1. ## Maximum Rate of Change of f(x,y,z). Gradient problem.

I'm hoping someone could check my work. My answers are not the same as the text gives.

The question is, "Find the maximum rate of change of f at the given point and the direction in which it occurs."

So I think the problem boils down to finding $\|\vec\nabla f\|$

Given $f(x,y,z)=\sqrt{x^2+y^2+z^2}$ and $P(3,6,-2)$

$\vec\nabla f = $

Differentiating $f(x,y,z)$ with respect to x and applying the chain rule;

$f_x=\frac{1}{2}(f(x,y,z))^{\frac{-1}{2}}*2x=(f(x,y,z))^{\frac{-1}{2}}*x$

Replacing x with y yields $f_y$. Replacing x with z yields $f_z$.

Since f(x,y,z) evaluated at P(3,6,-2) is 7 and I need $\|\vec\nabla f\|$ evaluated at P I get the following expression;

$\vec\nabla f=<(7)^{\frac{-1}{2}}*x,(7)^{\frac{-1}{2}}*y,(7)^{\frac{-1}{2}}*z>$

$\vec\nabla f$ evaluated at P is;
$\vec\nabla f=<(7)^{\frac{-1}{2}}*3,(7)^{\frac{-1}{2}}*6,(7)^{\frac{-1}{2}}*-2>=\frac{1}{\sqrt7}<3,6,-2>$

The text says the answer for this part of the question is $<3,6,-2>$ is my answer of $\frac{1}{\sqrt7}<3,6,-2>$wrong?

Also
$\|\vec\nabla f\|=\sqrt{ \frac{9}{7} + \frac{36}{7}+\frac{4}{7}}=\sqrt{\frac{49}{7}}$

the text says the solution for this part of the problem is 1 so i'm pretty sure something is wrong.

2. ## Re: Maximum Rate of Change of f(x,y,z). Gradient problem.

Hi bkbowser!

Originally Posted by bkbowser
I'm hoping someone could check my work. My answers are not the same as the text gives.

The question is, "Find the maximum rate of change of f at the given point and the direction in which it occurs."

So I think the problem boils down to finding $\|\vec\nabla f\|$

Given $f(x,y,z)=\sqrt{x^2+y^2+z^2}$ and $P(3,6,-2)$

$\vec\nabla f = $

Differentiating $f(x,y,z)$ with respect to x and applying the chain rule;

$f_x=\frac{1}{2}(f(x,y,z))^{\frac{-1}{2}}*2x=(f(x,y,z))^{\frac{-1}{2}}*x$
I think you've been doing a bit much at the same time here.

\begin{aligned}f_x &= \frac {\partial}{\partial x} \sqrt{x^2+y^2+z^2} \\ &= \frac 1 {2 \sqrt{x^2+y^2+z^2}} \cdot 2x \\ &= \frac x {\sqrt{x^2+y^2+z^2}} \\ &= f(x,y,z)^{-1} \cdot x \end{aligned}

Replacing x with y yields $f_y$. Replacing x with z yields $f_z$.

Since f(x,y,z) evaluated at P(3,6,-2) is 7 and I need $\|\vec\nabla f\|$ evaluated at P I get the following expression;

$\vec\nabla f=<(7)^{\frac{-1}{2}}*x,(7)^{\frac{-1}{2}}*y,(7)^{\frac{-1}{2}}*z>$

$\vec\nabla f$ evaluated at P is;
$\vec\nabla f=<(7)^{\frac{-1}{2}}*3,(7)^{\frac{-1}{2}}*6,(7)^{\frac{-1}{2}}*-2>=\frac{1}{\sqrt7}<3,6,-2>$

The text says the answer for this part of the question is $<3,6,-2>$ is my answer of $\frac{1}{\sqrt7}<3,6,-2>$wrong?

Also
$\|\vec\nabla f\|=\sqrt{ \frac{9}{7} + \frac{36}{7}+\frac{4}{7}}=\sqrt{\frac{49}{7}}$

the text says the solution for this part of the problem is 1 so i'm pretty sure something is wrong.

3. ## Re: Maximum Rate of Change of f(x,y,z). Gradient problem.

OOOOOOOOOOOOOOOOOOOOOOOOOhhhhhhhhhhhhhhhhhhhhhhhhh hhhhhh............................ OK Thanks!

4. ## Re: Maximum Rate of Change of f(x,y,z). Gradient problem.

Can I assume my answer $\frac{1}{7}<3,6,-2>$ instead of $<3,6,-2>$ is correct?

5. ## Re: Maximum Rate of Change of f(x,y,z). Gradient problem.

Originally Posted by bkbowser
Can I assume my answer $\frac{1}{7}<3,6,-2>$ instead of $<3,6,-2>$ is correct?
They are both correct.
The problem asks for the direction - both vectors point in the same direction.
Separately it asks for the maximum rate of change.

6. ## Re: Maximum Rate of Change of f(x,y,z). Gradient problem.

OK. I had kind of just assumed they'd just want the direction. But I was a bit put off by the lack of the scalar. I guess they just want to save printing costs...

Thanks!