Can anyone show me how to adjust my work below so that it is a correct answer? This is question number 14.6.28 in the 7th edition of Stewart Calculus.
Find the directions in which the directional derivative of $f(x,y)=ye^{-xy}$ at the point $(0,2)$ has the value 1.
My work is:
$\nabla f(x,y) = <-y^2 e^{-xy}, e^{-xy}(1-xy)>$
$\nabla f(0,2) = <-4,1>$
$|\nabla f(0,2)| = \sqrt{17}$
$D_v f(x,y)=|\nabla f|\cos{\theta}=\sqrt{17}\cos{\theta}$
$\sqrt{17} \cos{\theta}=1$ when $\cos{\theta}={{\sqrt{17}}/17}$
$\theta =\arccos{{\sqrt{17}}/17}\approx +1.326$ and $-1.326$
EDIT:
I tried the following, based on suggestions below:
$\vec{u}=<\cos{\theta},\sin{\theta}>$
$D_u f(x,y)=\nabla f(x,y)\cdot \vec{u}$
$D_u f(0,2)=\nabla f(0,2)\cdot \vec{u}=-4\cos{\theta}+\sin{\theta}=1$
I then plugged this into a spreadsheet and found that $-4\cos{\theta}+\sin{\theta}=1$ when $\theta = \pi/2 , 5\pi/2 , 9\pi/2 , ...$ and when $\theta = 4\pi/3 , 10\pi/3 , 16\pi/3 , ...$
Can anyone check the correctness of this approach?
Also, I found this result experimentally. If it is correct, I would rather be able to find it using calculus.