# How to find the parametric equation of a cycloid?

"A cycloid is the curve traced by a point on the rim of a circular wheel as the wheel rolls along a straight line." - Wikipedia

In many calculus books I have, the cycloid, in parametric form, is used in examples to find arc length of parametric equations. This is the parametric equation for the cycloid:

\begin{align*}x &= r(t - \sin t)\\ y &= r(1 - \cos t)\end{align*}

How are these equations found in the first place?

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Nice picture! Decompose the motion into two parts: (i) steady motion of the centre along the line $y=r$ and (ii) rotation. –  André Nicolas Apr 18 '12 at 19:56
I gave a general approach for deriving roulettes (of which your cycloid is a special case) in an answer to this question. –  Ｊ. Ｍ. Apr 28 '12 at 8:21

$t$ measures the angle through which the wheel has rotated, starting with your point in the "down" position. Since the wheel is rolling, the distance it has rolled is the distance along the circumference of the wheel from your point to the "down" position, which (since the wheel has radius $r$) is $rt$. So the centre of the wheel, which was initially at $(0,r)$, is now at $(rt,r)$. Your point is displaced from this by $-r\sin(t)$ horizontally and $-r\cos(t)$ vertically, so it is at $(rt - r\sin(t), r - r\cos(t))$.

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The center of the circle moves along a horizontal line at constant velocity. If we want the cusps to be at $y=0$, that means the center should be $(x_c,y_c)=(rt,r)$. Then we add on the location of the point on the rim relative to the center. This will be something like $(r\cos t, r\sin t)$ but we still need to get the phase right. If we start with the point on the rim at $(0,0)$ at $t=0$ the rim point is at an angle of $\frac {-\pi}2$ at $t=0$, that is, pointing straight down. A little fiddling with the phases gets the expression you quote. The scale between the center motion and rotation is set by the requirement that there be no slippage, which means the velocity of the point on the road must be $0$.