# Gradient as normal to a surface [duplicate]

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

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.