University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 4 - Section 4.8 - Antiderivatives - Exercises - Page 273: 107



Work Step by Step

We need to find the anti-derivative for $\dfrac{d^2r}{dt^2}=\dfrac{2}{t^3}$ We have: $\dfrac{dr}{dt}=\dfrac{-1}{t^2}+c$; $r=\dfrac{1}{t}+ct+c'$ Apply the initial conditions $r'(1)=1 ; r(1)=1$ Thus, $-1+c=1 \implies c=2 $ and $1+c+c'=1 \implies c'=-2$ Hence, $r=\dfrac{1}{t}+2t-2$
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