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?

You shouldn't assume that the naive parametrization $(\text{cos}(t), \text{sin}(t))$ will solve the equation. It doesn't, as it clearly doesn't satisfy even the first differential equation. However, if you assume a more general form $x(t) = \text{cos}(f(t))$, $y(t)= \text{sin}(f(t))$, which still traces out a semicircle (but with a different speed), then it should be possible to find an $f(t)$ namely a reparametrization which makes the geodesic equations satisfied. If you plug in the above ansatz, both equations tell you that you need to solve $$f''(t) = \text{cot}(f(t)) (f'(t))^2,$$ which is solved by
$$f(t) = 2\text{cot}^{-1}(e^t)$$
