Let $A \subset \mathbb{R}^n$, with $A$ open, and let $\emptyset\ne S \subset A$. Let $f\in C^1 (A, \mathbb{R} )$ be a function. Let $\vec{a}$ be a point in $S$, and suppose that $\vec{a}$ is a local extremum point for $f \mid S$. Prove that the gradient vector $( \nabla f ) ( \vec{a} )$ is orthogonal to every vector $\vec{v}$ which is tangent to $S$ at $\vec{a}$.
Aside (definition of a vector tangent to a set):
Let $\emptyset\ne S\subset\mathbb{R}^n$ and let $\mathbb{a}\in S$. A vector $\vec{v} \in \mathbb{R}^n$ is said to be tangent to $S$ at $\vec{a}$ when there exists a differentiable path $\gamma : I \to \mathbb{R}^n$, where $I \subseteq \mathbb{R}$ is an open interval containing $0$, such that the following conditions are fulfilled:
(i) $\gamma (t) \in S$ for every $t \in I$;
(ii) $\gamma (0) = \vec{a}$;
(iii) $\gamma ' (0) = \vec{v}$.
I thought that I easily proved this result, but now I'm wondering: if $\vec{a}$ is a point of local extremum for $f$ on $S$, does this mean that $\nabla f(\vec{a})=\vec{0}$? Also, is a stationary point the same as a point of local extremum?
Would appreciate your clarifications.
The approach I used is as follows:
Let $u:I\to\mathbb{R}$ be a function, such that $u(t) = f(\gamma(t)), t\in I$. Then $u$ is differentiable, with
$$u'(t)=\langle (\nabla f(\gamma(t)),\gamma'(t) \rangle$$ $$u'(0)=\langle (\nabla f(\gamma(0)),\gamma'(0) \rangle=\langle \nabla f(\vec{a}), \vec{v} \rangle=\langle \vec{0}, \vec{v} \rangle =0$$
But now, I think, it's not so clear.
If $\vec{a}$ is a point of local extremum for $f$ on $S$, this does not mean that $\nabla f(\vec{a})=\vec{0}$. Let $S$ be a singleton, $\{\vec{a}\}$, where $\nabla f(\vec{a})\neq\vec{0}$
A stationary point is not the same as a point of local extremum. Try $f(x_1, ... x_n)=x_1^3$. It has no extrema, but it has stationary points.