How to write the coordinates of a point on a parabola as a function of the distance traveled on the parabola to reach that point? Assume the parabola $y=kx^2$ in the 2D Cartesian coordinate system.
If I walk a distance $s$ on that curve from origin, what will my coordinates be?
It is easy to find the distance traveled on the parabola if I know my $x$ coordinate at the destination. But, here the reverse is needed.
It is evident that immediately after finding $x(s)$, $y(s)$ can be found using $y=kx^2$. So the goal is to find $x(s)$.
$$ds=\sqrt{(f'(x)dx)^2+dx^2}=\sqrt{1+(f'(x))^2}\ dx=\sqrt{1+4k^2x^2}\ dx$$
So we are facing the boundary value problem $ds=\sqrt{1+4k^2x^2}\ dx$, $x(0)=0$, $x(s)=?\,$.
This is separable. After integrating, we have:
$$s=\int_0^x\sqrt{1+4k^2X^2}\ dX$$
To calculate the integral, we use the known formula:
$$\int\sqrt{1+x^2}\ dx=\frac12x\sqrt{1+x^2}+\frac12\ln|x+\sqrt{1+x^2}|+\mathbb{constant}$$
and the substitution $x=2kX$, so $dX=\frac{dx}{2k}$. Finally, assuming $x\ge0,$
$$kx\sqrt{1+4k^2x^2}+\frac12\ln(2kx+\sqrt{1+4k^2x^2})=2ks$$
Up to here, the formal way was used. From now on, These are my own ideas (which may not be the best methods).
Using the substitution $2kx=\tan\theta$, we can go further simplifying the relation:
$$\tan\theta\sec\theta+\ln(\tan\theta+\sec\theta)=4ks$$
The goal was to find $x(s)$ which is equivalent to finding $\theta(x)$. But, I am stuck here.
Mathematica says it can't solve it with available methods. Can anyone help me?

EDIT:
I could achieve more simplifications. Denoting $\tan\theta$ by $T$ we can have a better look...

$$T\sqrt{1+T^2}+\ln(T+\sqrt{1+T^2})=4ks$$
That $\ln(T+\sqrt{1+T^2})$ is saying "write me in terms of hyperbolic functions!"
$$T\sqrt{1+T^2}+\sinh^{-1}(T)=4ks$$
This means there exists a $z$ such that $T=\sinh(z)$. Rewriting anything so that we have only $z$:
$$\sinh(z)\sqrt{1+\sinh(z)^2}+z=4ks$$
$$\sinh(z)\cosh(z)+z=4ks$$
$$\frac12\sinh(2z)+z=4ks$$
$$\sinh(2z)+2z=8ks$$
Now say $\psi=2z$, we have:
$$\psi+\sinh(\psi)=8ks$$
This implies we are very close to an explicit expression of $\psi(s)$!
But I again have to leave the rest up to you...
 A: EDIT 1/2
In 3D it is easier,  arc length $s$ does not change even when cone is developed. For this purpose semi vertical cone angle $\alpha $ is brought in:
$$ r= \tan \,\alpha z $$
$$ z{\prime } = \cot \alpha\, r{\prime}  $$
$$ \dfrac{ds}{d \theta} = \sqrt{ r^2+ r^{\prime^2}+z^{\prime ^2} } $$
which can be integrated in closed form using relations for polar forms:
$$ r= \dfrac{a}{1-\cos \theta}  $$
$$ r{\prime }= \dfrac{a\sin \theta}{(1-\cos \theta)^2}  $$
$$ \theta_{dev} = \theta \sin \alpha $$
$$ r_{dev}  = \rho = r/ \sin \alpha $$
$ \theta$  is projected angle $ \rho$ true radius in development.
EDIT 3
2- dimensional straight-forward calculation:
$$ x = 2 f t, y= f t^2, \,  \frac{dy}{dx} = t = \tan \phi \tag1 $$
$$   y= \frac{x^2}{4f},\,  y^{\prime \prime} = \frac{1}{2 f }, \quad \kappa = \frac{1/2f}{(1+\tan ^2 \phi)^{3/2}} \tag2 $$ 
$$ 2f \kappa = 2f \frac{ d\phi}{ds}= \cos^3\phi \tag3 $$
$$ ds = 2f \frac{d \phi} {\cos ^3 \phi} \tag4 $$
$$ \frac{s}{ 2f}= \int \sec^3 \phi \, d \phi \tag5  $$
$$\frac{s}{ f} =  log\, \frac{1+\tan \phi/2}{1-\tan \phi/2} + \sec\phi \tan \phi \tag6 $$ 
as arbitrary constant zero at parabola vertex. This can be expressed in terms of $x,t, x = 2 f \tan \phi, $ hyperbolic functions etc., please cross verify.
