00:01
So the rate of change of temperature at p in the direction of 3 -9 -3 -3 is going to be this directional unit vector of t to negative -1 -2, where the unit vector u is from the point p to 3 -3 -3 -3.
00:28
And we know that p is 2 -9 -3.
00:34
So first we want to find the unit vector.
00:37
So we see that we subtract this point right here by this point, and we end up getting our position vector 1 -2 -1, right, 1 negative 2, 1.
00:47
And then we want to find the magnitude of that, which we know is the square root of 6.
00:53
So that tells us that our unit vector is 1 over root 6, negative 2 over root 6, 1 over root 6.
01:02
So unit vector is good, checked off our list.
01:05
Now we want to find the gradient of t.
01:09
So that's going to be equal to all of our partial derivatives.
01:15
So we have a negative, our x partial derivative is negative 400xe to the negative x squared minus 3y squared minus 9 z squared.
01:27
So it's quite a mouthful...