Orientability of the level set surface

Question

Let $S$ be a regular surface in $R^{3}$ given as the set of solutions to the equation $F(x,y,z)=a$, where $F:U\in R^{3}\rightarrow R$ is differentiable and $a$ is a regular value of $F$. Prove $S$ is orientable.

A necessary and sufficient condition for orientability is that if we can find a non-vanishing continuous function from $S$ to normal vectors at $S$, our surface is orientable. My thought is to construct the normal vector by finding the gradient of $F$. But I don't know how to prove that the gradient is indeed orthogonal with the level set and that this gradient is non-vanishing on the level set.

Answer 1

The gradient is indeed the solution. Take a curve $\gamma$ that lies in $S$. Then $F\circ \gamma$ is $a$ everywhere.

So its derivative is zero. But its derivative is just $$ \nabla F(\gamma(s)) \cdot \gamma'(s)\,, $$ so the gradient is orthogonal to $\gamma'(s)$. All vectors tangent to $S$ arise in this way so you are done.

(Regular value means that the gradient is non-zero so that bit follows by definition.)