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: 23


$$\frac{d}{dt}(r(g(t)))=e^t\langle 2e^t, -1\rangle .$$

Work Step by Step

Since $ r(t)=\langle t^2, 1-t\rangle, \quad g(t)=e^t $, then we have $$\frac{d}{dt}(r(g(t)))=\frac{d}{dt}\langle e^{2t}, 1-e^t\rangle=\langle 2e^{2t}, -e^t\rangle \\=e^t\langle 2e^t, -1\rangle=r'(g(t)) g'(t).$$
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.