Directional Derivative and Gradient Help

**Alterah** · October 10th 2009, 01:35 PM

I am having some difficulty with the following problems:
1. For the function
$f(x,y)=\left\{\begin{array}{cc}\frac{x|y|}{\sqrt(x ^2+y^2)},&\mbox{ if }<br /> (x,y) != (0,0)\\0, & \mbox{ if } (x,y) = (0,0)\end{array}\right.$

use definition $D_vf(a) = \lim_{h \to 0}\frac{f(a + hv) - f(a)}{h}$ to determine for which unit vectors v = vi + wj the direction derivative $D_vf(0,0)$ exists.

I am able to get the function down to the following:
$\lim_{h \to 0}(\frac{hv|hw|}{\sqrt(h^2v^2+h^2w^2)} - 0)/h$

Simplifying I get:
$\lim_{h \to 0}(\frac{h^2v|w|}{\sqrt(v^2+w^2)}$
This is where I get lost. The answer is supposed to be v|w|. The limit I have goes to 0.
------------------------
Problem 2
The depth of a lake is given by $f(x,y) = 400 - 3x^2y^2$ find the direction a swimmer should swim so that there is no change in the depth. The swimmer begins at (1,-2).

This is part 2 of the problem. Part one has me find the direction the depth increases most rapidly and I got $\frac{1}{\sqrt5}(-2i + j)$

To be honest I am not quite sure where to begin. Would I set the gradient equal to zero at that point and go from there?

Thanks for any and all help.

**Danny** · October 11th 2009, 06:52 AM

Originally Posted by Alterah

I am having some difficulty with the following problems:
1. For the function
$f(x,y)=\left\{\begin{array}{cc}\frac{x|y|}{\sqrt(x ^2+y^2)},&\mbox{ if }<br /> (x,y) != (0,0)\\0, & \mbox{ if } (x,y) = (0,0)\end{array}\right.$

use definition $D_vf(a) = \lim_{h \to 0}\frac{f(a + hv) - f(a)}{h}$ to determine for which unit vectors v = vi + wj the direction derivative [tex] D_vf(0,0) exists.

I am able to get the function down to the following:
$\lim_{h \to 0}(\frac{hv|hw|}{\sqrt(h^2v^2+h^2w^2)} - 0)/h$

Simplifying I get:
$\lim_{h \to 0}\frac{h^2v|w|}{\sqrt(v^2+w^2)}$
This is where I get lost. The answer is supposed to be v|w|. The limit I have goes to 0.
------------------------
Problem 2
The depth of a lake is given by $f(x,y) = 400 - 3x^2y^2$ find the direction a swimmer should swim so that there is no change in the depth. The swimmer begins at (1,-2).

This is part 2 of the problem. Part one has me find the direction the depth increases most rapidly and I got $\frac{1}{\sqrt5}(-2i + j)$

To be honest I am not quite sure where to begin. Would I set the gradient equal to zero at that point and go from there?

Thanks for any and all help.

In the first part, you dropped an $h$ so you should have

$\lim_{h \to 0}\frac{h^2v|w|}{h^2\sqrt{v^2+w^2}} = v|w|$ since $u^2+v^2 = 1$ (it's a unit vector)

Second part. $\nabla f = <-6x y^2, - 6x^2 y>$ which at your point is $\frac{<-2,1>}{\sqrt{5}}$ . If the direction you move in is ${\bf u} = <u,v>$ , then $D_u f = \nabla f \cdot {\bf u} = \frac{<-2,1>}{\sqrt{5}} \cdot <u,v>$ and if you want this to be zero then $-2u + v = 0.$ So pick a $u$ and $v$ that satisfies this then make it a unit vector.

**Alterah** · October 11th 2009, 10:21 AM

For the first problem, in the first problem on this part:

$\lim_{h \to 0}(\frac{hv|hw|}{\sqrt(h^2v^2+h^2w^2)} - 0)/h<br />$

The square root symbol is over the entire line. So, if we factor out $h^2$ then we get $\sqrt(h^2)\sqrt(v^2+w^2)$ in that line. Then shouldn't $\sqrt(h^2) = |h|$ ? We can then move the h in the very bottom up to the top. I suppose I am not seeing where I am losing an h

**Danny** · October 11th 2009, 10:39 AM

Originally Posted by Alterah

For the first problem, in the first problem on this part:

$\lim_{h \to 0}(\frac{hv|hw|}{\sqrt(h^2v^2+h^2w^2)} - 0)/\color{red}{h}<br />$

The square root symbol is over the entire line. So, if we factor out $h^2$ then we get $\sqrt(h^2)\sqrt(v^2+w^2)$ in that line. Then shouldn't $\sqrt(h^2) = |h|$ ? We can then move the h in the very bottom up to the top. I suppose I am not seeing where I am losing an h

See above in red.

**Alterah** · October 11th 2009, 11:03 AM

Originally Posted by Danny

See above in red.

I suppose I am figuring that I didn't lose that h because we can move it to the numerator. Below is all the steps I am going through.

$\lim_{h \to 0}(\frac{hv|hw|}{\sqrt(h^2v^2+h^2w^2)} - 0)/h$

is the same as:
$\lim_{h \to 0}\frac{\frac{hv|hw|}{\sqrt(h^2v^2+h^2w^2)}- 0} {h}$

$\lim_{h \to 0} \frac{\frac{hv|h||w|}{|h|\sqrt{v^2+w^2}}}{h}$

Becomes:

$\lim_{h \to 0} \frac{h^2v|h||w|}{|h|\sqrt{v^2+w^2}}$

$\lim_{h \to 0} \frac{h^2v|w|}{\sqrt{v^2+w^2}}$

But, I am obviously forgetting something or doing something wrong.

**Danny** · October 11th 2009, 11:27 AM

Originally Posted by Alterah

I suppose I am figuring that I didn't lose that h because we can move it to the numerator. Below is all the steps I am going through.

$\lim_{h \to 0}(\frac{hv|hw|}{\sqrt(h^2v^2+h^2w^2)} - 0)/h$

is the same as:
$\lim_{h \to 0}\frac{\frac{hv|hw|}{\sqrt(h^2v^2+h^2w^2)}- 0} {h}$

$\lim_{h \to 0} \frac{\frac{hv|h||w|}{|h|\sqrt{v^2+w^2}}}{h}$

Becomes: It's this step

$\lim_{h \to 0} \frac{h^2v|h||w|}{|h|\sqrt{v^2+w^2}}$

$\lim_{h \to 0} \frac{h^2v|w|}{\sqrt{v^2+w^2}}$

But, I am obviously forgetting something or doing something wrong.

Above in red. Note that $\frac{\frac{1}{a}}{b} = \frac{1}{ab}.$

**Alterah** · October 11th 2009, 11:36 AM

Originally Posted by Danny

Above in red. Note that $\frac{\frac{1}{a}}{b} = \frac{1}{ab}.$

Thanks. Yeah, I am finally seeing where I went wrong.

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