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(e^rt - 1) can be used, where P(t) is the infected population t days after the first infection, 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 a virus 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 draw, with or without technology.

b. When will the infected population equal to the uninfected population?

Question

sicabeth sicabeth

23-06-2019
Mathematics

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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(e^rt - 1) can be used, where P(t) is the infected population t days after the first infection, 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 a virus 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 draw, with or without technology.

b. When will the infected population equal to the uninfected population?

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sqdancefan sqdancefan · Answer 1 · 2019-06-24T07:16:25+02:00

Answer:

a) growth will reach a peak and begin declining after about 42.6 days. 5000 people will be infected at that point

b) the infected an uninfected populations will be the same after about 42.6 days

Step-by-step explanation:

We have assumed you intend the function to match the form of a logistic function:

[tex]P(t)=\dfrac{Kpe^{rt}}{K+p(e^{rt}-1}[/tex]

This function is symmetrical about its point of inflection, when half the population is infected. That is, up to that point, it is concave upward, increasing at an increasing rate. After that point, it is concave downward, decreasing at a decreasing rate.

a) The growth rate starts to decline at the point of inflection, when half the population is infected. That time is about 42.6 days after the start of the infection. 5000 people will be infected at that point

b) The infected and uninfected populations will be equal about 42.6 days after the start of the infection.