## Calculus (3rd Edition)

The speed function: $f\left( t \right) = \frac{{ds}}{{dt}} = \sqrt {3 + 2\cos t - 2\sin t}$. The maximal speed is $f\left( { - \frac{\pi }{4}} \right) = \sqrt {3 + 2\sqrt 2 } \simeq 2.414$
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 $f\left( { - \frac{\pi }{4}} \right) = \frac{{ds}}{{dt}}{|_{t = - \pi /4}} = \sqrt {3 + 2\sqrt 2 } \simeq 2.414$