Gradient as normal to a surface

Question

Why the normal to a surface is given by gradient? How to see this intuitively?

Answer 1

Typically a surface is given by an equation like $$ g(x,y,z) = 0 $$ A path on the surface given by $g$ will be of the form $\vec{r}(t) = (x(t), y(t), z(t))$ where $$ g(x(t), y(t), z(t)) = 0 $$ Define $$ f(t) = g(x(t), y(t), z(t)) = 0 $$ Then $$ 0 = f'(t) = \frac{\partial g}{\partial x} x'(t) + \frac{\partial g}{\partial y} y'(t) + \frac{\partial g}{\partial z} z'(t) = (\nabla g ) \cdot \vec{v} $$ where $\vec{v}(t) = \vec{r}'(t)$. What this shows is that any curve on the surface defined by $g = 0$ has velocity perpendicular to the gradient of $g$. Being perpendicular to the velocity of any curve on the surface is exactly what we mean when we say that a vector is perpendicular to a surface.