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Show that surface $z = y \sin x$ has infinitely many saddle points.

Can someone show me the step-by-step solution for that statement? Detailed explanations will be appreciated. Thank you very much!

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Note that if you find that $(x,y)$ is a saddle point for some specific $x,y$ then since $z$ is periodic in $x$ there are infinitely many saddle points of the form $(x+2\pi n,y).$ – coffeemath Nov 19 '12 at 7:12
up vote 2 down vote accepted

You first find the critical points of $z=y \sin x$ by finding the points $(x,y)$ for which both partial derivatives are zero. Denoting the partials as $z_x$ and $z_y$, you have $$z_x=y \cos x,$$ $$z_y=\sin x.$$ If these are both zero then $\sin x = 0$ implying $x=n \pi$ for some integer $n$. But then $\cos x=1,-1$ so from $z_x=0$ you see that $y=0$ at any critical point.

So your critical points are all points of the form $(x,y)=(n \pi, 0).$

The next thing one does for checking for saddle points is to compute the following value at each critical point $(x,y)$: $$J(x,y)=z_{x,x} \cdot z_{y,y} - z_{x,y} z_{y,x},$$ where the double subscripts denote second partials (partials of partials). If this calculation comes out negative, you know you have a saddle point.

I'll leave the rest to you --- you should find that all the above critical points are saddle points.

Note: $J(x,y)$ above is actually the determinant of the "jacobian" at $(x,y)$, where the jacobian is really a matrix. In general one sees whether the jacobian is positive definite, negative definite, or indefinite, or none of these. That is, if the value $J(x,y)$ happens to be $0$ at a critical point, then nothing can be concluded about whether one has a local max, local min, or saddle point.

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Thank you very much! :D – Cossette Nov 20 '12 at 7:31
@coffeemath: The matrix of second partials of a scalar function is usually called the Hessian and denoted by $H$. The notion of Jacobian (and the letter $J$) is relevant for the first derivative of a map $f:\ {\mathbb R}^n\to{\mathbb R}^m$. – Christian Blatter Nov 21 '12 at 9:32

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