Calculating the "time down" different curves (Brachiostone problem)? I am interested in calculating the time taken to "go down" a variety of curves, such as quadratics and exponential curves etc. However, I am unsure of how to calculate these values. Off the internet, I have found formulae for the cycloid (The Brachiostone Problem) but I am not interested in this type of curve.
I will demonstrate what I am intending to find for the curve which I know how to solve; the linear line going from $(100,100)$ to the origin $(0,0)$.

As acceleration will be constant, we can use:
$$ mg\Delta h = \frac{mv^2}{2} $$
$$ v^2=2g\Delta h $$
$$ v = \sqrt{2g\Delta h} $$
As $ \Delta h = 100$:
$$v=\sqrt{200g}=44.3\space m\space s^{-1}$$
And as $time=\frac{s}{v}$:
$$T=\frac{\sqrt{100^2+100^2}}{44.3}=3.19\space s$$
Therefore, the question becomes: How do I solve this problem for different types of curves, i.e. when acceleration is not constant?
 A: time=distance /velocity.
$\frac{1}{2}mv^2=mg(y-y_0)\implies v=\sqrt{2g(y-y_0)}$
$\tau=\int \frac{ds/dx}{\sqrt{2g|y-y_0|}}dx= \int_{x_1}^{x_2} \frac{\sqrt{1+y'^2} dx}{\sqrt{2g|y(x)-y_0|}}$
Suppose you have a circular ramp, $(x-r)^2+(y-r)^2=r^2$ and a block slides from $(0,r)$ without friction. How long does it take to reach the bottom?
$2(x-r)+2y'(y-r)=0\implies y'=\frac{-(x-r)}{(y-r)}\implies y'^2=\frac{(x-r)^2}{(y-r)^2}$
$\sqrt{1+y'^2}=\sqrt{\frac{r^2}{(y-r)^2}}=\frac{r}{|y-r|}$
$\tau=\int_0^r\frac{rdx}{\sqrt{2g}|y-r|^{3/2}}$
Let $x=r-r\cos{\theta},y=r-r\sin{\theta}$
$\tau=\int_0^{\pi/2} \frac{r^2\sin{\theta}d\theta}{\sqrt{2g}\cdot r^{3/2}\sin^{3/2}{(\theta)}}=\sqrt{\frac{r}{2g}}\int_0^{\pi/2}\frac{d\theta}{\sqrt{\sin{\theta}}}\approx\sqrt{\frac{r}{2g}}(2.62)$ according to wolfram alpha: https://www.wolframalpha.com/input/?i=integrate%5B1%2Fsqrt%28sin%28x%29%29%5D
To answer your question more directly, you need some way of measuring distance along the curve which is given by $\sqrt{1+(dy/dx)^2}dx$ and you need to divide that by the instantaneous velocity which because of the conservation of energy can be given by $\sqrt{2g(y-y_0)}$ essentially the same as your $v=\sqrt{2g\Delta h}$ above.
$ds=\sqrt{1+y'^2}dx$. $v=ds/dt$, so $dt=ds/v$. The you have the effect of non-constant acceleration indirectly by keeping track of how the velocity changes. To get a sensible integral, you'll need to re-express everything in terms of the same variable. I introduced $\theta$ as a parameter to simplify the integral which I don't think can be solved with elementary functions.
