# Is it possible to express a cycloid in polar coordinates

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.

• 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? – Ted Shifrin May 1 '17 at 22:08
• You may not be able to make a cycloid, but you can make something similar: en.wikipedia.org/wiki/Epicycloid – Franklin Pezzuti Dyer May 1 '17 at 22:09
• You can use $t=cos^{-1} (r \sin \theta \pm 1)$ – N74 May 1 '17 at 22:13
• @N74: That is not giving the curve in polar coordinates (i.e., in the form $r=f(\theta)$), is it? – Ted Shifrin May 1 '17 at 22:38
• It answers your last question about how to eliminate the parameter $t$ – 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.
$$\rho\cos\theta=\arccos(1\pm\rho\sin\theta)-\pi-\sin(\arccos(1\pm\rho\sin\theta)),$$ which is a nasty implicit polar equation.