Finding Christoffel Symbols using via variational method. I'm trying to find the Christoffel Symbols for the Lorentz metric $${\rm d}s^2 = \cos(2\pi x)({\rm d}x^2-{\rm d}y^2) - 2\sin(2\pi x)\,{\rm d}x\,{\rm d}y$$by looking at the Euler-Lagrange equations for $$L(x,\dot{x},y,\dot{y}) = \cos(2\pi x)(\dot{x}^2-\dot{y}^2) - 2\sin(2\pi x)\,\dot{x}\,\dot{y}.$$I have already done my fair share of computations like this, but I must be making some algebraic mistake that I cannot find for the life of me. If we write $$\begin{align}\frac{\partial L}{\partial x} - \frac{{\rm d}}{{\rm d}t}&\left(\frac{\partial L}{\partial \dot{x}}\right) = -2\pi\sin(2\pi x)(\dot{x}^2-\dot{y}^2)-4\pi\cos(2\pi x)\dot{x}\dot{y} \\ &\qquad - \frac{{\rm d}}{{\rm d}t}\left(2\dot{x}\cos(2\pi x) - 2\dot{y}\sin(2\pi x)\right),\end{align}  $$and we will have a term with $\ddot{y}$. This is a problem, since as far as I understand the geodesic equation corresponding to the $x$-coordinate should have the form $$\ddot{x} + \Gamma(\dot{x},\dot{y})=0,$$maybe after dividing by something. What is going on?
 A: *

*The infinitesimal variation of the Lagrangian
$$ L(x,\dot{x})~=~ g_{ij}(x)~ \dot{x}^i\dot{x}^j \tag{1} $$
is
$$ \frac{1}{2}\delta L~=~ -\left\{ \color{blue}{ g_{k\ell}\ddot{x}^{\ell}} +\color{red}{\Gamma_{k,ij} \dot{x}^i\dot{x}^j}\right\}\delta x^k 
+\frac{\mathrm{d}}{\mathrm{d}t}\left\{ \color{green}{ g_{k\ell}\dot{x}^{\ell} \delta x^k}\right\},\tag{2} $$
where we have introduced the lowered Levi-Civita Christoffel symbols
$$\Gamma_{k,ij}~:=~g_{k\ell}\Gamma^{\ell}_{ij}. \tag{3} $$
Note that eq. (2) contains three different types of terms (displayed in different colors), which are uniquely characterized by how the $t$-derivatives are distributed. 
In particular we see that the geodesic equations are multiplied with the metric, cf. OP's last question.

*Example.  OP's Lagrangian reads
$$L~=~c(\dot{x}^2-\dot{y}^2) - 2s\dot{x}\dot{y}, \qquad c~:=~\cos(2\pi x), \qquad s~:=~\sin(2\pi x) ,\tag{4}$$
corresponding to the metric
$$ \begin{pmatrix} g_{xx} & g_{xy} \cr g_{yx} & g_{yy} \end{pmatrix} ~=~\begin{pmatrix} c & -s \cr -s & -c \end{pmatrix}. \tag{5}$$
We calculate the infinitesimal variation:
$$\begin{align}\frac{1}{2} \delta L
~=~&\left\{ \color{blue}{s\ddot{y}-c\ddot{x}}
+ \color{red}{\pi s (\dot{x}^2+\dot{y}^2)} \right\} \delta x 
+ \left\{  \color{blue}{s\ddot{x}+c\ddot{y}}
+ \color{red}{2\pi (c\dot{x}^2-s\dot{x}\dot{y})} \right\} \delta y \cr
& + \frac{\mathrm{d}}{\mathrm{d}t}\left\{ \color{green}{(c\dot{x}-s\dot{y})\delta x -(s\dot{x}+c\dot{y})\delta y}\right\},
\end{align}\tag{6}$$
which should be compared with the general formula (2).
From the red terms in eq. (6) we can read off the non-zero lowered Christoffel symbols 
$$ \Gamma_{x,xx}~=~-\pi s ~=~\Gamma_{x,yy}~=~-\Gamma_{y,xy}, \qquad \Gamma_{y,xx}~=~-2\pi c,\tag{7} $$
cf. OP's title.
