I'm having trouble with a math problem. I'm supposed to find the directional derivative for $f(x,y) = x^2cos(y)$ at $(x,y) = \big(0, \frac{\pi}{2}\big)$ in the direction $w = (1,-1)$ and also find the direction vector for which directional derivative is maximal.
What I've done so far is this:
$f_x(x,y) = 2x\cos{y}$
$f_y(x,y) = -x^2\sin{y}$
$|w| = \sqrt{1^2+(-1)^2}=\sqrt{2}$
$u=\frac{w}{|w|}=(1/\sqrt{2}, -1/\sqrt{2})$
So $D_u\,f(x,y) = \nabla f(x,y) \cdot u$
$\nabla f(x,y) = (2x\cos{y}, -x^2\sin{y})$
$\nabla f(0,\pi/2) = (2(0)\cos{\frac{\pi}{2}}, -(0)^2\sin{\frac{\pi}{2}} = (0, 0)$
$D_u\,f(x,y) = (0,0) \cdot (1/\sqrt{2}, -1/\sqrt{2}) = 0$
So the directional derivative is 0, meaning the function does not increase or decrease. Am I doing this correctly? How am I supposed to do the second part if the first part is 0? Thanks!
Best Answer
Yes, your answer is right.
For the second part, you can just say that the direction vector for which directional derivative is maximal does not exists.