6.5 logistic growth model years bears greg kelly, hanford high school, richland, washington
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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
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