Drill
• Find the integral using the given substitution
 2x
4x  3
2
 3x  1
dx
u  2 x  3x  1
2
1
du   4 x  3 dx
ln u  C
 u du
ln( 2 x  3 x  1)  C
2
Lesson 6.5: Logistic Growth
Day #1: P. 369: 1-17 (odd)
Day #2: P. 369/70: 19-29; odd, 37
Partial Fraction Decomposition with
Distinct Linear Denominators
• If f(x) = P(x)/Q(x) where P and Q are
polynomials with the degree of P less than the
degree of Q, and if Q(x) can be written as a
product of distinct linear factors, then f(x) can
be written as a sum of rational functions with
distinct linear denominators.
Example: Finding a Partial Fraction
Decomposition
• Write the function f(x)= (x-13)/(2x2 -7x+3) as a sum of rational
functions with linear denominators.
f ( x) 
x  13
( 2 x  1)( x  3 )
A

2x 1

B
f ( x) 
x3
A ( x  3 )  B ( 2 x  1)  x  13
1
, A(
2
A(
5
)  B (0) 
2
5 B   10 , B   2
1
1
1
 3 )  B ( 2 ( )  1)   13
2
2
2
 25
A(
2
f ( x) 
( 2 x  1)( x  3 )
If x  3, A ( 3  3 )  B ( 2 ( 3 )  1)  3  13
A ( 0 )  B ( 5 )   10
If x 
A ( x  3 )  B ( 2 x  1)
5
)
 25
2
x  13
( 2 x  1)( x  3 )

5
2x 1
2

2
x3
,A  5
Write the function f(x)= (2x+16)/(x2 +x-6) as a sum of
rational functions with linear denominators.
f ( x) 
2 x  16
( x  2 )( x  3 )

A
x2

B
f ( x) 
x3
A ( x  3 )  B ( x  2 )  2 x  16
A ( x  3)  B ( x  2 )
If x   3, A (  3  3 )  B (  3  2 )  2 (  3 )  16
A ( 0 )  B (  5 )  10
 5 B  10 , B   2
If x  2 , A ( 2  3 )  B ( 2  2 )  2 ( 2 )  16
A ( 5 )  B ( 0 )  20
f ( x) 
( x  2 )( x  3 )
A ( 5 )  20 , A  4
2 x  16
( x  2 )( x  3 )

4
x2

2
x3
Example: Antidifferentiating with
Partial Fractions
3x  1
4

x 1
2
Because the numerator’s degree is
larger than the denominator’s degree,
we will need to use polynomial division.
dx
3x2
+3
x  1 3x  0 x  0 x  0x  1
2
4
3x4
3
2
-3x2
3x2
3x2
+1
-3
4 
 2
  3 x  3  x 2  1 dx
4 
 2
  3 x  3  x 2  1 dx
 4 
x  3x    2
dx
 x 1
B 
 A
x  3x   

dx
 x 1 x 1
 A ( x  1)  B ( x  1) 
dx
x  3 x   
( x  1)( x  1)


3
3
3
A ( x  1)  B ( x  1)  4
When x = -1
A(0) + B(-2) = 4; B = - 2
When x = 1
A(2) + B(0) = 4; A= 2
2 
 2
x  3x   

dx
 x 1 x 1
3
x  3 x  2 ln( x  1)  2 ln( x  1)  C
3
x  3 x  2 ln ( x  1)  ln ( x  1)   C
3
x  3 x  2 ln
3
x 1
x 1
C
Finding Three Partial Fractions
6x  8x  4
2
 x
2

 4 ( x  1)
6x  8x  4
2
dx
x
2

 4 ( x  1)

A
x2

B
x2

C
x 1
A ( x  2 )( x  1)  B ( x  2 )( x  1)  C ( x  2 )( x  2 )
( x  2 )( x  2 )( x  1)
A ( x  2 )( x  1)  B ( x  2 )( x  1)  C ( x  2 )( x  2 )  6 x  8 x  4
2
When x = -2
B(-4)(-3)=36
12B = 36
B=3
When x = 1
C(-1)(3)=-6
-3C = -6
C=2
When x = 2
A(4)(1)=4
4A=4
A=1
3
2 
 1
  x  2  x  2  x  1  dx
ln x  2  3 ln x  2  2 ln x  1  C
3
ln x  2  ln x  2
3
 ln x  1
2
2
C
ln x  2 x  2 ln x  1  C
Drill: Evaluate the Integral
5 x  14
x
A
x

2
 7x
dx
B
A( x  7)  B ( x)
x7
x( x  7)
A ( x  7 )  B ( x )  5 x  14
3 
2
  x  x  7  dx
When x = -7, B(-7) = -21, B = 3
When x = 0, 7A = 14, A = 2
2 ln x  3 ln x  7  C 
ln x
ln x
2
 ln x  7
2
x7
3
3
C
C 
The Logistic Differential Equation
If we want the growth rate
to approach 0 as P approaches
a maximal carrying capacity M,
we can introduce a limiting
factor of M – P.
Remember that exponential
growth can be modeled by
dP/dt = kP, for some k >0
Logistic Differential
Equation
dP/dt = kP(M-P)
General Logistic Formula
• The solution of the general logistic differential
equation dP
 kP ( M  P )
dt
is
P 
M
1  Ae
 ( Mk ) t
Where A is a constant determined by an
appropriate initial condition. The carrying
capacity M and the growth constant k are
positive constants.
Example: Using Logistic Regression
• The population of Aurora, CO for selected
years between 1950 and 2003
Years after 1950
Population
0
11,421
20
74, 974
30
158,588
40
222,103
50
275,923
53
290,418
Use logistic regression to find a model.
• In order to this with a table, enter the data into
the calculator (L1 = year, L2 = population)
• Stat, Calc, B: Logistic
P 
316440 . 7
1  23 . 577 e
 . 1026 t
• Based on the regression equation, what will the
Aurora population approach in the long run?
 316,441 people (note that the carrying capacity is the
numerator of the equation!)
When will the population first
exceed 300,000?
• Use the calculator to
intersect.
y1 
316440 . 7
1  23 . 577 e
Write a logistic equation in
the form of dP/dt = kP(M-P)
that models the data.
P 
M
1  Ae
 ( Mk ) t
 . 1026 t
y 2  300 , 000
• When x = 59.12 or in the
59th year: 1950 + 59 = 2009
• We know that M = 316440.7 and
Mk =.1026
• Therefore 316440.7k=.1026,
making k = 3.24 X 10-7
• dP/dt= 3.24 X 10-7 P(316440.7-P)
6.5, day #2
• Review the following notes, ON YOUR OWN
• You can get extra credit if you complete the
homework for day 2. (Will replace a missing
homework, OR if you do have no missing
assignments, it will push your homework
grade over 100%)
• Correct answers + work will give you extra
credit on Friday’s quiz.
• No work = No extra credit
Example
• The growth rate of a population P of bears in a
newly established wildlife preserve is modeled by
the differential equation dP/dt = .008P(100-P),
where t is measured in years.
A. What is the carrying capacity for bears in this
wildlife preserve?
Remember that M is the carrying capacity, and
according to the equation, M = 100 bears
B. What is the bear
population when the
population is growing
the fastest?
C. What is the rate of
change of population
when it is growing the
fastest?
The growth rate is always
maximized when the
population reaches half
the carrying capacity.
dP/dt = .008P(100-P)
dP/dt = .008(50)(100-50)
dP/dt =20 bears per year
In this case, it is 50 bears.
Example
• In 1985 and 1987, the
• What is the carrying
Michigan Department of
capacity?
Natural Resources
1000 moose
airlifted 61 moose from
Algonquin Park, Ontario
Generate a slope field. The
to Marquette County in
window should be x: [0, 25]
the Upper Peninsula. It
and y: [0, 1000]
was originally hoped that
the population P would
reach carrying capacity in
about 25 years with a
growth rate of dP/dt =
.0003P(1000-P)
dP
• Solve the differential
equation with the initial
condition P(0) = 61
dP
P (1000  P )
 . 0003 dt
B
 A



  P 1000  P dP 
 . 0003 dt
A (1000  P )  B ( P )  1
 . 0003 P (1000  P )
dt
We first need to separate the variables.
Then we will integrate one side by using
partial fractions.
dP
 P (1000
 P)

 .0003 dt
 A (1000  P )  B ( P ) 
  P (1000  P ) dP 


P  0
P  1000
1000 A  1
1000 B  1
A  . 001
B  . 001
. 001
 . 001



  P 1000  P dP 
 . 0003 dt


1

  P (1000  P ) dP


. 001
 . 001



  P 1000  P dP 
1
1



  P 1000  P dP 
 . 0003 dt
 .3 dt
ln P   1 ln( 1000  P )  . 3 t  C
ln( 1000  P )  ln P  .  3t   C
ln
1000  P
P
  .3t  C
Solve the differential equation with the initial condition P(0) = 61
ln
1000  P
  .3t  C
P
1000  P
1000
 e
 .3 t  C
P
1000
P
e
 .3 t
e
C
1
1  e
 .3 t
e
C
P
1000
e
 .3 ( 0 )
61
15 . 39344  e
C
e
C
1
1000
e
 .3 t
e
C
1
P
P
1000
P 

1
15 . 3944 e
 .3 t
1
1000
15 . 3944 e
 .3 t
1