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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.)

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    @Michael: I guess that's fair and I'm just not feeling charitable today...2011-11-29

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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?

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    It's true that for this problem you don't need the fundament theorem of primitives, because the function being integrated is constant. That's as trivial as an integral can be. But you still need something other than purely discrete math in order to see that it's constant.2017-06-05
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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.

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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.