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Population Models (Autonomic Differential Equations) Population Models (Autonomic Differential Equations)

Population Models (Autonomic Differential Equations) - PowerPoint Presentation

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Population Models (Autonomic Differential Equations) - PPT Presentation

MAT 275 The first type of population model is called the unconstrained model In words it is typically stated as The rate of change of a population at time t is directly proportional to the population at time ID: 673596

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Slide1

Population Models (Autonomic Differential Equations)

MAT 275Slide2

The first type of population model is called the

unconstrained model

. In words, it is typically stated as:The rate of change of a population at time t is directly proportional to the population at time t.If is the population at time t, then the above statement can be parsed out as follows:The phrase “The rate of change of a population at time t” is written .The word “is” translates as =.The words “directly proportional” imply a proportion constant k.The words “population at time t” is .Combined, this is a differential equation .

 

(c) ASU Math - Scott Surgent. Report errors to surgent@asu.edu

2Slide3

The differential equation

can also be written

. It is autonomic.It can be solved by separating variables, integration factors, or using the general form .Example: The rate of change of a population of a city is directly proportional to the population of the city. If there were 25,000 people in the city in 2009, and 26,150 people in 2015, find the unconstrained population model and then find the population of the city in 2030, assuming growth continues at the same rate.Solution: The general form of the solution is . To find C, use the initial condition , where

t = 0 represents 2009.

 

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3Slide4

Example:

The rate of change of a population of a city is directly proportional to the population of the city. If there were 25,000 people in the city in 2009, and 26,150 people in 2015, find the unconstrained population model and then find the population of the city in 2030, assuming growth continues at the same rate

.From the last screen, we have . To find k, use the other known condition, . Here t = 6 represents six years since 2009.Thus, the particular solution is .In 2030, set t = 21, and there will be about

people.

This model is unconstrained since it assumes the same growth rate forever. In reality, other factors may inhibit growth. For example, actual physical space may be limited. The uninhibited model is an effective model for short-term modeling.

 

(c) ASU Math - Scott Surgent. Report errors to surgent@asu.edu

4Slide5

Constrained Growth Model (Version I):

Stated in words, it is

The rate of change in a quantity is directly proportional to the room for remaining growth of the quantity.Let be the quantity at time t, and let L be the limiting upper bound of P (that is, as , then ).Then the above statement can be translated as a differential equation of the formThe factor is the “room for growth”, the difference between P and its limiting upper bound L. 

(c) ASU Math - Scott Surgent. Report errors to surgent@asu.edu5Slide6

One way this can be solved is by using separation of variables:

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6Slide7

Example:

The value of a stock on week was $50, and analysts expected that the stock’s maximum value would be no higher than $75 in the future. After 5 weeks, the stock’s value was $62. Assume that the rate of change in the value of the stock is directly proportional with the remaining room for growth in the value of the stock. Find a function that models this stock’s value, and determine when the stock’s value will reach $70.

Solution: From the previous slide, we use the form . Here, , so we haveThe initial condition (0,50) is used to find C:

The known condition (5,62) is used to find

k

:

 

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7Slide8

Thus, the stock’s value is modeled by

We find when the stock reaches a value of $70:

 (c) ASU Math - Scott Surgent. Report errors to surgent@asu.edu8Slide9

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9

Example: A sample of 25 mg of a radioactive element decays such that after 48 hours, 22.5 mg remains. Assume the rate of change in the amount of the sample is proportional to the amount of the sample. Find a function that models this element’s decay, and determine when 90% of the sample has decayed away.Solution: The initial conditions are (0,25) and (48,22.5).The general model is . From the initial condition (0,25), we have C = 25, so now we have .From the other known condition (48,22.5), we have

 Slide10

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10

Thus, the element decays according to the model .When 90% of the element has decayed away, then 10% remains, so we are looking for t such that , which is 10% of 25: Slide11

Constrained Growth Model (Logistics Model):

In words, it is stated as

The rate of change in a quantity is directly proportional to the quantityand to the room for remaining growth of the quantity.Let be the quantity at time t, and let L be the limiting upper bound of P (that is, as , then ).Then the above statement can be translated as a differential equation of the formThe factor is the “room for growth”, the difference between P and its limiting upper bound L. 

(c) ASU Math - Scott Surgent. Report errors to surgent@asu.edu

11Slide12

To find

P

, separate variables and integrate:(c) ASU Math - Scott Surgent. Report errors to surgent@asu.edu12Slide13

Example:

In a dormitory of 1,000 students, 20 come down with the flu. It is a very contagious flu, and within an hour, the total number of students with the flu is 35. The authorities quarantine everyone in the dorm so no one can leave. Assume that the rate of change in the number of students with the flu is both proportional to the number of students with the flu, and the remaining room for growth of the spread of the flu. Find a function

that models the number of cases of flu after t hours, and determine when 750 of the students will have the flu.Solution: Using the form , we have and known points, (0, 20) and (1, 35).We find C first, using (0, 20):So now we have .We’ll find k on the next slide. 

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13Slide14

We have

and the other condition, (1, 35):

So now we have which simplifies to When will 750 of the students have the flu?

 

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14Slide15

Here is the graph:

The graph has an “S” shape. It is called a logistics graph. The max rate of change occurs at the inflection point.

(c) ASU Math - Scott Surgent. Report errors to surgent@asu.edu15Slide16

There are other growth models. For example, suppose the rate of change in the price of a commodity is inversely proportional to the square root of the price.

The differential equation that models this growth is

Its solution is (You should verify this). (c) ASU Math - Scott Surgent. Report errors to surgent@asu.edu16