Multivariable calculus

Gradient of x^3 - 3xy^2: Monkey Saddle Worked Out

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

At (0,0) every second partial is zero, so D = f_{xx}f_{yy} - f_{xy}^2 = 0 and the test is inconclusive. You classify the point directly instead: f(x,0) = x^3 changes sign through the origin, so it is neither a maximum nor a minimum.

Q: Where does the gradient of x^3 - 3xy^2 point purely along the y-axis?

You need 3x^2 - 3y^2 = 0 with -6xy \ne 0, so y = \pm x with x \ne 0. Along the diagonals the gradient is \langle 0, -6xy\rangle, which is vertical.

For f(x, y) = x^3 - 3xy^2 the gradient is (3x^2 - 3y^2, -6xy). At the point (2, 1) it equals (9, -12), a vector of length 15. The gradient is zero only at the origin, where the surface forms a monkey saddle with three rising and three falling directions.

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

\nabla f(x,y) = \left\langle 3x^2 - 3y^2,\ -6xy \right\rangle

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

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

Take the mixed term apart one variable at a time

The term $-3xy^2$ contains both variables, so it contributes to both partials, just in different ways. Holding $y$ fixed makes $-3y^2$ a coefficient in front of $x$ .

\frac{\partial f}{\partial x} = \frac{\partial}{\partial x}\left(x^3\right) - 3y^2 \cdot \frac{\partial}{\partial x}\left(x\right) = 3x^2 - 3y^2

Now hold $x$ fixed. The $x^3$ term is a constant and disappears, while $-3x$ becomes the coefficient in front of $y^2$ .

\frac{\partial f}{\partial y} = 0 - 3x \cdot \frac{\partial}{\partial y}\left(y^2\right) = -6xy

\nabla f(x,y) = \left\langle 3x^2 - 3y^2,\ -6xy \right\rangle

At $(2,1)$ : $3(4) - 3(1) = 9$ and $-6(2)(1) = -12$ , so $\nabla f(2,1) = \langle 9, -12\rangle$ with $\left|\nabla f\right| = \sqrt{81 + 144} = 15$ .

The mistake: letting the mixed term vanish from one partial

The most common wrong answer is $\nabla f = \langle 3x^2, -6xy\rangle$ , dropping $-3y^2$ from the $x$ -component. The reasoning behind the slip is that $y^2$ is a constant when $y$ is frozen, so surely it differentiates to zero.

It does not, because $y^2$ is not standing alone. It is multiplied by $x$ , and $x$ is the live variable. A constant times $x$ differentiates to that constant, not to zero. Compare $\frac{d}{dx}(7) = 0$ with $\frac{d}{dx}(7x) = 7$ .

A term dies under $\frac{\partial}{\partial x}$ only if it contains no $x$ at all.
$-3xy^2$ contains $x$ , so it survives in $\frac{\partial f}{\partial x}$ as $-3y^2$ .
Sanity check at $(0,1)$ : $f(x,1) = x^3 - 3x$ has derivative $3x^2 - 3$ , which is $-3$ at $x = 0$ . The formula gives $3(0) - 3(1) = -3$ , so it agrees.

One critical point, and a surface that goes up three ways

Set both components to zero. From $-6xy = 0$ you get $x = 0$ or $y = 0$ . If $x = 0$ , the first equation gives $-3y^2 = 0$ , so $y = 0$ . If $y = 0$ , it gives $3x^2 = 0$ , so $x = 0$ . The origin is the only critical point.

In polar coordinates $f = r^3\cos 3\theta$ , which oscillates three times as you circle the origin. That is the monkey saddle: three directions where the surface rises and three where it falls, so the animal gets grooves for two legs and a tail.

This function is also harmonic, meaning its two pure second partials cancel.

\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = 6x + (-6x) = 0

Frequently asked questions

Why does the second derivative test fail at the origin here?

At $(0,0)$ every second partial is zero, so $D = f_{xx}f_{yy} - f_{xy}^2 = 0$ and the test is inconclusive. You classify the point directly instead: $f(x,0) = x^3$ changes sign through the origin, so it is neither a maximum nor a minimum.

Where does the gradient of x^3 - 3xy^2 point purely along the y-axis?

You need $3x^2 - 3y^2 = 0$ with $-6xy \ne 0$ , so $y = \pm x$ with $x \ne 0$ . Along the diagonals the gradient is $\langle 0, -6xy\rangle$ , which is vertical.