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# Computing a Jacobian matrix

Transkrypcja filmu video (w języku angielskim)
- [Teacher] So, just as a reminder of where we are, we've got this very non-linear transformation and we showed that if you zoom in on a specific point while that transformation is happening, it looks a lot like something linear and we reason 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. And 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 sin of y. Sin of y and then y plus sin 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 and I'll go ahead and kind of redraw the matrix here. So, 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 one times x plus something that has nothing to do with x and 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, and the derivative of sin of x becomes cosine of x. And then up here, we're taking the partial derivative with respect to y of the first component; that upper one here, and for that, partial derivative of x, with respect to y, is zero and partial derivative of sin of y, with respect to y, is cosine of y. And then, finally, the partial derivative of the second component with respect to y looks like one because it's just one times y plus some constant. And this is the general Jacobian as a function of x and y, but if we want to understand what happens around this specific point that started off at, well, I think I recorded it here at negative two, one, we plug that in to each one of these values. So, when we plug in negative two, one. So go ahead and just kind of again, rewrite it to remember we're plugging in negative two, one as our specific point, that matrix as a function, kind of a matrix valued function, becomes one, and then next we have cosine, but we're plugging in negative two for x, cosine of negative two, and if you're curious, that is approximately equal to, I calculated this earlier. Negative zero point four two, 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 one is approximately equal to zero point five four; and then bottom right, that's just another constant: one. So, that is the matrix, just as a matrix full of numbers, and 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 factor, the thing it got turned into, which is this factor here, does look like it has coordinates one and negative zero point four two, 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 pretty believable that that's negative zero point four two. And then, likewise, this second column is telling us what happened to that second basis factor, which is the one that looks like this. And again, its y component is about as long as how it started, right, the 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.