Computing a Jacobian matrix

So just as a reminder of where we are, we've got this very nonlinear transformation. We showed that if you zoom in on a specific point while that transformation is happening, it looks a lot like something linear. We reasoned that you can figure out what linear transformation that looks like by taking the partial derivatives of your given function, the one that I defined up here, and then turning that into a matrix.

What I want to do here is basically just finish up what I was talking about by computing all of those partial derivatives. So first of all, let me just rewrite the function back on the screen so we have it in a convenient place to look at. The first component is X plus s of y, s of Y, and then y plus s of X was the second component.

So what I want to do here is just compute all of those partial derivatives to show what kind of thing this looks like. So let's go ahead and get rid of this word, then I'll go ahead and kind of redraw the matrix here. For that upper left component, we're taking the partial derivative with respect to X of the first component.

So we look up at this first component, and the partial derivative with respect to X is just one since there's 1 * X plus something that has nothing to do with X. Then below that, we take the partial derivative of the second component with respect to X down here, and that guy, the Y, well that looks like a constant, so nothing happens. The derivative of s of X becomes cosine of x.

Then up here, we're taking the partial derivative with respect to Y of the first component, that upper one here, and for that, you know, the partial derivative of x with respect to Y is zero. The partial derivative of s of Y with respect to Y is cosine of Y. Finally, the partial derivative of the second component with respect to Y looks like 1 because it's just 1 * y plus some constant.

This is the general Jacobian as a function of X and Y. But if we want to understand what happens around the specific point that started off at, well I think I recorded it here at -21. We plug that into each one of these values. When we plug in -21, so go ahead and just kind of again rewrite it to remember we're plugging in -21 as our specific point, that matrix as a function, kind of a matrix-valued function, becomes one.

Then next we have cosine, but we're plugging in -2 for x. Cosine of -2, and if you're curious, that is approximately equal to, I calculated this earlier, 0.42 if you just want to think in terms of a number there. Then for the upper right, we have cosine again, but now we're plugging in the value for Y, which is one, and cosine of 1 is approximately equal to 0.54.

Then bottom right, that's just another constant, one. So that is the matrix just as a matrix full of numbers. Just as kind of a gut check, we can take a look at the linear transformation this was supposed to look like and notice how the first basis vector, the thing it got turned into, which is this vector here, does look like it has coordinates 1 and 0.42.

Right, it's got this rightward component that's about as long as the vector itself started, and then this downward component, which I think that's, you know, pretty believable that that's 0.42. Likewise, this second column is telling us what happened to that second basis vector, which is the one that looks like this, and again, its y component is about as long as how it started, right? A length of one, and then the rightward component is around half of that, and we actually see that in the diagram.

But this is something you compute. Again, it's pretty straightforward; you just take all of the possible partial derivatives and you organize them into a grid like this. So with that, I'll see you guys next video.