Question: 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.

Answer: 
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.