Orthogonality of $\nabla f(x_0)$ and $\gamma'(0)$ where $\gamma(0)=x_0$

Question

let

$$f:\mathbb R^n\rightarrow \mathbb R$$ be differentiable, $c\in \mathbb R,\epsilon >0,x_0\in \mathbb R^n$ and $\gamma:(-\epsilon,\epsilon)\rightarrow \mathbb R^n$ a continuous differentiable curve with $\text{Image}(\gamma)\subseteq f^{-1}(\{c\})$ and $\gamma(0)=x_0$.

I want to show that $\nabla f(x_0)$ and $\gamma'(0)$ are orthogonal to eachother.

I dont know if its needed in order to solve this problem...but I am not even understanding the meaning of this problem..

Answer 1

You have, using the chain rule, $$ f(\gamma(t))=f(x)\Rightarrow 0=\frac{d}{dt}f(\gamma(t))=\nabla f(\gamma(t))\cdot \gamma(t)'. $$ (The derivative is vanishing because $f(\gamma(t))$ actually does not depend on $t$). In particular, at $t=0$, $$ \nabla f(x_0)\cdot \gamma(0)'=0. $$