Title: A Mathematical View of Our World
1A Mathematical View of Our World
- 1st ed.
- Parks, Musser, Trimpe, Maurer, and Maurer
2Chapter 12
3Section 12.1Malthusian Population Growth
- Goals
- Study Malthusian population growth
- Use the Malthusian growth formula
- Study Ponzi schemes
- Study chain letters
412.1 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.
5Population 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.
6Example 1
- Suppose a population grows by 5 each year.
- If the initial population is 20,000, what is the
approximate population after 3 years?
7Example 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.
8Example 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.
9Malthusian 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.
10Example 2
- Suppose a population grows by 5 each year.
- If the initial population is 20,000, what is the
population after 20 years?
11Example 2, contd
- Solution Use the Malthusian growth formula with
P0 20,000, r 0.05 and m 20. -
12Malthusian Growth, contd
- Note Even with low growth rates, a Malthusian
model of population growth always leads to very
large population estimates.
13Question
- 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
14Annual 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
15Example 3
- The table summarizes the population information
from the previous example. - Calculate the growth rate using 2 different pairs
of years.
16Example 3, contd
- Solution Suppose we choose P
20,000 and Q 53,066. - We have m 20.
-
17Example 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.
18Example 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?
19Example 4, contd
- Solution We have P 5000, Q
8000, and m 10. -
- The annual growth rate is about 4.8.
20Example 4, contd
- Solution Use the Malthusian growth formula to
find the population after 20 years. -
- The estimated population is about 12,800.
21Malthusian 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
22Example 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.
23Example 5, contd
- Solution Use the formula with
- P 6100, Q 9300, m 11, and
- n 25.
-
24Example 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?
25Example 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. -
26Example 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. -
27Ponzi 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.
28Ponzi 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).
29Example 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?
30Example 7, contd
- Solution We know that S1 18 and we need to
find S49. - Since and and so on, we find that
31Example 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.
32Chain 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.
33Chain 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
34Question
- 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
35Example 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?
36Example 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.
37Example 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.
3812.1 Initial Problem Solution
- Does it appear that the worlds population
follows a Malthusian model?
39Initial Problem Solution, contd
40Initial Problem Solution, contd
- The growth rates vary widely, so the Malthusian
model would not be a good predictor for this
population.
41Section 12.2Population Decrease,Radioactive
Decay
- Goals
- Study radioactive decay
- Study half-life
- Study carbon-14 dating
4212.2 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.
43Example 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.
44Example 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.
45Example 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.
46Example 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.
47Radioactive 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
48Example 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?
49Example 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.
50Question
- 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
51Half-Life
- The half-life of a radioactive substance is the
time at which exactly half of the initial amount
remains.
52Decay Rate Formula
- If d is the annual decay rate of a substance and
h is the half-life of the substance, then
53Example 3
- Suppose the half-life of a particular radioactive
substance is 25 years. - What is the annual decay rate?
54Example 3, contd
- Solution We have h 25.
-
- The decay rate is about 2.73.
55Example 4
- Suppose in 1995 you had 128 grams of the
substance from the previous example. - How much will remain in 2095?
56Example 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.
57Radioactive 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
58Example 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?
59Example 5, contd
- Solution We have h 300 years,
- m 1000 years, and A0 150 grams.
-
- After 1000 years, just under 15 grams will remain.
60Half-Life, contd
61Question
- 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
62Half-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.
63Example 6
- Suppose a radioactive substance has an annual
decay rate of 1. - What is the approximate half-life of this
substance?
64Example 6, contd
- Solution Because d 0.01, the approximation
formula can be used. -
- The approximate half-life is 69 years.
65Carbon-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.
66Carbon-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.
67Example 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.
68Example 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.
6912.2 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?
70Initial Problem Solution, contd
- Use the decay rate formula, with the time unit of
hours. - For h 24 hours
71Initial Problem Solution, contd
- For h 48 hours
- The drug leaves the bloodstream at a rate of 1.4
to 2.8 per hour.
72Section 12.3Logistic Population Models
- Goals
- Study logistic growth models
- Use the logistic growth law
- Study steady-state populations
- Study steady-state population fractions
7312.3 Initial Problem
- Does the world population appear to follow a
logistic model? - The solution will be given at the end of the
section.
74Logistic 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.
75Logistic Growth Model, contd
- Graphs comparing the Malthusian model and
logistic model are shown below.
76Logistic 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.
77Logistic 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
78Example 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.
79Example 1, contd
- Solution Using the logistic law we get the
formula
80Example 1, contd
- Solution, contd Use the formula to compute the
population after the next 15 breeding seasons. - For example
81Example 1, contd
82Example 1, contd
- Solution, contd A graph of the population
figures is shown below.
83Steady-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.
84Steady-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
85Steady-State Population, contd
86Question
- 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
87Example 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?
88Example 2, contd
- Solution We have r 0.47 and
- c 500,000.
- Calculate the steady-state population
89Population Fraction
- The population fraction is the fraction of the
maximum population that is present at a given
time. -
90Example 3
- For the population from a previous example with r
0.5 and c 6000, find the population fractions
p5 and p10.
91Example 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.
92Example 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.
93Verhulsts 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
94Example 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.
95Example 4, contd
- Solution We have and p0 0.7.
-
-
96Example 4, contd
- Solution Use Verhulsts equation to find the
rest of the population fractions. -
-
-
-
-
97Steady-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
98Example 5
- Suppose a population is governed by Verhulsts
equation with -
- Compute the steady-state population fraction.
99Example 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.
100Example 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.
101Example 6, contd
- Solution The population has r 0.5 and c
6000. - We find
- So
- This matches the steady-state population of 2000.
102Example 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.
103Steady-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.
104Example 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.
105Example 7, contd
- Solution The 5 population fractions are computed
106Example 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.
107Example 8, contd
- Solution Calculate the population fractions
-
-
-
-
10812.3 Initial Problem Solution
- Does the world population appear to follow a
logistic model?
109Initial Problem Solution, contd
110Initial 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.