Answer
The speed function:
$f\left( t \right) = \frac{{ds}}{{dt}} = \sqrt {3 + 2\cos t - 2\sin t} $.
The maximal speed is
$f\left( { - \frac{\pi }{4}} \right) = \sqrt {3 + 2\sqrt 2 } \simeq 2.414$
Work Step by Step
We have
$x\left( t \right) = \sin t + t$, ${\ \ }$ $x'\left( t \right) = \cos t + 1$,
$y\left( t \right) = \cos t + t$, ${\ \ }$ $y'\left( t \right) = - \sin t + 1$.
By Theorem 2 of Section 12.2, the speed along $c\left(t\right)$ is
$\frac{{ds}}{{dt}} = \sqrt {{{\left( {\cos t + 1} \right)}^2} + {{\left( { - \sin t + 1} \right)}^2}} $
$\frac{{ds}}{{dt}} = \sqrt {{{\cos }^2}t + 2\cos t + 1 + {{\sin }^2}t - 2\sin t + 1} $
$\frac{{ds}}{{dt}} = \sqrt {3 + 2\cos t - 2\sin t} $
Write $f\left( t \right) = \frac{{ds}}{{dt}} = \sqrt {3 + 2\cos t - 2\sin t} $. To find the maximal speed we take the derivatives of $f\left(t\right)$:
$f'\left( t \right) = \frac{{ - 2\cos t - 2\sin t}}{{2\sqrt {3 + 2\cos t - 2\sin t} }}$
$f{\rm{''}}\left( t \right) = \frac{{ - 3 - 3\cos t + 3\sin t + \sin 2t}}{{{{\left( {3 + 2\cos t - 2\sin t} \right)}^{3/2}}}}$
We find the critical point of $f\left(t\right)$ by solving $f'\left( t \right) = 0$:
$f'\left( t \right) = \frac{{ - 2\cos t - 2\sin t}}{{2\sqrt {3 + 2\cos t - 2\sin t} }} = 0$
This occurs when $ - 2\cos t - 2\sin t = 0$. So,
$\sin t = - \cos t$, ${\ \ \ }$ $\tan t = - 1$.
The solutions are $t = - \frac{\pi }{4} + \pi n$, for $n = 0, \pm 1. \pm 2, \pm 3,...$
$f\left(t\right)$ is maximal if
$f{\rm{''}}\left( t \right) = \frac{{ - 3 - 3\cos t + 3\sin t + \sin 2t}}{{{{\left( {3 + 2\cos t - 2\sin t} \right)}^{3/2}}}} < 0$
At critical points, the denominator is always positive, so we solve the equation:
$ - 3 - 3\cos t + 3\sin t + \sin 2t < 0$
Substituting $t = - \frac{\pi }{4} + \pi n$ in the equation we get
$ - 3 - 3\cos \left( { - \frac{\pi }{4} + \pi n} \right) + 3\sin \left( { - \frac{\pi }{4} + \pi n} \right) + \sin \left( { - \frac{\pi }{2} + 2\pi n} \right) < 0$
$ - 3 - 3\cos \left( { - \frac{\pi }{4}} \right)\cos \left( {\pi n} \right) + 3\sin \left( { - \frac{\pi }{4}} \right)\cos \left( {\pi n} \right) + \sin \left( { - \frac{\pi }{2}} \right) < 0$
$ - 3 - \frac{3}{2}\sqrt 2 \cos \left( {\pi n} \right) - \frac{3}{2}\sqrt 2 \cos \left( {\pi n} \right) - 1 < 0$
The left-hand side is negative if $n$ is even. So, the the particle's speed is maximal when
$t = - \frac{\pi }{4} + \pi k$ ${\ \ }$ for $k = 0, \pm 2, \pm 4,...$
We choose $t = - \frac{\pi }{4}$ and substitute it in the speed function $f\left( t \right) = \frac{{ds}}{{dt}} = \sqrt {3 + 2\cos t - 2\sin t} $. The maximal speed is
$f\left( { - \frac{\pi }{4}} \right) = \frac{{ds}}{{dt}}{|_{t = - \pi /4}} = \sqrt {3 + 2\sqrt 2 } \simeq 2.414$
