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# Nossi ch 12

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Population growth and decay

Population growth and decay

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### Transcript

• 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
• 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 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.
• 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.
• 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
• 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 .