Singularity of a surface

Question

$\DeclareMathOperator{\Grad}{grad}$I know a surface is singular at a point when gradient vanishes at that point. If $f(x,y) = x^2 + y^2$ then $\Grad f = (2x, 2y)$ and $\Grad f(0,0) = (0,0)$. Does this mean $f(x,y) = x^2 + y^2$ is singular at $(0,0)$? I don't see a cusp at this point and surface $f(x,y)$ seems to me having a tangent at $(0,0)$ which is the $x$-$y$ plane. Should I consider $\Grad f$ as normal to the tangent of the level curve at $(0,0)$ and as level curve $x^2 + y^2 = 0$ is indeed not a curve but only a point then the normal is not defined at this point? Thank you in advance

Answer 1

I know a surface is singular at a point when gradient vanishes at that point.

$\newcommand{\Reals}{\mathbf{R}}\newcommand{\grad}{\nabla}$To clarify: If $F:U \to \Reals$ is a real-valued function on some non-empty open subset $U$ of $\Reals^{3}$, and if $\grad F(p) \neq 0$ at some point $p$ of $U$, the implicit function theorem guarantees that the level set of $F$ through $p$ is a regular surface in some neighborhood of $p$.

There are at least two snags with your proposed example:

1. Your surface is expressed as a graph, not as a level surface. If you express the defining equation $z = f(x, y)$ in "level set form" $$ 0 = F(x, y, z) := z - f(x, y), $$ you find that $\grad F = (-f_{x}, -f_{y}, 1) \neq (0, 0, 0)$.

2. The converse of the implicit function theorem is not true. The level surface $$ 0 = F(x, y, z) = z^{3} $$ is the image of a regular surface, but $\grad F \equiv 0$ everywhere on the level surface.