Q: Consider an one sheet hyperboloid $S$ sitting in $\mathbb{R}^3$ which defined by $x^2+y^2-z^2 =1$. Show that there is a straight line in $S$ through every point of $S$. Also, deduce without any calculation, that the Gaussian curvature of $S$ is less than or equal to $0$ at each point.
What I done so far was, I know that the surface $S$ has regular parametrization:
$$r(u,v)=(\cos u-v\sin u, \sin u+v \cos u,v)$$
$$r(u,v)=(\cos u,\sin u,0)+v(-\sin u,\cos u,1)$$
by looking at the second equation, if we keep u fixed, it is the standard straight line equation, and by varying u from 0 to 2$\pi$, we obtain the whole surface $S$. Hence, for every point in $S$, there is a straight line that go through it and lie in $S$.
However I don't know how to deduce the curvature of $S$ is $\le 0$ from the fact that one sheet hyperboloid is a ruled surface.
Best Answer
Assume that the Gaussian curvature is positive. It means that the second fundamental form is positive (or negative, depends on the choice of orientation). It then follows, that the curvature of every curve on the surface is also positive, as it is at least equal to the second fundamental form of the derivative at each point. This contradicts, obviously, the existence of a straight line at every point.