Download Logistics

Math Project
June 2, 2004
Logistics
Logistics is similar to simple exponential growth, explained below:
Simple Exponential Growth
dy
 ky
dt
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.
y  Cekt
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:
dy
y
dy
 ky (1  ) or
 ky ( L  y ) (Important note: Do not use the same k value in both of
dt
L
dt
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:
dy
0
dy
 ky (1  )  ky . Note that
 ky is the formula
1. If y is equal to or close to 0:
dt
L
dt
for exponential growth. This means that logistic growth is initially exponential in
nature.
dy
L
 ky (1  )  ky (0)  0 . The growth approaches 0
2. If y is equal to or close to L:
dt
L
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:
L
y   kt
Ce  1
Math Project
June 2, 2004
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:
dy
 0.0003 y (18000  y )
dt
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:
3, 000, 000
y
1  2,500, 000e0.7t
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?
e. Find a differential equation to explain this situation.
Math Project
June 2, 2004
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
dy
 kn(3000  n)
and to (3,000 – n)
dt
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
*The goal is to find t at y = 2500
reach 2,500?
*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
L
logistics (in this case, use y   kt
).
Ce  1
Since L is known and two y, t pairs are known, the following system exists:
3000

500  Ce k *0  1

2500  3000

Ce k *1  1
The solution to the system is C = 5 and k  0.916
Using all variables known so far, the following equation can be written to express
population y in terms of t:
3000
y  0.916t
5e
1
Using y = 2500 (we want to know when the population will be 2500), we find that
t  3.514 , so the population will be 2500 about 2.514 years from now (subtract 1 year
because t = 0 represents the previous year.
Math Project
June 2, 2004
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:
dy
 0.0003 y (18000  y )
dt
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)
dy
 0.0003 y (18000  y )
dt
L
The goal is to substitute constants into the formula y   kt
to produce a
Ce  1
formula for y in terms of t.
Since the formula
dy
dy
 0.0003 y (18000  y ) is of the form
 ky ( L  y ) , one
dt
dt
can infer that k = 0.0003.
The only other constant to solve for is C. To find this, substitute all known
L
constants as well as the specific y, t pair known (1000, 0) into y   kt
. The result is
Ce  1
18000
1000  0.0003*0
, which gives a C value of 17.
Ce
1
The solution, then, is y 
18000
.
17e 0.0003t  1
Math Project
June 2, 2004
Problem 3
The population (y) of Lauxes in the state of New Jersey is modeled by:
3, 000, 000
y
1  2,500, 000e0.7t
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?
e. Find a differential equation to explain this situation.
a. 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.
L
b. Since the formula provided is of the form y   kt
, the value of L can be
Ce  1
simply taken as the number in the position of L, 3,000,000.
c. Using the same method used in b, k is 0.7.
d. The maximum is L. The question, then, is to find when y = 90% of L, or 0.9L.
3, 000, 000
Using substitution, one can say that (0.9*3, 000, 000) 
,
1  2,500, 000e0.7t
resulting in a t value of 20.895.
dy
 ky ( L  y ) , substitute the values
e. To find a differential equation of the form
dt
L
for the constants in the given equation of the form y   kt
. From the
Ce  1
formula provided, L = 3,000,000 and k = 0.7, so the differential equation is
dy
 0.7 y (3, 000, 000  y ) .
dt
y = 3000000/(1+2500000e^(-0.7x))