The argument presented here requires some familiarity with the properties of eigenvalues and eigenvectors of symmetric matrices.

Curvature

Recall that the curvature of a parametric curve $\rr(t)$ is given by \begin{equation} \kappa = \frac{|\vv\times\aa|}{|\vv|^3} \end{equation} This expression is independent of the parameter used, so choose an arclength parameterization, in which case $\vv$ is just the unit tangent vector $\TT$ and $|\vv|=1$, so that \begin{equation} \kappa = |\TT\times\aa| \end{equation}

Parametric curve on a graph

Now consider the graph of $z=f(x,y)$. The position vector from the origin to any point on this surface takes the form \begin{equation} \rr(x,y) = x\,\xhat + y\,\yhat + f(x,y)\,\zhat \end{equation} We can obtain a curve on this surface by specifying a relationship between $x$ and $y$. In particular, suppose that \begin{eqnarray} x &=& a+t\cos\alpha \\ y &=& b+t\sin\alpha \end{eqnarray} so that \begin{equation} \rr(t) = \rr_0 + t\cos\alpha\,\xhat + t\sin\alpha\,\yhat + f(x,y)\,\zhat \end{equation} which leads to \begin{eqnarray} \vv(t) &=& \cos\alpha\,\xhat + \sin\alpha\,\yhat + \frac{df}{dt}\,\zhat \\ \aa(t) &=& \frac{d^2f}{dt^2}\,\zhat \end{eqnarray} Using the master formula, we have \begin{equation} df = \grad f\cdot d\rr \end{equation} so that \begin{equation} \frac{df}{dt} = \grad f\cdot\vv = \Partial{f}{x}\,\frac{dx}{dt} + \Partial{f}{y}\,\frac{dy}{dt} = \Partial{f}{x}\,\cos\alpha + \Partial{f}{y}\,\sin\alpha \end{equation} Taking another derivative yields \begin{eqnarray} \frac{d^2f}{dt^2} &=& \frac{d}{dt}\Partial{f}{x}\,\cos\alpha + \frac{d}{dt}\Partial{f}{y}\,\sin\alpha \nonumber\\ &=& \Partial{}{x}\Partial{f}{x}\,\cos^2\alpha + \Partial{}{y}\Partial{f}{x}\,\sin\alpha\cos\alpha + \Partial{}{x}\Partial{f}{y}\,\sin\alpha\cos\alpha + \Partial{}{y}\Partial{f}{y}\,\sin^2\alpha \nonumber\\ &=& \frac{\partial^2 f}{\partial x^2} \cos^2\alpha + 2\frac{\partial^2 f}{\partial x\partial y} \sin\alpha\cos\alpha + \frac{\partial^2 f}{\partial y^2} \sin^2\alpha \nonumber\\ &=& \pmatrix{\cos\alpha & \sin\alpha} \pmatrix{\frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x\partial y} \cr \frac{\partial^2 f}{\partial y\partial x} & \frac{\partial^2 f}{\partial y^2}} \pmatrix{\cos\alpha \cr \sin\alpha} \nonumber\\ &=& v^T J v \end{eqnarray} where the column vector $v$ and the matrix $J$ are defined by the last equality.

Curvature at a critical point

Now suppose $(a,b)$ is a critical point, so that \begin{equation} \frac{\partial f}{\partial x}\bigg|_{(a,b)} = 0 = \frac{\partial f}{\partial y}\bigg|_{(a,b)} \end{equation} Then the velocity reduces to \begin{equation} \vv(0) = \cos\alpha\,\xhat + \sin\alpha\,\yhat \end{equation} which is a unit vector. Furthermore, $\vv\perp\aa$, so that the curvature becomes simply \begin{equation} \kappa = |\aa| = \left|\frac{d^2f}{dt^2}\right| \end{equation} If we drop the absolute value sign, we obtained a notion of “signed curvature”, which allows us to distinguish concave up (positive) from concave down (negative).

Thus, at a critical point, $v^T J v$ measures the concavity in the direction $v$.

Maximum and minimum concavity

We can now ask for the maximum and minimum values of the concavity. Using the product rule, we have \begin{equation} \frac{d}{dt} (v^T J v) = \frac{dv^T}{dt} J v + v^T J \frac{dv}{dt} \end{equation} But the last two terms are the same, since $J$ is symmetric. Thus, at a max or min, we must have $\frac{dv^T}{dt} Jv = 0$, or equivalently $\frac{dv}{dt}\perp Jv$.

On the other hand, since $v$ is a unit vector, $v^T v=1$, and a similar argument now shows that $\frac{dv^T}{dt} v=0$, or equivalently $\frac{dv}{dt}\perp v$.

Thus, both $v$ and $Jv$ are perpendicular to $\frac{dv}{dt}$, and in two dimensions this can only happen if $Jv\parallel v$. In other words, at a max or min of the concavity, $v$ must be an eigenvector of $J$!

But $J$ is a symmetric matrix, which implies first of all that $J$ admits two real eigenvalues, $\lambda_\pm$, and second that the corresponding eigenvectors are perpendicular. When classifying a critical point, it is enough to consider these two directions. If the graph is concave down/up in both directions, the critical point corresponds to a local max/min, respectively. If it is concave up in one direction, but concave down in the other, then it is a saddle point. If one or both concavities are zero, anything can happen.

Thus, in order to classify a critical point, one needs to know the product of the concavities, or equivalently the product of the eigenvalues. But the product of the eigenvalues of a matrix is precisely its determinant.

Second Derivative Test

This, finally, brings us to the second derivative test for classifying critical points. Determine the matrix $J$ of partial derivatives, evaluated at the critical point. Compute the determinant $D=\det(J)$ of this matrix.

• If $D>0$, the critical point is either a a local min or a local max, depending on the sign of the concavity in any direction.
• If $D<0$, the critical point is a saddle point.
• If $D=0$, anything can happen.

In the first case, the standard method for distinguishing between a max and a min is to examine the sign of the $(1,1)$ element of $J$, which gives the concavity in the $x$-direction.