Suppose that a given population can be divided into two parts: those who have a

Question

Suppose that a given population can be divided into two parts: those who have a given disease and can infect others, and those who don%u2019t have the disease but are susceptible. Let x be the percentage of susceptible individuals of the population and y be the percentage of infected individuals of the population. Assume that the disease spreads by contact between sick and well members, and that the rate of spread is directly proportional to the number of such contacts. Further, assume that the members of both groups move about freely among each other, so that the number of contacts is directly proportional to the product of the percentage of infected individuals and the percentage of non-infected individuals. Suppose that initially 5 people are infected of a town with 50,000 people. After 3 days 75 people are infected. How long will it take before 85% of the population is infected? Sketch the graph (using software or by hand) of the percentage of infected people as time increases

Explanation / Answer

dy/dt = k x y

so

in this case

dy/dt = k*(50000-y)y
with y(0) = 5

this has solution

y(t) = (50000 e^(50000 k t))/(e^(50000 k t)+9999)

use condition that y(3) = 75

75=(50000 e^(50000 k t))/(e^(50000 k t)+9999)

50000k=0.90315

so

y=
(50000 e^(0.90315 t))/(e^(0.90315 t)+9999)

we want to know when y = .85*50000

so

.85*50000=(50000 e^(0.90315 t))/(e^(0.90315 t)+9999)

t=12.12 days

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