# Directional Derivatives

• Aug 3rd 2009, 01:10 PM
Directional Derivatives
I need to find the directions in which the directional derivative $\displaystyle D_u f(x,y)$ of $\displaystyle f(x,y)=ye^{-xy}$ at the point $\displaystyle (0,2)$ has the value 1.

I use the dot product formula for directional derivatives where $\displaystyle u$ is a unit vector; I set equal to one:

$\displaystyle D_u f(x,y)=u\cdot\nabla f(x,y)=1$

The components of the gradient vector:

$\displaystyle \frac{\partial f}{\partial x}i=-y^2e^{-xy}i$

$\displaystyle \frac{\partial f}{\partial y}j=-xye^{-xy}j$

I plug in the coordinates:

$\displaystyle \nabla f(0,2)=<-4,0>$

$\displaystyle u\cdot<-4,0>$=1 , which is $\displaystyle u\cdot(-4i)=1$ This tells me that the $\displaystyle a4=1$ where $\displaystyle a$ is the component of the unit vector $\displaystyle u$ in $\displaystyle i$ direction. If I write the unit vector as $\displaystyle u=ai + bj$ I have $\displaystyle \mid u \mid= \sqrt{a^2+b^2}=1$.

Now I have the elementary system of equations:

$\displaystyle a4=1$

$\displaystyle \sqrt{a^2+b^2}=1$

I not sure if the solutions for a,b the correct components of the unit vector givng the direction required. It makes sense to me, but I don't have the answer in my book, so I can't check this.
• Aug 3rd 2009, 01:15 PM
Calculus26
Du can also be written |f||u|cos(t) = 1

so 4cos(t) =1

t = arccos(1/4)

u = cos(t) i +sin(t) j