Let $M = \mathbb{R}\times \mathbb{R}_{>0}$ with the metric $ds^2 = dx^2 + ydy^2$
a. Calculate the Gaussian curvature of the abstract surface $M$
b. Determine all geodesics in $M$ going through $(x_0, y_0)$ with tangent vector $(1, 0)$
My attempt
a. First we note that the matrix of the first fundamental form is $I =\begin{bmatrix} 1 & 0 \\ 0 & y \end{bmatrix}$.
From this, we determine that the only non-zero Christoffel symbol is $\Gamma_{vv}^v = \frac{1}{2v}$.
If we denote $R(x, y)z = \nabla_x\nabla_yz - \nabla_y\nabla_xz$, then the formula for the Gaussian curvature is $$K = \frac{\left< R(\mathbf{x}_u,\mathbf{x}_v)\mathbf{x}_v,\mathbf{x}_u\right>}{\left< \mathbf{x}_u,\mathbf{x}_u \right >\left< \mathbf{x}_v,\mathbf{x}_v \right > - \left< \mathbf{x}_u,\mathbf{x}_v \right >}$$
Now, using the formula for Gaussian curvature, we get that $$K = \frac{\left< \frac{-1}{2v^2}\mathbf{x}_v, \mathbf{x}_u\right>}{v}$$
b. We have that geodesics are the curves that satisfy $\nabla_{\alpha^\prime} \alpha^{\prime} = 0$
The question
First of all, is a correct? I think it is, but I may have been wrong in many places. Also, does $K$ simplify to $0$? It seems like it does, but I'm not too sure either...
Finally, how can I do b? I wrote down the equation, but I don't really know how to continue it. I think I should write $\alpha = \mathbf{x}(u(t), v(t))$, but I'm not very familiar with the context of abstract surfaces.
Any help is appreciated! Thanks in advance!
Hint for part b
(IMPORTANT: in the following context, direct superscripts on symbols do not imply raising to powers.)
The geodesics equations are $$ \frac{\partial^2 x^i}{\partial s^2}+\Gamma_{\alpha\beta}^i\frac{\partial x^\alpha}{\partial s}\frac{\partial x^\beta}{\partial s}=0, $$ where $i,\alpha,\beta \in \{1,2\}$. In the case of your question, the geodesics are found among $$ {\frac{\partial^2 x}{\partial s^2}=0, \\ \frac{\partial^2 y}{\partial s^2}+\frac{1}{2y}(\frac{\partial y}{\partial s})^2=0. } $$
Additional Remark
I couldn't fully conceive what $\Gamma_{vv}^v = \frac{1}{2v}$ means. Unless I'm missing something, the expression $\Gamma_{vv}^v = \frac{1}{2v}$ means $$ { \Gamma_{11}^1 = \frac{1}{2}, \\\Gamma_{22}^2 = \frac{1}{4}, } $$ which is indeed incorrect. Actually, you should write the only non-zero Christoffel symbol as $$ \Gamma_{22}^2 = \frac{1}{2x^2}=\frac{1}{2y}. $$