Question

If f:R^n-->R is a smooth function, define hypersurface f(x)=c for some constant c, then prove velocity vector is orthogonal to the normal vector i.e <\nabla f(p),v_{\gamma}(p)>=0

Answer 1

We have been provided that $f:\R^n\rightarrow \R$ is $\mathscr{C}^{\infty}$ . let us define the level set by

$$M_f:=\{\vec x\in \R^n:f(\vec x)=c\}$$

Let $\gamma:(a,b)\rightarrow \R^n$ is smooth path such that $t\mapsto\gamma(t)$ , for $t\in (a,b),a,b\in \R$ which passes through $M_f$ . let $\gamma(\theta)=p\in\R^n$ for $\theta\in (a,b)$ and denote $v_{\gamma}(p)=\gamma'(\theta)$ is velocity vector passes through $p$ .

Thus, by chain rule we get,

$$\frac{d}{dt}[f(\gamma(t))]=f'(\gamma(t))v_{\gamma}(p)$$

Since, $\nabla f:U\rightarrow \R^n$ is smooth map from open set $U$ to $\R^n$ such that

$$\nabla f(\vec x)=(D_1f(\vec x),...,D_nf(\vec x))$$

Thus, $\nabla f(\gamma(t))^T=f'(\gamma(t))$ , Hence, we get

$$\frac{d}{dt}[f(\gamma(t))]=f'(\gamma(t))v_{\gamma}(p)=\frac{d}{dt}(c)=0\\ \implies <\nabla f(p),v_{\gamma}(p)>=0$$

Therefore $\nabla f(p)\&v_{\gamma}(p)$ orthogonal to each other.

We are done.