Question

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 knows the rumor at time $t$, and $a$ and $k$ are positive constants. (a) find $lim_{t
ightarrowinfty}p(t)$. (b) find the rate of speed of the rumor. (c) how long will it take for 80% of the population to hear the rumor? here, take $a = 10$, $k = 0.5$ with $t$ measured in hours. (a) limit = (b) $p(t)=$ (c) hours

Explanation:

Step1: Find $\lim_{t

ightarrow\infty}p(t)$
Given $p(t)=\frac{1}{1 + ae^{-kt}}$. As $t
ightarrow\infty$, $e^{-kt}
ightarrow0$ since $k = 0.5>0$. So, $\lim_{t
ightarrow\infty}p(t)=\lim_{t
ightarrow\infty}\frac{1}{1 + ae^{-kt}}=\frac{1}{1+0}=1$.

Step2: Find the rate of spread $p^{\prime}(t)$

First, rewrite $p(t)=(1 + ae^{-kt})^{-1}$. Using the chain - rule, if $u = 1+ae^{-kt}$, then $p(t)=u^{-1}$. The derivative of $u$ with respect to $t$ is $u^{\prime}=-kae^{-kt}$, and the derivative of $p$ with respect to $u$ is $p^{\prime}(u)=-u^{-2}$. So, $p^{\prime}(t)=\frac{kae^{-kt}}{(1 + ae^{-kt})^{2}}$.

Step3: Find the time when 80% of the population hears the rumor

Set $p(t)=0.8=\frac{1}{1 + ae^{-kt}}$. Cross - multiply to get $0.8(1 + ae^{-kt}) = 1$. Then $1+ae^{-kt}=\frac{1}{0.8}=1.25$, and $ae^{-kt}=0.25$. Given $a = 10$ and $k = 0.5$, we have $10e^{-0.5t}=0.25$. So, $e^{-0.5t}=\frac{0.25}{10}=0.025$. Take the natural logarithm of both sides: $-0.5t=\ln(0.025)$. Then $t=\frac{\ln(0.025)}{- 0.5}=\frac{\ln(\frac{1}{40})}{-0.5}=\frac{-\ln(40)}{-0.5}=\frac{\ln(40)}{0.5}\approx\frac{3.6889}{0.5}=7.3778\approx7.38$ hours.

Answer:

(a) 1
(b) $\frac{kae^{-kt}}{(1 + ae^{-kt})^{2}}$
(c) 7.38 hours