Frederica's Cycloidal Film

For her photography show, Frederica Fiddleria attaches a light to a point on the circumference of a circular wheel with a radius of 1 meter. She points a camera at the wheel and, during a single long exposure, rolls the wheel for one revolution along the ground.

When she develops the film, she is curious about the path the light took as the wheel rolled. What is the length of this path?

Lo, and behold! Frederica's light source is traveling along a cycloidal path, which has some fairly well defined properties, but let's try to derive them from scratch. Let's assume that the center of the wheel starts at the point $(0,1).$ Since the circumference of the wheel is $2\pi,$ after one full revolution of the wheel, the center would end up at the point $(2\pi, 1).$ Without loss of generality, let's assume that the light was added to a point $P$ on the circumference of the circle such that the angle between $P$, the center of the circle and the positive $y$-axis is $\theta \in [0,\pi].$ In this case, we see that the point of light is at $x(0) = \sin \theta,$ and $y(0) = \cos \theta.$ After rotating for some time $t \in [0,2\pi),$ the center of the circle will be at the point $(t,1),$ while the angle that the point of light now makes as measured with respect to the positive $y$-axis, will now be $\theta + t.$ So the point of light will be given by the parametric equations \begin{align*} x(t) &= t + \sin(\theta + t),\\ y(t) & = 1 + \cos (\theta + t).\end{align*} In particular, taking derivatives with respect to time, we get \begin{align*} \frac{dx}{dt} &= 1 + \cos (\theta + t) \\ \frac{dy}{dt} &= - \sin (\theta + t).\end{align*}

Therefore, integrating the arclength formula and appealing to the half-angle formula, we see that no matter what the value of $\theta$ is, the length of the path the light took is given by \begin{align*}\ell &= \int_0^{2\pi} \sqrt{ \left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 } \,dt \\ &= \int_0^{2\pi} \sqrt{ \left(1 + \cos (\theta + t)\right)^2 + \left( - \sin (\theta + t) \right)^2 } \,dt \\ &= \int_0^{2\pi} \sqrt{ 2 + 2 \cos (\theta + t) } \,dt \\ &= 2 \int_0^{2\pi} \sqrt{ \frac{ 1 + \cos (\theta + t) }{2} } \,dt \\ &= 2 \int_0^{2\pi} \left|\cos \left( \frac{\theta + t}{2}\right)\right| \,dt \\ &= 2 \int_0^{\pi - \theta} \cos \left(\frac{\theta + t}{2}\right) \,dt + \int_{\pi-\theta}^{2\pi} - \cos \left(\frac{\theta + t}{2}\right) \,dt \\ &= 4 \left. \sin \left(\frac{\theta + t}{2}\right) \right|_0^{\pi - \theta} - 4 \left. \sin \left(\frac{\theta + t}{2}\right) \right|_{\pi - \theta}^{2\pi} \\ &= 4 \left( \sin \frac{\pi}{2} - \sin \frac{\theta}{2} \right) - 4 \left( \sin \left(\pi + \frac{\theta}{2}\right) - \sin \frac{\pi}{2} \right) \\ &= 4 + 4 \sin \frac{\theta}{2} + 4 - 4 \sin \frac{\theta}{2} = 8\end{align*}