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?
On
On each point of a ruled surface obviously there is a direction of $0$ curvature (along the line). The gaussian curvature at a point is the product of the principal curvatures, which are the maximal and minimal curvatures at the point, so in the case of a ruled surface either the minimal one is $0,$ in which case the gaussian curvature is $0,$ or it is negative and the maximal one is $\geq 0,$ from which follows that the gaussian curvature is $\leq0.$
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.