00:01
So for this problem, we have f of x, y being equal to 1 ,000 minus 0 .005x squared minus 0 .01y squared.
00:18
And we're given the point equal to 6040.
00:23
And we want to go, we're trying to find the directional derivative at a point p along v.
00:32
So we have our directional derivative equation.
00:36
We know all that.
00:37
What we ultimately want to find, though, is our unit vector.
00:41
So since we're moving due south, we know that our unit vector is just going to be 0 -9 -1, because if we think about a compass where this is north, this is south.
00:52
Due -south is this directional derivative of 0 -1, and that's a unit vector already.
01:01
So then our gradient of f, xy is going to be equal to negative 0 .01x, negative 0 .02y.
01:17
And then we want to evaluate it at the point 6040.
01:22
So the gradient of f at 6040 equals negative 0 .0 .6, negative 0 .6, negative 0 .8.
01:36
Then from the equation 9, we can write our directional derivative as being equal to this dot our unit vector.
01:49
So what we end up getting is just 0 .8 as our directional derivative right here...