Multivariable calculus

Partial Derivatives of x^3 - 3xy^2 (Monkey Saddle)

Q: Why does the second derivative test fail at the origin?

At (0,0) every second partial is zero, so the discriminant f_{xx}f_{yy} - f_{xy}^2 is 0 and the test is inconclusive. Restricting to the line y = 0 gives x^3, which changes sign through the origin, so the point is neither a maximum nor a minimum.

Q: Where does f_x vanish?

Setting 3x^2 - 3y^2 = 0 gives y = \pm x, the two diagonal lines. Along those lines the surface is momentarily flat in the x direction, though it is still changing in y unless you are at the origin.

For f(x, y) = x^3 - 3xy^2 the partial with respect to x is 3x^2 - 3y^2, and the partial with respect to y is -6xy. This is the monkey saddle. Its second partials are 6x and -6x, which sum to zero, so the function satisfies Laplace's equation and is harmonic everywhere.

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

\frac{\partial f}{\partial x} = 3x^2 - 3y^2

\frac{\partial f}{\partial y} = -6xy

Both terms contain x, only one contains y

For $\frac{\partial f}{\partial x}$ , both terms are in play. The first gives $3x^2$ ; in the second, $-3y^2$ is the constant multiplier on $x$ , so it contributes $-3y^2$ .

\frac{\partial f}{\partial x} = 3x^2 - 3y^2

For $\frac{\partial f}{\partial y}$ , the term $x^3$ has no $y$ in it and contributes nothing. Only $-3xy^2$ survives, with $-3x$ as the constant multiplier.

\frac{\partial f}{\partial y} = -3x \cdot 2y = -6xy

At $(2,1)$ the partials are $12 - 3 = 9$ and $-6(2)(1) = -12$ , so the surface climbs in $x$ and drops in $y$ .

The mistake: giving x^3 a y derivative

Writing $\frac{\partial f}{\partial y} = 3x^2 - 6xy$ is the standard slip. It differentiates $x^3$ as though $x$ were the active variable, in the middle of a $y$ derivative.

Reset before each partial. Say out loud which letter is moving, then scan the terms once and cross out anything without that letter.

For $\frac{\partial f}{\partial y}$ : cross out $x^3$ entirely, since $\frac{\partial}{\partial y}\left(x^3\right) = 0$ .
What remains is $-3xy^2$ , and only the $y^2$ is differentiated.
The coefficient $-3$ and the factor $x$ both ride along unchanged.

A second trap is the exponent on $y$ : $\frac{\partial}{\partial y}\left(y^2\right) = 2y$ , so the coefficient becomes $-6$ , not $-3$ .

Why this surface is called the monkey saddle

Setting both partials to zero gives $-6xy = 0$ and $3x^2 = 3y^2$ , whose only solution is the origin. There the surface has three directions going down and three going up, room for two legs and a tail, which is where the name comes from.

The second partials are worth computing here.

f_{xx} = 6x, \qquad f_{yy} = -6x, \qquad f_{xx} + f_{yy} = 0

A function whose second partials cancel like this satisfies Laplace's equation and is called harmonic. This one is the real part of the complex cube $(x + iy)^3$ , which is why the cancellation is exact at every point rather than a coincidence at one.

Frequently asked questions

Why does the second derivative test fail at the origin?

At $(0,0)$ every second partial is zero, so the discriminant $f_{xx}f_{yy} - f_{xy}^2$ is $0$ and the test is inconclusive. Restricting to the line $y = 0$ gives $x^3$ , which changes sign through the origin, so the point is neither a maximum nor a minimum.

Where does f_x vanish?

Setting $3x^2 - 3y^2 = 0$ gives $y = \pm x$ , the two diagonal lines. Along those lines the surface is momentarily flat in the $x$ direction, though it is still changing in $y$ unless you are at the origin.