Question

Under certain circumstances, a rumor spreads according to the equation p(t)=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 limt→[infinity]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.

Answer 1

Answer:

(a) $\lim_{t \to \infty} \frac{1}{1+ae^{-kt}}=1$

(b) $\frac{d}{dt}\left(\frac{1}{1+ae^{-kt}}\right)=\frac{ake^{-kt}}{\left(1+ae^{-kt}\right)^2}$

(c) 80% of the population will have heard the rumor in about 7.4 hrs.

Step-by-step explanation:

We know that 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) To find $\lim_{t \to \infty} p(t)$ you must:

Use this fact,

$\lim _{x\to a}\left[\frac{f\left(x\right)}{g\left(x\right)}\right]=\frac{\lim _{x\to a}f\left(x\right)}{\lim _{x\to a}g\left(x\right)},\:\quad \lim _{x\to a}g\left(x\right)\ne 0$

$\lim_{t \to \infty} \frac{1}{1+ae^{-kt}}\\\\\frac{\lim _{t\to \infty \:}\left(1\right)}{\lim _{t\to \infty \:}\left(1+ae^{-kt}\right)}$

Apply this identity, to find $\lim _{t\to \infty \:}\left(1)$

$\lim _{x\to a}c=c$

$\lim _{t\to \infty \:}\left(1\right)=1$

$\lim _{t\to \infty \:}\left(1+ae^{-kt}\right)=\lim _{t\to \infty \:}\left\left(1\right)+\lim _{t\to \infty \:}\left(ae^{-kt}\right)\\\\\lim _{t\to \infty \:}\left(1+ae^{-kt}\right)=1+a\cdot \lim _{t\to \infty \:}\left(e^{-kt}\right)\\\\\lim _{t\to \infty \:}\left(1+ae^{-kt}\right)=1+0$

$\frac{\lim _{t\to \infty \:}\left(1\right)}{\lim _{t\to \infty \:}\left(1+ae^{-kt}\right)}=\frac{1}{1+0} =1$

(b) To find the rate of speed of the rumor you must find the derivative $\frac{dp}{dt}$

$\frac{d}{dt}\left(\frac{1}{1+ae^{-kt}}\right)\\\\\frac{d}{dt}\left(\left(1+ae^{-kt}\right)^{-1}\right)\\\\\mathrm{Apply\:the\:chain\:rule}:\quad \frac{df\left(u\right)}{dx}=\frac{df}{du}\cdot \frac{du}{dx}\\\\\frac{d}{du}\left(u^{-1}\right)\frac{d}{dt}\left(1+ae^{-kt}\right)\\\\\left(-\frac{1}{u^2}\right)\left(-ake^{-kt}\right)\\\\\mathrm{Substitute\:back}\:u=\left(1+ae^{-kt}\right)\\\\\left(-\frac{1}{\left(1+ae^{-kt}\right)^2}\right)\left(-ake^{-kt}\right)$

$\frac{d}{dt}\left(\frac{1}{1+ae^{-kt}}\right)=\frac{ake^{-kt}}{\left(1+ae^{-kt}\right)^2}$

(c) To find the time that will take for 80% of the population to hear the rumor, you must substitute a = 10, k = 0.5, and p(t) = 0.8 into $p(t)=\frac{1}{1+ae^{-kt}}$ and solve for t

$\frac{1}{1+10e^{-0.5t}}=0.8\\\\\frac{1}{1+10e^{-0.5t}}\left(1+10e^{-0.5t}\right)=0.8\left(1+10e^{-0.5t}\right)\\\\1=0.8\left(1+10e^{-0.5t}\right)\\\\0.8\left(1+10e^{-0.5t}\right)=1\\\\0.8\left(1+10e^{-0.5t}\right)\cdot \:10=1\cdot \:10\\\\8\left(1+10e^{-0.5t}\right)=10\\\\\frac{8\left(1+10e^{-0.5t}\right)}{8}=\frac{10}{8}\\\\1+10e^{-0.5t}=\frac{5}{4}\\\\1+10e^{-0.5t}-1=\frac{5}{4}-1\\\\10e^{-0.5t}=\frac{1}{4}\\\\\frac{10e^{-0.5t}}{10}=\frac{\frac{1}{4}}{10}$

$e^{-0.5t}=\frac{1}{40}\\\\\ln \left(e^{-0.5t}\right)=\ln \left(\frac{1}{40}\right)\\\\-0.5t\ln \left(e\right)=\ln \left(\frac{1}{40}\right)\\\\-0.5t=\ln \left(\frac{1}{40}\right)\\\\t=2\ln \left(40\right) \approx 7.377 \:hrs$