Directional derivative at a point

[catch22]

Homework Statement

Homework Equations

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

 

[andrewkirk]

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

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

 

[Ray Vickson]

Homework Statement

View attachment 91055

Homework Equations

The Attempt at a Solution

part a) finding partial derivatives:
View attachment 91058

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):

View attachment 91056

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

View attachment 91057

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

View attachment 91059

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

Homework Statement

View attachment 91055

Homework Equations

The Attempt at a Solution

part a) finding partial derivatives:
View attachment 91058

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):

View attachment 91056

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

View attachment 91057

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

View attachment 91059

One of your partial derivatives is wrong.

 

[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.

 

[catch22]

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

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

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

 

[catch22]

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

 

[catch22]

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.

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

 

[geoffrey159]

Not at all and I wonder why you ask that ?

 

[Ray Vickson]

Not at all and I wonder why you ask that ?

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.