I have a problem in solving mathematical problem.
Take a ball with radius 60 cm. A creature walk from the southpole to northpole by following the spiral curve that goes once around the ball every time the creature has walked the distance that corresponds 5 cm rise on the diameter of the sphere. How long is the trip that creature walks?
I had problems even to find the parametrization of the curve.
Let $r = 60{\rm cm}$ and $\ell = 5{\rm cm}$. Express everything in polar coorindates
$$(x,y,z) = r(\sin\theta\cos\phi, \sin\theta\sin\phi, \cos\theta)$$ The condition
the spiral curve that goes once around the ball every time the creature has walked the distance that corresponds 5 cm rise on the diagonal of the sphere
translates to $$ | \phi(t) - \phi(0) | = \frac{2\pi r(1 + \cos\theta(t))}{\ell} = 24\pi (1 + \cos\theta(t)) \quad\implies\quad \left|\frac{d\phi}{d\theta}\right| = 24\pi \sin\theta $$ The metric on the sphere is given by
$$ds^2 = r^2 ( d\theta^2 + \sin\theta^2 d\phi^2 )$$ This means the total distance traveled is given by:
$$r \int_0^\pi \sqrt{1 + \left( \sin\theta \frac{d\phi}{d\theta}\right)^2} d\theta = r \int_0^\pi \sqrt{ 1 + ( 24\pi )^2 ( \sin\theta )^4 } d\theta $$
I am unable to evaluate this integral analytically. Numerically, it is about $7123.16587{\rm cm}$.
I assume that the radius of the ball is $1$. Choose the geographical longitude $\phi$ as parameter and assume that the creature crosses the equator ($\theta=0$) at the point $(1,0,0)$. Then its path has a parametrization of the form $$\eqalign{x(\phi)&=\cos\phi\cos\theta(\phi) \cr y(\phi)&=\sin\phi\cos\theta(\phi)\cr z(\phi)&=\sin\theta(\phi)\cr}\tag{1}$$ with an unknown increasing function $\phi\mapsto\theta(\phi)$, and $\theta(0)=0$. Denote by $\phi\mapsto s(\phi)$ the arc length along this curve, with $s(0)=0$. From $(1)$ we obtain $$s'^2(\phi)=x'^2(\phi)+y'^2(\phi)+z'^2(\phi)=\cos^2\theta(\phi)+\theta'^2(\phi)\ .\tag{2}$$ So far we have not yet used the decisive information about the nature of this curve. The simplest interpretation of the text would be that $$z'(\phi)\equiv p:={1\over24\pi}$$ which means that $z$ increases linearly with $\phi$ such to $\Delta\phi=2\pi$ corresponds $\Delta z={1\over12}$. This gives $$\sin\theta(\phi)=z(\phi)=p\phi\qquad\biggl(0\leq\phi\leq{1\over p}\biggr)$$ and similarly $$\cos\theta(\phi)\theta'(\phi)=p\ .$$ Plugging this into $(2)$ we then obtain $$s'^2(\phi)=1-p^2\phi^2+{p^2\over 1-p^2\phi^2}\ .$$ Now we should take the square root and integrate from $0$ to ${1\over p}$; but I'm afraid the resulting integral is not elementary. Numerical integration gives $$\int_0^{1/p}s'(\phi)\>d\phi\doteq 59.3597\ ,$$ which leads to a total length $$L\doteq7123.17\ {\rm cm}$$ (from south pole to north pole) when the radius of the sphere is $60$ cm.
