Answer
The arc length of the involute is $s = 2{\pi ^2}R$
Work Step by Step
Observe that OQP is a right triangle and that $\angle QPS = \frac{\pi }{2} - \theta $. So, from the figure we see that the coordinates of $P$ is
$x = R\cos \theta + ||\overline {PQ} ||\cos \left( {\frac{\pi }{2} - \theta } \right) = R\cos \theta + ||\overline {PQ} ||\sin \theta $,
$y = R\sin \theta - ||\overline {PQ} ||\sin \left( {\frac{\pi }{2} - \theta } \right) = R\sin \theta - ||\overline {PQ} ||\cos \theta $.
Since the length of $\overline {PQ} $ is $R\theta $, that is, $||\overline {PQ} || = R\theta $, we get
$x = R\cos \theta + R\theta \sin \theta $, ${\ \ \ }$ $y = R\sin \theta - R\theta \cos \theta $.
So,
$P = \left( {R\left( {\cos \theta + \theta \sin \theta } \right),R\left( {\sin \theta - \theta \cos \theta } \right)} \right)$
Since $P$ is any point on the involute, we obtain the parametrization
$c\left( \theta \right) = \left( {R\left( {\cos \theta + \theta \sin \theta } \right),R\left( {\sin \theta - \theta \cos \theta } \right)} \right)$
We have
$x\left( \theta \right) = R\left( {\cos \theta + \theta \sin \theta } \right)$,
$x'\left( \theta \right) = R\left( { - \sin \theta + \sin \theta + \theta \cos \theta } \right) = R\theta \cos \theta $,
$y\left( \theta \right) = R\left( {\sin \theta - \theta \cos \theta } \right)$,
$y'\left( \theta \right) = R\left( {\cos \theta - \cos \theta + \theta \sin \theta } \right) = R\theta \sin \theta $.
Using Eq. (3) of Theorem 1, the arc length for $0 \le t \le 2\pi $ is
$s = \mathop \smallint \limits_0^{2\pi } \sqrt {{{\left( {R\theta \cos \theta } \right)}^2} + {{\left( {R\theta \sin \theta } \right)}^2}} {\rm{d}}t = R\mathop \smallint \limits_0^{2\pi } \theta {\rm{d}}\theta $
$s = \frac{1}{2}R{\theta ^2}|_0^{2\pi } = 2{\pi ^2}R$.
