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.