steepest ascent.mp3 |
You know, if you imagine dotting this together with, I don't know, let's say it was a vector that's like one, two. Really, you're thinking this vector represents one step in the x direction, two steps in the y direction, so the amount that it changes things should be one times the change caused by a pure step in the x ... | Why the gradient is the direction of steepest ascent.mp3 |
You can see the directional derivative video if you want a little bit more discussion on that. And this is the formula that you have, but this starts to give us the key for how we could choose the direction of steepest descent, because now what we're really asking, when we say which one of these changes things the most... | Why the gradient is the direction of steepest ascent.mp3 |
What we're doing is we're saying, find the maximum for all unit vectors, so for all vectors v that satisfy the property that their length is one, find the maximum of the dot product between f, evaluated at that point, right, evaluated at whatever point we care about, and v. Find that maximum. Well, let's just think abo... | Why the gradient is the direction of steepest ascent.mp3 |
So if you imagine some vector v, you know, some unit vector v, let's say it was sticking off in this direction, the way that you interpret this dot product, the dot product between the gradient f and this new vector v, is you would project that vector directly, kind of a perpendicular projection onto your gradient vect... | Why the gradient is the direction of steepest ascent.mp3 |
And then you'd multiply that by the length of the gradient itself, of the vector against which you're dotting, and maybe that guy, maybe the length of the entire gradient vector, just again as an example, maybe that's two. It doesn't have to be, it could be anything. But the way that you interpret this whole dot produc... | Why the gradient is the direction of steepest ascent.mp3 |
You would take 0.7, the length of your projection, times the length of the original vector. And the question is, when is this maximized? What unit vector maximizes this? And if you start to imagine maybe swinging that unit vector around, so if instead of that guy, you were to use one that pointed a little bit more clos... | Why the gradient is the direction of steepest ascent.mp3 |
And if you start to imagine maybe swinging that unit vector around, so if instead of that guy, you were to use one that pointed a little bit more closely in the direction, then its projection would be a little bit longer. Maybe that projection would be like 0.75 or something. If you take the unit vector that points dir... | Why the gradient is the direction of steepest ascent.mp3 |
It would be one. Because projecting it doesn't change what it is at all. So it shouldn't be too hard to convince yourself. And if you have shaky intuitions on the dot product, I'd suggest finding the videos we have on Khan Academy for those. Sal does a great job giving that deep intuition. But it should kind of make se... | Why the gradient is the direction of steepest ascent.mp3 |
And if you have shaky intuitions on the dot product, I'd suggest finding the videos we have on Khan Academy for those. Sal does a great job giving that deep intuition. But it should kind of make sense why the unit vector that points in the same direction as your gradient is gonna be what maximizes it. So the answer her... | Why the gradient is the direction of steepest ascent.mp3 |
So the answer here, the answer to what vector maximizes this is gonna be, well, it's the gradient itself. It is that gradient vector evaluated at the point we care about, except you'd normalize it because we're only considering unit vectors. So to do that, you just divide it by whatever its magnitude is. If its magnitu... | Why the gradient is the direction of steepest ascent.mp3 |
If its magnitude was already one, it stays one. If its magnitude was two, you're dividing it down by a half. So this is your answer. This is the direction of steepest descent. So I think one thing to notice here is the most fundamental fact is that the gradient is this tool for computing directional derivatives. You ca... | Why the gradient is the direction of steepest ascent.mp3 |
This is the direction of steepest descent. So I think one thing to notice here is the most fundamental fact is that the gradient is this tool for computing directional derivatives. You can think of that vector as something that you really want to dot against. And that's actually a pretty powerful thought. It's that the... | Why the gradient is the direction of steepest ascent.mp3 |
And that's actually a pretty powerful thought. It's that the gradient, it's not just a vector. It's a vector that loves to be dotted together with other things. That's the fundamental. And as a consequence of this, as a consequence of that, the direction of steepest descent is that vector itself because anything, if yo... | Why the gradient is the direction of steepest ascent.mp3 |
That's the fundamental. And as a consequence of this, as a consequence of that, the direction of steepest descent is that vector itself because anything, if you're saying what maximizes the dot product with that thing, it's, well, the vector that points in the same direction as that thing. And this can also give us an ... | Why the gradient is the direction of steepest ascent.mp3 |
We know the direction is the direction of steepest ascent, but what does the length mean? So let's give this guy a name. Let's give this normalized version of it a name. I'm just gonna call it W. So W will be the unit vector that points in the direction of the gradient. If you take the directional derivative in the dir... | Why the gradient is the direction of steepest ascent.mp3 |
I'm just gonna call it W. So W will be the unit vector that points in the direction of the gradient. If you take the directional derivative in the direction of W of F, what that means is the gradient of F dotted with that W. And if you kind of spell out what W means here, that means you're taking the gradient of the ve... | Why the gradient is the direction of steepest ascent.mp3 |
And all of these, I'm just writing gradient of F, but maybe you should be thinking about gradient of F evaluated at AB, but I'm just being kind of lazy and just writing gradient of F. And the top, when you take the dot product with itself, what that means is the square of its magnitude. But the whole thing is divided b... | Why the gradient is the direction of steepest ascent.mp3 |
You can say this doesn't need to be there. That exponent doesn't need to be there. And basically, the directional derivative, the directional derivative in the direction of the gradient itself has a value equal to the magnitude of the gradient. So this tells you when you're moving in that direction, in the direction of... | Why the gradient is the direction of steepest ascent.mp3 |
Okay, so we are finally ready to express the quadratic approximation of a multivariable function in vector form. So I have the whole thing written out here where f is the function that we are trying to approximate, x-naught, y-naught is the constant point about which we are approximating, and then this entire expressio... | Vector form of multivariable quadratic approximation.mp3 |
Quadratic. Now to vectorize things, first of all, let's write down the input, the input variable x, y as a vector, and typically we'll do that with a bold-faced x to indicate that it's a vector, and its components are just gonna be the single variables x and y, the non-bold-faced, so this is the vector representing the... | Vector form of multivariable quadratic approximation.mp3 |
And more specifically, it's the Jacobian matrix, or sometimes the associated determinant. And here, I just want to talk about some of the background knowledge that I'm assuming. Because to understand the Jacobian, you do have to have a little bit of a background in linear algebra. And in particular, I want to make sure... | Jacobian prerequisite knowledge.mp3 |
And in particular, I want to make sure that everyone here understands how to think about matrices as transformations of space. And when I say transformations, let me just get kind of a matrix on here. I'll call it 2, 1, and negative 3, 1. You'll see why I'm coloring it like this in just a moment. And when I say how to ... | Jacobian prerequisite knowledge.mp3 |
You'll see why I'm coloring it like this in just a moment. And when I say how to think about this as a transformation of space, I mean you can multiply a matrix by some kind of two-dimensional vector, some kind of two-dimensional x, y. And this is gonna give us a new two-dimensional vector. This is gonna bring us to, l... | Jacobian prerequisite knowledge.mp3 |
This is gonna bring us to, let's see, in this case, it'll be, I'll write kind of 2, 1, negative 3, 1, where what it gives us is 2x plus negative 3 times y, and then 1x plus 1 times y. Right, this is a new two-dimensional vector somewhere else in space. And even if you know how to compute it, there's still room for a de... | Jacobian prerequisite knowledge.mp3 |
And there's also still a deeper understanding in what we mean when we call this a linear transformation, a linear transformation. So what I'm gonna do is just show you what this particular transformation looks like on the left here, where every single point on this blue grid, I'm gonna tell the computer, hey, if that p... | Jacobian prerequisite knowledge.mp3 |
So let me just kind of play it out here. All of the points in space move, and you end up in some final state here. And there are a couple important things to note. First of all, all of the grid lines remain parallel and evenly spaced. And they're still lines, they didn't get curved in some way. And that's very, very sp... | Jacobian prerequisite knowledge.mp3 |
First of all, all of the grid lines remain parallel and evenly spaced. And they're still lines, they didn't get curved in some way. And that's very, very special. That is the geometric way that you can think about this term, this idea of a linear transformation. I kind of like to think about it that lines stay lines. A... | Jacobian prerequisite knowledge.mp3 |
That is the geometric way that you can think about this term, this idea of a linear transformation. I kind of like to think about it that lines stay lines. And in particular, the grid lines here, the ones that started off as, you know, kind of vertical and horizontal, they still remain parallel and they still remain ev... | Jacobian prerequisite knowledge.mp3 |
And the other thing to notice here is I have these two vectors highlighted, the green vector and the red vector. And these are the ones that started off, if we kind of back things up, these are the ones that started off as the basis vectors, right? Let me kind of make a little bit more room here. The green vector is 1,... | Jacobian prerequisite knowledge.mp3 |
The green vector is 1, 0, 1 in the x direction, 0 in the y direction. And then that red vertical vector here is 0, 1. Right, 0, 1. And if we notice where they land under this transformation, when the matrix is multiplied by every single vector in space, the place where the green vector lands, the one that started off a... | Jacobian prerequisite knowledge.mp3 |
And if we notice where they land under this transformation, when the matrix is multiplied by every single vector in space, the place where the green vector lands, the one that started off as 1, 0, has coordinates 2, 1. And that corresponds very directly with the fact that the first column of our matrix is 2, 1. And the... | Jacobian prerequisite knowledge.mp3 |
And that's what corresponds with the fact that the next column is negative 3, 1. And it's actually relatively simple to see why that's going to be true. Here, I'll go ahead and multiply this matrix that we had that was... See, now it's kind of easy to remember what the matrix is, right? I can just kind of read it off h... | Jacobian prerequisite knowledge.mp3 |
I can just kind of read it off here as 2, 1, negative 3, 1. But just to see why it's actually taking the basis vectors to the columns like this, if we do the multiplication by 1, 0, notice how it's going to take us to... So it's 2 times 1, that'll be 2, and then negative 3 times 0, so that'll just be 0. And over here, ... | Jacobian prerequisite knowledge.mp3 |
And over here, it's 1 times 1, so that's 1, and then 1 times 0. So again, we're adding 0. So the only terms that actually mattered because of the 0 down here was everything in that first column. But if we take that same matrix, 2, 1, negative 3, 1, and we multiply it by 0, 1 over here, by the second basis vector, what ... | Jacobian prerequisite knowledge.mp3 |
But if we take that same matrix, 2, 1, negative 3, 1, and we multiply it by 0, 1 over here, by the second basis vector, what you're going to get is 2 times 0, so 0 plus that element in that second column, and then 1 times 0, so another 0, plus 1 times 1, plus that 1. It's kind of like that 0 knocks out all of the terms... | Jacobian prerequisite knowledge.mp3 |
And when you start to think about it a little bit, if you can know where this green vector lands and where this red vector lands, that's going to lock into place where the entire grid has to go. And let me show you what I mean and how this corresponds with maybe a different definition that you've heard for what linear ... | Jacobian prerequisite knowledge.mp3 |
So here you're applying the transformation to a scaled vector, and evidently that's the same as scaling the transformation of the vector. And similarly, second property of linearity is that if you add two vectors, it doesn't really matter if you add them before or after the transformation. If you take the sum of the ve... | Jacobian prerequisite knowledge.mp3 |
And one of the most important consequences of this formal definition of linearity is that it means if you take your function and apply it to some vector, x, y, well it can split up that vector as x times the first basis vector, x times one, zero, plus y, let's see, y, times that second basis vector, zero, one. And beca... | Jacobian prerequisite knowledge.mp3 |
So to be concrete, let's actually put in a value for x and y here and try to think about that specific vector geometrically. So maybe I'll put in something like vector two, one. So if we look over on the grid, we're going to be focusing on the point that's over here, two, one, and this particular point. And I'm going t... | Jacobian prerequisite knowledge.mp3 |
And I'm going to play the transformation, and I want you to follow this point to see where it lands, and it's going to end up over here. Okay, so in terms of the old grid, right, the original one that we started with, it's now at the point one, three. This is where we've ended up. But importantly, I want you to notice ... | Jacobian prerequisite knowledge.mp3 |
But importantly, I want you to notice how it's still two times that green vector, plus one times that red vector. So it's satisfying that property that it's still x times whatever the transformed version of that first basis vector is, plus y times the transformed version of that second basis vector. So that's all just ... | Jacobian prerequisite knowledge.mp3 |
So continuing on with where we were in the last video, we're looking for this unit tangent vector function given the parameterization. So the specific example that I have is a function that parameterizes a circle with radius capital R, but I also kind of want to show in parallel what this looks like more abstractly. So... | Curvature formula, part 3.mp3 |
So we actually have the same thing over here where the unit tangent vector should be the derivative function, which we know gives a tangent, right, it's just it might not be unit, but then we normalize it. We take the magnitude of that tangent vector function. And in our specific case with the circle, once we did this ... | Curvature formula, part 3.mp3 |
It's this, the X component of the derivative plus Y prime of T squared, just taking the magnitude of a vector here. So when we take the entire function and divide it by that, what we get doesn't simplify as it did in the case of a circle. Instead, we have that X prime of T, which is the X component of our S prime of T,... | Curvature formula, part 3.mp3 |
You have to divide it by that whole square root expression. I'm just gonna kind of write dot, dot, dot with the understanding that this square root expression is what goes up there. And similarly, over here, we'd have Y prime of T divided by that entire expression again, right? So simplification doesn't always happen. ... | Curvature formula, part 3.mp3 |
So simplification doesn't always happen. That was just kind of a lucky happenstance of our circle example. And then now what we want, once we have the unit tangent vector as a function of that same parameter, what we're hoping to find is the derivative of that unit tangent vector with respect to arc length, the arc len... | Curvature formula, part 3.mp3 |
That's gonna be what curvature is. But the way to do this is to take the derivative with respect to the parameter T, so D big T, D little t, and then divide it out by the derivative of our function S with respect to T, which we already found. And the reason I'm doing this, loosely, if you're just thinking of the notati... | Curvature formula, part 3.mp3 |
But another way to think about this is to say, when we have our tangent vector function as a function of T, the parameter T, we're not sure of what its change is with respect to S, right? That's something we don't know directly. But we do directly know its change with respect to a tiny change in that parameter. So then... | Curvature formula, part 3.mp3 |
So then if we just kind of correct that by saying, hey, how much does the length of the curve change? How far do you move along the curve as you change that parameter? And maybe if I go back up to the picture here, this Ds Dt is saying, for a tiny nudge in time, right, what is the ratio of the size of the movement ther... | Curvature formula, part 3.mp3 |
So the reason that this comes out to be a very large vector, right, it's not a tiny thing, is because you're taking the ratio. Maybe this tiny change was a just itty bitty smidgen vector, but you're dividing it by like one one millionth, or whatever the size of Dt that you're thinking. And in this specific case for our... | Curvature formula, part 3.mp3 |
And what this means in our specific case, if we wanna apply this to our circle example, we take Dt, D big T, the tangent vector function, and I'll go ahead and write it here. We have the derivative of our tangent vector with respect to the parameter, and we go up and we look here, we say, okay, the unit tangent vector ... | Curvature formula, part 3.mp3 |
You've got a cosine, you've got a sine. There's nothing else in there. You're gonna end up with cosine squared plus sine squared. So this magnitude just equals one. And when we do what we're supposed to over here and divide it by the magnitude of the derivative, right, we take this and we divide it by the magnitude of ... | Curvature formula, part 3.mp3 |
So this magnitude just equals one. And when we do what we're supposed to over here and divide it by the magnitude of the derivative, right, we take this and we divide it by the magnitude of the derivative, the S, Dt. Well, we've already computed the magnitude of the derivative. That was R. That's how we got this R, is ... | Curvature formula, part 3.mp3 |
That was R. That's how we got this R, is we took the derivative here and took its magnitude and found it. So we find that in the specific case of the circle, the curvature function that we want is just constantly equal to one over R, which is good and hopeful, right, because I said in the original video on curvature th... | Curvature formula, part 3.mp3 |
So I should hope that its curvature ends up being one divided by R. And in the more general case, if we take a look at what this ought to be, you can maybe imagine just how horrifying it's gonna be to compute this, right? We've got our tangent vector function, which itself, you know, is almost too long for me to write ... | Curvature formula, part 3.mp3 |
And you're gonna have to take this, take its derivative with respect to t, right? It's not gonna get any simpler when you take its derivative. Take the magnitude of that and divide all of that by the magnitude of the derivative of your original function. And I think what I'll do, I'm not gonna go through all of that he... | Curvature formula, part 3.mp3 |
And I think what I'll do, I'm not gonna go through all of that here. It's a little bit much, and I'm not sure how helpful it is to walk through all those steps. But for the sake of having it, for anyone who's curious, I think I'll put that into an article, and you can kind of go through the steps at your own pace and s... | Curvature formula, part 3.mp3 |
And I'll just tell you right now, maybe kind of a spoiler alert, what that formula comes out to be is x prime, the derivative of that first component, multiplied by y double prime, the second derivative of that second component, minus y prime, first derivative of that second component, multiplied by x double prime. And... | Curvature formula, part 3.mp3 |
And you can maybe see why you're gonna get terms like this, right, because when you're taking, when you're taking the derivative of, when you're taking the derivative of the unit tangent vector function, you have the square root term in it, the square root that has x primes and y primes. So that's where you're gonna ge... | Curvature formula, part 3.mp3 |
And it turns out it comes in here at a 3 halves power. And what I'm gonna do in the next video, I'm gonna go ahead and describe kind of an intuition for why this formula isn't random. Why if you break down what this is saying, it really does give a feeling for the curvature, the amount that a curve curves, that we want... | Curvature formula, part 3.mp3 |
So it's almost like this is a third way of thinking about it, right? The first one, I said you have whatever circle most closely hugs your curve, and you take one divided by its radius. And then the second way, you're thinking of this DTDS, the change in the unit tangent vector with respect to arc length and taking its... | Curvature formula, part 3.mp3 |
And what that means is we're starting to allow ourselves to use terms like x squared, x times y, and y squared. And quadratic basically just means anytime you have two variables multiplied together. So here you have two x's multiplied together, here it's an x multiplied with a y, and here, y squared, that kind of thing... | Quadratic approximation formula, part 1.mp3 |
So let's take a look at this local linearization. It seems like a lot, but once you actually kind of go through term by term, you realize it's a relatively simple function, and if you were to plug in numbers for the constant terms, it would come out as something relatively simple. Because this right here, where you're ... | Quadratic approximation formula, part 1.mp3 |
That's just gonna output some kind of number. And similarly, when you do that to the partial derivative, this little f sub x means partial derivative at that point, you're just getting another number. And over here, this is also just another number, but we've written it in the abstract form so that you can see what you... | Quadratic approximation formula, part 1.mp3 |
And the reason for having it like this, the reason that it comes out to this form is because of a few important properties that this linearization has. Let me move this stuff out of the way. We'll get back to it in a moment. But I just wanna emphasize a few properties that this has, because it's gonna be properties tha... | Quadratic approximation formula, part 1.mp3 |
But I just wanna emphasize a few properties that this has, because it's gonna be properties that we want our quadratic approximation to have as well. First of all, when you actually evaluate this function at the desired point, at x naught, y naught, what do you get? Well, this constant term isn't influenced by the vari... | Quadratic approximation formula, part 1.mp3 |
And now the rest of the terms, when we plug in x here, this is the only place where you actually see the variable. Maybe that's worth pointing out, right? We've got two variables here, and there's a lot going on, but the only places where you actually see those variables show up, where you have to plug in anything, is ... | Quadratic approximation formula, part 1.mp3 |
When you plug in x naught for our initial input, this entire term goes to zero, right? And then similarly, when you plug in y naught over here, this entire term is gonna go to zero, which multiplies out to zero for everything. So what you end up with, you don't have to add anything else. This is just a fact. And this i... | Quadratic approximation formula, part 1.mp3 |
This is just a fact. And this is an important fact, because it tells you your approximation for the function at the point about which you are approximating actually equals the value of the function at that point, so that's very good. But we have a couple other important facts also, because this isn't just a constant ap... | Quadratic approximation formula, part 1.mp3 |
If you were to take the partial derivative of this linearization with respect to x, what do you get? What do you get when you actually take this partial derivative? Well, if you look up at the original function, this constant term is nothing, so that just corresponds to a zero. Over here, this entire thing looks like a... | Quadratic approximation formula, part 1.mp3 |
Over here, this entire thing looks like a constant multiplied by x minus something, and if you differentiate this with respect to x, what you're gonna get is that constant term, which is the partial derivative of f evaluated at our specific point. And then the other term has no x's in it, it's just a y, which as far as... | Quadratic approximation formula, part 1.mp3 |
Now notice, this is not saying that our linearization has the same partial derivative as f everywhere, it's just saying that its partial derivative happens to be a constant, and the constant that it is is the value of the partial derivative of f at that specific input point. And you can do pretty much the same thing, a... | Quadratic approximation formula, part 1.mp3 |
You know the value of the linearization at the point, and the value of its two different partial derivatives, and these kind of define the linearization itself. Now what we're gonna do for the quadratic approximation is take this entire formula, and I'm just literally gonna copy it here, and then we're gonna add to it ... | Quadratic approximation formula, part 1.mp3 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.