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LOGISTIC GROWTH

Section 6.3A

Calculus BC AP/Dual, Revised ©2017

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7/30/2018 11:40 AM §6.3A: Logistic Growth 1

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RECALL

Solve the logistic differential equation 𝒅𝑷

𝒅𝒕= 𝒌𝑷 𝟏 −

𝑷

𝑳.


1dP P

kPdt L

= −

1

1

dP kdtP

PL

= −

1 1dP kdt

P L P

+ =

−

RECALL

Solve the logistic differential equation 𝒅𝒚


𝑷

𝑳.


ln lnP L P kt C+ − = +

lnL P

kt CP

−= − −

kt CL Pe

P

− −−=

C ktL Pe e

P

− −−=

ktL PBe

P

−−=

1 kt

LP

Be−=

+

LOGISTIC CURVES

A. Logistics Differential Equation: 𝒅𝑷


𝑷

𝑳or

𝒅𝑷

𝒅𝒕= 𝒌𝑷 𝑳 − 𝑷 where

𝒌 will be multiplied and the inside is divided

B. The 𝒌 can be different depending on the use of the equation

C. Logical Growth Model: 𝑷 =𝑳

𝟏+𝑩𝒆−𝒌𝒕

1. 𝑳 = Carrying Capacity (Upper Horizontal LIMIT)

2. 𝑲 = Proportionality Constant

3. 𝑩 = Beginning Amount (arbitrary), use the equation: 𝑩 =𝑳−𝑷𝟎

𝑷𝟎where P0 is the initial

population

C. Point of Inflection: 𝒚 =𝑳

𝟐


LOGISTIC GROWTH


EXAMPLE 1

Given 𝒅𝑷

𝒅𝒕= 𝑷 𝟐 −

𝑷

𝟓𝟎𝟎𝟎identify the 𝒌 and 𝑳 of the equation.


1dy y

kydt L

= −

1dP P

kPdt L

= −

25000

dP PP

dt

= −

EXAMPLE 1

Given 𝒅𝑷

𝒅𝒕= 𝑷 𝟐 −

𝑷

𝟓𝟎𝟎𝟎identify the 𝒌 and 𝑳 of the equation.


25000

dP PP

dt

= −

( )2 500022 2

P

dPP

dt

= −

2 110000

dP PP

dt

= −

2

10,000

k

L

=

=

YOUR TURN

Given 𝒅𝑷

𝒅𝒕= 𝟎. 𝟎𝟏𝑷 𝟏𝟎𝟎 − 𝑷 identify the 𝒌 and 𝑳 of the equation.


1

100

k

L

=

=

EXAMPLE 2

Using the equation, 𝒚 =𝟒

𝟏+𝟐𝒆−𝟑𝒕identify the point of inflection and

sketch a graph.


( )1

34 1 2 ty e−

−= +

( ) ( ) ( )2

3 3' 4 1 1 2 6t ty e e−

− −= − + −

4, 2, 3L B and k= = =

3

3 3

4 2' 3

1 2 1 2

t

t t

ey

e e

−

− −

=

+ +

3

1' 3 1

1 2

using Long Division

ty y

e−

= −

+

EXAMPLE 2



sketch a graph.


' 3 14

yy y

= −

( )3

4' 3 1

4 1 2 ty y

e−

= −

+

( ) ( )3 0

4 4; 0

31 2y y

e−

= =+

3

1' 3 1

1 2

using Long Division

ty y

e−

= −

+

EXAMPLE 2



sketch a graph.


2

Ly =

4

2y =

2y =

3

4

1 2 ty

e−=

+

3

42

1 2 te−=

+

( )32 1 2 4te−+ =

31 2 2te−+ =32 1te− =

3 1

2

te− =

3

1 1

2te=

3 2te =3ln ln 2te =

3 ln 2t =

ln 2

3t =ln 2

: ,23

POI

EXAMPLE 2



sketch a graph.


ln 2: ,2

3POI

YOUR TURN

Using the equation, 𝒚 =𝟖

𝟏+𝟐𝒆−𝟐𝒕identify the point of inflection and

sketch a graph.


ln 2: ,4

2POI

EXAMPLE 3

A state game commission releases 40 elk into a game refuge. After 5 years, the elk population is 104. The commission believes that the environment can support no

more than 4000 elk. The growth rate of the elk population, 𝒑, is: 𝒅𝑷

𝒅𝑻=

𝒌𝑷 𝟏 −𝑷

𝟒𝟎𝟎𝟎, 𝟒𝟎 ≤ 𝒑 ≤ 𝟒𝟎𝟎𝟎 where 𝒕 is the number of years.

A. Write a model for the elk population in terms of 𝒕.

B. Estimate the elk population in 15 years.

C. Find the limit of the model as 𝒕 → ∞.


EXAMPLE 3A

A state game commission releases 40 elk into a game refuge. After 5 years, the elk population is 104. The commission believes that the environment can support no

more than 4000 elk. The growth rate of the elk population, 𝒑, is: 𝒅𝑷

𝒅𝑻=

𝒌𝑷 𝟏 −𝑷

𝟒𝟎𝟎𝟎, 𝟒𝟎 ≤ 𝒑 ≤ 𝟒𝟎𝟎𝟎 where 𝒕 is the number of years.



1 ,40 40004000

4000, ??, B ??

dP Pkp P

dt

L k

= −

= = =

1 kt

Ly

Be−=

+

4000

1 ktP

Be−=

+

( )0

400040

1k

Be−

=+

( )( )040 1 4000

kBe

−+ =

40 40 4000B+ =

40 3960B =

99B =

4000

1 99 ktP

e−=

+

1 ,40 40004000

4000, ??,B 99

dP PkP P

dt

L k

= −

= = =

4000

1 99 ktP

e−=

+

EXAMPLE 3A

A state game commission releases 40 elk into a game refuge. After 5 years, the elk population is 104. The commission believes that the environment can support no more than 4000 elk. The growth

rate of the elk population, p, is: 𝒅𝑷

𝒅𝑻= 𝒌𝑷 𝟏 −

𝑷

𝟒𝟎𝟎𝟎, 𝟒𝟎 ≤ 𝒑 ≤ 𝟒𝟎𝟎𝟎 where t is the number of

years.



4000

1 99 kty

e−=

+

( )5

4000104

1 99k

e−

=+

( )5104 1 99 4000ke−+ =

5 40001 99

104

ke−+ =

5 400099 1

104

ke− = −

599 37.5

99 99

ke−

=

0.1942k 0.1942

4000

1 99 tP

e−=

+

1 ,40 40004000

4000, ??, B 99

dP Pkp P

dt

L k

= −

= = =

5 37.5

99

ke− =

5

1 37.5

99ke=

( ) 537.5 99ke =

5 2.64ke =

5ln ln 2.64ke =

5 ln 2.64

5 5

k=

1 ,40 40004000

4000, 0.1942,B 99

dP PkP P

dt

L k

= −

= =

EXAMPLE 3B




𝑷


years.

B. Estimate the elk population in 15 years.


1 ,40 40004000

4000, ??, B ??

dP Pkp P

dt

L k

= −

= = =( )0.1942 15

4000

1 99P

e−

=+

627.2598 elk0.1942

4000

1 99 tP

e−=

+

1 ,40 40004000

4000, ??, B 99

dP Pkp P

dt

L k

= −

= = =

EXAMPLE 3C




𝑷


years.

C. Find the limit of the model as 𝒕 → ∞.


1 ,40 40004000

4000, ??, B ??

dP Pkp P

dt

L k

= −

= = =

( )0.1942

4000lim

1 99tt e

−→ +

4000

0.1942

4000

1 99 tP

e−=

+1 ,40 4000

4000

4000, ??,B 99

dP PkP P

dt

L k

= −

= = =

EXAMPLE 4

Suppose the population of bears in a national park grows according to

the logistic differential equation, 𝒅𝑷

𝒅𝑻= 𝟓𝑷 − 𝟎. 𝟎𝟎𝟐𝑷𝟐 , where 𝑷 is

the number of bears at time 𝒕 in years. Find the limit of the model as 𝒕 → ∞ where 𝑷 𝟎 = 𝟏𝟎𝟎.


1dP P

kPdt L

= −

( )0.002

2500lim

1 24tx e

−→ +2500

25 0.002dP

P Pdt

= −( )0.002

2500100

1 24

24

te

B

−=

+

=

5 12500

dP PP

dt

= −

25 0.002

5 5

dP P P

dt= −

YOUR TURN

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?


1 kt

LP

Be−=

+

( )100 0

10010

1 Be−

=+

( )0,10

( )10,23

10010

1 B=

+

10 10 100B+ =

10 90B =

9B =

100

1 9 ktP

e−=

+

YOUR TURN

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?


10 1001 9

23

ke−+ =

10 779

23

ke− =

10 0.371981ke− =

( )0.09889

100

1 9t

Pe−

=+

( )0,10 ( )10,23

( )10

10023

1 9k

e−

=+

10 ln 0.371981

10 10

k−=

− −

YOUR TURN


0

20

40

60

80

100

20 40 60 80 100Years

Bears

𝒀 = 𝟓𝟎 𝒂𝒕 𝟐𝟐 𝒚𝒆𝒂𝒓𝒔

𝒀 = 𝟕𝟓 𝒂𝒕 𝟑𝟑 𝒚𝒆𝒂𝒓𝒔

𝒀 = 𝟏𝟎𝟎 𝒂𝒕 𝟕𝟓 𝒚𝒆𝒂𝒓𝒔

( )0.09889

100

1 9t

Pe−

=+

EXAMPLE 5

In a particular town of 100,000 residents, 20,000 watched a viral video on the Internet. The rate of growth of the spread of information was jointly proportional to the amount of people who had not watched it. If 50% watched it after one hour, how long does 80% of the population watched the viral video?


1 kt

LP

Be−=

+

100,000 20,000

20,000B

−=

( )

( )

100,000

?

0,20000

1,50000

L

B

=

=

4B =

( )100,000

1 4 ktP t

e−=

+

EXAMPLE 5



( )1

100,00050,000

1 4k

e−

=+

( )50,000 1 4 100,000ke+ =

1 4 2ke−+ =

( )

( )

100,000

?

0,20000

1,50000

L

B

=

=( )100,000

1 4 ktP t

e−=

+

EXAMPLE 5



1 4 2ke−+ =

4 1ke− =1

4

ke− =

( )

( )

100,000

?

0,20000

1,50000

L

B

=

=

1 1

4ke=

4ke =

EXAMPLE 5



( )

( )

100,000

?

0,20000

1,50000

L

B

=

=4ke =

( ) ( )ln 4

100,000

1 4t

P te−

=+

ln 4k =

ln 4

100,00080,000

1 4 te−=

+

EXAMPLE 5



ln 4

100,00080,000

1 4 te−=

+

( )ln 480,000 1 4 100,000te−+ =

ln 4 51 4

4

te−+ =

( )

( )

100,000

?

0,20000

1,50000

L

B

=

=

EXAMPLE 5



ln 4 1

16

te− =

ln 4 1ln ln

16

1ln

16ln 4ln 4

te

t

− =

= −

2 t hours=

ln 4 14

4

te− =

ln 4 1

16

te− =

( )( )ln 4 1

16

te

−= ( )

( )

100,000

?

0,20000

1,50000

L

B

=

=ln 4 51 4

4

te−+ =

AP MULTIPLE CHOICE PRACTICE QUESTION 1 (NON-CALCULATOR)

Which of the following differential equations for a population could model the logistic growth equation for the graph below?

(A) 𝒅𝑷

𝒅𝒕= 𝟎. 𝟐𝑷 − 𝟎. 𝟎𝟎𝟏𝑷𝟐

(B) 𝒅𝑷

𝒅𝒕= 𝟎. 𝟏𝑷 − 𝟎. 𝟎𝟎𝟏𝑷𝟐

(C) 𝒅𝑷

𝒅𝒕= 𝟎. 𝟐𝑷𝟐 − 𝟎. 𝟎𝟎𝟏𝑷

(D) 𝒅𝑷

𝒅𝒕= 𝟎. 𝟏𝑷𝟐 − 𝟎. 𝟎𝟎𝟏𝑷





Vocabulary Connections and Process

Logistic Growth

Equation( )1 ; 200

200

dP P dPkP kP P

dt dt

= − = −

2

200

dP kPkP

dt= −

2

)

0.2 0.001

0.2

10.2 1 0.2

200 1000

A

dPP P

dt

k

dP PP P P

dt

= −

=

= − = −

2

)

0.1 0.001

0.1

10.1 1 0.1

200 2000

B

dPP P

dt

k

dP PP P P

dt

= −

=

= − −




Connections and Process Answer

( )1 ; 200200

dP P dPkP kP P

dt dt

= − = −

2

200

dP kPkP

dt= −

2

)

0.2 0.001

0.001

10.001 1 0.001

200 200000

C

dPP P

dt

k

dP PP P P

dt

= −

=

= − −

2

)

0.1 0.001

0.001

10.001 1 0.001

200 200000

D

dPP P

dt

k

dP PP P P

dt

= −

=

= − −

A

ASSIGNMENT

Worksheet


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