Answer
(a) the speed is at a maximum when $t=\pi$.
(b) one arch of the cycloid has length $8$.
Work Step by Step
(a) We have ${\bf{r}}\left( t \right) = \left( {t - \sin t,1 - \cos t} \right)$. So, the derivative is
${\bf{r}}'\left( t \right) = \left( {1 - \cos t,\sin t} \right)$
The speed is
$||{\bf{r}}'\left( t \right)|| = \sqrt {\left( {1 - \cos t,\sin t} \right)\cdot\left( {1 - \cos t,\sin t} \right)} $
$||{\bf{r}}'\left( t \right)|| = \sqrt {{{\left( {1 - \cos t} \right)}^2} + {{\sin }^2}t} $
$||{\bf{r}}'\left( t \right)|| = \sqrt {1 - 2\cos t + {{\cos }^2}t + {{\sin }^2}t} $
$||{\bf{r}}'\left( t \right)|| = \sqrt {2\left( {1 - \cos t} \right)} $
Recall the identity ${\sin ^2}\left( {\frac{t}{2}} \right) = \frac{1}{2}\left( {1 - \cos t} \right)$. So,
$||{\bf{r}}'\left( t \right)|| = \sqrt {4{{\sin }^2}\left( {\frac{t}{2}} \right)} = 2\sin \left( {\frac{t}{2}} \right)$
Let $f\left( t \right)$ denote the speed such that $f\left( t \right) = ||{\bf{r}}'\left( t \right)|| = 2\sin \left( {\frac{t}{2}} \right)$.
To find the maximum speed we take the derivative of $f$ and find the critical point by solving the equation:
$f'\left( t \right) = \cos \left( {\frac{t}{2}} \right) = 0$
For $t$ in $\left[ {0,2\pi } \right]$, the solution is $t = \pi $. So, the critical point occurs at $t=\pi$.
Taking the second derivative gives $f{\rm{''}}\left( t \right) = - \frac{1}{2}\sin \left( {\frac{t}{2}} \right)$. Since $f{\rm{''}}\left( \pi \right) = - \frac{1}{2} < 0$, therefore the speed is at a maximum when $t=\pi$.
(b) From part (a) we obtain the speed is $||{\bf{r}}'\left( t \right)|| = 2\sin \left( {\frac{t}{2}} \right)$
Recall from the parametrization of a cycloid on page 599, an arch of the cycloid corresponds to the interval $0 \le t \le 2\pi $. Thus, the length of one arch of the cycloid is
$s = \mathop \smallint \limits_0^{2\pi } ||{\bf{r}}'\left( t \right)||{\rm{d}}t = \mathop \smallint \limits_0^{2\pi } 2\sin \left( {\frac{t}{2}} \right){\rm{d}}t$
$s = - 2\cdot2\cdot\cos \left( {\frac{t}{2}} \right)|_0^{2\pi } = - 4\left( { - 1 - 1} \right) = 8$
Hence, one arch of the cycloid has length $8$.
