Length of loxodrome On a sphere with radius $R$, find the length of a loxodrome which starts at the equator and makes an angle $\gamma$ with all the meridians.
(No equations for such a loxodrome are given, and should be derived.)
 A: It can be done without calculus. Here is a hint: Consider two latitude circles at latitudes $\theta$ and $\theta+\Delta\theta$ with $0<\Delta\theta \ll 1$. How long is a piece of meridian between these two circles, and how long is a piece of your loxodrome between these two circles?
A: It looks to me as if no "equation" of the loxodrome is needed.
Suppose $d\theta$ is an infinitely small increment of latitude $\theta$.  Going from a point at latitude $\theta$ to a point straight north of it at latitude $\theta+d\theta$ means going northward by a distance $R\;d\theta$.  Now suppose we are heading $\gamma$ east of north.  Thus we go northward by $R\;d\theta$ (the "adjacent" side of a right triangle) and eastward by $R\tan\gamma\;d\theta$ the ("opposite" side), covering a distance of $ds=R\sec\gamma\;d\theta$, the length of the hypotenuse (I'm using "$\sec = \mathrm{hyp}/\mathrm{adj}$").
Then the total length of the loxodrome, from the south pole to the north pole, is
$$
\int_{\theta=-\pi/2}^{\theta=\pi/2} ds = \int_{-\pi/2}^{\pi/2} R\sec\gamma\;d\theta.
$$
The quantity $R\sec\gamma$ is a constant, i.e. it does not change as $\theta$ changes, so this is
$$
R\sec\gamma \int_{-\pi/2}^{\pi/2}\;d\theta.
$$
That's a trivial integral.
A: Given the change in latitude is $dt$ for a loxodrome the corresponding change in longitude will definitely be $\tan(g)\,dt$.  However, the longitudinal contribution to arclength depends on the latitude -- circles of constant latitude are smaller near the poles by $\cos(t)$.
So, the actual arclength would be given by $ds^2 = R^2\big(1+\tan^2(g)\cos^2(t)\big)dt^2$.
The corresponding integral is elliptic which, I am afraid, has no elementary expression.
