If we have an implicit surface $g(x, y, z)=…$, the gradient is a normal vector. But why? Do I have to start with tangent vectors and then show the gradient is perpendicular? Not sure how to approach this.
Differential Geometry – Why the Gradient of Implicit Surface is the Normal Vector
differential-geometry
Best Answer
Fix a point $p$ on the surface and take a curve $\gamma=\gamma(t)$ lying on the surface and passing through $p$, say $\gamma(0)=p$. That implies $$ g(\gamma(t))=0 $$ for all $t$. Differentiate this relation termwise.
This argument shows that $\nabla g(p)$ is orthogonal to the velocity vector $\dot{\gamma}(0)$ for all curves on the surface that pass through $p$. By definition, the plane tangent to the surface at $p$ is the one that is composed of those vectors. Therefore, $\nabla g(p)$ is orthogonal to the tangent plane.