I wanted to prove that a helix $h(t)$ on a circular cylinder $f(u,v)$ always is a geodesic by showing that the covariant derivative $\nabla_{h(t)}h'(t)$ vanishes everywhere. One way to solve this problem is to parametrize the cylinder as $f(u,v) = (r\cdot \cos u,r\cdot \sin u, v)$ for $r \neq 0$ and the helix as $h(t) = (r \cdot \cos t, r \cdot \sin t, bt)$, where $b \neq 0$. It is easy to see, that a normal field for the cylinder is given by $N(x,y,z) = (x,y,0)$. And by computing the second derivative of $h$ as
$$h''(t) = (-r\cdot \cos t, -r \cdot \sin t, 0) = - N(h(t))$$
it follows that
$$\nabla_{h(t)}h'(t) = h''(t) - \langle h''(t), N(h(t)) \rangle N(h(t)) = 0.$$
This proves the proposition. From the definition it's clear that $h$ is indeed a curve on the cylinder. However, I think it's much more elegant to express the helix with respect to $f$ as $H(t) := f(\alpha(t), \beta(t)) = (r \cdot cos t, r \cdot sin t, bt)$, that is by using the plane curve $\hbar(t) = (\alpha(t), \beta(t)) = (t, bt)$. It should be possible to reproduce the steps in the proof above with this parametrization. But I'm confused as to what is the derivative of $H(t)$? What is the proper way to derive this function?