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)$
As $t
ightarrow\infty$, $e^{-kt}
ightarrow0$ since $k = 0.5>0$. Given $p(t)=\frac{1}{1 + ae^{-kt}}$ and $a = 10$, we have $\lim_{t
ightarrow\infty}p(t)=\lim_{t
ightarrow\infty}\frac{1}{1 + 10e^{-0.5t}}=\frac{1}{1+0}=1$.

Step2: Find $p^{\prime}(t)$

First, rewrite $p(t)=(1 + ae^{-kt})^{-1}$. Using the chain - rule, if $y=(1 + ae^{-kt})^{-1}$ and $u = 1+ae^{-kt}$, then $y = u^{-1}$ and $\frac{dy}{du}=-u^{-2}$, $\frac{du}{dt}=-kae^{-kt}$. So $p^{\prime}(t)=\frac{dy}{du}\cdot\frac{du}{dt}=\frac{kae^{-kt}}{(1 + ae^{-kt})^{2}}$. Substituting $a = 10$ and $k = 0.5$, we get $p^{\prime}(t)=\frac{5e^{-0.5t}}{(1 + 10e^{-0.5t})^{2}}$.

Step3: Find the time when $p(t)=0.8$

Set $p(t)=0.8=\frac{1}{1 + 10e^{-0.5t}}$. Then $1 + 10e^{-0.5t}=\frac{1}{0.8}=1.25$. So $10e^{-0.5t}=1.25 - 1=0.25$, and $e^{-0.5t}=0.025$. Taking the natural logarithm of both sides, $-0.5t=\ln(0.025)$. Solving for $t$, we have $t=\frac{\ln(0.025)}{- 0.5}=\frac{\ln(\frac{1}{40})}{-0.5}=\frac{-\ln(40)}{-0.5}=\frac{\ln(40)}{0.5}\approx\frac{3.6889}{0.5}=7.3778\approx7.38$ hours.

Answer:

(a) $\lim_{t
ightarrow\infty}p(t)=1$
(b) $p^{\prime}(t)=\frac{5e^{-0.5t}}{(1 + 10e^{-0.5t})^{2}}$
(c) $7.38$ hours