When will the infected population equal the uninfected population?

Question

The spread of a virus can be modeled by exponential growth, but its growth is limited by the number of individuals that can be infected. For such situations the function, P(t)=Kpe^rt/K+p(ert+1), can be used, where P(t) is the infected populations t days after the first infection and p is the initial infected population, K is the total population that can be infected, and r is the rate the virus spreads written as a decimal.
a.A town of 10,000 people starts with 2 infected people and the virus has a growth rate of 20%. When will the growth of the infected population start to level off, and how many people will be infected at that point? Explain your reasoning and include any graphs you use to help you arrive at your conclusion.
b. When will the infected population equal the uninfected population.

Answer 1

(a)

 

 

 

p = 2, K = 10000, and r = 1.2 (since the infected population for a given day is 120% the previous day's infected population). So:

 

P(t) = (10000(2)e^1.2t) / (10000 + 2(e^1.2t + 1))

 

 

 

That's pretty difficult to think through without graphing. So, go to wolframalpha.com and plug in:

 

"y = (10000(2)e^(1.2t)) / (10000 + 2(e^(1.2t) + 1)) from t = 0 to 20"

 

From that graph, it seems to me like the infected population will start leveling off around 9 or 10 days, when 9500 or so people are infected.

 

Make sure you come to your own conclusion about what the problem means by "leveling off." That way you can explain your own reasoning behind what time and population you picked. Include the graph from wolframalpha.com in your answer.

 

 

 

(b)

 

The infected population and the uninfected population will be equal when the infected population is 5000. So, plug in P(t) = 5000 and solve for t.

 

5000 = (10000(2)e^1.2t) / (10000 + 2(e^1.2t + 1))

5000(10000 + 2(e^1.2t + 1)) = 10000(2)e^1.2t

50000000 + 10000(e^1.2t + 1) = 20000e^1.2t

50000000 + 10000e^1.2t + 10000 = 20000e^1.2t

50010000 + 10000e^1.2t = 20000e^1.2t

50010000 = 20000e^1.2t - 10000e^1.2t

50010000 = 10000e^1.2t

5001 = e^1.2t

ln(5001) = 1.2t

ln(5001)/1.2 = t

7.092 = t

 

The populations will be equal after approximately 7.092 days. This looks right on the graph.