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Length functional. On the unit sphere, write the position vector in spherical coordinates asIts two coordinate tangent vectors satisfy $r_{\theta }\cdot r_{\varphi }=0$, $|r_{\theta }|=1$ and $|r_{\varphi }|=\sin \theta$. Consequently the induced Riemannian metric is $d{s}^2=d{\theta }^2+{\sin }^2\theta d{\varphi }^2$. On a segment for which $\theta$ increases and $\varphi =\varphi (\theta )$,Segments where $\theta$ is not monotone require another parameter or a subdivision. Endpoint turning points can be handled by a limit.

First integral. With $F=\sqrt{1+{\sin }^2\theta ({\varphi }^{\prime }{)}^2}$, there is no explicit $\varphi$ dependence. The Euler-Lagrange equation therefore says thatis constant. For $c\ne 0$, put $k=1/|c|$ and solve:The radicand must be nonnegative along the chosen segment. The printed expression with both limits fixed at ${\theta }_1,{\theta }_2$ describes a longitude difference between endpoints; a formula for $\varphi (\theta )$ needs the variable upper limit and the additive constant shown here. The case $c=0$ gives a meridian with constant longitude. The equator is also a stationary route, although it cannot be parameterized using changing $\theta$.

Integration and geometric meaning. Put $t=\cot \theta$, so $d\theta =-dt/(1+t^2)$ andFor a nonconstant segment $k>1$, and writing $K=\sqrt{k^2-1}$, the integral becomesRearranging, and absorbing phase and sign into two constants, givesMultiplying by $\sin \theta$ shows that the corresponding Cartesian coordinates satisfy $z=Ax+By$. Thus every such stationary route lies in a plane through the sphere's center: it is a great circle, the geodesic of the sphere. The meridians and equator are the limiting or separately parameterized cases of the same description.

Rome to Singapore. Use exactly the approximate coordinates supplied in the question: $(\theta ,\varphi )=(\pi /4,\pi /12)$ at departure and $(\pi /2,7\pi /12)$ at arrival. Inserting them into the preceding relation determinesIndeed the right side is one at Rome and zero at Singapore. This corresponds to $K=1$, $k=\sqrt{2}$. In latitude $\ell =\pi /2-\theta$, the route is $\ell =\arctan [\cos (\varphi -\pi /12)]$: it starts at $45^{\circ }$ north with an initially eastward heading and then curves southward to the equator. The first-integral expression in $\theta$ has an integrable endpoint singularity at departure, where $\theta$ has a turning point; the longitude parameter is regular there.

For an explicit regular arc-length parametrization, let $a=r(\pi /4,\pi /12)$ and $b=r(\pi /2,7\pi /12)$. Their dot product is $\sin (\pi /4)\cos (\pi /2)=0$. HenceBoth $|r(s)|$ and $|r^{\prime }(s)|$ equal one. The shorter great circle arc is the global shortest path: for any unit-speed curve from $a$, its central angle $\alpha =\arccos (a\cdot r)$ satisfies $|d\alpha /ds|\le 1$ by the Cauchy-Schwarz inequality, since the tangential projection of $a$ has length $\sin \alpha$. Integrating gives length at least the endpoint angle $\pi /2$, which this route attains. For a sphere of radius $R$, the corresponding shortest distance is $\pi R/2$.