The parametric equations for a cycloid of radius 1 centered at the pole are $$ x(t) = t - \pi - \sin t \\ y(t) = \pm (1- \cos t) $$ where the plus sign is a cycloid above the x-axis and the minus sign a cycloid below the x-axis. Is it possible to convert these equations into polar coordinates so that the cycloid is expressed as a angle dependent radius $r(\theta)$ ? I tried to convert them to polar coordinates by using $$ r = \sqrt{x(t)^2 + y(t)^2} \\ \tan \theta = \frac{y(t)}{x(t)} $$ But I don't understand how I can eliminate the parameter $t$ so that I get a function $r(\theta)$ in the end.

  • $\begingroup$ I don't think it's possible. Why are you trying to do so? Are you perchance trying to find the area under one arch of a cycloid? $\endgroup$ – Ted Shifrin May 1 '17 at 22:08
  • $\begingroup$ You may not be able to make a cycloid, but you can make something similar: en.wikipedia.org/wiki/Epicycloid $\endgroup$ – Franklin Pezzuti Dyer May 1 '17 at 22:09
  • $\begingroup$ You can use $t=cos^{-1} (r \sin \theta \pm 1)$ $\endgroup$ – N74 May 1 '17 at 22:13
  • $\begingroup$ @N74: That is not giving the curve in polar coordinates (i.e., in the form $r=f(\theta)$), is it? $\endgroup$ – Ted Shifrin May 1 '17 at 22:38
  • $\begingroup$ It answers your last question about how to eliminate the parameter $t$ $\endgroup$ – N74 May 1 '17 at 22:40

To express $\rho$ as a function of $\theta$, you could express $t$ as a function of $\theta$ by inverting

$$\tan\theta=\pm\frac{1-\cos t}{t-\pi-\sin t}$$and plug the result in the expression of $\rho=\sqrt{x^2+y^2}$.

But there is no closed-form expression for $t(\theta)$ because $t$ appears inside and outside of a trigonometric function, making a transcendental equation.

Following @N74's approach, you can write

$$\rho\cos\theta=\arccos(1\pm\rho\sin\theta)-\pi-\sin(\arccos(1\pm\rho\sin\theta)),$$ which is a nasty implicit polar equation.


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