Answer
The length of one arch of a cycloid generated by a circle of radius $R$ is $8R$.
Work Step by Step
The parametrization of a cycloid generated by a circle of radius $R$ is given by (Exercise 97 of Section 12.1)
$c\left( t \right) = \left( {Rt - R\sin t,R - R\cos t} \right)$.
So, we have
$x\left( t \right) = Rt - R\sin t$, ${\ \ \ }$ $x'\left( t \right) = R - R\cos t$,
$y\left( t \right) = R - R\cos t$, ${\ \ \ }$ $y'\left( t \right) = R\sin t$.
The $x$-interval for one arch of the cycloid is $0 \le x \le 2\pi R$ as is shown in Figure 25 of Exercise 97 of Section 12.1. The corresponding $t$-interval is $0 \le t \le 2\pi $.
Using Eq. (3) of Theorem 1, the arc length is
$s = \mathop \smallint \limits_0^{2\pi } \sqrt {{{\left( {R - R\cos t} \right)}^2} + {{\left( {R\sin t} \right)}^2}} {\rm{d}}t = \mathop \smallint \limits_0^{2\pi } \sqrt {2{R^2}\left( {1 - \cos t} \right)} {\rm{d}}t$
$s = \sqrt 2 R\mathop \smallint \limits_0^{2\pi } \sqrt {1 - \cos t} {\rm{d}}t$
Since $2{\sin ^2}\frac{t}{2} = 1 - \cos t$, we get
$s = 2R\mathop \smallint \limits_0^{2\pi } \sin \frac{t}{2}{\rm{d}}t = - 4R\cos \frac{t}{2}|_0^{2\pi } = 8R$.
