Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 3 - Review - Exercises - Page 271: 91

Answer

$v = -Ae^{-ct}~[c~cos(\omega t+\delta)+\omega~sin(\omega t+\delta)~]$ $a = Ae^{-ct}~[(c^2-\omega^2)~cos(\omega t+\delta)+2c\omega~sin(\omega t+\delta)~]$

Work Step by Step

$s = Ae^{-ct}~cos(\omega t+\delta)$ $v = \frac{ds}{dt}$ We can find an expression for $v$: $v = Ae^{-ct}(-c)~cos(\omega t+\delta)-Ae^{-ct}~\omega~sin(\omega t+\delta)$ $v = -Ae^{-ct}~[c~cos(\omega t+\delta)+\omega~sin(\omega t+\delta)~]$ $a = \frac{dv}{dt}$ We can find an expression for $a$: $a = -Ae^{-ct}(-c)~[c~cos(\omega t+\delta)+\omega~sin(\omega t+\delta)~]-Ae^{-ct}~[-c~\omega~sin(\omega t+\delta)+\omega^2~cos(\omega t+\delta)~]$ $a = Ae^{-ct}c^2~cos(\omega t+\delta)+Ae^{-ct}c\omega~sin(\omega t+\delta)+Ae^{-ct}~c~\omega~sin(\omega t+\delta)-Ae^{-ct}\omega^2~cos(\omega t+\delta)$ $a = Ae^{-ct}~[(c^2-\omega^2)~cos(\omega t+\delta)+2c\omega~sin(\omega t+\delta)~]$
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