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.

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>?

Re: Directional Derivative