Question #213964

3.A single individual starts a rumour in a community of 200 people. the rumour spread at a rate proportional to the number of people who have not yet heard the rumour. After 2days 10 people have heard the rumour

  1. how many people will have heard the rumour after 5days
  2. how long will it take for the rumour to spread to 100 people
1
Expert's answer
2021-07-07T13:40:52-0400

If NN denotes the number of people who have heard the rumor, then 200N200-N represents the number of people who haven’t heard the rumor.

Then


dNdt=k(200N)\dfrac{dN}{dt}=k(200-N)

dN200N=kdt\dfrac{dN}{200-N}=kdt

Integrate


ln200N=kt+lnC\ln|200-N|=-kt+\ln C

N=200CektN=200-Ce^{-kt}

N(0)=200C=0=>C=200N(0)=200-C=0=>C=200

N=200200ektN=200-200e^{-kt}

N(2)=200200ek(2)=10N(2)=200-200e^{-k(2)}=10

ek(2)=0.95e^{-k(2)}=0.95

2k=ln0.952k=-\ln0.95

k=12ln0.95k=-\dfrac{1}{2}\ln0.95

N(t)=200200e(ln0.95)t/2N(t)=200-200e^{(\ln0.95)t/2}

N(t)=200200(0.95)t/2N(t)=200-200(0.95)^{t/2}



1.


N(5)=200200(0.95)5/2=24N(5)=200-200(0.95)^{5/2}=24

2.

N(t)=200200e(ln0.95)t/2=100N(t)=200-200e^{(\ln0.95)t/2}=100

e(ln0.95)t/2=0.5e^{(\ln0.95)t/2}=0.5

(ln0.95)t=2ln0.5(\ln0.95)t=2\ln0.5

t=2ln0.5ln0.95t=\dfrac{2\ln0.5}{\ln0.95}

t=27 dayst=27\ days


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