Finding the equation for a (inverted) cycloid given two points

Question

If I have two points on a Cartesian plane, and I know that they are connected by a cycloid, then how do I find the equation for that cycloid?

For background information, I have been playing around with the brachistochrone problem, and I've found that the solution curve, an inverted cycloid, has parameters $x = \frac C2(t - \sin\;t)$ and $y = C - \frac C2(1 - \cos\;t)$, where $\frac C2$ is the radius of the rolling circle and $t$ is measured in radians.

Actually, I found $y = \frac C2(1 - \cos\;t)$; I would not know why the true solution would be an inverted cycloid as opposed to an upright one and how I would derive that from my calculations.

In any case, suppose we had two points, $(0,A)$ and $(B,0)$, where the former corresponds to the rolling circle having passed through an angle of $0$; the latter, an angle of $\pi$. The parametric equations for this example are then $x = \frac A2(t - \sin\;t)$ and $y = A - \frac A2(1 - \cos\;t)$.

Now suppose we have a similar pair of coordinates, specifically $(0,A)$ and $(2B,0)$. The angles corresponding to those angles are no longer $0$ and $\pi$. How would I then find the parametric equations for the curve connecting those two points?

Regards.

Answer 1

A cycloid can also be interpreted the equation of motion of a point in a rolling-circle. You can check here if you are not convinced. Or even prove it mathematically.

Therefore, you have two parameters: the radius of the circle $r$, and the angular speed of the circle $\omega$. The angle of the point in the circle is $t$. Then:

\begin{array}{} x = r(\omega t + \cos\omega t) \\ y = r\sin\omega t \end{array}

Those equations are the rolling-circle I describe, in function of parameter $t$. I'm not sure if it is an cycloid or an inverted cycloid. But for your problem, I guess it is irrelevant.

Watching closely, we can notice that those equations are describing a circle $(\cos\omega t, \sin\omega t)$, but adding $r\omega t$ in the variable $x$, which says circle center is moving in $x$ direction with speed $r\omega$.

You can center the circle in arbitrary position $(x_0, y_0)$. Or you can define a initial position for the point in the circle. If you do so, you will get:

\begin{array}{} \Delta x = r(\omega t + \cos\omega \Delta t) \\ \Delta y = r\sin\omega \Delta t \end{array}

Now you have $r, \omega, x_0, y_0, t_0$ to chose. Fixing $r=1$, you would have to solve a system of 4 variables and 4 equations to get the solution to your problem.

Basically in your derivation, you just forgot that a point in a circle can rotate with $\omega$ different than 1.

Answer 2

I drew from the information provided by Semiclassical and Physicist137 (thank you for helping!) to draw out a direct solution to finding the curve connecting two points.

Suppose we wanted the cycloid connecting an initial, known point $A$ and a second, arbitrary point $B$. For simplicity, set $A=(0,0)$; a different initial point means a simple translation. Using the parametric equations,

\begin{array}{} x = a(t-\sin t) \\ y = -a(1-\cos t), \end{array}

where $a$ is the unknown constant or the radius of the rolling circle, we can see that the slope connecting $A$ and $B$ is,

\begin{array}{} \frac yx = \frac{\cos t+1}{t-\sin t}. \end{array}

Notice that the slope is independent of the radius $a$. One unknown; one equation. In theory, we can then solve for $t$, $0 \leq t \leq 2\pi$, and then use $t$ to solve for the unknown $a$. From there, you have your parametric equations that would describe the cycloid you're looking for.

To answer my note about the "inverted cycloid" and the inclusion of a negative sign, I realize that in my derivation of the brachistochrone curve, I had defined the $y$-direction -- for gravity -- to be positive downwards. Adjusting for the conventional Cartesian plane where $y$ downwards is negative, you would then swap the signs for $y$; $x$ is unaffected.

I suppose the next question to ask would be about the number of arches you could have connecting both points.