
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.

Du can also be written fucos(t) = 1
so 4cos(t) =1
t = arccos(1/4)
u = cos(t) i +sin(t) j