Different approaches to calculating the Christoffel symbols I would be very grateful to whoever can debug the following calculations...
We have the metric for static spacetime: $$ds^2 = -\exp(2U(\vec x))dt^2+h_{ij}(\vec x) d x^i d x^j$$
I want to find the associated Christoffel symbols.
I know of 2 ways:
The first is to use the formula $$\Gamma^a_{bc}={1\over 2}g^{ad}(g_{dc,b}+g_{db,c}-g_{bc,d})$$
Using this formula, I get $$\Gamma^0_{0i}={\partial U\over \partial x^i}$$ for $i=1,2,3$.
However, by using a second way -- by reading off the Euler-Lagrange equation ,I got an extra factor of $2$ and I can't spot where I have made a mistake. Here is what I did:
The Lagrangian $$L=\exp(2U(\vec x))\dot t^2-h_{ij}(\vec x) \dot x^i \dot x^j$$
So by the Euler-Lagrange equation for the $t$ component, $${d\over d\tau}(\exp(2U(\vec x))\dot t)=0$$
which leads to $$\ddot t +2{\partial U\over \partial x^i} \dot x^i \dot t=0$$
Therefore giving $$\Gamma^0_{0i}=2{\partial U\over \partial x^i}$$
What went wrong, where? Are my approaches and formulae correct?
Thanks.
 A: The first one is probably correct, and you are likely not counting correctly in the second method. You should compare you result
$\ddot{t}+ 2 \frac{\partial U}{\partial x^{i}} \dot{x}_i \dot{t}=0$
to 
$\ddot{t}+ \Gamma^0_{\mu \nu} \dot{x}^\mu \dot{x}^{\nu}=0  $
or
$\ddot{t}+ \Gamma^0_{i 0} \dot{x}^i \dot{x}^{0}+ \Gamma^0_{0 i} \dot{x}^0 \dot{x}^{i}=0  $
which is
$\ddot{t}+ \Gamma^0_{i 0} \dot{x}^i \dot{t} + \Gamma^0_{0 i} \dot{t} \dot{x}^{i}=
\ddot{t}+ 2\Gamma^0_{i 0} \dot{x}^i \dot{t} = 0$.
Which yields 
$\Gamma^0_{i 0} = \frac{\partial U}{\partial x^{i}}$.
A: Couple mistakes:


*

*The second term in the metric should read $h_{ij}(\vec x)dx^idx^j$ not $h_{ij}(\vec x) \dot x^i\dot x^j$

*The first term in the Langrangian should have $\dot t^2$ instead of $dt^2$ which means that in the equation after the Lagrangian you should have a $\dot t^2$, and (I redid the calc.) this gives an answer that matches the first method without the extra factor of $2$.
Edit. My #2 doesn't solve the problem of the factor of two; BJBunk's answer does.
