Ruled surface with negative/zero gaussian curvature. 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?
 A: 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.
