# Ruled surface out of lines of curvatures

I'm trying to proof the following statement:

Let $c$ be a curve inside a surface element $f:U\rightarrow\mathbb{R}^3$ (i.e $c=f\circ\gamma$ where $\gamma:I\rightarrow U$). Then $c$ is a line of curvature iff the ruled surface $g$ defined by $f$'s surface normal $\nu$ along the directix c is developable.

To make it clear- $g$ is the ruled surface created out of the normal vector of $f$ sliding through the curve $c$.

Both conditions have many equivalent definitions. For the first condition I thought using that $c'$ is an eigenvector of the shape operator $L$. For the second condition I thought of either using that the normal vector of $g$ doesn't change along $\nu$ or equivalently that the Gaussian curavture of $g$ vanishes.

Any ideas for solving this? Also hints are welcome, but please make them helpful enough!

## 1 Answer

You should prove that the ruled surface $x(s,t) = \alpha(s)+t\beta(s)$ (say, with $\|\alpha'\|=\|\beta\|=1$) is developable if and only if $\alpha', \beta, \beta'$ are everywhere linearly dependent.