00:01
So for this problem, we are attempting to find the gradient and the norm of the gradient.
00:05
As a reminder, what the gradient is, the gradient always gives you the direction of the maximum rate of change of your function.
00:13
So the steepest descent in a way.
00:16
In part a, we are attempting to use this function to find the maximum rate of change.
00:24
And the maximum rate of change is always the magnitude of the gradient.
00:28
So we must find the gradient.
00:30
As a reminder what the gradient is, the gradient is just the vector of partial derivatives.
00:39
So we want to take the derivative with respect to x1 and the derivative of respect to x2.
00:44
And this function is actually relatively simple.
00:48
The first one, the derivative with respect to x1 is just e to the x2, right, because x1 just x1e to the x2.
00:54
And secondly, this one is just x1e to the x2 because the derivative e to the x2 is just e to the x2.
01:02
And then what we do is we actually plug in the gradient of f at our point of interest, which is 1 comma 0.
01:09
And we would find this vector, which is just 1, and then we would have 1 as again.
01:16
So if we go in the direction of 1 -1 while standing at 1 .0, we would be moving in the fastest rate of change.
01:22
How fast is that? well, we just find the magnitude of the gradient, which is just equal to the square root of 1 squared plus 1 squared, which is equal to.
01:32
The square root of 2.
01:34
And that's our answer.
01:36
This is the magnitude of the fastest rate of change.
01:39
It's the fastest rate of change, actually, at the point 1 .0.
01:43
You can increasing your height at a rate of square root of 2.
01:48
Secondly, here, in this next one, we have another function of x1 and x2, and we are asked to find the gradient of this one, because we're asked to find the direction of the steepest ascent...