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# Drill - PowerPoint PPT Presentation

Drill. Find the integral using the given substitution. 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.

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Presentation Transcript
Drill
• Find the integral using the given substitution

### Lesson 6.5: Logistic Growth

Day #1: P. 369: 1-17 (odd)

Day #2: P. 369/70: 19-29; odd, 37

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

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

3x2

+3

-

3x4 -3x2

-

3x2 +1

3x2 -3

4

When x = -1

A(0) + B(-2) = 4; B = - 2

When x = 1

A(2) + B(0) = 4; A= 2

Finding Three Partial Fractions

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

Drill: Evaluate the Integral

When x = -7, B(-7) = -21, B = 3

When x = 0, 7A = 14, A = 2

The Logistic Differential Equation

Remember that exponential

growth can be modeled by

dP/dt = kP, for some k >0

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.

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.

Example: Using Logistic Regression
• The population of Aurora, CO for selected years between 1950 and 2003
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!)
When will the population first exceed 300,000?

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)

• Use the calculator to intersect.
• When x = 59.12 or in the 59th year: 1950 + 59 = 2009
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.
• 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

The growth rate is always

maximized when the

population reaches half

the carrying capacity.

In this case, it is 50 bears.

• 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]

We first need to separate the variables.

Then we will integrate one side by using partial fractions.

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

Multiply both sides by 1000 to clear decimals.

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

Remember that ln(x) =y

Can be rewritten as ey = x