# Parametrization of hyperbolic geodesic line on sphere

A curve in $\mathbb R^3$ starts at equator of sphere radius $a$ being inclined at $\alpha$ to longitude goes to North pole along a hyperbolic geodesic.

Find its radius $r(\theta)$ as function of longitude $\theta$ in cylindrical/polar coordinates.

EDITS 1-3:

Considering differentials on a sphere meridian (arc of curve makes angle $\psi$ to longitude or sphere meridian) we have

$$\sin \phi = \frac{ dr \tan \psi } { r d\theta};$$

on the sphere of DE ($\phi$ is slope angle between arc and symmetry axis)

$$\cos \phi = r/a ;$$

but to proceed in finding parametric equations of the curve I am stuck not knowing dependence of $\psi$ on $r$... or $\theta$

Normally we have geodesics on a sphere as great circles, hyperbolic geodesics on a Beltrami pseudosphere as asymptotic lines of zero normal curvature.

But when seen interchangeably in the nature of geodesic, is it not possible making sense of it to parametrically define hyperbolic geodesics on a Riemann sphere and common geodesics on a pseudosphere?

At least we know geodesics cannot cross cuspidal equator of a Beltrami pseudosphere.

Now what about any restrictions that may exist in the present ( hyperbolic geodesics on sphere) case?

If not possible to define it, I appreciate your help to comment on how it fails to make enough sense. Regards

• You've left out too many words for this to make sense. For example: a unit sphere does not have any "hyperbolic geodesics", unless you have more words to explain what you mean by a "unit sphere" and a "hyperbolic geodesic". – Lee Mosher Mar 2 '18 at 16:16
• I think line curvture is determined by chosen geometry. Is the simpler question alright now? – Narasimham Mar 2 '18 at 18:36
• Are you talking about a rhumb line, a path that always makes the same angle to meridians? What “radius” are you talking about in your second paragraph? – Lubin Mar 3 '18 at 4:12
• I mentioned cylindrical coordinates in the same line... where radius $r$ and z- coordinate satisfy $r^2+z^2 = a^2$ – Narasimham Mar 3 '18 at 6:16