# Directional Derivative

• Sep 14th 2011, 06:38 AM
dbakeg00
Directional Derivative
Ok, so I kept getting an Latex Error:Unknown Error when I tried to enter my stuff in LaTex. Then I scanned my handwritten work and tried to attach it...it just timed out on me everytime. Sorry this is so primitive, but I just typed everything out. I'm looking for someone to check my work and help me with the last part of #20. Here are the two problems:

19. Suppose that z=e^{xy+x-y}. How fast is z changing when we move away from the origin toward (2,1)?

20. In problem 19 in what direction should we move away from the origin for z to change most rapidly? What is the maximum rate of change? In what directions is the derivative zero at the origin?

Here is my work:

19. z=e^{xy+x-y}
P=(0,0)
v=<2,1>
u={1/sqrt(5)}<2,1>
DZ/DX=(y+1)*e^{xy+x-y}
DZ/DY=(x-1)e^{xy+x-y}
DZ/DX(0,0)=1
DZ/DY(0,0)=-1
Directional Derivative of z at (0,0)=<1,-1> . (1/sqrt{5})<2,1>=(1/sqrt{5})

20. The gradient of z(0,0)=<1,-1> is the direction we should move away from (0,0) for z to change most rapidly.

The maximum rate is sqrt{2}

-I'm stuck on the last part of question 20: "In what directions is the derivative zero at the origin?" Can someone help me with this part? Thanks.
• Sep 14th 2011, 09:36 AM
dbakeg00
Re: Directional Derivative
I finally clued into the fact that Latex is only working with [tex] tags, so I retyped everything to make it easier.

19. $\displaystyle z=e^{xy+x-y}$

P: (0,0)
$\displaystyle \vec{v}=<2,1>$

$\displaystyle \vec{u}=\frac{1}{\sqrt{5}}<2,1>$

$\displaystyle \frac{\delta z}{\delta x}=(y+1)*e^{xy+x-y}=1$

$\displaystyle \frac{\delta z}{\delta y}=(x-1)*e^{xy+x-y}=-1$

$\displaystyle D_{\vec{u}}f(0,0)=<1,-1> \cdot\ \frac{1}{\sqrt{5}}<2,1>$

$\displaystyle D_{\vec{u}}f(0,0)=\frac{1}{\sqrt{5}}$

20.

<1,-1> is the direction we should move away from the origin for z to change most rapidly.

The max rate of change is $\displaystyle \sqrt{2}$

I still have yet to figure out how to solve the last question in problem #20.

Any help with where I'm stumped or in just reviewing the work I have completed so far would be appreciated. Thanks.
• Sep 14th 2011, 12:22 PM
HallsofIvy
Re: Directional Derivative
The "directional derivative" in the direction of unit vector, v, is the dot product of grad f and v. In particular, the derivative will be 0 is that dot product is 0 which means v must be perpendicular to grad f. What vectors are perpendicular to <1, -1>?
• Sep 15th 2011, 03:55 AM
dbakeg00
Re: Directional Derivative
<-1,-1> & <1,1>