2
$\begingroup$

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 ?

$\endgroup$
  • 1
    $\begingroup$ If you want the gradient to vanish along $g(x,y)=c$, take $f(x,y) = (g(x,y)-c)^2$ $\endgroup$ – ronno Jun 3 '13 at 13:00
1
$\begingroup$

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 $-y^2$.

$\endgroup$
  • $\begingroup$ 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 ? $\endgroup$ – Teddy Jun 3 '13 at 17:51
  • $\begingroup$ 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. $\endgroup$ – Ted Shifrin Jun 3 '13 at 18:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.