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: 26


$\langle 16t\cos 2t^2, -24\sin2t^2\rangle $

Work Step by Step

Since $ r(t)=\langle 4\sin 2t, 6\cos 2t \rangle, \quad g(t)=t^2 $, then we have $$ \frac{d}{dt}(r(g(t)))=r'(g(t)) g'(t)=2t \langle 8\cos 2t^2, -12\sin2t^2\rangle= \langle 16t\cos 2t^2, -24\sin2t^2\rangle.$$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.