Frederica's Trochoidal Film

For her next show, Frederica wants to mix things up. Instead of placing the light on the circumference of the wheel, she will pick a random point inside the circle. (Before you ask, let me clarify what “random” means here: Any two regions with the same area are equally likely to contain the point.)

As before, she will roll the wheel for one revolution along the ground and capture the motion with a single long exposure on her camera. On average, what can she expect the length of the path to be?

As before we will compute for any given starting point $$(x_0, y_0) \in \{ (x,y) \mid \sqrt{x^2 + (y-1)^2} \leq 1 \}$$ and calculate the path that it traces out as the wheel makes a full rotation. If the point $(x_0,y_0)$ is on the circumference of the circle, the path is known as a cycloid; however, if $(x_0,y_0)$ is in the interior of the wheel it is known as a curtate trochoid. Anywhoozle, let's again define our parametric equations ...

Let's assume that the point of light is at $x(0) = x_0, y(0) = y_0$, where without loss of generality let's assume that $x_0 \gt 0,$ and of course $r = \sqrt{x_0^2 + (y_0-1)^2} \leq 1.$ Then we see that we can write $x(0) = r \sin \theta, y(0) = 1 + r \cos \theta,$ for $\theta = \tan^{-1} \frac{x_0}{y_0 - 1} \in [0, \pi].$ Since after a rotation of $t \in [0,2\pi)$ we will have the wheel centered at $(t,1),$ 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 at time $t$ the point of light would be at \begin{align*} x(t) &= t + r \sin (\theta + t) \\ y(t) &= 1 + r \cos (\theta + t).\end{align*} Differentiating with respect to time we get \begin{align*} \frac{dx}{dt} &= 1 + r \cos (\theta + t) \\ \frac{dy}{dt} &= -r \sin (\theta + t), \end{align*} so we see that the arclength of the path the light takes is \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 + r \cos (\theta + t)\right)^2 + \left( - r \sin (\theta + t) \right)^2 } \,dt \\ &= \int_0^{2\pi} \sqrt{ 1 + r^2 + 2r \cos (\theta + t) } \,dt.\end{align*} We can easily set $u = \theta + t$ and then through periodicity recognize that we have \begin{align*}\ell = \ell(r) &= \int_0^{2\pi} \sqrt{ 1 + r^2 + 2r \cos(\theta+t) } \,dt\\ &= \int_\theta^{\theta + 2\pi} \sqrt{ 1 + r^2 + 2r \cos u} \,du\\& = \int_0^{2\pi} \sqrt{ 1 + r^2 + 2r \cos u } \,du,\end{align*} regardless of the the value of $\theta.$ Now at this point, if we wanted to we could do some more algebra and appeal to the complete elliptic integrals of the second kind to find that $$\ell(r) = 4(1+r)E\left(\frac{2 \sqrt{r}}{1+r}\right),$$ where $$E(k) = \int_0^{\pi/2} \sqrt{1-k^2\sin^2t}\,dt;$$ however, since we would still need to plug this into another integral, this doesn't really do us any good.

However, we can us the integral definition of $\ell(r)$ and the definition of the randomly chosen starting point to get an expression for the expected length of the path. If we want to get the probability that some starting point $(x_0,y_0)$ is in the annulus centered at $(0,1)$ between radii $r$ and $r + dr,$ we get that the area of the annulus is $2\pi r \,dr,$ while the total probability of the unit disk centered at $(0,1)$ is $\pi,$ then we see that the conditional probability of the starting point being in the annulus between radii $r$ and $r + dr$ is $2r\, dr.$ Therefore, we get from the law of total expectation that $$\mathcal{L} = \mathbb{E} \left[ \ell \right] = \int_0^1 \ell(r) 2r \,dr.$$ Plugging back in the integral form of $\ell(r),$ we get $$\mathcal{L} = \int_0^1 \ell(r) 2r\,dr = \int_0^1 \int_0^{2\pi} 2r \sqrt{ 1 + r^2 + 2r \cos u } \,du \,dr,$$ which we can also recognize as the polar area integral of the function $$f(x,y) = 2\sqrt{ (1-x)^2 + y^2 }$$ over the unit disk. That is, $$\mathcal{L} = \int_{-1}^1 \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} 2\sqrt{(1-x)^2 + y^2} \,dy \,dx.$$

Since $$\int \sqrt{a^2 + t^2} \,dt = \frac{t}{2} \sqrt{a^2 + t^2} + \frac{a^2}{2} \ln \left| t + \sqrt{a^2 + t^2} \right| + C,$$ we can do this integration directly. We first get that \begin{align*}g(x) &= \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} 2 \sqrt{ (1-x)^2 + y^2} \,dy \\ &= 4 \int_0^{\sqrt{1-x^2}} \sqrt{ (1-x)^2 + y^2} \,dy \\ &= 4 \left[ \frac{y}{2} \sqrt{ (1-x)^2 + y^2 } + \frac{(1-x)^2}{2} \ln \left| y + \sqrt{ (1-x)^2 + y^2 } \right| \right]_0^{\sqrt{1-x^2}} \\ &= 4 \left[ \frac{\sqrt{1-x^2} \sqrt{2 - 2x}}{2} + \frac{(1-x)^2}{2} \ln \left| \sqrt{1-x^2} + \sqrt{2-2x} \right| - \frac{(1-x)^2}{2} \ln |1-x| \right] \\ &= 2\sqrt{2} (1-x) \sqrt{1+x} + 2(1-x)^2 \ln \left( \frac{\sqrt{1+x} + \sqrt{2}}{\sqrt{1-x}} \right).\end{align*} So we now have \begin{align*}\mathcal{L} &= \int_{-1}^1 g(x) \,dx \\ &= \int_{-1}^1 2\sqrt{2} (1-x)\sqrt{1+x} \,dx + \int_{-1}^1 2(1-x)^2 \ln \left( \frac{\sqrt{1+x} + \sqrt{2}}{\sqrt{1-x}} \right) \,dx.\end{align*} Using integration by parts, we have first integral is \begin{align*}I_1 &= \int_{-1}^1 2\sqrt{2}(1-x) \sqrt{1+x} \,dx \\ &= 2 \sqrt{2} \left( \left[ \frac{2}{3} (1-x) (1+x)^{3/2} \right]_{-1}^1 + \frac{2}{3} \int_{-1}^1 (1+x)^{3/2} \,dx \right) \\ &= 2 \sqrt{2} \left.\frac{4}{15} (1+x)^{5/2} \right|_{-1}^1 = 2\sqrt{2} \frac{4}{15} 2^{5/2} = \frac{64}{15}.\end{align*} We can break the second integral further into two integrals, namely $$I_2 = \int_{-1}^1 2(1-x)^2\ln \left( \sqrt{1+x} + \sqrt{2} \right) \,dx$$ and $$I_3 = \int_{-1}^1 2(1-x)^2 \ln \sqrt{1-x} \,dx = \int_{-1}^1 (1-x)^2 \ln (1-x) \,dx.$$

If we set $u = \ln \left( \sqrt{1+x} + \sqrt{2} \right)$ then we see that $$du = \frac{dx}{2 \sqrt{1+ x} ( \sqrt{1+x} + \sqrt{2} )} = \frac{ \sqrt{2} - \sqrt{1+x} }{2 \sqrt{1+x} (1-x)} dx,$$ so then if we pursue integration by parts we get to obtain \begin{align*}I_2 &= \int_{-1}^1 2(1-x)^2 \ln \left(\sqrt{1+x} + \sqrt{2}\right) \,dx \\ &= 2 \left( \left[ -\frac{(1-x)^3}{3} \ln \left( \sqrt{1+x} + \sqrt{2} \right) \right]_{-1}^1 + \int_{-1}^1 \frac{(1-x)^3}{3} \frac{\sqrt{2} - \sqrt{1+x}}{2 \sqrt{1+x} (1-x) } \,dx \right) \\ &= 2 \left( -\frac{4}{3} \ln 2 + \int_{-1}^1 \frac{\sqrt{2}}{6} \frac{(1-x)^2}{\sqrt{1+x}} \,dx - \int_{-1}^1 \frac{(1-x)^2}{6} \,dx \right) \\ &= 2 \left( -\frac{4}{3} \ln 2 + \left[ \frac{\sqrt{2}}{3} (1-x)^2 \sqrt{1+x} \right]_{-1}^1 + \frac{2\sqrt{2}}{3} \int_{-1}^1 (1-x) \sqrt{1+x} \,dx + \left[ \frac{(1-x)^3}{18} \right]_{-1}^1 \right) \\ &= 2 \left( -\frac{4}{3} \ln 2 + \frac{4}{9} + \frac{1}{3} \left( \frac{64}{15} \right) \right) = \frac{56}{15} -\frac{8}{3} \ln 2, \end{align*} where we are using our precalculated knowledge of $I_1$ in the second to last step.

Using integration by parts gives $$\int t^2 \ln t \,dt = \frac{t^3}{3} \ln t - \int \frac{t^2}{3} \,dt = \frac{t^3}{3} \ln t - \frac{t^3}{9} + C.$$ With the u = 1-x substituion we get $$I_3 = \int_{-1}^1 (1-x)^2 \ln (1-x) \,dx = \int_{0}^2 u^2 \ln u \,du = \left[\frac{u^3}{3} \ln u - \frac{u^3}{9} \right]_{0}^2 = \frac{8}{3} \ln 2 - \frac{8}{9}.$$ So putting it altogether we get the expected length of the uniformly random point of light to be \begin{align*}\mathcal{L} &= I_1 + I_2 + I_3 \\ &= \frac{64}{15} + \left( \frac{56}{15} - \frac{8}{3} \ln 2 \right) + \left( \frac{8}{3} \ln 2 - \frac{8}{9} \right) \\ &= 8 - \frac{8}{9} = \frac{64}{9} = 7.111111\dots.\end{align*}

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*}

Fiddler: The Helical Ball?

Instead of requiring a vertical line down the cylinder’s surface, now any helix down the surface is allowed. An example of such a helix passing through all three gaps is shown below.

What is the probability that there exists at least one such helix that can pass through each ring’s gap?

Again, let's assume that we have cylinder with radius one, that the broken rings are embedded within the planes $z=2,$ $z=1$ and $z=0$ and finally that the gap on the ring at $z=2$ is from $\theta \in [0, \pi / 4],$ while the gaps at $z=1$ and $z=0$ are from $\theta \in [\alpha, \alpha+\pi/4]$ and $\theta \in [\beta, \beta + \pi / 4]$ for some $\alpha, \beta \sim U(0,2\pi),$ respectively. Here again we will play a little fast and loose with notation and say that $[\psi_1, \psi_2] \subseteq [0,2\pi]$ should mean the set $[0, \psi_2 - 2\pi] \cup [\psi_1, 2\pi]$ in the case that $\psi_1 \lt 2\pi \lt \psi_2.$

We also see that the helical path from any point $(2, \theta_2)$ to $(1, \theta_1)$ for any $\theta_1, \theta_2 \in [0,2\pi)$ is given by $z(t) = 2-t,$ $\theta(t) = \theta_2 + t(\theta_1 - \theta_2).$ All we need to do is understand where this helical path will hit the plane $z = 0,$ in this case at $\theta(2) = 2 \theta_1 - \theta_2.$ Since any helical path that gets through the upper and middle gaps will have $\theta_2 \in [0, \pi/4]$ and $\theta_1 \in [\alpha, \alpha + \pi/4],$ we see that $\theta_0 = \theta(2) = 2 \theta_1 - \theta_2 \in [2 \alpha - \pi/4, 2 \alpha + \pi/2].$ See the figure below, for instance for the choice of $\alpha = 4.613846199712232.$

So in order to fit through all of the gaps, then we would need to have $$[\beta, \beta + \pi/4] \cap [2 \alpha - \pi/4, 2 \alpha + \pi/2 ] \ne \emptyset,$$ that is, we would need to have $\beta \in [ 2 \alpha - \pi/2, 2\alpha + \pi/2].$ Therefore, the conditional probability of there being a helical path conditional on choice of $\alpha$ is $$p(\alpha) = \mathbb{P} \{ \beta \in [ 2 \alpha - \pi/2, 2 \alpha + \pi/2 ] \mid \alpha \} = \frac{\pi}{2\pi} \equiv \frac{1}{2}.$$ Therefore, the probability of there being at least one such helix that can pass through each of the gaps is $p = \int_0^{2\pi} p(\alpha) \,d\alpha = \frac{1}{2}.$

Ring around the cylinder

A long vertical cylinder has three narrow open rings, each of which wraps around seven-eighths of the cylinder (leaving a one-eighth “gap”). The rings are evenly spaced vertically, but are otherwise randomly rotated about the cylinder’s central axis. For some orientations of the rings, there exists at least one vertical line down the cylinder’s surface that passes through each ring’s gap, as illustrated below.

What is the probability that at least one such vertical line exists?

Let's define some terms. Let's assume we have a unit cylinder and that the broken rings are embedded within the planes $z=2,$ $z=1$ and $z=0.$ Let's, without loss of generality, assume that the opening at the topmost ring is one-eighth of the way around and oriented such that it is from $\theta \in [0, \pi/4].$ Let's further assume that for $\alpha, \beta \sim U(0,2\pi)$ that the middle and bottom rings at z=1 and z=0 are given by $\theta \in [\alpha, \alpha+\pi/4]$ and $\theta \in [\beta, \beta+\pi/4],$ respectively. Here we shall abuse notation slightly and assume that if $\psi_1 \lt 2\pi \lt \psi_2$ then by $[\psi_1, \psi_2] \subseteq [0,2\pi]$ we mean $[0, \psi_2 - 2\pi] \cup [ \psi_2, 2\pi].$

Any vertical line on this cylinder can be written in cylindrical coordinates as $\theta = \theta_0$ for some $\theta_0 \in [0,2\pi).$ If a vertical line can pass through the top and middle gaps, then we must have $[0, \pi/4] \cap [\alpha, \alpha+\pi/4] \ne \emptyset,$ or equivalently, $\alpha \leq \pi/4$ and $\alpha + \pi/4 \geq 0,$ that is $\alpha \in [-\pi/4, \pi/4].$ In order for this line to pass through all three gaps, we must have $$[ \beta, \beta + \pi/4 ] \cap [ \alpha, \alpha + \pi/4 ] \cap [0, \pi/4] \ne \emptyset,$$ or equivalently $\beta \in [ \max \{ 0, \alpha \} - \pi/4, \min \{0, \alpha\} + \pi/4 ].$

Therefore, the probability that a vertical line can pass through all gaps is \begin{align*}p &= \int_{-\pi/4}^{\pi/4} \int_{\max \{0, \alpha\} - \pi/4}^{\min \{0, \alpha\} + \pi/4} \frac{d\beta}{2\pi} \, \frac{d\alpha}{2\pi} \\ &= \int_{-\pi/4}^{\pi/4} \left( \frac{\pi}{2} - |\alpha| \right) \,\frac{d\alpha}{4\pi^2}\\ &= \frac{1}{2\pi^2} \int_0^{\pi/4} \left( \frac{\pi}{2} - \alpha \right), d\alpha\\ &= \frac{1}{2\pi^2} \int_{\pi/4}^{\pi/2} u \,du = \frac{1}{4\pi^2} \left[ \left(\frac{\pi}{2}\right)^2 - \left(\frac{\pi}{4}\right)^2 \right] \\ &= \frac{1}{4\pi^2} \frac{3\pi^2}{16} = \frac{3}{64} = 4.6875\%\end{align*}