Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 4: Applications of Derivatives - Section 4.2 - The Mean Value Theorem - Exercises 4.2 - Page 198: 44


$s(t)=\sin (\dfrac{2t}{\pi})+1$

Work Step by Step

Since, the derivative of the function, $v= \dfrac{ds}{dt}=\dfrac{2}{\pi} \cos (\dfrac{2t}{\pi})$ is $s(t)=\sin \dfrac{2t}{\pi}+C$ As we are given that $s=1, t=\pi^2$ This implies that $s=1, t=\pi^2$ and $1=\sin (\dfrac{2 \pi^2}{\pi})+C \implies C=1$ Hence, $s(t)=\sin (\dfrac{2t}{\pi})+1$
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