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.
Best Answer
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.