Drill

• Find the integral using the given substitution

dxxx

x

132

342 132 2 xxu

dxxdu 34 duu

1

Cu ln Cxx )132ln( 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.

312)3)(12(

13)(

x

B

x

A

xx

xxf

)3)(12(

)12()3()(

xx

xBxAxf

13)12()3( xxBxA 133)1)3(2()33(,3 BAxIf

10)5()0( BA 2,105 BB

132

1)1)

2

1(2()3

2

1(,

2

1 BAxIf

2

25)0()

2

5(

BA 5,

2

25)

2

5(

AA

3

2

12

5

)3)(12(

13)(

xxxx

xxf

Write the function f(x)= (2x+16)/(x2 +x-6) as a sum of rational functions with linear denominators.

32)3)(2(

162)(

x

B

x

A

xx

xxf

)3)(2(

)2()3()(

xx

xBxAxf

162)2()3( xxBxA 16)3(2)23()33(,3 BAxIf

10)5()0( BA 2,105 BB

16)2(2)22()32(,2 BAxIf

20)0()5( BA 4,20)5( AA

3

2

2

4

)3)(2(

162)(

xxxx

xxf

Example: Antidifferentiating with Partial Fractions

dxx

x

1

132

4 Because the numerator’s degree is larger than the denominator’s degree, we will need to use polynomial division.

100031 2342 xxxxx

3x2

3x4 -3x2

3x2 +1

+3

3x2 -3

--

4

dxx

x

1

433

22

dxx

x

1

433

22 dx

xxx

1

43

23

dxx

B

x

Axx

1133 dx

xx

xBxAxx

)1)(1(

)1()1(33

4)1()1( xBxA

When x = -1A(0) + B(-2) = 4; B = - 2

When x = 1A(2) + B(0) = 4; A= 2

dxxx

xx

1

2

1

233

Cxxxx )1ln(2)1ln(233

Cxxxx )1(ln)1(ln233 Cx

xxx

1

1ln233

Finding Three Partial Fractions

dxxx

xx

)1(4

4862

2

122)1(4

4862

2

x

C

x

B

x

A

xx

xx

)1)(2)(2(

)2)(2()1)(2()1)(2(

xxx

xxCxxBxxA

486)2)(2()1)(2()1)(2( 2 xxxxCxxBxxA

When x = -2B(-4)(-3)=36

12B = 36B = 3

When x = 1C(-1)(3)=-6

-3C = -6C = 2

When x = 2A(4)(1)=4

4A=4A=1

dx

xxx 1

2

2

3

2

1

Cxxx 1ln22ln32ln

Cxxx 231ln2ln2ln

Cxxx 231ln22ln

Drill: Evaluate the Integraldxxx

x

7

1452

7x

B

x

A

)7(

)()7(

xx

xBxA

145)()7( xxBxA When x = -7, B(-7) = -21, B = 3When x = 0, 7A = 14, A = 2

dxxx 7

32

Cxx

Cxx

Cxx

32

32

7ln

7lnln

7ln3ln2

The Logistic Differential Equation

Remember that exponentialgrowth can be modeled bydP/dt = kP, for some k >0

If we want the growth rateto approach 0 as P approachesa maximal carrying capacity M,we can introduce a limitingfactor of M – P.

Logistic Differential Equation

dP/dt = kP(M-P)

General Logistic Formula

• The solution of the general logistic differential equation

is

Where A is a constant determined by an appropriate initial condition. The carrying capacity M and the growth constant k are positive constants.

)( PMkPdt

dP

tMkAe

MP

)(1

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

• 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!)

teP

1026.577.231

7.316440

When will the population first exceed 300,000?

• Use the calculator to intersect.

• When x = 59.12 or in the 59th year: 1950 + 59 = 2009

Write a logistic equation in the form of dP/dt = kP(M-P) that models the data.

• 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)

tey

1026.1 577.231

7.316440

000,3002 y

tMkAe

MP

)(1

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?

The growth rate is alwaysmaximized when thepopulation reaches halfthe carrying capacity.

In this case, it is 50 bears.

C. What is the rate of change of population when it is growing the fastest?

dP/dt = .008P(100-P)dP/dt = .008(50)(100-50)dP/dt =20 bears per year

Example

• In 1985 and 1987, the Michigan Department of Natural Resources airlifted 61 moose from Algonquin Park, Ontario to Marquette County in the Upper Peninsula. It 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)

• What is the carrying capacity?

1000 moose

Generate a slope field. The window should be x: [0, 25] and y: [0, 1000]

• Solve the differential equation with the initial condition P(0) = 61

)1000(0003. PPdt

dP

dtPP

dP0003.

)1000(

dtPP

dP0003.

)1000(

dtdP

P

B

P

A0003.

1000

1)()1000( PBPA

dPPP

dPPP

PBPA

)1000(

1

)1000(

)()1000(

001.

11000

0

A

A

P

001.

11000

1000

B

B

P

dtdP

PP0003.

1000

001.001.

We first need to separate the variables.Then we will integrate one side by using partial fractions.

dtdP

PP0003.

1000

001.001.

Multiply both sides by 1000 to clear decimals.

dtdP

PP3.

1000

11

CtPP 3.)1000ln(1ln

Multiply both sides by -1 to put 1000-P on the top

CtPP 3.ln)1000ln(

CtP

P

3.

1000ln

CtP

P

3.

1000ln

CteP

P 3.1000

Remember that ln(x) =y Can be rewritten as ey = x

CteeP

3.11000

11000 3. CteeP

Solve the differential equation with the initial condition P(0) = 61

161

1000 )0(3. Cee

Ce 39344.15

11000 3. CteeP

Solve for P

13944.15

1

1000 3. te

P

13944.15

10003.

teP