Calculus: Early Transcendentals 9th Edition

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

Chapter 9 - Section 9.3 - Separable Equations - 9.3 Exercises - Page 626: 15

Answer

$A = {b^3}{e^{b\sin br}}$

Work Step by Step

$$\eqalign{ & \frac{{dA}}{{dr}} = A{b^2}\cos br,{\text{ }}A\left( 0 \right) = {b^3} \cr & {\text{Separating the variables}} \cr & \frac{{dA}}{A} = {b^2}\cos brdr \cr & {\text{Integrate both sides of the equation}} \cr & \int {\frac{{dA}}{A}} = \int {A{b^2}\cos brdr} \cr & \int {\frac{{dA}}{A}} = b\int {\cos br\left( b \right)dr} \cr & \ln \left| A \right| = b\sin br + C{\text{ }}\left( {\bf{1}} \right) \cr & {\text{Use the initial condition }}A\left( 0 \right) = {b^3} \cr & \ln \left| {{b^3}} \right| = b\sin b\left( 0 \right) + C \cr & C = \ln \left| {{b^3}} \right| \cr & {\text{Substitute }}C{\text{ into }}\left( {\bf{1}} \right) \cr & \ln \left| A \right| = b\sin br + \ln \left| {{b^3}} \right| \cr & {\text{Solve for }}A \cr & {e^{\ln \left| A \right|}} = {e^{b\sin br + \ln \left| {{b^3}} \right|}} \cr & A = {e^{\ln \left| {{b^3}} \right|}}{e^{b\sin br}} \cr & A = {b^3}{e^{b\sin br}} \cr} $$
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