Directional Derivatives

**adkinsjr** · August 3rd 2009, 01:10 PM

I need to find the directions in which the directional derivative $D_u f(x,y)$ of $f(x,y)=ye^{-xy}$ at the point $(0,2)$ has the value 1.

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

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

The components of the gradient vector:

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

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

I plug in the coordinates:

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

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

Now I have the elementary system of equations:

$a4=1$

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

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.

**Calculus26** · August 3rd 2009, 01:15 PM

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

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