Geodesics in a right cylinder

Consider the right cylinder $X(u,v)=\alpha(u)+v\vec{a}$, where $\alpha$ is a regular curve contained in a orthogonal plane to the vector $\vec{a}$. A curve $\alpha(t)=X(u(t),v(t))$ is a geodesic if the tangent part of $\alpha''(t)$ vanishes for every $t$. We have

$$\alpha''(t)=u''X_u+v''X_v+u'^2 X_{uu}+2u'v' X_{uv}+v'^2X_{vv},$$

where

$$X_u=\alpha'(u),\ X_v= \vec{a}$$

$$X_{uu}=\alpha''(u),\ X_{uv}=X_{vv}=0.$$

Thus the tangent part of $\alpha''$ is 

$$\alpha''(t)^T=u''\alpha'(u)+v''\vec{a}+u'^2\alpha''(t)^T.$$

Since $\alpha''(t)=\kappa(t)n(t)$ is a vector orthogonal to the surface, $\alpha''(t)^T=0$. Thus $\alpha$ is a geodesic if and only if $$u''=0,\ v''=0\Leftrightarrow u(t)=at+b, v(t)=ct+d,$$

for some constant $a,b,c,d$. Then the preimage of $\alpha$ in the domain of the parametrization is a straight-line so when we carry into the surface by $X$ we obtain a helix. For example, when $u=ct$, the geodesic is a vertical line in the cylinder and when $v=ct$, the geodesic is a vertical translation of the base curve $\alpha$. In the picture, and a for a circular cylinder $x^2+y^2=1$,  we have the lines in the domain of $X$, the geodesics and the cylinder with the geodesics.

Ruled surfaces

A ruled surface is a surface constructed moving a straight-line along a given curve. If $\alpha=\alpha(s)$ is this curve and $w(s)$ is the direction of the straight-line at $\alpha(s)$, the straight-line is the set $\{\alpha(s)+t w(s):t\in{\mathbb R}\}$. Thus the surface $S=\{\alpha(s)+t w(s): s\in I,t\in{\mathbb R}\}$ and the parametrization is $X(s,t)=\alpha(s)+t w(s)$. Since $X_s=\alpha'(s)+tw'(s)$ and $X_t=w(s)$, then we have to assume that they are independent linearly. Then $X$ is a parametrized surface and so, locally, $S$ is a surface.   

We show some examples. Suppose $\alpha$ is a planar curve included in the plane $z=0$. If we take $w(s)=a=(0,0,1)$, we have the right cylinder over the curve $\alpha$. In the next figure, $\alpha$ is the parabola $\alpha(s)=(s,s^2,0)$ and $w(s)=(0,0,1)$.

We can take $w(s)$ to be tilted at each point of $s$. If $\{T(s), N(s), B(s)\}$ is the Frenet trihedron along $\alpha$,  then up to reparametrizations, $B(s)=(0,0,1)$. If we take $w(s)=\cos(m) N(s)+\sin(m) B(s)$, with $m\in{\mathbb R}$,  we obtain a cone along $\alpha$. In the next picture,  $\alpha$ is the parabola again.

If we replace the constant $m$ by a function $\theta(s)$, then $w(s)$ goes changing at each point. Here we take $\alpha$ the circle $\alpha(s)=(\cos(s),\sin(s),0)$. If $w(s)$ is a $2\pi$-periodic function, then $w$ is also $2\pi$-periodic. This occurs for example if $\theta(s)=s$. The parametrization is $X(s,t)=(\cos (s)-t \cos ^2(s),\sin (s)-t \sin (s) \cos (s),t \sin (s))$ and the surface is:

But if $w$ is $4\pi$-periodic, then we obtain a Möbius strip. For this, we take $\theta(s)=s/2$. Then $$X(s,t)=\left(\cos (s) \left(1-t \cos \left(\frac{s}{2}\right)\right),\sin (s) \left(1-t \cos \left(\frac{s}{2}\right)\right),t \sin \left(\frac{s}{2}\right)\right).$$