University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 4 - Section 4.2 - The Mean Value Theorem - Exercises - Page 223: 45


$s(t)=-\dfrac{\cos \pi t}{\pi}+\dfrac{1}{\pi}$

Work Step by Step

Here, $v= \sin \pi t$ and $\dfrac{d}{dt}(-\dfrac{\cos \pi t}{\pi}) =\sin \pi t$ and thus, the general form of the function is: $s(t)=-\dfrac{\cos \pi t}{\pi}+C$ Since we have $s(0)=0$ Therefore, $s(0)=0 \implies 0=-\dfrac{\cos \pi (0)}{\pi}+C$ Thus, $C=\dfrac{1}{\pi}$ Hence, $s(t)=-\dfrac{\cos \pi t}{\pi}+\dfrac{1}{\pi}$
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.