If an embedded surface $S \subset \mathbf{R}^{3}$ contains a line $L$, show that the Gaussian curvature is non-positive at each point of $L$. Give an example where the Gaussian curvature is zero at each point of $L$.

Consider the helicoid $S$ given as the image of $\mathbf{R}^{2}$ in $\mathbf{R}^{3}$ under the map

$$\phi(u, v)=(\sinh v \cos u, \sinh v \sin u, u) .$$

What is the image of the corresponding Gauss map? Show that the Gaussian curvature at a point $\phi(u, v) \in S$ is given by $-1 / \cosh ^{4} v$, and hence is strictly negative everywhere. Show moreover that there is a line in $S$ passing through any point of $S$.

[General results concerning the first and second fundamental forms on an oriented embedded surface $S \subset \mathbf{R}^{3}$ and the Gauss map may be used without proof in this question.]