Geodesics on the cylinder and Levi-Civita connection $\newcommand{\Reals}{\mathbb{R}}\newcommand{\Vec}{\mathcal{T}}\newcommand{\Brak}[1]{\left\langle #1\right\rangle}$
Just a question about the following exercise.
Exercise
Let $C=\{(x,y,z)\in\mathbb{R}^3|x^2+y^2=1\}\subset\mathbb{R}^3$. The cylinder $C$ can be identified with $M=\mathbb{R}^2\diagdown\{(0,0)\}$ through the diffeomorphism
$$ \psi\colon M\to C,\qquad (\theta,\rho)\mapsto(x=\cos\theta, y=\sin\theta, z=\log\rho)$$
where $(\theta,\rho)$ are the usual polar coordinates in $M$.
The euclidean metric $g$ of $\mathbb{R}^3$ induces a metric $g_C$ on the cylinder $C$.
$(a)$ Write explicitly the matrix of the metric $g_M$ on $M$ (in the $(\theta,\rho)$ coordinates) wich corresponds to the metric $g_C$ on $C$ through the diffeomorphism $\psi$.
$(b)$ Taking into account the formula
$$ \Gamma_{ij}^k=\frac{1}{2}\sum_{l=1}^2g^{kl}\Big(\frac{\partial g_{lj}}{\partial x^i}+\frac{\partial g_{il}}{\partial x^j}+\frac{\partial g_{ij}}{\partial x^l}\Big)\quad i,j,k=1,2$$
determine explicitly the Christoffel symbols of the Levi-Civita connection on $M$ and write down the equations of the geodesics.
$(c)$ Given $\alpha_0,\alpha_1,\beta_0,\beta_1\in\mathbb{R}$, $\beta_0>0$, verify that the curves
$$t\mapsto (\rho(t)=\beta_0 e^{\beta_1 t}, \theta(t)=\alpha_0+\alpha_1 t)  $$
are geodesics in $M$.
Solution
$(a)$ I get $g_M=d\theta^2+\frac{1}{\rho^2}d\rho^2$ and so
$$ g_{M_{ij}}=\begin{bmatrix}1 & 0 \\0 & \frac{1}{\rho^2}\end{bmatrix}$$
and
$$ g_{M}^{ij}=\begin{bmatrix}1 & 0 \\0 & \rho^2\end{bmatrix}.$$
So for the eight Christoffel coefficients I get that they are all $0$ except for 
$$\Gamma_{22}^2=-\frac{1}{\rho}.$$
But since the equations for the geodesics are given in terms of the Christoffel coefficients by
$$
\left\{
\begin{aligned}
\ddot{\theta}(t)+\sum_{i,j=1}^2\Gamma^1_{ij}\dot{\sigma}^i(t)\dot{\sigma}^j(t) &= 0\\
\ddot{\rho}(t)+\sum_{i,j=1}^2\Gamma^2_{ij}\dot{\sigma}^i(t)\dot{\sigma}^j(t) &= 0\\
\end{aligned}
\right.
$$
where $\sigma(t)=(\sigma^1(t),\sigma^2(t))=(\theta(t),\rho(t))$,
this would mean that the only equation which matters in determining the geodesics on the given cylinder is the one involving the $\rho$ coordinate of the given parametrization. One can actually verify (I did it) that the curves given in $(c)$ are geodesics in $M$ (this amounts only to verify that the first component of those curves solves the second order differential equation involving $\rho$ of the system). 
I'm just wondering if such a thing is possible, I mean the fact that the only thing that matters is "what $\rho$ does" in order for a curve parametrized by $t$ of that form to be a geodesic on the cylinder.
 A: Secretly, the induced metric on the cylinder $C$ is the Euclidean (!) metric
$$
g_{C} = d\theta^{2} + d(\log \rho)^{2}
  = d\theta^{2} + dz^{2}
$$
in the $(\theta, z)$-plane. Consequently, the geodesics are "straight lines parametrized at constant speed". (On the cylinder, these paths trace meridians, helices, or latitude circles.)
Just as your calculations predict,
$$
\ddot{\theta}(t) = 0,\qquad
\ddot{z}(t)
  = \frac{\ddot{\rho}(t)}{\rho(t)} - \left(\frac{\dot{\rho}(t)}{\rho(t)}\right)^{2}
  = \frac{1}{\rho(t)}\left(\ddot{\rho}(t) - \frac{\dot{\rho}(t)^{2}}{\rho(t)}\right)
  = 0.
$$
Note, incidentally, that the geodesic equations decouple, but they're not independent of $\theta$, which must be an affine function of $t$.
A: As a set, $C$ is same to flat cylinder we already know In further in problem $C$ has induced metric from canonical metric on $\mathbb{R}^3$ So it is in fact flat cylinder 
And since we give a new coordination which is usually not used we obtain nontrivial geodesic equation 
