# A continuum of critical points in multi-variable calculus

In multivariable calculus, we learn to find local extrema by identifying the critical points, and deciding (using the second derivative test, or otherwise) the type of the point - a local max, a local min, or a saddle point.

What if we get a continuum of critical points ? say the gradient of $f(x,y)$ vanishes along a curve in the $xy$ plane. Is this situation possible ? what can we say about classifying these points into min/max/saddle ?

• If you want the gradient to vanish along $g(x,y)=c$, take $f(x,y) = (g(x,y)-c)^2$ – ronno Jun 3 '13 at 13:00

The second-derivative test will fail at all such points, but you can take a generalized cylinder and get a curve of any of these: Consider $f(x,y)=y^2$, $y^3$, or $-y^2$.
• The second derivative test certainly fails in the examples you provide, since the function is $x$-independent, and the determinant of the Hessian is zero. Do you claim that the fact that the first derivative vanishes on a plane curve implies that the second derivative test fails at each of the curve's points ? – Teddy Jun 3 '13 at 17:51
• Yup. The fact that $\partial f/\partial x = \partial f/\partial y = 0$ along a curve means that if $\vec v$ is the tangent vector to this curve, then $\vec w^\top H\vec v = 0$ for every $\vec w$, where $H$ is the Hessian at any point of the curve. – Ted Shifrin Jun 3 '13 at 18:14