Can we solve for $x(t)$ and $y(t)$ in closed-form with respect to time? I would like to ask a question I was discussing the other day with a friend of mine. Suppose you have a point mass m, sliding on the friction-free curve $y = e^{-x}$ starting from position $x(0) = 0$ and $y(0) = 1$ with zero initial velocity. Can we find in closed-form the exact position $(x,y)$ with respect to time?
Going down the usual path, from conservation of energy I found that $v = \sqrt{2g(1-y)}$. Assuming $φ$ is the angle between $\vec{v}$ and the downward vertical direction (the direction of negative ordinates), and using that $\dot{x} = v \sin(φ)$, $\dot{y} = v \cos(φ)$, $φ = \tan^{-1}(e^{x})$, I reached
$$\dot{x} = \sqrt{2g}\sqrt{\frac{e^{2x}-e^{x}}{e^{2x}+1}}$$
$$\dot{y} = y\sqrt{2g}\sqrt{\frac{1-y}{1+y^2}}$$
From these two, I need to solve only one. Choosing the first, separating $dx$ and $dy$ and integrating yields:
$$\int_0^{x(t)}\sqrt{\frac{e^{2x} + 1}{e^{2x} - e^{x} }}dx = \sqrt{2g}t$$
However, plugging the integral in the LHS in Mathematica doesn't compute anything, most likely meaning that this integral is not solvable in closed form, let alone invertible in closed-form.
Choosing now the second equation and doing the same yields:
$$\int_1^{y(t)}\sqrt{\frac{1+y^2}{1 - y}}\frac{dy}{y} = - \sqrt{2g}t$$
Again, trying to compute the integral in the LHS, gives an assortment of complex elliptic integrals that are gain not invertible in some closed-form. I find it very strange that the only method I could find involved integration and then inversion of the result. At best, shouldn't there be a way to acquire the solution at least in some integral form?
 A: I am fairly certain it is not possible. 
I used the convention opposite to yours for angle (i.e. $\theta= tan^{-1}(e^{-x})$) which yields the following integral for x
$$
\int_{0}^{x(t)} dx' \frac{e^{x'}}{1-e^{-x'}} \sqrt{(1-e^{-x'})(1+e^{-2x})} = \frac{t}{\sqrt{2g}}
$$
Then, substituting $u=e^{-x'}$ one gets
$$
-\int_{1}^{\ln(x)} du \frac{e^{x'}}{1-u} \sqrt{1-u+u^2-u^3} = \frac{t}{\sqrt{2g}}
$$
This integral does not appear to have a closed form solution (at least according to mathematica).
The integral does have the indefinite value of 
$$\frac{2 \sqrt{-u^3+u^2-u+1} \left(\left(u^3+u^2+u+1\right) \sqrt{1-u}+2 \sqrt{-1-i} \sqrt{\frac{u-i}{u-1}} \sqrt{\frac{u+i}{u-1}} (u-1)^2 F\left(\left.i \sinh ^{-1}\left(\frac{\sqrt{-1-i}}{\sqrt{1-u}}\right)\right|-i\right)+2 i \sqrt{-1-i} \sqrt{\frac{u-i}{u-1}} \sqrt{\frac{u+i}{u-1}} (u-1)^2 E\left(\left.i \sinh ^{-1}\left(\frac{\sqrt{-1-i}}{\sqrt{1-u}}\right)\right|-i\right)\right)}{3 (1-u)^{3/2} \left(u^2+1\right)}$$
Which seems unlikely to give you a closed form answer for x(t) or y(t)
Hope this helps!
