Transcript Logistic Growth - Northland Preparatory Academy

Do Now: #1-8, p.346
Let f
x 
50
1  5e
 0.1 x
Check the graph first?
1. Continuous for all real numbers
2. lim f
x 
 x   50
lim f
x  
x  0
3. H.A.: y = 0, y = 50
4. In both the first and second derivatives, the denominator
will be a power of 1  5 e
 0.1 x
, which is never 0. Thus, the
domains of both are all real numbers.
Do Now: #1-8, p.346
Let f
x 
50
1  5e
 0.1 x
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].
Conc. up:   ,1 6 .0 9 4  Conc. down:  1 6 .0 9 4,  
8. Point of inflection:  1 6 .0 9 4, 2 5 
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
P
k
is constant,
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
dP
k
P 


P M  P
 k 1 
 or
dt
M
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
dt

k
M
P M  P 
0.1
100
P 100  P 
 0 .0 0 1 P  1 0 0  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
P  0   10
dt
 0.001 P  100  P 
Rewrite
1
dP
P  100  P  dt
 0.001
1  1
1
 dP
Partial Fractions
 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
1
 1

Rewrite  
 dP  0.1dt
 P 100  P 
Integrate ln P  ln 1 0 0  P  0 .1t  C
Prop. of Logs ln
P
100  P
 0.1t  C
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.
ln
P
100  P
Prop. of Logs ln
 0.1t  C
100  P
  0.1t  C
P
Exponentiate
100  P
e
 0.1 t  C
P
Rewrite
100  P
P
 e
C
e
 0.1 t
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
 e
C
P
 0.1 t
–c 100
Let A = +– e
 1  Ae
P
100
Solve for P P 
 0.1 t
1  Ae
Initial Condition 10 
100
1  Ae
0
e
 0.1 t
The Model:
P 
100
1  9e
 0.1 t
Graph this on top
of our slope field!
 A9
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:
50 
1  9e
1  9e
 0.1 t
e
 0.1 t
e
t
100
ln 9
0.1
0.1 t
 0.1 t
2
1 9
9
 21.972 yr
Note: As illustrated in this example,
the solution to the general logistic
differential equation
dP

dt
k
M
P M  P
is always
P 
M
1  Ae
 kt
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
dt
 0.0083 P
P  60, 500 e
How does the graph look???
0.0083 t
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
dt
P 

k
M
P  M  P   0.00008 P  500  P 
M
1  Ae
 kt

500
1  11.5 e
 0.04 t
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.
P t  
200
1 e
5.3  t

200
1 e e
5.3
t

M
1  Ae
This is a logistic growth model
with k = 1 and M = 200.
 kt
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.
P 0 
200
1 e
5.3
 0 .9 9 3
1
Initially (t = 0), 1 student has the measles.