Logistic
Growth
10.4
A question on the 2008 AP exam:
dy
If
 ky and k is a nonzero constant, then y could be
dt
a) 2e
kty
b) 2e
kt
kt
c)e +3
d) kty  5
1 2 1
e) ky 
2
2
A question on the 2008 AP exam:
dy
Population y grows according to the equation
 ky, where k is
dt
a nonzero constant and t is measured in years. If the population
doubles every 10 years, then the value of k is
a) 0.069
b) 0.200
c)0.301
d) 3.322
e)5.000
If the growth rate (or decay rate) of a population, P, is
proportional to the population itself, we say :
dP
 kP
dt
In other words, the larger the population, the faster it
grows. The smaller the population, the slower it grows.
Solving this differential equation results in the population
growth model :
P  P0ekt

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, L.
A more realistic model is the logistic growth model where
growth rate is proportional to both the amount present and
the carrying capacity.

The equation then becomes:
Logistics Differential Equation
 y
y  ky 1  
 L
We can solve this differential equation to find the logistics
growth model.

Logistics Growth Model
L  y0
L
y
where A 
 kt
1  Ae
y0
Logistics Differential Equation
 y
y  ky 1  
 L

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?
L  p0
L
P
where A 
 kt
1  Ae
p0
L  100
P(0)  10
P(10)  23

L  p0
L
P
where A 
 kt
1  Ae
p0
100  10
A
9
10
100
P
1  9e  kt
Solve for 100
k using P(10)  23.
P
1  9100
e .099t
23 
1  9e  k (10)
1  9e
 k (10)
100

23
L  100 P(0)  10 P(10)  23
77
Note:
10 k
9e
 also be
The value of A can
23
found algebraically
77  1by

10 k
e
P(0)=10.
substituting
 
23  9 
ln e
10 k
 77  1  
 ln    
 23  9  
 77  1 
10k  ln   
 23  9 
k  .099

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?
100
P
1  9e .099t

100
P
1  9e 0.099t
We can graph
this equation
and use “trace”
to find the
solutions.
100
80
60
Bears
40
20
0
We can also solve
algebraically.
20
40
Years
60
80
100
y=50 at 22 years
y=75 at 33 years
y=100 at 75 years
p
Solve the initial value problem.
y  0.0004 y(100  y), y (0)  40
Factor out the 100.
 
y 
y  0.0004 y 100 1 

100

 
y 

y  0.04 y 1 

 100 
k  0.04 L  100
Find A.
100  40
A
 1.5
40
Logistics Growth Model
y
L  y0
L
where
A

1  Ae kt
y0
Logistics Differential Equation
 y
y  ky 1  
 L
L
y
1  Ae  kt
100
y
1  1.5e 0.04t
Match the differential equation with the corresponding
slope field. The window is [-20,100]x[-10,200].
1.
y 

y  .06 y 1 

 180 
a.
2.
y 

y  .06 y 1 

 100 
b.
y 

3. y  .06 y 1 

 100 
c.
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