University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.6 - The Chain Rule - Exercises: 26

Answer

$\frac{ds}{dt} = \frac{3 \pi}{2}(cos(\frac{3 \pi t}{2}) -sin (\frac{3 \pi t}{2}))$

Work Step by Step

$s = sin(u) + cos(u)$ where, $u = \frac{3 \pi t}{2}$ Now, $\frac{ds}{du} = cos(u) -sin (u)$ and, $\frac{du}{dt} = \frac{3 \pi}{2}$ So, $\frac{ds}{dt} = \frac{ds}{du} \times \frac{du}{dt}$ $\frac{ds}{dt} = \frac{3 \pi}{2}(cos(u) -sin (u))$ $\frac{ds}{dt} = \frac{3 \pi}{2}(cos(\frac{3 \pi t}{2}) -sin (\frac{3 \pi t}{2}))$
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.