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How to Find the Logistic Growth Using the Carrying Capacity of a Population

The carrying capacity K is the maximum population of a particular group that may be sustained in a region. Logistic growth is when the growth factor N is proportional to the population itself N and proportional to the difference between the population N and the carrying capacity K. You can therefore write logistic growth as a separable differential equation. The differential equation is:

N = k N (K N)

The general solution is given by the formula:

N = K CeKkt + 1

If you are asked to solve the integral, this equation is useful:

1 N (K N)dN = 1 K ln N K N + C,0 < N(0) < K

1 N (K N)dN = 1 K ln N K N + C,0 < N(0) < K.

Example 1

The number of bacteria in a polluted drinking water reservoir was originally 800. In the first hour afterwards, the number of bacteria increased by 320. Assume that the number of bacteria follows a model of logistic growth and that the carrying capacity is 7500.

Let N be the number of bacteria after t hours. Find k, determine the growth model and find how many bacteria there were after 5 and 13 hours.

From the text you see that N(0) = 800, N = 320, K = 7500. You find k by entering the values into the equation and solving for k: 320 = k 800 (7500 800) k = 0.00006

You can now enter k straight into the differential equation and solve this to find the growth model:

N = 0.00006 N (7500 N)

Solve the differential equation by entering into the formula:

N = 7500 Ce75000.00006t + 1 = 7500 Ce0.4478t + 1

N = 7500 Ce75000.00006t + 1 = 7500 Ce0.4478t + 1

To find the growth model in this case you need to find C. From the text you know the initial condition N(0) = 800. Enter that into the equation for N and you get
800 = 7500 Ce0.44780 + 1 = 7500 C + 1| (C + 1) 800 (C + 1) = 7500 800C + 800 = 7500 C = 7500 800 800 = 8.375

800 = 7500 Ce0.44780 + 1 = 7500 C + 1| (C + 1) 800 (C + 1) = 7500 800C + 800 = 7500 C = 7500 800 800 = 8.375

This gives the growth model:
N(t) = 7500 8.375e0.4478t + 1

You can now find the number of bacteria for t = 5 and t = 13: N(5) = 7500 8.375e0.44785 + 1 = 3963 N(13) = 7500 8.375e0.447813 + 1 = 7318

The number of bacteria after 5 and 13 hours is 3963 and 7318 respectively.

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