In what instances is the **magnitude** of $∇f$ at a point equal to $0$?

Question

Is it just when it's at a critical point?

I am trying to determine graphically when the directional derivative at a point is $>0$, $<0$ or $=0$.

For a particular question, I need to justify that it is $>0$ (not $=0$). Therefore, using $D_f(a,b)=\lVert∇f(a,b)\rVert\cos\theta$, I know that $\cos\theta>0$ but then also need to justify that $\lVert∇f(a,b)\rVert\neq0$.

Answer 1

Given $f=(x, y)$, let $|∇f| = 0$.

By definition of gradient, we have

$$|<f_{x}, f_{y}>| = 0.$$

Hence,

$$\sqrt{f_{x}^2 + f_{y}^2} = 0$$

$$\Leftrightarrow f_{x}^2 + f_{y}^2 = 0$$

$$\Leftrightarrow f_{x}^2 = -f_{y}^2$$

Note that a number squared is always positive. Hence, the only means by which the above equation is true is when $f_{x} = f_{y} = 0.$ We can also show the converse.

Thus, given $f=(x, y)$, $|∇f| = 0$ if and only if $\frac{\partial f}{\partial x} = \frac{\partial f}{\partial y} = 0$

Answer 2

By definition of the norm to be positive definite we have $$ \forall x \in \mathbb{R}^n: \Vert x \Vert = 0 \Leftrightarrow x = 0 ~~.$$

Hence for any $(a,b) \in \mathbb{R}^2$ we have $$\Vert \nabla f(a,b) \Vert = 0 \Leftrightarrow \nabla f(a,b) = 0 \Leftrightarrow (\partial_1 f)(a) = (\partial_2 f)(b) = 0$$

(Depending on your definition of a critical point) this does indeed happen exactly when $(a,b)$ is a critical point.