Vertical curve in a Riemannian warped product is a geodesic Let $(N,g_N)$ be Riemannian manifold, $I \subset \mathbb{R}$ open with coordinate $r$ and $M:= I \times N$ with metric $g=dr^2+f^2 g_N$. 
Let $c_p: I \rightarrow M, (t \mapsto (t,p))$ for a fixed $p$ be a curve in $M$. 
I want to show that $c$ is a geodesic. 
To do that, I first look at local coordinates for $M$, define $(y^1,....,y^{n+1})=((t,q) \mapsto t, x^1,...,x^n)$ where $x^i$ are local coordinates in $N$. Now:
$y^1 \circ c_p= (t \mapsto t)$ and $y^i \circ c_p=(t \mapsto x^i(p))$ for $i>1$. I can conclude that $$\dfrac{\partial (y^1 \circ c_p)}{\partial t} =1,\quad \dfrac{\partial^2 (y^i \circ c_p))}{\partial t^2}=0 \ \forall i,\quad \dfrac{\partial (y^i \circ c_p)}{\partial t} =0 \ \forall i>1.$$ 
Now the equation for $\nabla_{c^{'}} c^{'}$ reduces to:
$$\nabla_{c^{'}} c^{'}= \sum\limits_{i=1}^{n+1} \Gamma^i_{11} \dfrac{\partial}{\partial y_i} = \nabla_{\partial_1} \partial_1$$
Is it correct, that $ \nabla_{\partial_1} \partial_1=0$ (then my proof would be complete)? And if yes, why?
 A: $F(t,x)=(t,\phi (x))$ is chart for $M$ where
 $\phi$ is a chart for $N$.
Hence $E_t=\partial_t,\ E_\alpha=
 d\phi\ e_\alpha$ are coordinate field. Note that $g(E_t,E_t)=1 ,\ g(
 E_t,E_\alpha)=0,\ g(E_\alpha,E_{
 \alpha'})=f^2 g_N(E_\alpha,E_{\alpha'})$
So we must show that $\nabla_t E_t=0$ : $g(\nabla_tE_t,E_t)=0$ and $
 g(\nabla_tE_t,E_\alpha)=E_tg(E_t,E_\alpha)
 -\frac{1}{2}E_\alpha g(E_t,E_t) =0$ 
A: Since your metric is Riemannian, we can just show that given two instants $t_0<t_1$ in $I$, $c_p|_{[t_0,t_1]}$ minimizes arc-length among all other curves joining $c_p(t_0)$ to $c_p(t_1)$. This show that $c_p|_{[t_0,t_1]}$ is a geodesic, and since $t_0<t_1$ were arbitrary, we will conclude that $c_p$ itself is a geodesic all along $I$. So, let $\gamma = (\gamma_1,\gamma_2)\colon [t_0,t_1] \to M$ be a smooth curve joining $(t_0,p)$ and $(t_1,p)$. We have that $$\gamma'(t) = \gamma_1'(t)\partial_r|_{\gamma(t)} + \gamma_2'(t),$$ Thus we compute $$\begin{align} L[\gamma] &= \int_{t_0}^{t_1} \sqrt{\gamma_1'(t)^2 + f^2(\gamma_1(t)) g_N(\gamma_2'(t),\gamma_2'(t))}\,{\rm d}t \\ & \geq \int_{t_0}^{t_1} \sqrt{\gamma_1'(t)^2}\,{\rm d}t \\ &= \int_{t_0}^{t_1} \gamma_1'(t)\,{\rm d}t \\ &= \gamma_1(t_1) - \gamma_1(t_0) \\ &= t_1-t_0 \\ &= L[c_p|_{[t_0,t_1]}]. \end{align}$$This concludes the proof.
