Answer
$v(\pi/2)= 5$.
Work Step by Step
Since $r(t) =\lt \sin3t, \cos4t, \cos5t\gt$, using the facts that $(\cos at)'=-a\sin at$ and $(\sin at)'=a\cos at$, we have $r'(t) =\lt 3\cos 3t, -4\sin 4t, -5\sin 5t\gt$. Hence the speed $v(t)$ is given by $$v(t)=\|r'(t)\|=\sqrt{9\cos^23t+16\sin^24t+25\sin^25t } . $$ At $t=\pi/2$, $v(\pi/2)=\sqrt{0+0+25}=5$.
