Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 13: Vector-Valued Functions and Motion in Space - Practice Exercises - Page 777: 24


$t=0; t=\dfrac{\pi}{2}; t =\pi$

Work Step by Step

$v(t)=\dfrac{dr}{dt}=3 \cos t \ k -5 \sin t j$ and $a(t)=\dfrac{dv(t)}{dt}=-3 \sin t \ k -5j \cos t $ Now, $v(t) \times a(t)=(3 \cos t \ k -5 \sin t j) \times (-3 \sin t \ k -5j \cos t) =16 \sin t \ \cos t $ When $v \times a=0$ This implies that $16 \sin t \ \cos t =0$ or, $\sin t =0$ or, $\cos t=0$ This implies that $t=0; t=\dfrac{\pi}{2}; t =\pi$
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