Geometric interpretation of $\frac {\partial^2} {\partial x \partial y} f(x,y)$

Question

Are there any geometric interpretation for the second partial derivative? i.e.

$$f_{xy} = \frac {\partial^2 f} {\partial x \partial y}$$

In particular, I'm trying to understand the determinant from second partial derivative test for determining whether a critical point is a minima/maxima/saddle points:

$$D(a, b) = f_{xx}(a,b) f_{yy}(a,b) - f_{xy}(a,b)^2$$

I have no trouble understanding $f_{xx}(x,y)$ and $f_{yy}(x,y)$ as the of measure of concavity/convexity of f in the direction of x and y axis. But what does $f_{xy}(x,y)$ means?

Answer 1

The object that truly has geometric meaning is the Hessian, i.e. the matrix consisting of the second order partial derivatives: $$ H(x,y) := \begin{pmatrix} f_{xx} & f_{xy} \\ f_{xy} & f_{yy} \end{pmatrix}. $$

(In the following, I will denote the dot/scalar product by $\langle(u_1, u_2), (v_1, v_2)\rangle = u_1 v_1 + u_2 v_2$.)

Write $\mathbf x = (x, y)$. Taylor's theorem says that the best second order approximation to the (smooth) function $f$ is given by $$ f(\mathbf x) = f(0) + \langle \nabla f(0), \mathbf x \rangle + \langle H(0) \mathbf x, \mathbf x\rangle + O( \| \mathbf x \|^3 ).$$

If you are at a critical point, the relevant term is the quadratic term $\langle H(0) \mathbf x, \mathbf x\rangle$. Level sets of a quadratic expression like this are conic sections. The determinant of $H$ (which is given by the formula you wrote above) allows you to determine what the level sets are, whether this quadratic function is positive definite, negative definite or indefinite. If you think of the graph of the function as a mountain range, the eigenvalues of $H$ tell you how spiky the mountain is, and the eigenvectors tell you the directions of steepest ascent/gentlest ascent (or descent, as the case may be).

Answer 2

Think of the mixed partial as an abstract tendency of the curve to twist like DNA, (but just because it's positive in absolute value doesn't mean the surface actually twists since other tendencies may overwhelm it), and the straight second derivatives $f_{xx}$ and $f_{yy}$ as an abstract tendency of the curve to bulge up or down in the $xz$- and $yz$- cross-sections.

To tease out the individual roles of the partials, let's assume for the moment that the mixed partials are identically zero, the better to isolate their effect later. If $f_{xx}$ and $f_{yy}$ are of opposite signs, then the curve has a saddle tendency (think of two curves, one in the $yz$-plane and one in the $xz$-plane, intersecting at right angles, one opening up and the other opening down: this will produce a negative intrinsic curvature or saddle shape like a hyperbolic paraboloid). Now think of the same parabolas but both opening say down, with $f_{xx}$ and $f_{yy}$ tending to have the same signs: that will tend to produce an intrinsic positive curvature (bulging) like an ellipsoid. I'm sure you knew this.

Now let's add in the mixed partials to show what their effect is. Think of the mixed partials as a pure twisting factor, also tending to produce a negative or saddle curvature, but rotated $45$ degrees! Yes, a twist is the same as a saddle tendency, but we think of it differently. Imagining your hand riding down along the wall of a saddle shape, like an airplane that dives while rotating, may help you see this.

Now: ever seen a diagram of a saddle showing a kind of "X" shape over it, diagonal lines coming out of the saddle-point, where the tendency to go up in say $y$ is canceled by the tendency to go down in $x$, and the thing just holds steady along the diagonal line, with $f_{xx}$ and $f_{yy}$ both zero here? Well, if you had a properly rotated saddle shape it might not go up or down in the $x$- or $y$- directions at all--and yet it would be negatively curved or twisting nonetheless as would be plainly visible from the other directions. It is THIS twist that the mixed partials measure (just as an $xy$-term rotates a conic, by the way).

If the curvature from the two ordinary second partials is negative, forget it--the mixed partials will make it even more negative. In other words, if $f_{xx}f_{yy}$ is negative because these differ in sign, the intrinsic curvature is already negative, and subtracting $f_{xy}^2$ will make it even more so. But if the curvature from the straight second partials is positive (bulging up or down), because $f_{xx}$ and $f_{yy}$ agree in sign, then possibly this positive tendency can still be overwhelmed by the independent negative-curvature twisting action of the mixed partials. That is why $f_{xy}$ is squared and subtracted: its sign doesn't matter to the saddle-ness and it is an inherently negatively-curved factor. The contest between these is the discriminant test you mention.

Answer 3

Try thinking about the Hessian as expressing the curvature of the surface at a particular point. Negative curvature corresponds to a saddle point, positive with a relative extrema, etc. If you want a real geometric interpretation you should just pick up a differential geometry book (Do Carmo's is a good one) and look up sectional curvature. I tend to think of the $f_{xy}$ as expressing how much the curves obtained by intersecting the graphs with planes parallel to the xz,y-planes deviate from having maximum acceleration with respect to curves tangent to the surface intersecting your point of interest. Also the determinant is independant of your choice of coordinates of the tangent space-hence coordinate free.

Answer 4

Roughly, the mixed partial represents how fast (and in what direction) a tangent line "spins" as you "drag" the tangent point across a surface. At least this is how I think of it. Consider a surface such as $z = xy$, which is a fairly simple case. Its mixed partial is identically $1$, so the discriminant is identically $-1$ and the critical point at $(0, 0)$ is a saddle point (as expected). If you draw a tangent line at $(-1, 0)$ parallel to the y-z plane, you get the line $(-1, 0, 0) + t(0, 1, -1)$. Now drag this line toward the origin, and it spins around to meet the y axis, then farther until you reach $(1, 0, 0) + t(0, 1, 1)$. In fact this tangent coincides with the surface at all points, but in general this will not be the case; try $z = x^2 + 3xy + y^2 = (x+y)^2 + xy$, which has an extra confounding term but still has the same basic behavior (and the same saddle point since its discriminant is identically $-5$).