Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 6 - Differential Equations - 6.2 Exercises - Page 412: 24

Answer

$y = \frac{1}{{2000}}{e^{2.3025t}}$

Work Step by Step

$$\eqalign{ & {\text{The exponential function is in the form }}y = C{e^{kt}}{\text{ }}\left( {\bf{1}} \right) \cr & {\text{Where }}C{\text{ and }}k{\text{ are constants}} \cr & \cr & {\text{From the graph we know the points }}\left( {3,\frac{1}{2}} \right){\text{ and }}\left( {4,5} \right) \cr & {\text{*Substituing the point }}\underbrace {\left( {3,\frac{1}{2}} \right)}_{\left( {t,y} \right)}{\text{ in the equation }}\left( {\bf{1}} \right){\text{we obtain:}} \cr & \frac{1}{2} = C{e^{k\left( 3 \right)}} \to \frac{1}{2} = C{e^{3k}}{\text{, }}\left( {\bf{2}} \right) \cr & {\text{*Substituing the point }}\underbrace {\left( {4,5} \right)}_{\left( {t,y} \right)}{\text{ in the equation }}\left( {\bf{1}} \right){\text{we obtain:}} \cr & 5 = C{e^{k\left( 4 \right)}} \to 5 = C{e^{4k}}{\text{, }}\left( {\bf{3}} \right) \cr & \cr & {\text{*Solve the equation }}\left( {\bf{2}} \right){\text{ for }}C \cr & C = \frac{1}{{2{e^{3k}}}} \cr & {\text{Substitute the previous result in the equation }}\left( {\bf{3}} \right) \cr & 5 = \left( {\frac{1}{{2{e^{3k}}}}} \right){e^{4k}} \cr & 10 = {e^k} \cr & {\text{Solve for }}k \cr & k = \ln \left( {10} \right) \cr & {\text{Substitute }}k\,{\text{into the equation }}\left( {\bf{2}} \right){\text{ or }}\left( {\bf{3}} \right) \cr & \frac{1}{2} = C{e^{3\left( {\ln \left( {10} \right)} \right)}} \cr & {\text{Solve for }}C \cr & C = \frac{1}{2}{e^{ - 3\left( {\ln \left( {10} \right)} \right)}} \cr & \cr & {\text{Substitute }}k = \ln \left( {10} \right){\text{ and }}C = \frac{1}{2}{e^{ - 3\left( {\ln \left( {10} \right)} \right)}}{\text{ into }}\left( {\bf{1}} \right) \cr & y = \frac{1}{2}{e^{ - 3\left( {\ln \left( {10} \right)} \right)}}{e^{\ln \left( {10} \right)t}} \cr & or \cr & y = \frac{1}{{2000}}{e^{2.3025t}} \cr} $$
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