Consider the hyperbolic space as the upper half space of $\mathbb{R}^3$ $$\mathbb{H}^3=\{(x,y,z)\in\mathbb{R}^3:z>0\}$$ equipped with the metric $$ds^2=\frac{1}{z^2}(dx^2+dy^2+dz^2)$$ Prove that straight lines perpendicular to the plane $z=0$ are geodesics if they are arc-length parameterized.
Of course, such lines can be parameterized by $$\alpha(t)=(x,y,t)$$ where $(x,y)$ is an arbitrary fixed point of the plane. But, even though this is arc-length parameterized under the usual metric, it's not under the defined metric, and indeed it's not geodesic.
My problem is conceptually serious too, as I don't know how to even check it. For example, assume the parameterization given by $$\alpha(t)=(x,y,t^2+t^3)$$ Then we would have $$\alpha'(t)=(0,0,2t+3t^2)$$ Then the module is given by $$|\alpha'(t)|^2=\frac{1}{z^2}(2t+3t^2)^2$$ Now is this arc-length parameterized? I don't know, since different variables appear.
To arc-length parametrize you may integrate: $ds = dz/z$ to get $s-s_0=\pm \log(z/z_0)$ or $z= z_0 \exp( \pm (s-s_0))$.
In order to see that a vertical segment minimizes a distance (i.e. is a geodesic) consider a $C^1$ path $t\in [a,b]\mapsto \gamma(t)=(x(t),y(t),z(t))$ from $(x,y,z_0)$ to $(x,y,z_1)$ with $0<z_0<z_1$. Since $ds\geq dz/z$, with equality iff $x(t),y(t)$ are constant, you deduce that the projection of the path to the vertical segment gives the minimal length (and provided $z_0\leq z(t)\leq z_1$ for $t\in [a,b]$).