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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
Drill
  • Find the integral using the given substitution
lesson 6 5 logistic growth

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
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
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 x 2 x 6 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
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

slide7

When x = -1

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

When x = 1

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

finding three partial fractions
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
Drill: Evaluate the Integral

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

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

the logistic differential equation
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
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
Example: Using Logistic Regression
  • The population of Aurora, CO for selected years between 1950 and 2003
use logistic regression to find a model
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!)
slide15
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
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
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

slide18

What is the bear population when the population is growing the fastest?

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

example1
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]

slide20

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
slide21

Multiply both sides by 1000 to clear decimals.

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

slide22

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

Remember that ln(x) =y

Can be rewritten as ey = x

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