Here is the question:
If g is a function differentiable at $(a, b)$ such that $\nabla g(a, b) = (2, 3)$, then find all the vectors $\vec{u} = (x, y)$ with $x^2 + y^2 = 1 $ such that $D_\vec{u} g(a, b) = \vec{0}.$
Is the following correct?
$D_\vec{u} g(a, b) = \nabla g(a, b)\cdot \vec{u}$
$g_x(a,b) = 2$ and $g_y(a,b) = 3$
$\therefore \vec{0} = \langle2,3\rangle\cdot\langle u_1,u_2\rangle$
$\therefore \vec{0} = 2u_1 + 3u_2$
$\therefore u_2 = \frac{2}{3}u_1$
In $x^2 + y^2 = 1$ :
$u_1^2 + u_2^2 = 1$
$\therefore u_1^2 = \frac 9{13}$
$\therefore u_1 = \pm \sqrt{\frac {9}{13}}$
In $u_2 = \frac{2}{3}u_1$ :
$u_2 = -\frac 23(\pm \sqrt{\frac {9}{13}})$
These are all my own calculations. Is this correct and final?