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Section 11.9 Curvature and the Second Derivative Test

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}\tag{11.9.1} \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 \(|\vv|=1\text{,}\) so that

\begin{equation} \kappa = |\TT\times\aa|\tag{11.9.2} \end{equation}
Parametric curve on a graph.

Now consider the graph of \(z=f(x,y)\text{.}\) 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\tag{11.9.3} \end{equation}

We can obtain a curve on this surface by specifying a relationship between \(x\) and \(y\text{.}\) In particular, suppose that

\begin{align*} x \amp= a+t\cos\alpha ,\\ y \amp= b+t\sin\alpha , \end{align*}

so that

\begin{equation} \rr(t) = \rr_0 + t\cos\alpha\,\xhat + t\sin\alpha\,\yhat + f(x,y)\,\zhat\tag{11.9.4} \end{equation}

which leads to

\begin{align*} \vv(t) \amp= \cos\alpha\,\xhat + \sin\alpha\,\yhat + \frac{df}{dt}\,\zhat ,\\ \aa(t) \amp= \frac{d^2f}{dt^2}\,\zhat . \end{align*}

Using the master formula, we have

\begin{equation} df = \grad f\cdot d\rr\tag{11.9.5} \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 .\tag{11.9.6} \end{equation}

Taking another derivative yields

\begin{align*} \frac{d^2f}{dt^2} \amp= \frac{d}{dt}\Partial{f}{x}\,\cos\alpha + \frac{d}{dt}\Partial{f}{y}\,\sin\alpha\\ \amp= \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\\ \amp= \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\\ \amp= \begin{pmatrix}\cos\alpha \amp \sin\alpha\end{pmatrix} \begin{pmatrix} \frac{\partial^2 f}{\partial x^2} \amp \frac{\partial^2 f}{\partial x\partial y}\\ \frac{\partial^2 f}{\partial y\partial x} \amp \frac{\partial^2 f}{\partial y^2}\end{pmatrix} \begin{pmatrix}\cos\alpha\\ \sin\alpha \end{pmatrix}\\ \amp= v^T J v \end{align*}

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)} .\tag{11.9.7} \end{equation}

Then the velocity reduces to

\begin{equation} \vv(0) = \cos\alpha\,\xhat + \sin\alpha\,\yhat\tag{11.9.8} \end{equation}

which is a unit vector. Furthermore, \(\vv\perp\aa\text{,}\) so that the curvature becomes simply

\begin{equation} \kappa = |\aa| = \left|\frac{d^2f}{dt^2}\right|\tag{11.9.9} \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 \(v\text{.}\)

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}\tag{11.9.10} \end{equation}

But the last two terms are the same, since \(J\) is symmetric. Thus, at a max or min, we must have \(\frac{dv^T}{dt} Jv = 0\text{,}\) or equivalently \(\frac{dv}{dt}\perp Jv\text{.}\)

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

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

But \(J\) is a symmetric matrix, which implies first of all that \(J\) admits two real eigenvalues, \(\lambda_\pm\text{,}\) 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 \(D>0\text{,}\) the critical point is either a local min or a local max, depending on the sign of the concavity in any direction.

  • If \(D\lt 0\text{,}\) the critical point is a saddle point.

  • If \(D=0\text{,}\) 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 \(J\text{,}\) which gives the concavity in the \(x\)-direction.

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