The following simulation try to illustrate the properties of cyclodial curves.

\theta=\omega t, x=r(\theta-\sin\theta), y=r(1+\cos\theta)
so x'=r(1-\cos\theta), y'=r\sin\theta
From conservation of energy: \tfrac{1}{2}mv^2=mgy, v=\tfrac{ds}{dt}=\sqrt{2gy}
dt=\frac{ds}{v}=\frac{\sqrt{dx^2+dy^2}d\theta}{\sqrt{2gy}}=\frac{\sqrt{2r^2(1-\cos\theta)}}{\sqrt{2gr(1-\cos\theta)}}=\sqrt{\frac{r}{g}}d\theta, so T_{1/4}=\sqrt{\frac{r}{g}}\pi

However, from an intermediate point \theta_0, v=\frac{ds}{dt}=\sqrt{2g(y-y_0)}
So T=\int_{\theta_0}^{\pi}\sqrt{\frac{2r^2(1-\cos\theta)}{2gr(\cos\theta_0-\cos\theta)}}d\theta

with \sin\frac{x}{2}=\sqrt{\frac{1-\cos x}{2}}, \cos\frac{x}{2}=\sqrt{\frac{1-\cos x}{2}}

T=\sqrt{r}{g}\int {\theta_0}^{\pi} \sqrt{\frac{\sin\tfrac{\theta}{2}d\theta}{\cos^2\tfrac{\theta_0}{2}-\cos^2\tfrac{\theta}{2}}}
define u=\frac{\cos\tfrac{\theta}{2}}{\cos\tfrac{\theta_0}{2}}, du=\frac{\sin\frac{\theta}{2}d\theta}{2\cos\tfrac{\theta_0}{2}}

T=\sqrt{r}{g}\int_1^0 \frac{du}{\sqrt{1-u^2}}=2\sqrt{r}{g}\sin^{-1}u|_0^1=\sqrt{r}{g} \pi
So the amount of time is the same from any point.

1. A cycloid is the curve defined by the path of a point on the edge of circular wheel as the wheel rolls along a straight line.
2. Cycloidal curves is the curve of fastest descent under gravity
3. the period of a ball rolling back and forth inside this curve does not depend on the ballís starting position.

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