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