Saddle points - Show that surface $z = y \sin x$ has infinitely many saddle points. 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!
 A: 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.
