finding the no change point with Directional Derivatives

Question

Let $f(x,y) = x^2-xy+y^2-y$ Find the directions $u$ for which

$D_uf(1,-1)=0$

I don't really have a method of solving this. I tried cross product, but that was completely off. Eventually after some plug and play, I came up with $u = 5/3i+5/4j$ which fits the equation of $D_uf(1,-1)=0$, but the answer doesn't match the back of my textbook.

Answer 1

The function is smooth, so $df(x,u) = Df(x)u$. Since $Df(x) = (2x-y, 2y-x-1 ) $, we have $Df((1,-1)^T) = (3,-4)$.

Now compute $u$ such that $(3,-4) u = 0$.

This is equivalent to finding $u_1,u_2$ such that $3 u_1-4 u_2 = 0$, which can be seen to be the subspace { $\lambda(4,3)^T \}_{\lambda \in \mathbb{R}}$.

Answer 2

If $u = (a,b)$ then $D_u f (1, -1) = \frac {\partial f} {\partial x} (1, -1) \cdot a + \frac {\partial f} {\partial y} (1, -1) \cdot b = 3a -5b$. If $D_u f (1, -1) = 0$ then $3a - 5b =0$, i.e. the solutions are of the form $(t, \frac 3 5 t) = t \Bbb i + \frac 3 5 t \Bbb j, \space t \in \Bbb R$ (geometrically this describes a straight line).