Homework Help: Directional derivative at a point

1. Oct 30, 2015

catch22

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
part a) finding partial derivatives:

and plugging in (2,0,1) into each, I get the gradient which is <0,-2,0>

to find the directional derivative, it is the dot product of the gradient and unit vector of (3,1,1):

part b) isn't the direction of the greatest increase just the gradient, which I found in part A?

and the greatest increase at (2,0,1) is the magnitude of the gradient there, which is

2. Oct 30, 2015

andrewkirk

Yes.
A pedant might insist that a direction should be given as a normalized vector. But we are not pedants.

3. Oct 30, 2015

Ray Vickson

Some of your partial derivatives are incorrect (although not by very much).
One of your partial derivatives is wrong.

4. Oct 30, 2015

geoffrey159

Also, you don't say anything about the continuity of the first partial derivatives in a neighborhood of $(2,0,1)$. It is under that condition that the formula $(D_{\vec v}f)(2,0,1) = (\nabla f. \vec v) (2,0,1)$ is valid.

5. Oct 30, 2015

catch22

wow, I can't believe I didn't spot that +x at the end.

it should be (-yz e^-xyz) + 1

6. Oct 30, 2015

catch22

the correct answers after fixing the partial derivative with respect to x

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7. Oct 30, 2015

catch22

do you mean like saying the domain is from negative infinity to positive infinity?

8. Oct 30, 2015

geoffrey159

Not at all and I wonder why you ask that ?

9. Oct 30, 2015

Ray Vickson

Well, it IS possible to have a well-defined directional derivatives in some directions $\vec{d}$ at a point $\vec{r_0}= (x_0,y_0,z_0)$, but to not have well-defined, finite partial derivatives at $\vec{r_0}$. However, the present example is not so pathological.