00:01
In this problem, we will compute the change in f from p to q, and then we will use the differential of f at p to approximate that change.
00:12
So let's start by computing delta f, which is the change in f from p to q, and that's simply f evaluated at q minus f evaluated at p.
00:43
And then from here we just simply plug in the values for x, y, and z.
00:59
So this is 2 times .99 times minus 1 .02 squared times 2 .02 cubed, minus 2 times 1 times minus 1 squared times 2 cubed.
01:38
And when we evaluate this, we get that delta f is approximately 0 .97 -929.
02:01
So let's write that over here.
02:21
And so now we're going to use the differential of f at p to approximate the change in f from p to q.
02:30
So the differential of f, and this time we have three variables, but that's equal to the partial derivative of f with respect to x at p times d x plus the partial derivative of f with respect to y evaluated at p times d y and then finally the partial derivative of f with respect to z evaluated at p times d z and as we can see right away to compute this, we need to find dx, dy, and dz.
03:50
So to do so, we can let dx be the change in x, d.
04:00
Y, the change in y, and then dz to be the change in z.
04:15
So we get that dx is .99 minus 1, which is minus 0 .01.
04:29
And we get that dx is minus 1 .02 minus negative 1.
04:37
So that's minus 0 .02 and then d z which is 2 .02 minus 2 which is 0 .02 and now from here we want to compute our partial derivatives.
05:02
So we have the partial derivative of f with respect to x and our function is 2.
05:16
X times y squared times z cubed so we'll treat x as the only variable and when we do that we get that this partial derivative is two times y squared times z cubed and we still have to evaluate this at p so we get that this is two times negative one squared which is still two times two cubed which is two to the fourth which is 16.
06:09
And now we want the partial derivative of f with respect to y...