A particle travels on a unit sphere from a point $(z, \theta) = (z_0, 0)$ to $(z_0, \theta_0)$ in cylindrical coordinates, so z0 = const. Find the distance travelled and compare with the "great circle" (G.C.) distance between those points.
My solution:
a. $Distance = r \theta = \sqrt{(1 - z_0^2)} \theta_0$, where r is the radius of a circle, cut from sphere by $z = z_0$.
b. The great circle arc distance can be found by a dot product of vectors to the beginning and the end of the journey:
$r_0 = (\sqrt{(1 - z_0^2)}, 0, z_0)$
$r1 = (\sqrt {(1 - z_0^2)} \cos \theta_0, \sqrt {(1 - z_0^2)} \sin \theta_0, z0)$
$G.C. Distance = \arccos (r_0 r_1) = \arccos((1-z_0^2) \cos \theta_0 + z_0^2) $
My question: how to show (from this calculation that $Distance >= G.C. Distance$. I can do it from geometric considerations, but would like to derive it analytically from the formulas above (or any alternative formulas that can be derived).
You want to show that $$ \arccos[(1-z_0^2)\cos\theta_0+z_0^2]\le\theta_0\sqrt{1-z_0^2}, $$ for every $\theta_0\in[0,\pi]$. As the cosine is decreasing on $[0,\pi]$, that is the same as proving that $$ (1-z_0^2)\cos\theta_0+z_0^2\ge\cos\left(\theta_0\sqrt{1-z_0^2}\right). $$ If we define $$ f(\theta)=(1-z_0^2)\cos\theta+z_0^2-\cos\left(\theta\sqrt{1-z_0^2}\right) $$ we must then prove that $f(\theta)\ge0$ for $\theta\in[0,\pi]$.
Notice that $f(0)=0$ and $f'(0)=0$, but on the other hand $$ f''(\theta)=(1-z_0^2)\left[\cos\left(\theta\sqrt{1-z_0^2}\right)-\cos\theta\right]\ge0 $$ for $\theta\in[0,\pi]$. It follows that $f'(\theta)$ is increasing, that is $f'(\theta)\ge0$ for $\theta\in[0,\pi]$, which in turn implies that $f(\theta)$ is also increasing, that is $f(\theta)\ge0$ for $\theta\in[0,\pi]$, as we were to prove.