Find a unit vector u in which the directional derivative of $f(x,y)=\ln(1-x^2-y^2)$ at ($\frac{1}{2} ,\frac{1}{2}$) is zero
$f_x=\frac{-2x}{1-x^2-y^2}$ and $f_y=\frac{-2y}{1-x^2-y^2}$
Giving us the gradient at $\frac{1}{2} ,\frac{1}{2}$ to be $-2,-2$
How do I now find the unit vector?
Best Answer
The directional derivative $\nabla_vf$ of a function $f$ in direction of a vector $v = (v_x, v_y)^\top$ (in $\mathbb{R}^2$) can be obtained by
$$\nabla_vf = \langle \operatorname{grad} f , v\rangle$$
So you have to find a vector $v$ such that
$$\langle \operatorname{grad} f , v\rangle = 0$$
Thus, in your case:
$$-2v_x -2v_y = 0$$
This is satisfied by all vectors of the form $v = (t,-t)^\top$ with $t \in \mathbb{R}$. So a corresponding unit vector is e.g. $u = \left({1 \over\sqrt{2}},-{1 \over \sqrt{2}}\right)^\top$, because $\Vert u\Vert = 1$.