Let $\alpha(s) = (x(s),0,z(s))$ a curve parametrized by arc length, with x>0. Let $S$ be the surface of revolution obtained by rotating $\alpha$ around the $z$ axis.
I'm being asked to give a necessary and sufficient condition for a parallel of $S$ to be geodesic. My problem is that I got the answer "That can't happen!". Here's my work:
I can parametrize S by $\phi(s,t) = (x(s) cos(t), x(s) sin(t), z(s))$
Then, $\phi_{s}(s_0,t)= (x'(s)cos(t), x'(s)sin(t), z'(s))$
For $\gamma(t)=(x(s_0)cos(t),x(s_0)sin(t),z(s_0))$ to be geodesic,
$\gamma'' \cdot \phi_s = 0$ (and the same with $\phi_t$)
But $\gamma''(t)=(-x(s_0)cos(t),-x(s_0)sin(t),0)$
And then $\gamma'' \cdot \phi_s = x(s) x'(s) cos^2(t) + x(s) x'(s) sin^2(t) = x(s) x'(s) > 0$ by hypothesis. So it can't be geodesic! I'm pretty sure that my approach is correct, but I think the exercise is given because...there are parallels that are geodesics.
Am I doing something wrong? Thanks!
Hint: You might consider $\alpha(s) = (1, 0, s)$, since the corresponding surface of revolution is a cylinder, on which every parallel is a geodesic, and see where your reasoning breaks down.