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: 48


$s(t)= -\cos (\dfrac{3t}{\pi})$

Work Step by Step

Given: $a=(\dfrac{9}{\pi^2}) \cos (\dfrac{3t}{\pi}), v(0)=0$ and $s(0)=-1$ Since, we know that the derivative of velocity is acceleration. and $v(t)=\dfrac{3}{\pi}\sin \dfrac{3t}{\pi}+C$ Since,we have: $v(0)=0$ So, $0=(\dfrac{3}{\pi})\sin (\dfrac{3(0)}{\pi})+C \implies C=0$ Therefore, $v(t)=\dfrac{3}{\pi}\sin \dfrac{3t}{\pi}$ Since, we know that the derivative of position $s(t)$ is velocity. So, $s(t)=-\cos \dfrac{3t}{\pi}+C'$ Thus, $s(0)=-1$ and $-1=\cos \dfrac{3(0)}{\pi}+C' $ or, $C'=0$ Hence, $s(t)= -\cos (\dfrac{3t}{\pi})$
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