6.5

Logistic Growth Model

Years

Bears

Greg Kelly, Hanford High School, Richland, Washington

We have used the exponential growth equationto represent population growth.

0kty y e

The exponential growth equation occurs when the rate of growth is proportional to the amount present.

If we use P to represent the population, the differential equation becomes: dP

kPdt

The constant k is called the relative growth rate.

/dP dtk

P

The population growth model becomes: 0ktP P e

However, real-life populations do not increase forever. There is some limiting factor such as food, living space or waste disposal.

There is a maximum population, or carrying capacity, M.

A more realistic model is the logistic growth model where

growth rate is proportional to both the amount present (P)

and the fraction of the carrying capacity that remains: M P

M

The equation then becomes:

dP M PkP

dt M

Our book writes it this way:

Logistics Differential Equation

dP kP M P

dt M

We can solve this differential equation to find the logistics growth model.

PartialFractions

Logistics Differential Equation

dP kP M P

dt M

1 k

dP dtP M P M

1 A B

P M P P M P

1 A M P BP

1 AM AP BP

1 AM

1A

M

0 AP BP AP BP

A B1

BM

1 1 1 kdP dt

M P M P M

ln lnP M P kt C

lnP

kt CM P

Logistics Differential Equation

dP kP M P

dt M

1 k

dP dtP M P M

1 1 1 kdP dt

M P M P M

ln lnP M P kt C

lnP

kt CM P

kt CPe

M P

kt CM Pe

P

1 kt CMe

P

1 kt CMe

P

Logistics Differential Equation

kt CPe

M P

kt CM Pe

P

1 kt CMe

P

1 kt CMe

P

1 kt C

MP

e

1 C kt

MP

e e

CLet A e

1 kt

MP

Ae

Logistics Growth Model

1 kt

MP

Ae

Logistics Differential Equation

dP kP M P

dt M

So the solution for:

is:

Example:

Logistic Growth Model

Ten grizzly bears were introduced to a national park 10 years ago. There are 23 bears in the park at the present time. The park can support a maximum of 100 bears.

Assuming a logistic growth model, when will the bear population reach 50? 75? 100?

Ten grizzly bears were introduced to a national park 10 years ago. There are 23 bears in the park at the present time. The park can support a maximum of 100 bears.

Assuming a logistic growth model, when will the bear population reach 50? 75? 100?

1 kt

MP

Ae 100M 0 10P 10 23P

1 kt

MP

Ae 100M 0 10P 10 23P

0

10010

1 Ae

10010

1 A

10 10 100A

10 90A

9A

At time zero, the population is 10.

100

1 9 ktP

e

1 kt

MP

Ae 100M 0 10P 10 23P

After 10 years, the population is 23.

100

1 9 ktP

e

10

10023

1 9 ke

10 1001 9

23ke

10 779

23ke

10 0.371981ke

10 0.988913k

0.098891k

100

1 9 ktP

e

Let’s leave k as is here…

100

1 9 ktP

e

Years

Bears

We can graph this equation and use “trace” to find the solutions.

y=50 at 22 years

y=75 at 33 years

y=100 at 75 years

Before graphing, use the store function on your calculator to assign

0.098891k

We’d rather not round k when plugging into the calculator so that our graph can be as accurate as possible

0.1

100

1 9 tP

e

Years

Bears

It appears too that the graph is steepest around 22 years. What does this mean in terms of ?

dt

dP

y=50 at 22 years

Before graphing, use the store function on your calculator to assign

0.098891k

)( PMPM

k

dt

dP

To find the maximum rate of change, we need to find when the second derivative equals zero.

)()(2

2

PPM

kPMP

M

k

dt

Pd

)(2

2

PPMPM

k

dt

Pd

)2(2

2

PMPM

k

dt

Pd

)2)((2

2

2

PMPMPM

k

dt

Pd

To find the maximum rate of change, we need to find when the second derivative equals zero.

0)2)((2

2

2

PMPMP

M

k

dt

Pd

Which happens

when P = 0

M – P = 0 which happens when P = M or when the population has reached its carrying capacity

M – 2P = 0 which

happens when 2P = M or when the population has reached half of its carrying capacity

So in any logistic growth model, the rate of increase of a population is always a maximum when

2

MP