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So a surface $S \in \mathbb{R}^3$ is ruled if through each point $p$ there is a line in $\mathbb{R}^3$ entirely contained in $S$.

Show that the line through $p$ lies along an asymptotic direction.

Prove that if a surface is ruled, then $K\leq0 $ at each point.

Can someone give a hint?

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    $\begingroup$ For the second part, I would proceed by showing the following: If $K>0$ at some point $p \in S$, then for a small neighbourhood of $p$ in S, say $N_{p}$, $N_{p} \setminus \{p\}$ is contained in only one of the connected components of $\mathbb{R}^{3} \setminus T_{p}(S)$. This clearly rules out the existence of a line in $S$ through $p$. $\endgroup$ – Nick L Nov 5 '18 at 18:27
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Approach 1: One can show that a ruled surface locally can be parametrised as $$ x(u,v) = c(u) + v X(u), $$ where $c$ is a curve and $X$ a vector field along $c$. A calculation shows that the Gauss curvature is non-positive and that the lines $v\mapsto x(u_0,v)$ are asymptotic lines. This is probably not the quickest approach for this problem, but you can look up the calculations in, e.g. Kühnel - Differential Geometry. Curves-Surfaces-Manifolds (p. 85).

Approach 2: Let $c\colon I \to S $ be a unit speed parametrisation of a line in the surface $S$. Then $T'=c'' = 0$. You can easily show that $S_N T$ vanishes. ($S_N$ stands for the shape operator.)

Next you can use Euler's formula $k_n = k_1 \cos \theta + k_2 \sin \theta$. Here $k_n$ is the normal curvature in the direction $\cos\theta e_1 + \sin\theta e_2$, $k_1$, $k_2$ are the principal curvatures and $e_1$, $e_2$ are the principal directions. If there is one asymptotic direction, then $k_1$ and $k_2$ must have different signs and hence $K$ cannot be positive.

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