10. Consider the function f(x, y, z) = xy^2z where x

10. Consider the function $f(x, y, z) = xy^2z$ where $x > 0$, $y > 0$, and $z > 0$. a. Find the derivative of $f$ at the point $(1,2,3)$ in the direction of $(4,5,6)$. b. In what direction is $f$ increasing the fastest at the point $(1,2,3)$? c. What is the maximum rate of change of $f$ at the point $(1,2,3)$?

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00:01 So in this question, i have a function, and it's f of x, y, z, f of x, y, z is equal to x times y squared times z cubed.

00:16 In part a, i want to find the gradient of f.

00:20 Remember what the gradient of f is.

00:23 It is the vector, having components partial of f with respect to x, partial of f with respect to y, partial of f with respect to z.

00:31 My partial with respect to x this time is x squared z cubed.

00:39 My partial with respect to y is 2xy z cubed, and my partial with respect to z is 3xy squared z squared.

00:54 And there's part a.

00:56 In part b, i want my directional derivative at the point 211 in the direction towards the point 0 -3 -5.

01:07 So how do i get a directional derivative? remember, a directional derivative is a derivative in the direction of some unit vector u.

01:20 And this is equal to the gradient of f dotted with you.

01:25 So i need to figure out my unit vector u since i already have my gradient vector.

01:31 Well, i am traveling in the direction from 211 towards the point 0 negative 3 .5.

01:41 So what direction is that? well, my x decreased by 2.

01:46 So negative 2.

01:47 My y, that decreased by 4.

01:51 So negative 4.

01:54 And my z, that increased by 4.

01:57 So that's up 4.

02:00 And so now i need the magnitude of that.

02:04 Vector so that i can turn it into a unit vector well this will be the square root of 4 plus 16 plus 16 this is actually nice the square root of 36 is 6 and so this means in order to get my unit vector i have v over my magnitude of v i have components negative one -third negative 2 thirds, positive 2 thirds...

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