Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Prentice Hall

6.5Logistic Growth

Slide 6- 2 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Prentice Hall

Partial Fraction Decomposition with Distinct Linear Denominators

( )If ( ) , where and are polynomials with the ( )

degree of less than the degree of , and if ( ) can be writtenas a product of distinct linear factors, then ( ) can be writtenas a sum of r

P xf x P QQ xP Q Q x

f x

ational functions with distinct linear denominators.

Slide 6- 3 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Prentice Hall

Example Finding a Partial Fraction Decomposition

2

2

6 8 4Write the function ( ) as a sum of rational 4 1

functions with linear denominators.

x xf xx x

2

2

6 8 4Since ( ) , we will find numbers A, B and C2 2 1

so that ( ) .- 2 2 1

2 1 2 1 2 2Note that ,

- 2 2 1 2 2 1

so it follows that 2 1 2 1 2 2 6 8 4.S

x xf xx x xA B Cf xx x x

A x x B x x C x xA B Cx x x x x x

A x x B x x C x x x x

2

etting 2 : (4)(1) (0) (0) 4, so 1.Setting -2 : (0) (-4)(-3) (0) 36, so 3.Setting 1: (0) (0) (-1)(3) -6, so 2.

6 8 4 1 3 2Therefore ( ) .2 2 1 2 2 1

x A B C Ax A B C Bx A B C C

x xf xx x x x x x

Slide 6- 4 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Prentice Hall

Example Finding a Partial Fraction Decomposition

2

2

6 8 4Write the function ( ) as a sum of rational 4 1

functions with linear denominators.

x xf xx x

2

2

6 8 4Since ( ) , we will find numbers A, B and C2 2 1

so that ( ) .- 2 2 1

2 1 2 1 2 2Note that ,

- 2 2 1 2 2 1

so it follows that 2 1 2 1 2 2 6 8 4.S

x xf xx x xA B Cf xx x x

A x x B x x C x xA B Cx x x x x x

A x x B x x C x x x x

2

etting 2 : (4)(1) (0) (0) 4, so 1.Setting -2 : (0) (-4)(-3) (0) 36, so 3.Setting 1: (0) (0) (-1)(3) -6, so 2.

6 8 4 1 3 2Therefore ( ) .2 2 1 2 2 1

x A B C Ax A B C Bx A B C C

x xf xx x x x x x

Slide 6- 5 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Prentice Hall

Example Finding a Partial Fraction Decomposition

2

2

6 8 4Write the function ( ) as a sum of rational 4 1

functions with linear denominators.

x xf xx x

2

2

6 8 4Since ( ) , we will find numbers A, B and C2 2 1

so that ( ) .- 2 2 1

2 1 2 1 2 2Note that ,

- 2 2 1 2 2 1

so it follows that 2 1 2 1 2 2 6 8 4.S

x xf xx x xA B Cf xx x x

A x x B x x C x xA B Cx x x x x x

A x x B x x C x x x x

2

etting 2 : (4)(1) (0) (0) 4, so 1.Setting -2 : (0) (-4)(-3) (0) 36, so 3.Setting 1: (0) (0) (-1)(3) -6, so 2.

6 8 4 1 3 2Therefore ( ) .2 2 1 2 2 1

x A B C Ax A B C Bx A B C C

x xf xx x x x x x

Slide 6- 6 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Prentice Hall

Example Finding a Partial Fraction Decomposition

2

2

6 8 4Write the function ( ) as a sum of rational 4 1

functions with linear denominators.

x xf xx x

2

2

6 8 4Since ( ) , we will find numbers A, B and C2 2 1

so that ( ) .- 2 2 1

2 1 2 1 2 2Note that ,

- 2 2 1 2 2 1

so it follows that 2 1 2 1 2 2 6 8 4.S

x xf xx x xA B Cf xx x x

A x x B x x C x xA B Cx x x x x x

A x x B x x C x x x x

2

etting 2 : (4)(1) (0) (0) 4, so 1.Setting -2 : (0) (-4)(-3) (0) 36, so 3.Setting 1: (0) (0) (-1)(3) -6, so 2.

6 8 4 1 3 2Therefore ( ) .2 2 1 2 2 1

x A B C Ax A B C Bx A B C C

x xf xx x x x x x

Slide 6- 7 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Prentice Hall

Example Finding a Partial Fraction Decomposition

2

2

6 8 4Write the function ( ) as a sum of rational 4 1

functions with linear denominators.

x xf xx x

2

2

6 8 4Since ( ) , we will find numbers A, B and C2 2 1

so that ( ) .- 2 2 1

2 1 2 1 2 2Note that ,

- 2 2 1 2 2 1

so it follows that 2 1 2 1 2 2 6 8 4.S

x xf xx x xA B Cf xx x x

A x x B x x C x xA B Cx x x x x x

A x x B x x C x x x x

2

etting 2 : (4)(1) (0) (0) 4, so 1.Setting -2 : (0) (-4)(-3) (0) 36, so 3.Setting 1: (0) (0) (-1)(3) -6, so 2.

6 8 4 1 3 2Therefore ( ) .2 2 1 2 2 1

x A B C Ax A B C Bx A B C C

x xf xx x x x x x

Slide 6- 8 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Prentice Hall

Example Finding a Partial Fraction Decomposition

2

2

6 8 4Write the function ( ) as a sum of rational 4 1

functions with linear denominators.

x xf xx x

2

2

6 8 4Since ( ) , we will find numbers A, B and C2 2 1

so that ( ) .- 2 2 1

2 1 2 1 2 2Note that ,

- 2 2 1 2 2 1

so it follows that 2 1 2 1 2 2 6 8 4.S

x xf xx x xA B Cf xx x x

A x x B x x C x xA B Cx x x x x x

A x x B x x C x x x x

2

etting 2 : (4)(1) (0) (0) 4, so 1.Setting -2 : (0) (-4)(-3) (0) 36, so 3.Setting 1: (0) (0) (-1)(3) -6, so 2.

6 8 4 1 3 2Therefore ( ) .2 2 1 2 2 1

x A B C Ax A B C Bx A B C C

x xf xx x x x x x

Slide 6- 9 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Prentice Hall

Example Finding a Partial Fraction Decomposition

2

2

6 8 4Write the function ( ) as a sum of rational 4 1

functions with linear denominators.

x xf xx x

2

2

6 8 4Since ( ) , we will find numbers A, B and C2 2 1

so that ( ) .- 2 2 1

2 1 2 1 2 2Note that ,

- 2 2 1 2 2 1

so it follows that 2 1 2 1 2 2 6 8 4.S

x xf xx x xA B Cf xx x x

A x x B x x C x xA B Cx x x x x x

A x x B x x C x x x x

2

etting 2 : (4)(1) (0) (0) 4, so 1.Setting -2 : (0) (-4)(-3) (0) 36, so 3.Setting 1: (0) (0) (-1)(3) -6, so 2.

6 8 4 1 3 2Therefore ( ) .2 2 1 2 2 1

x A B C Ax A B C Bx A B C C

x xf xx x x x x x

Slide 6- 10 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Prentice Hall

Example Finding a Partial Fraction Decomposition

2

2

6 8 4Write the function ( ) as a sum of rational 4 1

functions with linear denominators.

x xf xx x

2

2

6 8 4Since ( ) , we will find numbers A, B and C2 2 1

so that ( ) .- 2 2 1

2 1 2 1 2 2Note that ,

- 2 2 1 2 2 1

so it follows that 2 1 2 1 2 2 6 8 4.S

x xf xx x xA B Cf xx x x

A x x B x x C x xA B Cx x x x x x

A x x B x x C x x x x

2

etting 2 : (4)(1) (0) (0) 4, so 1.Setting -2 : (0) (-4)(-3) (0) 36, so 3.Setting 1: (0) (0) (-1)(3) -6, so 2.

6 8 4 1 3 2Therefore ( ) .2 2 1 2 2 1

x A B C Ax A B C Bx A B C C

x xf xx x x x x x

Slide 6- 11 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Prentice Hall

Example Finding a Partial Fraction Decomposition

2

2

6 8 4Write the function ( ) as a sum of rational 4 1

functions with linear denominators.

x xf xx x

2

2

6 8 4Since ( ) , we will find numbers A, B and C2 2 1

so that ( ) .- 2 2 1

2 1 2 1 2 2Note that ,

- 2 2 1 2 2 1

so it follows that 2 1 2 1 2 2 6 8 4.S

x xf xx x xA B Cf xx x x

A x x B x x C x xA B Cx x x x x x

A x x B x x C x x x x

2

etting 2 : (4)(1) (0) (0) 4, so 1.Setting -2 : (0) (-4)(-3) (0) 36, so 3.Setting 1: (0) (0) (-1)(3) -6, so 2.

6 8 4 1 3 2Therefore ( ) .2 2 1 2 2 1

x A B C Ax A B C Bx A B C C

x xf xx x x x x x

Slide 6- 12 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Prentice Hall

Example Antidifferentiating with Partial Fractions

26 8 4Find .

2 2 1x x dx

x x x

2

3 2

We know from the last example that 6 8 4 1 3 22 2 1 - 2 2 1

ln - 2 3ln 2 2ln 1

ln - 2 2 1

x x dx dxx x x x x x

x x x C

x x x C

Slide 6- 13 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Prentice Hall

Example Antidifferentiating with Partial Fractions

26 8 4Find .

2 2 1x x dx

x x x

2

3 2

We know from the last example that 6 8 4 1 3 22 2 1 - 2 2 1

ln - 2 3ln 2 2ln 1

ln - 2 2 1

x x dx dxx x x x x x

x x x C

x x x C

Slide 6- 14 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Prentice Hall

Example Antidifferentiating with Partial Fractions

26 8 4Find .

2 2 1x x dx

x x x

2

3 2

We know from the last example that 6 8 4 1 3 22 2 1 - 2 2 1

ln - 2 3ln 2 2ln 1

ln - 2 2 1

x x dx dxx x x x x x

x x x C

x x x C

Slide 6- 15 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Prentice Hall

Example Antidifferentiating with Partial Fractions

26 8 4Find .

2 2 1x x dx

x x x

2

3 2

We know from the last example that 6 8 4 1 3 22 2 1 - 2 2 1

ln - 2 3ln 2 2ln 1

ln - 2 2 1

x x dx dxx x x x x x

x x x C

x x x C

Slide 6- 16 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Prentice Hall

Example Antidifferentiating with Partial Fractions

26 8 4Find .

2 2 1x x dx

x x x

2

3 2

We know from the last example that 6 8 4 1 3 22 2 1 - 2 2 1

ln - 2 3ln 2 2ln 1

ln - 2 2 1

x x dx dxx x x x x x

x x x C

x x x C

Slide 6- 17 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Prentice Hall

Logistic Differential Equation

Exponential growth can be modeled by the differential equation

for some 0.

If we want the growth rate to approach zero as approaches a maximal carrying capacity , we can introduce a limitin

dP kP kdt

PM

g factor

of - : .

This is the .

dPM P kP M Pdt

logistic differential equation

Slide 6- 18 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Prentice Hall

Logistic Differential Equation

Exponential growth can be modeled by the differential equation

for some 0.

If we want the growth rate to approach zero as approaches a maximal carrying capacity , we can introduce a limitin

dP kP kdt

PM

g factor

of - : .

This is the .

dPM P kP M Pdt

logistic differential equation

Slide 6- 19 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Prentice Hall

Logistic Differential Equation

Exponential growth can be modeled by the differential equation

for some 0.

If we want the growth rate to approach zero as approaches a maximal carrying capacity , we can introduce a limitin

dP kP kdt

PM

g factor

of - : .

This is the .

dPM P kP M Pdt

logistic differential equation

Slide 6- 20 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Prentice Hall

Logistic Differential Equation

Exponential growth can be modeled by the differential equation

for some 0.

If we want the growth rate to approach zero as approaches a maximal carrying capacity , we can introduce a limitin

dP kP kdt

PM

g factor

of - : .

This is the .

dPM P kP M Pdt

logistic differential equation

Slide 6- 21 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Prentice Hall

Example Logistic Differential Equation

The growth rate of a population of bears in a newly establishedwildlife preserve is modeled by the differential equation

0.008 100 - , where is measured in years.

a. What is the carrying capac

P

dP P P tdt

ity for bears in this wildlife preserve?b. What is the bear population when the population is growing the fastest?c. What is the rate of change of the population when it is growing the fastest?

a. The carrying capacity is 100 bears.b. The bear population is growing the fastest when it is halfthe carrying capacity, 50 bears.

c. When 50, 0.008 50 100 50 20 bears per year.dPPdt

Slide 6- 22 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Prentice Hall

Example Logistic Differential Equation

The growth rate of a population of bears in a newly establishedwildlife preserve is modeled by the differential equation

0.008 100 - , where is measured in years.

a. What is the carrying capac

P

dP P P tdt

ity for bears in this wildlife preserve?b. What is the bear population when the population is growing the fastest?c. What is the rate of change of the population when it is growing the fastest?

a. The carrying capacity is 100 bears.b. The bear population is growing the fastest when it is halfthe carrying capacity, 50 bears.

c. When 50, 0.008 50 100 50 20 bears per year.dPPdt

Slide 6- 23 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Prentice Hall

Example Logistic Differential Equation

The growth rate of a population of bears in a newly establishedwildlife preserve is modeled by the differential equation

0.008 100 - , where is measured in years.

a. What is the carrying capac

P

dP P P tdt

ity for bears in this wildlife preserve?b. What is the bear population when the population is growing the fastest?c. What is the rate of change of the population when it is growing the fastest?

a. The carrying capacity is 100 bears.b. The bear population is growing the fastest when it is halfthe carrying capacity, 50 bears.

c. When 50, 0.008 50 100 50 20 bears per year.dPPdt

Slide 6- 24 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Prentice Hall

Example Logistic Differential Equation

The growth rate of a population of bears in a newly establishedwildlife preserve is modeled by the differential equation

0.008 100 - , where is measured in years.

a. What is the carrying capac

P

dP P P tdt

ity for bears in this wildlife preserve?b. What is the bear population when the population is growing the fastest?c. What is the rate of change of the population when it is growing the fastest?

a. The carrying capacity is 100 bears.b. The bear population is growing the fastest when it is halfthe carrying capacity, 50 bears.

c. When 50, 0.008 50 100 50 20 bears per year.dPPdt

Slide 6- 25 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Prentice Hall

Example Logistic Differential Equation

The growth rate of a population of bears in a newly establishedwildlife preserve is modeled by the differential equation

0.008 100 - , where is measured in years.

a. What is the carrying capac

P

dP P P tdt

ity for bears in this wildlife preserve?b. What is the bear population when the population is growing the fastest?c. What is the rate of change of the population when it is growing the fastest?

a. The carrying capacity is 100 bears.b. The bear population is growing the fastest when it is halfthe carrying capacity, 50 bears.

c. When 50, 0.008 50 100 50 20 bears per year.dPPdt

Slide 6- 26 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Prentice Hall

Example Logistic Differential Equation

The growth rate of a population of bears in a newly establishedwildlife preserve is modeled by the differential equation

0.008 100 - , where is measured in years.

a. What is the carrying capac

P

dP P P tdt

ity for bears in this wildlife preserve?b. What is the bear population when the population is growing the fastest?c. What is the rate of change of the population when it is growing the fastest?

a. The carrying capacity is 100 bears.b. The bear population is growing the fastest when it is halfthe carrying capacity, 50 bears.

c. When 50, 0.008 50 100 50 20 bears per year.dPPdt

Slide 6- 27 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Prentice Hall

The General Logistic Formula

-

The solution of the general logistic differential equation

is

1

where is a constant determined by an appropriateinitial condition. The and

Mk t

dP kP M Pdt

MPAe

AM

carrying capacity the are positive constants.kgrowth constant

tMkAeMP )(1

Slide 6- 28 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Prentice Hall

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