Answer
The speed at $t = \frac{\pi }{4}$ is approximately $4.53$
Work Step by Step
We have
$x\left( t \right) = \sin 4t$, ${\ \ \ }$ $x'\left( t \right) = 4\cos 4t$,
$y\left( t \right) = \cos 3t$, ${\ \ \ }$ $y'\left( t \right) = - 3\sin 3t$.
By Theorem 2 of Section 12.2, the speed along c(t) is
$\frac{{ds}}{{dt}} = \sqrt {{{\left( {4\cos 4t} \right)}^2} + {{\left( { - 3\sin 3t} \right)}^2}} = \sqrt {16{{\cos }^2}4t + 9{{\sin }^2}3t} $
The speed at $t = \frac{\pi }{4}$ is
$\frac{{ds}}{{dt}}{|_{t = \pi /4}} = \sqrt {16{{\cos }^2}\pi + 9{{\sin }^2}\frac{{3\pi }}{4}} = \sqrt {\frac{{41}}{2}} \simeq 4.53$
