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Nossi ch 12
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Nossi ch 12

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Population growth and decay

Population growth and decay

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  • 1. Chapter 12 Growth and Decay
  • 2. Section 12.1 Malthusian Population Growth
    • Goals
      • Study Malthusian population growth
        • Use the Malthusian growth formula
        • Study Ponzi schemes
        • Study chain letters
  • 3. Population Growth
    • The ratio of births to a population size is called the birth rate .
    • The ratio of deaths to a population size is called the death rate .
    • The difference between the birth and death rates is called the growth rate .
  • 4. Example 1
    • Suppose a population grows by 5% each year.
    • If the initial population is 20,000, what is the approximate population after 3 years?
  • 5. Example 1
    • Solution: The growth rate is 5% or 0.05.
    • The initial population is P 0 = 20,000.
    • The population after 1 year is found to be P 1 = P 0 + 0.05( P 0 ) =
    • 20,000 + 0.05(20,000) = 21,000 people.
  • 6. Example 1
    • The population after 1 year is P 1 = 21,000.
    • The population after 2 years is
    • P 2 = 21,000 + 21,000(.05) = 22,050.
    • The population after 3 years is approximately
    • P 3 =22,050 + 22,050(.05) = 23,153.
  • 7. Malthusian Growth Formula
    • If a population with growth rate r is initially P 0 , then after m years the population will be
      • The Malthusian population model assumes the population can be computed using this formula.
  • 8. Example 2
    • Suppose a population grows by 5% each year.
    • If the initial population is 20,000, what is the population after 20 years?
  • 9. Example 2
    • Solution: Use the Malthusian growth formula with P 0 = 20,000, r = 0.05 and m = 20.
  • 10. Malthusian Growth
    • Even with low growth rates, a Malthusian model of population growth always leads to very large population estimates.
  • 11. Annual Growth Rate
    • If the population changes from P to Q in m years and a Malthusian population model is assumed, then the annual growth rate is given by the formula:
  • 12. Example 3
    • The table summarizes the population information from the previous example.
    • Calculate the growth rate using 2 different pairs of years.
  • 13. Example 3
    • Solution: Suppose we choose the first and last dates.
    • P = 20,000 and Q = 53,066.
      • We have m = 20.
  • 14. Example 3
    • Suppose instead we choose
    • P = 21,000 and Q = 23,153.
      • We have m = 2.
      • The growth rate is constant = 5%.
  • 15. Example 4
    • Suppose a population is initially 5000 and increases to 8000 in 10 years.
      • Assume a Malthusian population model.
    • Estimate the annual growth rate.
    • What is the predicted population in 20 years?
  • 16. Example 4
    • We have P = 5000,
    • Q = 8000, and m = 10.
      • The annual growth rate is about 4.8%.
  • 17. Example 4
    • Solution: Use the Malthusian growth formula to find the population after 20 years.
      • The estimated population is about 12,800.
      • Note : there is an error in this slide r should = .048
  • 18. Malthusian Growth
    • For cases in which the value of r is not known, a different formula can be used instead of the steps in the previous example.
    • If the population changes from P to Q in m years, then after n years the population changes from P to
  • 19. Example 5
    • Suppose a population is initially 6100 and increases to 9300 in 11 years.
    • What is the predicted population in 25 years?
      • Assume a Malthusian population model.
  • 20. Example 5
    • Solution: Use the formula with
    • P = 6100, Q = 9300, m = 11, and
    • n = 25.
  • 21. Example 6
    • China’s population increased by 132,200,000 from 1990 to 2000, for a total population of 1,260,000,000 in 2000.
    • Find the annual growth rate between 1990 and 2000.
    • What is the predicted population for the year 2010?
  • 22. Example 6
    • We know
    • Q = 1,260,000,000.
      • Since there was an increase of 132,200,000, the population in 1990 must have been P = 1,260,000,000 - 132,200,000 = 1,127,800,000.
  • 23. Example 6
    • Solution: Predict the population of China in 2010.
      • We have P = 1,127,800,000, Q = 1,260,000,000, n = 20, and m = 10.
  • 24. Ponzi Schemes
    • A confidence man named Charles Ponzi developed a fraudulent investment scheme which became known as the Ponzi scheme .
      • Instead of offering a true investment, Ponzi just paid old investors with the money received from new investors.
  • 25. Ponzi Schemes
    • Suppose a Ponzi scheme offers an interest rate of 40% in 90 days (a quarter year).
    • Let S m be the number of investors in the m th quarter.
    • Ponzi must have enough investors in the next quarter to pay the interest to these current investors.
      • The number of investors in the m + 1 quarter, S m+1 , must be at least 1.4( S m ).
  • 26. Example 7
    • Suppose the number of investors in the first quarter of a Ponzi scheme is 18.
    • What is the minimum number of investors necessary after 12 years, or 48 quarters?
  • 27. Example 7
    • Solution: We know that S 1 = 18 and we need to find S 49 .
      • Since and and so on, we find that
  • 28. Example 7
    • We can use this formula to find S 49 .
      • The scheme started with 18 investors, but after 12 years he must have 185,959,435 investors to keep the scheme going.
  • 29. Chain Letters
    • A chain letter arrives with a list of m names.
      • The number of names on the list is called the number of levels ,
    • The recipient mails something valuable to the person at the top of the list.
      • The recipient then removes that name from the top of the list and adds his or her name to the bottom.
    • He or she sends the letter to n new people.
  • 30.  
  • 31. Chain Letters
    • For a chain letter with m levels, n new participants for each letter, and a price to participate of P , the payoff in rising from the bottom to the top of the letter is
    • The total number of people who must join for an individual to make it to the top of the list is
  • 32. Example 8
    • Suppose a chain letter has 7 levels, asks you to send $10 to the person at the top of the list, and requires you to send out 5 new letters.
    • How much is the payoff in going from the bottom to the top of the list?
    • How many people must participate?
  • 33. Example 8
    • Solution: We know that m = 7 levels,
    • n = 5 new participants, and P = $10.
      • The payoff in going from the bottom to the top of the list is $781,250.
  • 34. Example 8
    • Solution: Since we know that m = 7 and n = 5. Use:
      • A total of 97,656 people must participate for the payoff to be achieved.
  • 35. Section 12.2 Population Decrease, Radioactive Decay
    • Goals
      • Study radioactive decay
        • Study half-life
        • Study carbon-14 dating
  • 36. 12.2 Initial Problem
    • The tranquilizer Librium ® has a half-life of between 24 and 48 hours.
    • What is the hourly rate at which Librium ® leaves the bloodstream as the drug is metabolized?
      • The solution will be given at the end of the section.
  • 37. Example 1
    • The population of Lake County, Oregon was 7532 in 1980 and 7186 in 1990. Assume a Malthusian model.
    • Determine the annual rate of decline between 1980 and 1990.
    • Estimate the population in 2000.
  • 38. Example 1
    • Solution: We have P = 7532,
    • Q = 7186, and m = 10.
      • If the growth rate remained constant, the population decreased by 0.469% each year between 1980 and 1990.
  • 39. Example 1
    • Solution: We have P 0 = 7532,
    • r = -0.00469, and m = 20.
      • If the population continued to decrease by 0.469% each year, the population in 2000 would have been 6856 people.
  • 40. Example 1
    • The population of Lake County, Oregon in 2000 was actually 7422.
      • This indicates that the population did not continue to decrease at a constant rate.
      • Many factors affect population growth or decrease making reliable predictions difficult.
  • 41. Radioactive Decay
    • If a radioactive substance has an annual rate of decay of d and there are initially A 0 units of the substance present, then the amount of the radioactive substance present after m years will be: (This is the same formula as before except d is used in place of r and A is used in place of P )
  • 42. Example 2
    • Suppose a radioactive substance has an annual decay rate of 1%.
    • If a sample contains 100 grams of the substance, how much will remain after 25 years?
  • 43. Example 2
    • Solution: The decay rate is d = 0.01, the initial amount is A 0 = 100, and the length of time is m = 25.
      • After 25 years, 77.8 grams will remain.
  • 44. Half-Life
    • The half-life of a radioactive substance is the time at which exactly half of the initial amount remains.
  • 45. Decay Rate Formula
    • If d is the annual decay rate of a substance and h is the half-life of the substance, then:
  • 46. Example 3
    • Suppose the half-life of a particular radioactive substance is 25 years.
    • What is the annual decay rate?
  • 47. Example 3
    • Solution: We have h = 25.
      • The decay rate is about 2.73%.
  • 48. Example 4
    • Suppose in 1995 you had 128 grams of the substance from the previous example.
    • How much will remain in 2095?
  • 49. Example 4
    • Solution: We know h = 25 and
    • d = 0.0273.
      • The initial amount was A 0 = 128 and the length of time is m = 100.
      • You would have about 8 grams left in 2095.
  • 50. Radioactive Decay Formula
    • If a radioactive substance has a half-life of h and A 0 units of the substance are initially present, then the units of the substance present after m years will be:
  • 51. Example 5
    • Suppose a radioactive substance has a half-life of 300 years.
    • If a sample contains 150 grams of the substance, how much will remain after 1000 years?
  • 52. Example 5
    • Solution: We have h = 300 years,
    • m = 1000 years, and A 0 = 150 grams.
      • After 1000 years, just under 15 grams will remain.
  • 53. Half-Life
  • 54. Carbon-14 Dating
    • Carbon-14 dating uses the known half-life of the radioactive isotope of carbon to estimate the time since the death of living organisms.
      • Scientists assume the ratio of normal carbon, C-12, to radioactive carbon, C-14, has been constant for millions of years.
  • 55. Carbon-14 Dating
    • All living things have the same ratio of C-12 to C-14.
    • As soon as an organism dies, its amount of C-14 begins to decay.
    • Percentages of C-14 present are given in the table.
  • 56.  
  • 57. 12.2 Initial Problem Solution
    • The tranquilizer Librium ® has a half-life of between 24 and 48 hours. What is the hourly rate at which Librium ® leaves the bloodstream as the drug is metabolized?
  • 58. Initial Problem Solution, cont’d
    • Use the decay rate formula, with the time unit of hours.
    • For h = 24 hours:
  • 59. Initial Problem Solution, cont’d
    • For h = 48 hours:
    • The drug leaves the bloodstream at a rate of 1.4% to 2.8% per hour.
  • 60. Section 12.3 Logistic Population Models
    • Goals
      • Study logistic growth models
        • Use the logistic growth law
        • Study steady-state populations
        • Study steady-state population fractions
  • 61. Logistic Growth Model
    • For a population with limited resources, there may be a maximum population size, called the carrying capacity .
    • A model for population growth that takes into account the carrying capacity is called the logistic population model .
  • 62. Logistic Growth Model
    • Graphs comparing the Malthusian model and logistic model are shown below.
  • 63. Logistic Growth Model
    • The growth rate in a logistic model varies over time.
    • The natural growth rate , r , is the rate at which the population would grow if there were no limitations.
  • 64. Logistic Growth Law
    • If a population has a natural growth rate of r from 0 to 3, and the environment has a carrying capacity of c , then after m + 1 breeding seasons, the population will be: (You will NOT use this formula in a problem. I just want you to have seen it.)
  • 65. Chapter 12 Assignment Due Tues Aug 5
    • Section 12.1 pg 764 (5,7,11,15,19,23,31,33)
    • Section 12.2 pg 776 (1,5,9,13,21)
    • Section 12.3 pg 791 (1,2)
    • Next week Aug 5 class meets at 7:30 .

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