Intuitive explanation of second derivative test for functions of two variables. I will be teaching multivariable calculus again this semester, and I am not so happy with the explanation I have for the second derivatives test for functions of two variables.
QUESTION: What is a good intuitive explanation of what the quantity $D=f_{xx}f_{yy}-f_{xy}f_{yx}$ is really telling us about the function $f$ at a given point $(a,b)$?
Clearly the first term is telling us about the concavity of $f$ along the lines $x=a$ and $y=b$.  What is the second term $f_{xy}(a,b)f_{yx}(ab)$ telling us about $f$ at $(a,b)$?
 A: I don't teach that shortcut. Instead I teach the students to complete the square in the quadratic form
$$
h^2 f''_{xx} + 2hk f''_{xy} + k^2 f''_{yy}
$$
coming from the Taylor expansion of $f$ around the critical point. (Of course, the first order terms vanish.) If the quadratic form is positive definite, $f$ has a local minimum (since $f(x,y) > f(a,b)$ for nearby points) and so on. Now, $D$ is (a multiple of) the discriminant of the quadratic form, so we can think of it as a measure of definiteness.
A: First, the only way for $D(a,b)=D=f_{xx}f_{yy}-f_{xy}^2$ to be positive is if both $f_{xx}$ and $f_{yy}$ have the same sign. So as you observe, a necessary condition to obtain a local max/min is that the concavity in both the $x$-direction and $y$-direction are the same.
Now, the term $f_{xy}$ is giving the rate of change of $f_x$ in the $y$-direction (i.e. we are looking at how fast the slope of the surface is changing). If the magnitude of this rate of change is large (relative to the magnitude of both $f_{xx}$ and $f_{yy}$) it is reasonable to expect the surface to look hyperbolic near $(a,b)$, whereas when the relative magnitude is small, you would expect the surface to look spherical.
A: I'm not quite sure if this counts as intuitive, but this is the way i make it clear to myself:
Suppose our critial point is $0$, then the Taylor-expansion of $f$ at $0$ yields
$$f(x) = f(0) + x^t H x + O(||x||^3)$$
where $H$ is the hessian at $0$. Now if you are close to $0$, then the term $x^tHx$ dominates over the term $O(||x||^3)$ and therefore determines the local behaviour of the function. If $H$ is positive definite, then $f$ locally looks like a paraboloid that is open towards the top. This means that $f$ has a minimum at $0$. If $H$ is negative definite, then the paraboloid is open towards the bottom and the same argumentation yields that $f$ has a local maximum. If $H$ is indefinite, then it locally looks like a saddle and $0$ is neither a maximum nor minimum.
