Directional Derivatives and max rate of change

[Blkmage]

Homework Statement

See attachment.

2. Homework Equations /solution attempt

Part (a)
Well, the gradient evaluated at (1,2-1) will give the rate of change. If we want the maximum rate of change then we need the directional direction such that the unit vector $$\mathbf{u}$$ is in the same direction as $$\nabla f(1,2,-1)$$. So the unit vector would just be $$\frac{\nabla f(1,2,-1)}{\text{norm}(\nabla f(1,2,-1))}$$? And the max rate of change is just the direction derivative at this point and direction?

Anyways, part (b) is what I'm having trouble with:

Since we want it to be in the direction of the vector (1,-1,-1), does that mean that we want $$\nabla g(x_{Q},y_{Q},z_{Q}) = (1,-1,-1)$$? Since (1,-1,-1) isn't a unit vector, I didn't divide by the norm of the gradient...

 

[lanedance]

the gradient gives the direction of maximum rate of change, its magnitude is the rate of change moving along that direction

the magnitude of the gradient is equivalent to a dot product with a unit vector in the same direction, ie the value of the directional derivative in that direction

 

[Blkmage]

So for part (b), we have

$$\text{norm}(\nabla g(x_{Q},y_{Q},z_{Q})) = \nabla g(x_{Q},y_{Q},z_{Q}) \cdot \frac{1}{\sqrt{3}}\left(1,-1,-1\right)$$

This seems so tedious to solve though. It's on a practice exam my teacher gave us and his tests are very reasonable given the time limits, so I'm guessing there is something I'm missing?

edit: not to mention that would give me a single equation of 3 variables, which I wouldn't be able to solve

 

[lanedance]

$$g(x,y,z) = x^2yz$$
$$\nabla g(x,y,z) = (2xyz, x^2z, x^2y)$$

and you want to know when is in the direction (1,-1,-1), first clearly this will not be possible if any of the varibales are 0 as it will lead tio .

substituting in gives
2xyz = -x^2z _______(1)
2xyz = -x^2y _______(2)
x^2z = x^2y _______(3)

so you know y=z from (3), as x dne 0, which should simplify things