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Logistic Differential Equations for AP Calculus

based on 5 ratings
By — McGraw-Hill Professional
Updated on Apr 25, 2014

Practice problems for these concepts can be found at: Applications of Definite Integrals Practice Problems for AP Calculus

Often population may grow exponentially at first, but eventually slows as it nears a limit, called the carrying capacity. This patten is called logistic growth, and is represented by the differential equation , in which P is the population, K is the carrying capacity, and k is the proportional constant. The differential equation is separable so . This equation can be integrated using a partial fraction decomposition.

Exponention produces . Solving for P yields . Dividing numerator and denominator by C2e kt,. At t = 0, . Solving for C2 yields or , Let ,and the solution of this logistic differential equation with initial condition P(0)= P0 is where K is the carrying capacity and .

Example 1

The population of Great Britain was 57.1 million in 2001 and 60.6 million in 2006. Find a logistic model for the growth of the population, assuming a carrying capacity of 100 million. Use the model to predict the population in 2020.

Step 1:   Since the carrying capacity is K =100,.

Step 2:   The solution of the differential equation, if is .

Step 3:   Take 2006 as t =5, P (5)=60.6. Then. Solving gives k ≈ 0.0289 so

Step 1:   Since the year 2020 corresponds to t =19, Substitute and evaluate . The population of Great Britain in 2020 is predicted to be approximately 69.742 million.

Example 2

The spread of an infectious disease can often be modeled by a logistic equation with the total exposed population as the carrying capacity. In a community of 2000 individuals, the first case of a new virus is diagnosed on March 31, and by April 10, there are 500 individuals infected. Write a differential equation that models the rate at which the virus spread through the community and determine when 98% of the population will have contracted the virus.

Step 1:   The rate of spread is

Step 2:   The solution of the differential equation is , and with one person exposed, or .

Step 3:   Taking April 10 as day 10, . Solving the equation gives k ≈ .6502, so .

Step 4:   98% of the population of 2000 is 1960 people. To determine the day when 1960 people are infected, solve . This gives t ≈ 17.6749, so the 98% infection rate should be reached by April 18.

Practice problems for these concepts can be found at: Applications of Definite Integrals Practice Problems for AP Calculus

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