Multivariable calculus

Critical Points of x^3 - 3xy^2

The only critical point of f(x,y) = x^3 - 3xy^2 is the origin, where every second partial is zero and so D = 0. The test is inconclusive. In polar form the function is r^3 cos(3t), which is positive in three sectors and negative in three others, so the origin is a monkey saddle.

f(x,y) = x^{3} - 3xy^{2}

$\left(0, 0\right)$ degenerate, the test is inconclusivediscriminant D = 0

Finding the one critical point

\frac{\partial f}{\partial x} = 3x^{2} - 3y^{2}, \qquad \frac{\partial f}{\partial y} = -6xy

Start with the equation that factors. From $-6xy = 0$ you get $x = 0$ or $y = 0$ . If $x = 0$ , the first equation becomes $-3y^{2} = 0$ , so $y = 0$ . If $y = 0$ , the first equation becomes $3x^{2} = 0$ , so $x = 0$ . Either branch lands on the origin, and there are no other critical points.

A discriminant that is zero exactly where it matters

f_{xx} = 6x, \qquad f_{yy} = -6x, \qquad f_{xy} = -6y

D(x,y) = (6x)(-6x) - (-6y)^{2} = -36\left(x^{2} + y^{2}\right)

This discriminant is negative everywhere except at the origin, where it is exactly zero. The one point you actually need it at is the one point where it tells you nothing, so the test is inconclusive and you have to look at $f$ itself.

Switch to polar coordinates with $x = r\cos t$ and $y = r\sin t$ . Using $\cos 3t = 4\cos^{3} t - 3\cos t$ , the function collapses to a single term.

f = r^{3}\left(\cos^{3} t - 3\cos t \sin^{2} t\right) = r^{3}\cos 3t

As $t$ runs once around the origin, $\cos 3t$ changes sign six times. So on any small circle the surface rises in three sectors and falls in three others: a monkey saddle, not a maximum or a minimum.

The mistake students make

Students often expect a degenerate point to be flat or ambiguous, something in between. It is neither. $D = 0$ is a statement about the test, not about the surface, and this surface has very definite behaviour: three ridges and three valleys meeting at the origin.

The other slip is checking only the axes. Along the $y$ axis $f = 0$ identically, so that direction alone suggests nothing is happening. Along the $x$ axis $f = x^{3}$ , which already rules out an extremum. When the test fails, sample several directions, or convert to polar coordinates and read the whole circle at once.

Frequently asked questions

Why is it called a monkey saddle?

An ordinary saddle has two directions going down, one for each leg. This surface has three descending sectors, which leaves room for two legs and a tail, so it is the saddle a monkey could sit in. The three come from the factor $\cos 3t$ .

Could I classify it with third derivatives instead?

In effect, yes. The behaviour near the origin is governed by the first nonzero term of the Taylor expansion, and here that is the cubic $x^{3} - 3xy^{2}$ itself. Since a cubic form takes both signs arbitrarily close to the origin, the point cannot be a maximum or a minimum.