Geodesics on surface of revolution of regular curve I was recently presented with this in differential geometry stating the following:

Let us define the regular curve on the XZ plane as: $ \gamma (t) = (sin(t)+2,0,t) $ on XZ plane for $ t \in R $, and define S the regular surface obtained via rotation of the curve $ \gamma $ about the z axis. We are to determine for which real values of c the intersection curve of S with the plane $ z=c $ is a geodesic on the surface S

I do not know how to proceed all I know is the definition of a geodesic but cannot do this technically can someone please help on this? Thank you kindly 
 A: First, if $S$ is a regular surface and $c\colon I\subset \mathbb{R}\longrightarrow S$ is a geodesic then $$N (c(s))\parallel \vec{n}(s),\ \ \forall s\in I,\ \ \ \ \ \ \ \ (\dagger)$$ where $N$ is the unit normal of $S$ and $\vec{n}$ the first normal of the curve $c$. Thus, in your problem you are interested in finding the parallels of your surface of revolution that are also geodesics. Equivalently you want to find the parallels that do satisfy $(\dagger)$. Therefore, if $\alpha_t$ is the intersection curve of your $S$ with the plane $z=t$ you want to find the parallels $\alpha_t$ that do satisfy the following: $$N(a_t)\perp \vec{e}_3=(0,0,1)\ (\leftarrow \text{direction of the axis of rotation}).$$
To do that, consider the following parametrization of your surface of revolution  $$X(t,u):=(\underbrace{(\sin t+2)}_{f(t):=}\cos u,(\sin t+2)\sin u,t).$$
The unit normal is $$N(t,u)=\frac{X_t\times X_u}{||X_t\times X_u||}=\frac{1}{f(1+(f')^2)}(-\cos u,-\sin u,f').$$ Thus, $\alpha_t$ is a geodesic iff $f'(t)=0\Leftrightarrow t=k\pi+\frac{\pi}{2},\ k\in \mathbb{Z}$. Therefore you obtain that the only parallels which are also geodesics are $$\alpha_{k\pi+\frac{\pi}{2}}(u)=X\left(k\pi+\frac{\pi}{2}, u\right),\ \ k\in \mathbb{Z}.$$
