Chapter 12 Growth and Decay
Section 12.1 Malthusian Population Growth <ul><li>Goals </li></ul><ul><ul><li>Study Malthusian population growth </li></ul...
Population Growth <ul><li>The ratio of births to a population size is called the  birth rate . </li></ul><ul><li>The ratio...
Example 1 <ul><li>Suppose a population grows by 5% each year. </li></ul><ul><li>If the initial population is 20,000, what ...
Example 1 <ul><li>Solution: The growth rate is 5% or 0.05. </li></ul><ul><li>The initial population is  P 0  = 20,000. </l...
Example 1 <ul><li>The population after 1 year is  P 1  = 21,000. </li></ul><ul><li>The population after 2 years is  </li><...
Malthusian Growth Formula <ul><li>If a population with growth rate  r  is initially  P 0 , then after  m  years the popula...
Example 2 <ul><li>Suppose a population grows by 5% each year.  </li></ul><ul><li>If the initial population is 20,000, what...
Example 2 <ul><li>Solution: Use the Malthusian growth formula with  P 0  = 20,000,  r  = 0.05 and  m  = 20. </li></ul>
Malthusian Growth <ul><li>Even with low growth rates, a Malthusian model of population growth always leads to very large p...
Annual Growth Rate <ul><li>If the population changes from  P  to  Q  in  m  years and a Malthusian population model is ass...
Example 3 <ul><li>The table summarizes the population information from the previous example. </li></ul><ul><li>Calculate t...
Example 3 <ul><li>Solution: Suppose we choose the first and last dates. </li></ul><ul><li>P  = 20,000 and  Q  = 53,066. </...
Example 3 <ul><li>Suppose instead we choose  </li></ul><ul><li>P  = 21,000 and  Q  = 23,153. </li></ul><ul><ul><li>We have...
Example 4 <ul><li>Suppose a population is initially 5000 and increases to 8000 in 10 years. </li></ul><ul><ul><li>Assume a...
Example 4 <ul><li>We have  P  = 5000,  </li></ul><ul><li>  Q  = 8000, and  m  = 10. </li></ul><ul><ul><li>The annual growt...
Example 4 <ul><li>Solution: Use the Malthusian growth formula to find the population after 20 years. </li></ul><ul><ul><li...
Malthusian Growth <ul><li>For cases in which the value of  r  is not known, a different formula can be used instead of the...
Example 5 <ul><li>Suppose a population is initially 6100 and increases to 9300 in 11 years. </li></ul><ul><li>What is the ...
Example 5 <ul><li>Solution: Use the formula with  </li></ul><ul><li>P  = 6100,  Q  = 9300,  m  = 11, and </li></ul><ul><li...
Example 6 <ul><li>China’s population increased by 132,200,000 from 1990 to 2000, for a total population of 1,260,000,000 i...
Example 6 <ul><li>We know  </li></ul><ul><li>Q  = 1,260,000,000. </li></ul><ul><ul><li>Since there was an increase of 132,...
Example 6 <ul><li>Solution: Predict the population of China in 2010. </li></ul><ul><ul><li>We have  P  = 1,127,800,000,  Q...
Ponzi Schemes <ul><li>A confidence man named Charles Ponzi developed a fraudulent investment scheme which became known as ...
Ponzi Schemes <ul><li>Suppose a Ponzi scheme offers an interest rate of 40% in 90 days (a quarter year). </li></ul><ul><li...
Example 7 <ul><li>Suppose the number of investors in the first quarter of a Ponzi scheme is 18. </li></ul><ul><li>What is ...
Example 7 <ul><li>Solution: We know that  S 1  = 18 and we need to find  S 49 . </li></ul><ul><ul><li>Since     and    and...
Example 7 <ul><li>We can use this formula to find  S 49 . </li></ul><ul><ul><li>The scheme started with 18 investors, but ...
Chain Letters <ul><li>A chain letter arrives with a list of  m  names. </li></ul><ul><ul><li>The number of names on the li...
 
Chain Letters <ul><li>For a chain letter with  m  levels,  n  new participants for each letter, and a price to participate...
Example 8 <ul><li>Suppose a chain letter has 7 levels, asks you to send $10 to the person at the top of the list, and requ...
Example 8 <ul><li>Solution: We know that  m  = 7 levels,  </li></ul><ul><li>n  =  5 new participants, and  P  = $10. </li>...
Example 8 <ul><li>Solution: Since we know that  m  = 7 and  n  =  5. Use: </li></ul><ul><ul><li>A total of 97,656 people m...
Section 12.2 Population Decrease, Radioactive Decay <ul><li>Goals </li></ul><ul><ul><li>Study radioactive decay </li></ul>...
12.2 Initial Problem <ul><li>The tranquilizer Librium ®  has a half-life of between 24 and 48 hours. </li></ul><ul><li>Wha...
Example 1 <ul><li>The population of Lake County, Oregon was 7532 in 1980 and 7186 in 1990.  Assume a Malthusian model. </l...
Example 1 <ul><li>Solution: We have  P  = 7532,  </li></ul><ul><li>Q  = 7186, and  m  = 10. </li></ul><ul><ul><li>If the g...
Example 1 <ul><li>Solution: We have  P 0  = 7532,  </li></ul><ul><li>r  = -0.00469, and  m  = 20. </li></ul><ul><ul><li>If...
Example 1 <ul><li>The population of Lake County, Oregon in 2000 was actually 7422. </li></ul><ul><ul><li>This indicates th...
Radioactive Decay <ul><li>If a radioactive substance has an annual rate of decay of  d  and there are initially  A 0  unit...
Example 2 <ul><li>Suppose a radioactive substance has an annual decay rate of 1%. </li></ul><ul><li>If a sample contains 1...
Example 2 <ul><li>Solution: The decay rate is  d  = 0.01, the initial amount is  A 0  = 100, and the length of time is  m ...
Half-Life <ul><li>The  half-life  of a radioactive substance is the time at which exactly half of the initial amount remai...
Decay Rate Formula <ul><li>If  d  is the annual decay rate of a substance and  h  is the half-life of the substance, then:...
Example 3 <ul><li>Suppose the half-life of a particular radioactive substance is 25 years. </li></ul><ul><li>What is the a...
Example 3 <ul><li>Solution: We have  h  = 25. </li></ul><ul><ul><li>The decay rate is about 2.73%. </li></ul></ul>
Example 4 <ul><li>Suppose in 1995 you had 128 grams of the substance from the previous example.  </li></ul><ul><li>How muc...
Example 4 <ul><li>Solution: We know  h  = 25 and  </li></ul><ul><li>d  = 0.0273. </li></ul><ul><ul><li>The initial amount ...
Radioactive Decay Formula <ul><li>If a radioactive substance has a half-life of  h  and  A 0  units of the substance are i...
Example 5 <ul><li>Suppose a radioactive substance has a half-life of 300 years. </li></ul><ul><li>If a sample contains 150...
Example 5 <ul><li>Solution: We have  h  = 300 years,  </li></ul><ul><li>m  = 1000 years, and  A 0  = 150 grams. </li></ul>...
Half-Life
Carbon-14 Dating <ul><li>Carbon-14 dating uses the known half-life of the radioactive isotope of carbon to estimate the ti...
Carbon-14 Dating <ul><li>All living things have the same ratio of C-12 to C-14. </li></ul><ul><li>As soon as an organism d...
 
12.2 Initial Problem Solution <ul><li>The tranquilizer Librium ®  has a half-life of between 24 and 48 hours.  What is the...
Initial Problem Solution, cont’d <ul><li>Use the decay rate formula, with the time unit of hours. </li></ul><ul><li>For  h...
Initial Problem Solution, cont’d <ul><li>For  h  = 48 hours: </li></ul><ul><li>The drug leaves the bloodstream at a rate o...
Section 12.3 Logistic Population Models <ul><li>Goals </li></ul><ul><ul><li>Study logistic growth models </li></ul></ul><u...
Logistic Growth Model <ul><li>For a population with limited resources, there may be a maximum population size, called the ...
Logistic Growth Model <ul><li>Graphs comparing the Malthusian model and logistic model are shown below. </li></ul>
Logistic Growth Model <ul><li>The growth rate in a logistic model varies over time. </li></ul><ul><li>The  natural growth ...
Logistic Growth Law <ul><li>If a population has a natural growth rate of  r  from 0 to 3, and the environment has a carryi...
Chapter 12 Assignment Due Tues Aug 5 <ul><li>Section 12.1 pg 764 (5,7,11,15,19,23,31,33) </li></ul><ul><li>Section 12.2 pg...
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Population growth and decay

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Nossi ch 12

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