# Population Models (Autonomic Differential Equations)

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• 1.Population Models (Autonomic Differential Equations) MAT 275
• 2.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 𝑃(𝑡) 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 𝑑𝑃 𝑑𝑡 . The word “is” translates as =. The words “directly proportional” imply a proportion constant k. The words “population at time t” is 𝑃(𝑡). Combined, this is a differential equation 𝑑𝑃 𝑑𝑡 =𝑘𝑃(𝑡). (c) ASU Math - Scott Surgent. Report errors to surgent@asu.edu 2
• 3.The differential equation 𝑑𝑃 𝑑𝑡 =𝑘𝑃(𝑡) can also be written 𝑃 ′ =𝑘𝑃. It is autonomic. It can be solved by separating variables, integration factors, or using the general form 𝑃 𝑡 =𝐶 𝑒 𝑘𝑡 . 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 𝑃 𝑡 =𝐶 𝑒 𝑘𝑡 . To find C, use the initial condition 𝑃 0 =25,000, where t = 0 represents 2009. (c) ASU Math - Scott Surgent. Report errors to surgent@asu.edu 3
• 4.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 𝑃 𝑡 =25,000 𝑒 𝑘𝑡 . To find k, use the other known condition, 𝑃 6 =26,150. Here t = 6 represents six years since 2009. Thus, the particular solution is 𝑃 𝑡 =25,000 𝑒 0.0075𝑡 . In 2030, set t = 21, and there will be about 𝑃 21 =25,000 𝑒 0.0075 21 ≈29,265 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 4
• 5.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 𝑃(𝑡) be the quantity at time t, and let L be the limiting upper bound of P (that is, as 𝑡→+∞, then 𝑃 𝑡 →𝐿). Then the above statement can be translated as a differential equation of the form 𝑑𝑃 𝑑𝑡 =𝑘 𝐿−𝑃 . The factor 𝐿−𝑃 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 5
• 6.One way this can be solved is by using separation of variables: (c) ASU Math - Scott Surgent. Report errors to surgent@asu.edu 6
• 7.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 𝑃 𝑡 =𝐿+𝐶 𝑒 −𝑘𝑡 . Here, 𝐿=75, so we have 𝑃 𝑡 =75+𝐶 𝑒 −𝑘𝑡 . The initial condition (0,50) is used to find C: 50=75+𝐶 𝑒 −𝑘 0 →−25=𝐶 1 . Thus, 𝑃 𝑡 =75−25 𝑒 −𝑘𝑡 . The known condition (5,62) is used to find k: 62=75−25 𝑒 −𝑘 5 →−13=−25 𝑒 −5𝑘 →𝑘= ln 13 25 −5 ≈0.131. (c) ASU Math - Scott Surgent. Report errors to surgent@asu.edu 7
• 8.Thus, the stock’s value is modeled by 𝑃 𝑡 =75−25 𝑒 −0.131𝑡 . We find when the stock reaches a value of \$70: 70=75−25 𝑒 −0.131𝑡 −5=−25 𝑒 −0.131𝑡 0.2= 𝑒 −0.131𝑡 ln 0.2 −0.131 =𝑡 𝑡≈12.3 weeks. (c) ASU Math - Scott Surgent. Report errors to surgent@asu.edu 8
• 9.(c) ASU Math - Scott Surgent. Report errors to surgent@asu.edu 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 𝑃 𝑡 =𝐶 𝑒 𝑘𝑡 . From the initial condition (0,25), we have C = 25, so now we have 𝑃 𝑡 =25 𝑒 𝑘𝑡 . From the other known condition (48,22.5), we have 22.5=25 𝑒 𝑘 48 → 22.5 25 = 𝑒 48𝑘 → ln 22.5 25 =48𝑘→𝑘= ln 22.5 25 48 ≈−0.0022.
• 10.(c) ASU Math - Scott Surgent. Report errors to surgent@asu.edu 10 Thus, the element decays according to the model 𝑃 𝑡 =25 𝑒 −0.0022𝑡 . When 90% of the element has decayed away, then 10% remains, so we are looking for t such that 𝑃 𝑡 =2.5, which is 10% of 25:
• 11.Constrained Growth Model (Logistics Model): In words, it is stated as The rate of change in a quantity is directly proportional to the quantity and to the room for remaining growth of the quantity. Let 𝑃(𝑡) be the quantity at time t, and let L be the limiting upper bound of P (that is, as 𝑡→+∞, then 𝑃 𝑡 →𝐿). Then the above statement can be translated as a differential equation of the form 𝑑𝑃 𝑑𝑡 =𝑘𝑃 𝐿−𝑃 . The factor 𝐿−𝑃 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 11
• 12.To find P, separate variables and integrate: (c) ASU Math - Scott Surgent. Report errors to surgent@asu.edu 12
• 13.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 𝑃 𝑡 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 𝑃 𝑡 = 𝐿 1+𝐶 𝑒 −𝐿𝑘𝑡 , we have 𝐿=1000 and known points, (0, 20) and (1, 35). We find C first, using (0, 20): So now we have 𝑃 𝑡 = 1000 1+49 𝑒 −1000𝑘𝑡 . We’ll find k on the next slide. (c) ASU Math - Scott Surgent. Report errors to surgent@asu.edu 13
• 14.We have 𝑃 𝑡 = 1000 1+49 𝑒 −1000𝑘𝑡 and the other condition, (1, 35): So now we have 𝑃 𝑡 = 1000 1+49 𝑒 −1000(0.000575)𝑡 which simplifies to 𝑃 𝑡 = 1000 1+49 𝑒 −0.575𝑡 . When will 750 of the students have the flu? 750= 1000 1+49 𝑒 −0.575𝑡 →1+49 𝑒 −0.575𝑡 = 1000 750 → 𝑒 −0.575𝑡 = 4 3 −1 49 →𝑡= ln 0.0068… −0.575 ≈8.68 ℎ𝑜𝑢𝑟𝑠. (c) ASU Math - Scott Surgent. Report errors to surgent@asu.edu 14
• 15.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.edu 15
• 16.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 𝑑𝑃 𝑑𝑡 = 𝑘 𝑃 . Its solution is 𝑃 𝑡 = 3 3 2 𝑘𝑡+𝐶 2 . (You should verify this). (c) ASU Math - Scott Surgent. Report errors to surgent@asu.edu 16