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 $a = 10$, then $\lim_{t
ightarrow\infty}p(t)=\frac{1}{1+0}=1$.

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

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 $p^{\prime}(t)=\frac{kae^{-kt}}{(1 + ae^{-kt})^{2}}$. 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 $t$ when $p(t)=0.8$

Set $p(t)=0.8=\frac{1}{1 + 10e^{-0.5t}}$. Cross - multiply: $0.8(1 + 10e^{-0.5t})=1$. Expand: $0.8+8e^{-0.5t}=1$. Then $8e^{-0.5t}=1 - 0.8=0.2$. So $e^{-0.5t}=\frac{0.2}{8}=0.025$. Take the natural logarithm of both sides: $-0.5t=\ln(0.025)$. Solve for $t$: $t=\frac{\ln(0.025)}{-0.5}=\frac{-3.6888}{-0.5}=7.3776\approx7.38$ hours.

Answer:

(a) $\lim_{t
ightarrow\infty}p(t)=1$
(b) $p^{\prime}(t)=\frac{5e^{-0.5t}}{(1 + 10e^{-0.5t})^{2}}$
(c) $t\approx7.38$ hours