Do Now: #1-8, p.346
50
Let f  x  
1  5e 0.1x
Check the graph first?
1. Continuous for all real numbers
2.
lim f  x   50 lim f  x   0
x 
x 
3. H.A.: y = 0, y = 50
4. In both the first and second derivatives, the denominator
1  5e 0.1x, which is never 0.
will be a power of
domains of both are all real numbers.
Thus, the
Do Now: #1-8, p.346
50
Let f  x  
1  5e 0.1x
Check the graph first?
5. Graph f in [–30, 70] by [–10, 60]. f (x) has no zeros.
6. Graph the first derivative in [–30, 70] by [–0.5, 2].
Inc. interval:
 ,  
Dec. interval: None
7. Graph the second derivative in [–30, 70] by [–0.08, 0.08].
 ,16.094 Conc. down: 16.094,
8. Point of inflection: 16.094, 25
Conc. up:
LOGISTIC GROWTH
Section 6.5a
Review from last section…
Many populations grow at a rate proportional to the size of the
population. Thus, for some constant k,
dP
 kP
dt
Notice that
dP dt
 k is constant,
P and is called the relative growth rate.
Solution (from Sec. 6.4):
P  P0 e
kt
Logistic Growth Models
In reality, most populations are limited in growth. The maximum
population (M) is the carrying capacity.
The relative growth rate is proportional to 1 – (P/M), with
positive proportionality constant k:
dP dt
P

 k 1  
P
 M
or
dP k

P M  P
dt M
The solution to this logistic differential equation is called
the logistic growth model.
(What happens when P exceeds M???)
A national park is known to be capable of supporting no more
than 100 grizzly bears. Ten bears are in the park at present. We
model the population with a logistic differential eq. with k = 0.1.
(a) Draw and describe a slope field for the differential equation.
Carrying capacity = M = 100
k = 0.1
Differential Equation:
dP k
0.1

P M  P 
P 100  P 
dt M
100
 0.001P 100  P 
Use your calculator to get the slope field for this equation.
(Window: [0, 150] by [0, 150])
A national park is known to be capable of supporting no more
than 100 grizzly bears. Ten bears are in the park at present. We
model the population with a logistic differential eq. with k = 0.1.
(b) Find a logistic growth model P(t) for the population and draw
its graph.
Differential Equation:
Initial Condition:
dP
 0.001P 100  P  P  0  10
dt
1
dP
 0.001
Rewrite
P 100  P  dt
Partial Fractions
1 1
1  dP
 0.001
 

100  P 100  P  dt
A national park is known to be capable of supporting no more
than 100 grizzly bears. Ten bears are in the park at present. We
model the population with a logistic differential eq. with k = 0.1.
(b) Find a logistic growth model P(t) for the population and draw
its graph.
1 1
1  dP
 0.001
 

100  P 100  P  dt
Rewrite
1 
1
 
 dP  0.1dt
 P 100  P 
Integrate
Prop. of Logs
ln P  ln 100  P  0.1t  C
P
ln
 0.1t  C
100  P
A national park is known to be capable of supporting no more
than 100 grizzly bears. Ten bears are in the park at present. We
model the population with a logistic differential eq. with k = 0.1.
(b) Find a logistic growth model P(t) for the population and draw
its graph.
P
ln
 0.1t  C
100  P
100  P
Prop. of Logs ln
 0.1t  C
P
100  P
0.1t  C
Exponentiate
e
P
100  P
C
0.1t
Rewrite
 e e
P


A national park is known to be capable of supporting no more
than 100 grizzly bears. Ten bears are in the park at present. We
model the population with a logistic differential eq. with k = 0.1.
(b) Find a logistic growth model P(t) for the population and draw
its graph.
100  P
C
0.1t
  e  e
P
The Model:
0.1t
–c 100
Let A = +– e
 1  Ae
100
P
P
0.1t
1

9
e
100
Solve for P P 
0.1t
Graph this on top
1  Ae
of our slope field!
Initial Condition
100
10 
0  A9
1  Ae
A national park is known to be capable of supporting no more
than 100 grizzly bears. Ten bears are in the park at present. We
model the population with a logistic differential eq. with k = 0.1.
(c) When will the bear population reach 50?
Solve:
100
50 
0.1t
1  9e
1  9e
0.1t
e
0.1t
e
0.1t
2
1 9
9
ln 9
t
 21.972yr
0.1
Note: As illustrated in this example,
the solution to the general logistic
differential equation
dP k

P M  P
dt M
is always
M
P
 kt
1  Ae
More Practice Problems
For the population described, (a) write a diff. eq. for the
population, (b) find a formula for the population in terms of t, and
(c) superimpose the graph of the population function on a slope
field for the differential equation.
1. The relative growth rate of Flagstaff is 0.83% and its
current population is 60,500.
dP
 0.0083P
dt
P  60,500e
How does the graph look???
0.0083t
More Practice Problems
For the population described, (a) write a diff. eq. for the
population, (b) find a formula for the population in terms of t, and
(c) superimpose the graph of the population function on a slope
field for the differential equation.
2. A population of birds follows logistic growth with k = 0.04,
carrying capacity of 500, and initial population of 40.
dP k

P  M  P   0.00008 P  500  P 
dt M
M
500

P
0.04 t
 kt
1  11.5e
1  Ae
How does the
graph look???
More Practice Problems
The number of students infected by measles in a certain school
is given by the formula
200
P t  
1 e
5.3t
where t is the number of days after students are first exposed
to an infected student.
(a) Show that the function is a solution of a logistic differential
equation. Identify k and the carrying capacity.
200
200
M
P t  


5.3t
5.3  t
 kt
1 e
1 e e
1  Ae
This is a logistic growth model
with k = 1 and M = 200.
More Practice Problems
The number of students infected by measles in a certain school
is given by the formula
200
P t  
1 e
5.3t
where t is the number of days after students are first exposed
to an infected student.
(b) Estimate P(0). Explain its meaning in the context of the
problem.
200
P  0 
5.3  0.993  1
1 e
Initially (t = 0), 1 student has the measles.