Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 9 - Review - Exercises - Page 636: 20

Answer

$$ R=A S^{k} \quad\quad [\text { R as a function of S.}] $$ where $A=e^{C}$ is a positive constant.

Work Step by Step

$$ \frac{1}{R} \frac{d R}{d t}=\frac{k}{S} \frac{d S}{d t} \Rightarrow \frac{d}{d t}(\ln R)=\frac{d}{d t}(k \ln S) $$ $ \Rightarrow $ $$ \ln R=k \ln S+C \Rightarrow R=e^{k \ln S+C} = e^{C}\left(e^{\ln S}\right)^{k} $$ $$ \Rightarrow R=A S^{k} \quad\quad [\text { R as a function of S.}] $$ where $A=e^{C}$ is a positive constant.
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