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 - 14.2 Calculus of Vector-Valued Functions - Exercises - Page 720: 12

Answer

$$ a'(\theta) =(-3\sin \theta) i+( \sin2\theta ) j+(\sec^2 \theta) k.$$

Work Step by Step

Since $ a(\theta)=(\cos3 \theta) i+(\sin^2 \theta) j+(\tan \theta) k $, then by using the chain rule, the derivative $ a'(\theta)$ is given by $$ a'(\theta)=(-3\sin \theta) i+(2\sin \theta \cos \theta) j+(\sec^2 \theta) k\\ =(-3\sin \theta) i+( \sin2\theta ) j+(\sec^2 \theta) k.$$
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