A Mathematical View of Our World

1
A Mathematical View of Our World

• 1st ed.
• Parks, Musser, Trimpe, Maurer, and Maurer

2
Chapter 12

• Growth and Decay

3
Section 12.1Malthusian Population Growth

• Goals
• Study Malthusian population growth
• Use the Malthusian growth formula
• Study Ponzi schemes
• Study chain letters

4
12.1 Initial Problem

• Based on the data in the table, does it appear
  that the world population follows a Malthusian
  model?
• The solution will be given at the end of the
  section.

5
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.

6
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?

7
Example 1, contd

• Solution The growth rate is 5 or 0.05.
• The initial population is P0 20,000.
• The population after 1 year is found to be P1
  P0 0.05(P0)
• 20,000 0.05(20,000) 21,000 people.

8
Example 1, contd

• Solution, contd The population after 1 year is
  P1 21,000.
• The population after 2 years is
• P2 22,050.
• The population after 3 years is approximately P3
  23,153.

9
Malthusian Growth Formula

• If a population with growth rate r is initially
  P0, then after m years the population will be
• The Malthusian population model assumes the
  population can be computed using this formula.

10
Example 2

• Suppose a population grows by 5 each year.
• If the initial population is 20,000, what is the
  population after 20 years?

11
Example 2, contd

• Solution Use the Malthusian growth formula with
  P0 20,000, r 0.05 and m 20.

12
Malthusian Growth, contd

• Note Even with low growth rates, a Malthusian
  model of population growth always leads to very
  large population estimates.

13
Question

• Suppose a population grows by 3.5 each year.
  If the population is currently 6500, what will it
  be in 30 years? Round to the nearest whole
  number.
• a. 52,835,572 b. 6728
• c. 18,244 d. 19,169

14
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

15
Example 3

• The table summarizes the population information
  from the previous example.
• Calculate the growth rate using 2 different pairs
  of years.

16
Example 3, contd

• Solution Suppose we choose P
  20,000 and Q 53,066.
• We have m 20.

17
Example 3, contd

• Solution, contd Suppose instead we choose P
  21,000 and Q 23,153.
• We have m 2.
• The growth rate is constant, as stated.

18
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?

19
Example 4, contd

• Solution We have P 5000, Q
  8000, and m 10.
• The annual growth rate is about 4.8.

20
Example 4, contd

• Solution Use the Malthusian growth formula to
  find the population after 20 years.
• The estimated population is about 12,800.

21
Malthusian Growth, contd

• For cases in which the value of r is not known, a
  different formula can be used.
• If the population changes from P to Q in m years,
  then after n years the population changes from P
  to

22
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.

23
Example 5, contd

• Solution Use the formula with
• P 6100, Q 9300, m 11, and
• n 25.

24
Example 6

• Chinas 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?

25
Example 6, contd

• Solution 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.

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Example 6, contd

• 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.

27
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.

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Ponzi Schemes, contd

• Suppose a Ponzi scheme offers an interest rate of
  40 in 90 days (a quarter year).
• Let Sm be the number of investors in the mth
  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,
  Sm1 , must be at least 1.4(Sm).

29
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?

30
Example 7, contd

• Solution We know that S1 18 and we need to
  find S49.
• Since and and so on, we find that

31
Example 7, contd

• Solution, contd We can use this formula to find
  S49.
• The scheme started with 18 investors, but after
  12 years he must have 185,959,435 investors to
  keep the scheme going.

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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.

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Chain Letters, contd

• 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

34
Question

• Suppose you get a chain letter with a list of 10
  names which promises you will receive over 9
  million dollars if you send the letter on to 5
  new people. How many people must participate in
  the chain letter for you to move to the top of
  the list and receive your big payoff?

a. 2,441,406 b. 12,207,031 c. 9,765,625 d.
11,111
35
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?

36
Example 8, contd

• 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.

37
Example 8, contd

• Solution Since know that m 7 and
• n 5.
• A total of 97,656 people must participate for the
  payoff to be achieved.

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12.1 Initial Problem Solution

• Does it appear that the worlds population
  follows a Malthusian model?

39
Initial Problem Solution, contd

• Table 12.4

40
Initial Problem Solution, contd

• The growth rates vary widely, so the Malthusian
  model would not be a good predictor for this
  population.

41
Section 12.2Population Decrease,Radioactive
Decay

• Goals
• Study radioactive decay
• Study half-life
• Study carbon-14 dating

42
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.

43
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.

44
Example 1, contd

• 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.

45
Example 1, contd

• Solution We have P0 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.

46
Example 1, contd

• 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.

47
Radioactive Decay

• If a radioactive substance has an annual rate of
  decay of d and there are initially A0 units of
  the substance present, then the amount of the
  radioactive substance present after m years will
  be

48
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?

49
Example 2, contd

• Solution The decay rate is d 0.01, the initial
  amount is A0 100, and the length of time is m
  25.
• After 25 years, 77.8 grams will remain.

50
Question

• A radioactive substance has an annual decay rate
  of 5. If after 100 years there is 12 grams of
  the substance left, how much of the substance was
  present in the beginning? Round to the nearest
  whole gram.
• a. 2027 grams b. 20 grams
• c. 1689 grams d. 1.5 x 1031 grams

51
Half-Life

• The half-life of a radioactive substance is the
  time at which exactly half of the initial amount
  remains.

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Decay Rate Formula

• If d is the annual decay rate of a substance and
  h is the half-life of the substance, then

53
Example 3

• Suppose the half-life of a particular radioactive
  substance is 25 years.
• What is the annual decay rate?

54
Example 3, contd

• Solution We have h 25.
• The decay rate is about 2.73.

55
Example 4

• Suppose in 1995 you had 128 grams of the
  substance from the previous example.
• How much will remain in 2095?

56
Example 4, contd

• Solution We know h 25 and
• d 0.0273.
• The initial amount was A0 128 and the length of
  time is m 100.
• You would have about 8 grams left in 2095.

57
Radioactive Decay Formula

• If a radioactive substance has a half-life of h
  and A0 units of the substance are initially
  present, then the units of the substance present
  after m years will be

58
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?

59
Example 5, contd

• Solution We have h 300 years,
• m 1000 years, and A0 150 grams.
• After 1000 years, just under 15 grams will remain.

60
Half-Life, contd
61
Question

• If you had 10 grams of plutonium 241 in 1990,
  how much uranium would be left in 2010? Round to
  the nearest tenth of a gram.
• a. 6.4 grams
• b. 3.4 grams
• c. 7.3 grams
• d. 2.8 grams

62
Half-Life, contd

• If the exact half-life of a substance is unknown,
  the half-life can be approximated.
• If d is no greater than 1/6, the half-life
  approximation formula is
• Only the first 3 digits are meaningful.

63
Example 6

• Suppose a radioactive substance has an annual
  decay rate of 1.
• What is the approximate half-life of this
  substance?

64
Example 6, contd

• Solution Because d 0.01, the approximation
  formula can be used.
• The approximate half-life is 69 years.

65
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.

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Carbon-14 Dating, contd

• 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.

67
Example 7

• Archaeologists find bones from animals killed by
  ancient hunters. Testing of the bones reveals
  that 35 of the original C-14 remains undecayed.
• Estimate the age of the bones and, hence, the age
  of the village.

68
Example 7, contd

• Solution From the table we find that when 35 of
  the C-14 remains, the sample is 1.51 half-lives
  old.
• The half-life of C-14 is 5730 years.
• Calculate 1.51(5730) 8652.3
• The bones, and the village, are approximately
  8652 years old.

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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?

70
Initial Problem Solution, contd

• Use the decay rate formula, with the time unit of
  hours.
• For h 24 hours

71
Initial Problem Solution, contd

• For h 48 hours
• The drug leaves the bloodstream at a rate of 1.4
  to 2.8 per hour.

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Section 12.3Logistic Population Models

• Goals
• Study logistic growth models
• Use the logistic growth law
• Study steady-state populations
• Study steady-state population fractions

73
12.3 Initial Problem

• Does the world population appear to follow a
  logistic model?
• The solution will be given at the end of the
  section.

74
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.

75
Logistic Growth Model, contd

• Graphs comparing the Malthusian model and
  logistic model are shown below.

76
Logistic Growth Model, contd

• 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.

77
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

78
Example 1

• Suppose a population is governed by the logistic
  growth law, with r 0.5 and c 6000.
• Assume the initial population size is 500 and
  compute and graph the population after each of
  the next 15 breeding seasons.

79
Example 1, contd

• Solution Using the logistic law we get the
  formula

80
Example 1, contd

• Solution, contd Use the formula to compute the
  population after the next 15 breeding seasons.
• For example

81
Example 1, contd
82
Example 1, contd

• Solution, contd A graph of the population
  figures is shown below.

83
Steady-State Population

• A steady-state population is a population value
  at which the population would remain constant.
• Any population governed by the logistic growth
  law has a steady-state population, which is
  always smaller than the carrying capacity.

84
Steady-State Population, contd

• If a population has a natural growth rate of r
  and the environment has a carrying capacity of c,
  then the steady-state population size is

85
Steady-State Population, contd
86
Question

• An aquarium can hold a maximum of 8 goldfish.
  If the natural growth rate of the goldfish is
  35, what is the steady-state population of the
  goldfish in the aquarium? Round your answer to
  the nearest whole number.
• a. 3 fish b. 1 fish
• c. 5 fish d. 2 fish

87
Example 2

• Goats were first brought to Isabela Island in the
  Galapagos Islands in the 1970s and their
  population has increased ever since.
• Approximately 100,000 goats were on Isabela
  Island in 2003, with an estimated carrying
  capacity of 500,000.
• If the natural growth rate is r 0.47, what is
  the steady-state goat population?

88
Example 2, contd

• Solution We have r 0.47 and
• c 500,000.
• Calculate the steady-state population

89
Population Fraction

• The population fraction is the fraction of the
  maximum population that is present at a given
  time.

90
Example 3

• For the population from a previous example with r
  0.5 and c 6000, find the population fractions
  p5 and p10.

91
Example 3, contd

• Solution In the previous example we found that
  P5 1614.
• After 5 breeding seasons, the population is at
  about 26.9 of maximum.

92
Example 3, contd

• Solution, contd In the previous example we
  found that P10 1982.
• After 10 breeding seasons, the population is at
  about 33.0 of maximum.

93
Verhulsts Equation

• We call the growth parameter.
• Verhulsts equation gives a formula for the
  population fraction
• After m 1 breeding seasons, the population
  fraction will be

94
Example 4

• Suppose a population is governed by Verhulsts
  equation with growth parameter .
• If p0 0.7, find the population fraction for
  each of the next 5 breeding seasons.

95
Example 4, contd

• Solution We have and p0 0.7.

96
Example 4, contd

• Solution Use Verhulsts equation to find the
  rest of the population fractions.

97
Steady-State Population Fraction

• If a population has a growth parameter of ,
  where r is the natural growth rate, then the
  steady-state population fraction is

98
Example 5

• Suppose a population is governed by Verhulsts
  equation with
• Compute the steady-state population fraction.

99
Example 5, contd

• Solution The steady-state population fraction is
  calculated
• This means that over time the population would
  level off at about 70 of the carrying capacity.

100
Example 6

• Compute the steady-state population fractions for
  each of the populations.
• The population with r 0.5 and c 6000.
• The goat population with r 0.47 and c
  500,000.

101
Example 6, contd

• Solution The population has r 0.5 and c
  6000.
• We find
• So
• This matches the steady-state population of 2000.

102
Example 6, contd

• Solution The goat population has r 0.47 and c
  500,000.
• We find
• So
• This matches the steady-state population in the
  previous example.

103
Steady-State, contd

• For values of the growth parameter larger than 3,
  we cannot rely on the population fraction to
  return its steady-state value.
• This will be illustrated in the next examples.

104
Example 7

• Suppose a population is governed by Verhulsts
  equation with and the steady-state
  population fraction is known to equal 0.7.
• If the initial population fraction is 0.69, find
  the population fraction for each of the next 5
  breeding seasons.

105
Example 7, contd

• Solution The 5 population fractions are computed

106
Example 8

• Suppose a population is governed by Verhulsts
  equation with .
• If the initial population fraction is 0.4697,
  find the population fraction for each of the next
  4 breeding seasons.

107
Example 8, contd

• Solution Calculate the population fractions

108
12.3 Initial Problem Solution

• Does the world population appear to follow a
  logistic model?

109
Initial Problem Solution, contd
110
Initial Problem Solution, contd

• The logistic growth law does not apply in this
  case because the growth rate has increased,
  rather than decreased, as the population has
  increased.