Advanced Placement Calculus
Differential Equations and Mathematical Modeling
Chapter 7
Section 5
Logistic Growth
Essential Question: How are population growth equations determined? What are their
differences?
Student Objective: The student will be able to solve problems involving exponential or
logistic population growth.
Vocabulary:
Continuous growth
Differential equation
Logistic Growth
Partial Fractions
Relative growth rate
Key Ideas:
Exponential Model
Logistic Growth Model
Logistic Regression
Mathematical Concepts:
Exponential Model
This growth model has continuous non-bounded growth.
kt
1. P ( t ) = Ce
P ( t ) ! The population at a given time
C
! The beginning population
k
! The growth rate
t
! The amount of time
kt
2. P! ( t ) = k "Ce or P! ( t ) = kP! ( t )
Logistic Growth Model
The growth model has a maximum carrying capacity.
1.
P (t ) =
M
1+ Aekt
P ( t ) ! The population at a given time
M
! The maximum carrying capacity
A
! An arbitrary constant determined by the initial condition
k
! The growth rate
t
! The amount of time
P! (t ) =
2.
dp k
dp "kAMekt
=
P ( M " P ) or P! ( t ) =
=
dt M
dt (1+ Aekt )2
Exponential Regression
1. Place your time values in L1 .
2. Place your population values in L2 .
3. Make L3 = ln ( L2 ) .
4. Take the linear regression of L1 and L 3 .
5. Determine the value of C and store it in variable C. e ^ ( b ) ! C
The value of b is located at Vars ! Statistics ! EQ ! b
6. Place the equation into Y1 on your calculator.
Y1 = Ce ^ (ax)
Logistic Regression
1. Place your time values in L1 .
2. Place your population values in L2 .
3. Make L3 = ln ( L2 ) .
4. Take the linear regression of L1 and L 3 .
5. Determine the value of C and store it in variable C. e ^ ( b ) ! C
The value of b is located at Vars ! Statistics ! EQ ! b
6. Place the equation into Y1 on your calculator.
Y1 = Ce ^ (ax)
Sample Questions:
1. Perform the following integration by using partial fractions.
"x
2
2x + 5
dx
! 2x ! 15
2.
Determine the function that represents the given situations using a logistic growth model.
Determine the population after 5 years and determine the rate of growth after the 5th year.
The maximum sustainable amount is 100.
P ( 0 ) = 25 and P ( 2 ) = 36
3.
Determine the function that represents the given situations using a logistic growth model.
Determine the population after 50 years and determine the rate of growth after the 50th
year. The maximum sustainable amount is 10000.
P ( 3) = 1250 and P (15 ) = 5,000
Homework
Pages 373 - 375
Exercises: #3, 7, 11, 19, 23, 29, 31, 33, 37, 39, 41, and 43
Exercises: #2, 6, 14, 16, 26, 28, 32, 34, 38, 40, 42, and 44
Sample Questions:
1. Perform the following integration by using partial fractions.
2x + 5
2x + 5
=
2
x ! 2x ! 15 ( x ! 5 ) ( x + 3)
"x
2
2x + 5
dx
! 2x ! 15
2x + 5
A
B
=
+
( x ! 5 )( x + 3) x ! 5 x + 3
2x + 5
A " x + 3%
B " x ! 5%
=
$#
'& +
( x ! 5 )( x + 3) x ! 5 x + 3 x + 3 $# x ! 5 '&
A ( x + 3)
B( x ! 5)
2x + 5
=
+
( x ! 5 )( x + 3) ( x ! 5 )( x + 3) ( x ! 5 )( x + 3)
2x + 5
dx
! 2x ! 15
15 &
# 1
%
(
" % x +8 3 + x16! 5 ( dx
%$
('
"x
2x + 5
Ax + 3A + Bx ! 5B
=
( x ! 5 )( x + 3)
( x ! 5 )( x + 2 )
2x + 5
Ax + Bx + 3A ! 5B
=
( x ! 5 )( x + 3)
( x ! 5 )( x + 3)
2x + 5
( A + B ) x + ( 3A ! 5B )
=
( x ! 5 )( x + 3)
( x ! 5 )( x + 3)
So,
( A + B ) x = 2x ( 3A ! 5B ) = 5
I.
II.
!
I.
!
( !3)
II.
(1)
!
!
!
!
!
!
A
3A
!
A
+ B =
! 5B =
! ! !
+ B =
2
5
!
2
3A
! 5B =
5
!
!3A
3A
!
! ! ! ! !
! 3B = !6
! 5B = 5
! ! ! ! !
! 8B = !1
1
B =
8
! ! ! ! !
1
+
= 2
8
15
=
16
!
A
A
!
2
# 1 &
# 15 &
% 8 (
%
(
" % x + 3 ( dx + " % x16! 5 ( dx
%$
('
%$
('
1 # 1 &
15 # 1 &
%$
(' dx + " %$
( dx
"
8 x+3
16 x ! 5 '
1
15
ln x + 3 + ln x ! 5 + C
8
16
2
15
ln x + 3 + ln x ! 5 + C
16
16
1
( 2 ln x + 3 + 15 ln x ! 5 ) + C
16
1
2
15
ln ( x + 3) + ln x ! 5 + C
16
1
2
15
ln ( x + 3) + ln ( x ! 5 ) + C
16
1
2
15
ln ( x + 3) ( x ! 5 ) + C
16
(
(
)
(
)
1
15 16
ln ( x + 3) ( x ! 5 )
2
ln 16 ( x + 3) ( x ! 5 )
2
15
)
+C
+C
2.
Determine the function that represents the given situations using a logistic growth model.
Determine the population after 5 years and determine the rate of growth after the 5th year.
The maximum sustainable amount is 100.
P ( 0 ) = 25 and P ( 2 ) = 36
M
1+ Aekt
100
P (0) =
1+ Aek( 0 )
100
25 =
1+ Ae0
100
25 =
1+ A (1)
100
1+ 3ekt
100
P (2) =
1+ 3ek( 2 )
36 (1+ 3e2 k ) = 100
P (t ) =
P (t ) =
1+ 3e2 k = 2.7778
3e2 k = 1.7888
e2 k = 0.5926
100
25 =
1+ A
25 (1+ A ) = 100
ln ( e2 k ) = ln ( 0.5926 )
2k = !0.5232
k = !0.2616
1+ A = 4
A=3
" P (t ) =
100
1+ 3e!0.2616t
100
P (5) =
1+ 3e!0.2616( 5 )
100
P (5) =
1+ 3e!1.3081
100
P (5) =
1+ 3( 0.2703)
100
1+ 3e!0.2616t
P (t ) =
100
1+ 0.8110
100
P (5) =
1.8110
P ( 5 ) = 55.2186
P (5) =
" P ( 5 ) # 55
k
P( M ! P)
M
" !0.2616 %
P '( 5 ) = ! $
( 55.2186 )(100 ! 55.2186 )
# 100 '&
P '(t ) = !
P ' ( 5 ) = ! ( !0.0262 ) ( 55.2186 ) ( 44.7814 )
P ' ( 5 ) = 6.4694
3.
Determine the function that represents the given situations using a logistic growth model.
Determine the population after 50 years and determine the rate of growth after the 50th
year. The maximum sustainable amount is 10,000.
P ( 3) = 1250 and P (15 ) = 5,000
P (t ) =
M
1+ Aekt
10000
P ( 3) =
1+ Aek( 3)
10000
1250 =
1+ Ae3k
1250 (1+ Ae3k ) = 10000
M
1+ Aekt
10000
P (15 ) =
1+ Ae15 k
10000
5000 =
1+ Ae15 k
5000 (1+ Ae15 k ) = 10000
1+ Ae3k = 8
1+ Ae15 k = 2
Ae3k = 7
7
A = 3k
e
Ae15 k = 1
P (t ) =
! 7 $ 15 k
#" 3k &% e = 1
e
7e12 k = 1
A=
A=
7
e
3( !0.1622 )
7
e!0.4865
7
A=
0.6148
A = 11.3860
" P (t ) =
10000
1+ 11.3860e!0.1622t
e12 k = 0.1429
ln e12 k = ln ( 0.1429 )
12k = '1.9460
k = '0.1622
10000
1+ 11.3860e!0.1622t
10000
P ( 50 ) =
1+ 11.3860e!0.1622( 50 )
10000
P ( 50 ) =
1+ 11.3860e!8.1080
10000
P ( 50 ) =
1+ 11.3860 ( 0.0003011)
P (t ) =
10000
1+ 0.003429
10000
P ( 50 ) =
1.0034
P ( 50 ) = 9965.8301
P ( 50 ) =
" P ( 50 ) # 9,966
k
P( M ! P)
M
" !0.1622 %
P ' ( 50 ) = ! $
( 9965.8301)(10000 ! 9965.8301)
# 10000 '&
P '(t ) = !
P ' ( 50 ) = ! ( !0.00001622 ) ( 9965.8301) ( 34.1699 )
P ' ( 50 ) = 5.5220