Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 6 - Inverse Functions - 6.2 Exponential Functions and Their Derivatives - 6.2 Exercises: 104

Answer

41666.06 Kg

Work Step by Step

The rate of growth of a fish population was modeled by the equation $G(t)=\frac{60,000e^{-0.6t}}{(1+5e^{-0.6t})^{2}}$ Where, $t$ is measured in years and G in kilograms per year. Given: The biomass was 25,000 kg in the year 2000. Our aim is to calculate predicted biomass for the year 2020. For this, we will have to take integral of the function $G(t)=\frac{60,000e^{-0.6t}}{(1+5e^{-0.6t})^{2}}$ Integral $G(t)=\frac{60,000e^{-0.6t}}{(1+5e^{-0.6t})^{2}}$ with respect to t. $\int G(t) dt=\int \frac{60,000e^{-0.6t}}{(1+5e^{-0.6t})^{2}} dt$ Consider $(1+5e^{-0.6t})=x$ $dx=-3e^{-0.6t} dt$ Therefore, $\int G(t) dt=\frac{60,000}{3}\int- \frac{dx}{x^{2}}=\frac{20,000}{x}+C$ $=\frac{20,000}{(1+5e^{-0.6t})}+C$ The above expression is for biomass in the year of 2000. To calculate predicted value for biomass for the year 2020, we will put $t = 0 $ in above expression and equate this term to 25,000. Thus, $C=21666.67$ Let M be the predicted biomass for the year 2020. Then, $M==\frac{20,000}{(1+5e^{-0.6(20)})}+21666.67$ Hence, $M=19999.38+21666.67=41666.06$ kg The predicted biomass for the year 2020, M = 41666.06 Kg.
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