Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 14 - Calculus of Vector-Valued Functions - 14.3 Arc Length and Speed - Exercises - Page 725: 15


$v(\pi/2)= 5$.

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 $t=\pi/2$, $v(\pi/2)=\sqrt{0+0+25}=5$.
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