Cycloid: equation, length of arc, area

The curve traced by a point on the rim of the circle as the circle rolls along a straight line without skidding is called cycloid. There are many online resources you can read about cycloid. An interesting one includes the history of it which can be found here.
 

 

Below is the animated .gif of cycloid from Wikipedia

 

The equation of the cycloid can be written easily if expressed in terms of parameter θ. θ is the angle rotated by the rolling circle.
 

From the figure, line OB = arc AB.
$s = r\theta$
 

 

From right triangle ACO
$a = r \cos \theta$

$b = r \sin \theta$
 

Hence,
$x = s - b = r\theta - r \sin \theta$

$y = r - a = r - r \cos \theta$
 

Thus,
$x = r(\theta - \sin \theta)$

$y = r(1 - \cos \theta)$
 

Differentiate x and y
$dx = r(1 - \cos \theta) \, d\theta$

$dy = r\sin \theta \, d\theta$
 

Length of one arc
$\displaystyle s = \int_{x_1}^{x_2} \sqrt{1 + \left( \dfrac{dy}{dx} \right)^2} \, dx$
 

Where,
$\dfrac{dy}{dx} = \dfrac{r\sin \theta \, d\theta}{r(1 - \cos \theta) \, d\theta}$

$\dfrac{dy}{dx} = \dfrac{\sin \theta}{1 - \cos \theta}$
 

Thus,
$\displaystyle s = \int_0^{2\pi} \sqrt{1 + \left( \dfrac{\sin \theta}{1 - \cos \theta} \right)^2} \, r(1 - \cos \theta) \, d\theta$

$\displaystyle s = r\int_0^{2\pi} \sqrt{1 + \dfrac{\sin^2 \theta}{(1 - \cos \theta)^2}} \, (1 - \cos \theta) \, d\theta$

$\displaystyle s = r\int_0^{2\pi} \sqrt{1 + \dfrac{1 - \cos^2 \theta}{(1 - \cos \theta)^2}} \, (1 - \cos \theta) \, d\theta$

$\displaystyle s = r\int_0^{2\pi} \sqrt{1 + \dfrac{(1 - \cos \theta)(1 + \cos \theta)}{(1 - \cos \theta)^2}} \, (1 - \cos \theta) \, d\theta$

$\displaystyle s = r\int_0^{2\pi} \sqrt{1 + \dfrac{1 + \cos \theta}{1 - \cos \theta}} \, (1 - \cos \theta) \, d\theta$

$\displaystyle s = r\int_0^{2\pi} \sqrt{\dfrac{(1 - \cos \theta) + (1 + \cos \theta)}{1 - \cos \theta}} \, (1 - \cos \theta) \, d\theta$

$\displaystyle s = r\int_0^{2\pi} \sqrt{\dfrac{2}{1 - \cos \theta}} \, (1 - \cos \theta) \, d\theta$

$\displaystyle s = r\sqrt{2}\int_0^{2\pi} \dfrac{1}{(1 - \cos \theta)^{1/2}} \, (1 - \cos \theta) \, d\theta$

$\displaystyle s = r\sqrt{2}\int_0^{2\pi} (1 - \cos \theta)^{1/2} \, d\theta$
 

From
$\cos 2x = 1 - 2\sin^2 x$

$2\sin^2 x = 1 - \cos 2x$

$1 - \cos 2x = 2\sin^2 x$
 

Let x = ½ θ
$1 - \cos 2(\frac{1}{2}\theta) = 2\sin^2 (\frac{1}{2}\theta)$

$1 - \cos \theta = 2\sin^2 (\frac{1}{2}\theta)$

$(1 - \cos \theta)^{1/2} = \sqrt{2}\sin (\frac{1}{2}\theta)$
 

Thus,
$\displaystyle s = r\sqrt{2}\int_0^{2\pi} \sqrt{2}\sin \left( \frac{1}{2}\theta \right) \, d\theta$

$\displaystyle s = 2r\int_0^{2\pi} \sin \left( \frac{1}{2}\theta \right) \, d\theta$

$\displaystyle s = 4r\int_0^{2\pi} \sin \left( \frac{1}{2}\theta \right) \, \left( \frac{1}{2} d\theta \right)$

$s = 4r \left[ -\cos \left( \dfrac{1}{2}\theta \right) \right]_0^{2\pi}$

$s = -4r \left[ \cos \left( \dfrac{1}{2}\theta \right) \right]_0^{2\pi}$

$s = -4r(\cos \pi - \cos 0)$

$s = -4r(-1 - 1)$

$s = -4r(-2)$

$s = 8r$           answer
 

Area under one arc of cycloid
$\displaystyle A = \int_{x_1}^{x_2} y \, dx$
 

 

$\displaystyle A = \int_0^{2\pi} r(1 - \cos \theta) \times r(1 - \cos \theta) \, d\theta$

$\displaystyle A = r^2 \int_0^{2\pi} (1 - \cos \theta)^2 \, d\theta$

$\displaystyle A = r^2 \int_0^{2\pi} (1 - 2\cos \theta + \cos^2 \theta) \, d\theta$

$\displaystyle A = r^2 \int_0^{2\pi} \left[ 1 - 2\cos \theta + \dfrac{1}{2}(1 + \cos 2\theta) \right] \, d\theta$

$\displaystyle A = r^2 \int_0^{2\pi} \left(\dfrac{3}{2} - 2\cos \theta + \dfrac{1}{2}\cos 2\theta \right) \, d\theta$

$A = r^2 \left[ \frac{3}{2}\theta - 2\sin \theta + \frac{1}{4}\sin 2\theta \right]_0^{2\pi}$

$A = r^2 \left[ \frac{3}{2}(2\pi) - 2\sin (2\pi) + \frac{1}{4}\sin 2(2\pi) \right] - r^2 \left[ 0 - 2\sin 0 + \frac{1}{4}\sin 0 \right]$

$A = 3\pi r^2$           answer