Error when computing geodesics in hyperbolic half plane It is known that the geodesic equations for the upper half plane equipped with the hyperbolic metric are 
$$x''=\frac{2x'y'}{y},$$
$$y''=\frac{(y')^2 -(x')^2}{y}.$$
It is also well known that the geodesics are semi-circles lying on the x-axis. However, I am encountering difficulties when I attempt to verify this fact.

I am trying to follow the solution to part (b) of problem 5 on paper 1
  here. 

The path $\tilde \gamma(t)=(\cos(t),\sin(t))$, when re-parameterized, should be a geodesic. At the point $(\cos(t),\sin(t))$, the unit tangent vector is 
$$(-\sin^2(t), \sin(t)\cos(t)).$$
So we need
$$x''(t)=\frac{2(-\sin^2(t))(\sin(t)\cos(t)}{\sin(t)},$$
or $$x''(t)=-2\sin^2(t)\cos(t).$$
However, $(-\sin^2(t))'=-2\cos(t)\sin(t)$, which is not quite what we want. I'm confused why this does not line up. I suppose because the velocity field does not arise from a curve. 
How can this be fixed up?

Recall that $$ 0=E_1(E_j,E_k)= \frac{1}{y^2} \{ \Gamma_{1j}^k +
\Gamma_{1k}^j \} \Rightarrow \Gamma_{1j}^k =-\Gamma_{1k}^j  $$
$$ -2y^{-3}= E_2(E_2,E_2)= \frac{1}{y^2}
(2\Gamma_{22}^2) \Rightarrow \Gamma_{22}^2= - \frac{1}{y} $$
By definition,
$$ \Gamma_{12}^1 = -\frac{1}{y}$$
If $$ v= (-\sin^2 t, \sin\ t\cos\ t) :=fE_1+ g E_2$$ then $$ v (F(t))= y \frac{d}{dt} F(t) $$
and
$$
\nabla_vv = \nabla_v (fE_1) + \nabla_v (gE_2) $$ $$= v(f) E_1+
v(g)E_2 + f\nabla_vE_1 + g\nabla_vE_2 $$ $$ = y(f',g') + f ( f
\Gamma_{11}^i E_i + g \Gamma_{21}^i E_i) + g(  f \Gamma_{12}^i E_i +
g \Gamma_{22}^i E_i )
$$
$$ = y(f',g') + f^2 \Gamma_{11}^2E_2
+2f g \Gamma_{12}^1 E_1 + g^2  \Gamma_{22}^2 E_2 =0
$$
