Why is the normal vector to the surface $f$ given by $\text{grad}(f)$?

Question

According to my lecture notes, the normal vector to a surface $f$ is given by $\text{grad}(f) = \underline{\nabla} f$. However, surely the normal vector to the surface $f$ should be perpendicular to the vector $\underline{\nabla}f$?

Answer 1

If the surface, $M$, say, is given by $f^{-1}(c)$ for a regular value $c$ of $f$ then $f$ is constant along $M$. So for any curve $\gamma(t)$ in $M$, $f\circ \gamma= c$ and consequently $$\frac{d}{dt}f\circ \gamma = 0 =\langle \gamma^\prime, (\nabla f)\circ\gamma\rangle$$ Since any tangent vector to $M$ arises as a tangent to a curve in $M$, $\nabla f$ is perpendicular to $M$.