Math Project

June 2, 2004

Logistics

Logistics is similar to simple exponential growth, explained below:

Simple Exponential Growth

This shows that the growth is directly proportional (by constant k) to the amount that is present.

The resulting equation (can be proven by integration) in terms of y is as follows.

Logistic growth begins proportional to the amount present, but the growth approaches 0 as y approaches L.

Following are two formulas that could be used in logistic growth:

or (Important note: Do not use the same k value in both of these formulas to describe the same situation.)

L is the limit that y approaches. k is a constant that determines how rapidly y approaches the limit. It has no effect on the final height the graph reaches, which depends only on L.

Consider what happens at significant values of y:

  1. If y is equal to or close to 0: . Note that is the formula for exponential growth. This means that logistic growth is initially exponential in nature.
  2. If y is equal to or close to L: . The growth approaches 0 as y approaches L. (This explains why L is the limit. The decrease in growth as y approaches L prevents it from going past that value.)

The equation for logistic growth in terms of y, which can be proven by integration, is as follows:

Examples

1. The carrying capacity for frogs is 3,000, and the growth is proportional to population n and to (3,000 – n). Last year, there were 500 frogs. This year, the population is 1,000. How long will it take for the population to reach 2,500?

2. A colony of bacteria has a population limit of 18,000. There were 1,000 at t = 0 hours.

The rate of growth of the population (y) is:

Find the formula for bacteria population in terms of t.

3. The population (y) of Lauxes in the state of New Jersey is modeled by:

  1. What is the initial population?
  2. What is the population limit?
  3. What is the constant k?
  4. When will the population be 90% of its maximum?
  5. Find a differential equation to explain this situation.

Solutions

Problem 1

The carrying capacity for frogs is 3,000, and the growth is proportional to population n and to (3,000 – n). Last year, there were 500 frogs. This year, the population is 1,000. How long will it take for the population to reach 2,500?

From various phrases in the question, certain algebraic expressions can be obtained.

Phrase / Algebraic expression
The carrying capacity for frogs is 3,000 / L = 3000
The growth is proportional to population n and to (3,000 – n) /
Last year, there were 500 frogs. / At t = 0, y = 500 (Assume last year to be the starting point, since it is the earliest time mentioned in the problem.)
This year, the population is 1,000. / At t = 1, y = 1000
How long will it take for the population to reach 2,500? / *The goal is to find t at y = 2500

*A point to keep in mind: A common approach to solving a problem when given two t,y pairs is to use a system of equations derived from one of the general formulas for logistics (in this case, use ).

Since L is known and two y, t pairs are known, the following system exists:

The solution to the system is C = 5 and

Using all variables known so far, the following equation can be written to express population y in terms of t:

Using y = 2500 (we want to know when the population will be 2500), we find that , so the population will be 2500 about 2.514 years from now (subtract 1 year because t = 0 represents the previous year.

Problem 2

A colony of bacteria has a population limit of 18,000. There were 1,000 at t = 0 hours.

The rate of growth of the population (y) is:

Find the formula for bacteria population in terms of t.

Here are some algebraic expressions that can be derived from the problem

Phrase / Algebraic expression
population limit of 18,000 / L = 18000
There were 1,000 at t = 0 hours / y = 1000 at t = 0
(formula) /

The goal is to substitute constants into the formula to produce a formula for y in terms of t.

Since the formula is of the form , one can infer that k = 0.0003.

The only other constant to solve for is C. To find this, substitute all known constants as well as the specific y, t pair known (1000, 0) into . The result is , which gives a C value of 17.

The solution, then, is .

Problem 3

The population (y) of Lauxes in the state of New Jersey is modeled by:

a. What is the initial population?

b. What is the population limit?

c. What is the constant k?

d. When will the population be 90% of its maximum?

  1. Find a differential equation to explain this situation.
  1. The initial population is the population at t = 0. Setting t equal to 0 in the equation gives a y value of about 1.200.
  2. Since the formula provided is of the form , the value of L can be simply taken as the number in the position of L, 3,000,000.
  3. Using the same method used in b, k is 0.7.
  4. The maximum is L. The question, then, is to find when y = 90% of L, or 0.9L. Using substitution, one can say that , resulting in a t value of 20.895.
  5. To find a differential equation of the form , substitute the values for the constants in the given equation of the form . From the formula provided, L = 3,000,000 and k = 0.7, so the differential equation is .