## Calculus 8th Edition

Published by Cengage

# Chapter 6 - Inverse Functions - 6.2 Exponential Functions and Their Derivatives - 6.2 Exercises: 65

#### Answer

(a) 1 (b) $\frac{d{p(t)}}{dt} =\frac{ake^{-kt}}{ (1+ae^{-kt})^{2}}$ (c) Graph is depicted as below:

#### Work Step by Step

Under certain circumstances a rumor spreads according to the equation $p(t) =\frac{1}{1+ae^{-kt}}$ where $p(t)$ is the proportion of the population that has heard the rumor at time t and a and k are positive constants. (a) $\lim\limits_{t \to \infty} p(t)=\lim\limits_{t \to \infty} \frac{1}{1+ae^{-kt}}=\frac{1}{1+0}=1$ (b) Need to find the rate of spread of the rumor. $\frac{d{p(t)}}{dt} =\frac{d}{dt} (1+ae^{-kt})^{-1}$ Hence, $\frac{d{p(t)}}{dt} =\frac{ake^{-kt}}{ (1+ae^{-kt})^{2}}$ (c) Graph $p$ for the case $a= 10, k = 0.5$ with $t$ measured in hours. Use the graph to estimate how long it will take for 80% of the population to hear the rumor. $p(t) =\frac{1}{1+10e^{-0.5t}}$ This implies $p(t)=0.8$ when $t\approx 7.4$hours

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