Chapter 12Chapter 12

Growth and DecayGrowth and Decay

Section 12.1Section 12.1Malthusian Population GrowthMalthusian Population Growth

• GoalsGoals• Study Malthusian population growthStudy Malthusian population growth

• Use the Malthusian growth formulaUse the Malthusian growth formula• Study Ponzi schemesStudy Ponzi schemes• Study chain lettersStudy chain letters

Population GrowthPopulation Growth

• The ratio of births to a population size is The ratio of births to a population size is called the called the birth ratebirth rate..

• The ratio of deaths to a population size The ratio of deaths to a population size is called the is called the death ratedeath rate..

• The difference between the birth and The difference between the birth and death rates is called the death rates is called the growth rategrowth rate..

Example 1Example 1

• Suppose a population grows by 5% Suppose a population grows by 5% each year.each year.

• If the initial population is 20,000, what is If the initial population is 20,000, what is the approximate population after 3 the approximate population after 3 years? years?

Example 1Example 1

• Solution: The growth rate is 5% or 0.05.Solution: The growth rate is 5% or 0.05.• The initial population is The initial population is PP00 = 20,000. = 20,000.• The population after 1 year is found to The population after 1 year is found to

be be PP11 = = PP00 + 0.05( + 0.05(PP00) = ) =

20,000 + 0.05(20,000) = 21,000 people.20,000 + 0.05(20,000) = 21,000 people.

Example 1Example 1

• The population after 1 year is The population after 1 year is PP11 = 21,000. = 21,000.• The population after 2 years is The population after 2 years is

PP22 = 21,000 + 21,000(.05) = 22,050.= 21,000 + 21,000(.05) = 22,050.• The population after 3 years is The population after 3 years is

approximately approximately • PP33 =22,050 + 22,050(.05) = 23,153.=22,050 + 22,050(.05) = 23,153.

Malthusian Growth FormulaMalthusian Growth Formula

• If a population with growth rate If a population with growth rate rr is initially is initially PP00, then after , then after mm years the population will be years the population will be

• The Malthusian population model assumes the The Malthusian population model assumes the population can be computed using this formula.population can be computed using this formula.

( ) 01 mmP r P= + ⋅

Example 2Example 2

• Suppose a population grows by 5% Suppose a population grows by 5% each year. each year.

• If the initial population is 20,000, what If the initial population is 20,000, what is the population after 20 years? is the population after 20 years?

Example 2Example 2

• Solution: Use the Malthusian growth Solution: Use the Malthusian growth formula with formula with PP00 = 20,000, = 20,000, rr = 0.05 and = 0.05 and mm = 20. = 20.• ( )

( )20 0

20

1

1 .05 20000 53,066

mP r P= + ⋅

= + ⋅ ≈

Malthusian GrowthMalthusian Growth• Even with low Even with low

growth rates, a growth rates, a Malthusian model Malthusian model of population of population growth always growth always leads to very large leads to very large population population estimates.estimates.

Annual Growth RateAnnual Growth Rate• If the population changes from If the population changes from PP to to QQ in in

mm years and a Malthusian population years and a Malthusian population model is assumed, then the model is assumed, then the annual annual growth rategrowth rate is given by the formula: is given by the formula:

1

1mQr

P⎛ ⎞= −⎜ ⎟⎝ ⎠

Example 3Example 3• The table The table

summarizes the summarizes the population population information from the information from the previous example.previous example.

• Calculate the Calculate the growth rate using 2 growth rate using 2 different pairs of different pairs of years. years.

Example 3Example 3• Solution: Suppose we choose the first Solution: Suppose we choose the first

and last dates.and last dates.• PP = 20,000 and = 20,000 and QQ = 53,066. = 53,066.

• We have We have mm = 20. = 20.12053066 1 5%

20000r ⎛ ⎞= − ≈⎜ ⎟⎝ ⎠

Example 3Example 3• Suppose instead we choose Suppose instead we choose

PP = 21,000 and = 21,000 and QQ = 23,153. = 23,153.• We have We have mm = 2. = 2.

•

• The growth rate is constant = 5%.The growth rate is constant = 5%.

1223153 1 5%

21000r ⎛ ⎞= − ≈⎜ ⎟⎝ ⎠

Example 4Example 4• Suppose a population is initially 5000 Suppose a population is initially 5000

and increases to 8000 in 10 years.and increases to 8000 in 10 years.• Assume a Malthusian population model.Assume a Malthusian population model.

a)a) Estimate the annual growth rate.Estimate the annual growth rate.b)b) What is the predicted population in 20 What is the predicted population in 20

years?years?

Example 4Example 4a)a) We have We have PP = 5000, = 5000,

QQ = 8000, and = 8000, and mm = 10. = 10.

•

• The annual growth rate is about 4.8%.The annual growth rate is about 4.8%.

1108000 1 0.048122

5000r ⎛ ⎞= − ≈⎜ ⎟⎝ ⎠

Example 4Example 4b)b) Solution: Use the Malthusian growth formula to Solution: Use the Malthusian growth formula to

find the population after 20 years.find the population after 20 years.

•

• The estimated population is about 12,800.The estimated population is about 12,800.• NoteNote: there is an error in this slide r should = .048: there is an error in this slide r should = .048

( )201 0.48122 5000 12,800+ ⋅ ≈

Malthusian GrowthMalthusian Growth• For cases in which the value of For cases in which the value of r r is not known, a is not known, a

different formula can be used instead of the steps different formula can be used instead of the steps in the previous example.in the previous example.

• If the population changes from If the population changes from PP to to QQ in in mm years, years, then after then after nn years the population changes from years the population changes from PP to to

nmQP

P⎛ ⎞⎜ ⎟⎝ ⎠

Example 5Example 5

• Suppose a population is initially 6100 Suppose a population is initially 6100 and increases to 9300 in 11 years.and increases to 9300 in 11 years.

• What is the predicted population in 25 What is the predicted population in 25 years?years?

• Assume a Malthusian population model.Assume a Malthusian population model.

Example 5Example 5

• Solution: Use the formula with Solution: Use the formula with

PP = 6100, = 6100, QQ = 9300, = 9300, mm = 11, and = 11, and

nn = 25. = 25.

•

2511

2593006100 15,9006100

nmQP P

P⎛ ⎞ ⎛ ⎞= = ≈⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

Example 6Example 6• China’s population increased by China’s population increased by

132,200,000 from 1990 to 2000, for a total 132,200,000 from 1990 to 2000, for a total population of 1,260,000,000 in 2000.population of 1,260,000,000 in 2000.

a)a) Find the annual growth rate between 1990 Find the annual growth rate between 1990 and 2000.and 2000.

b)b) What is the predicted population for the What is the predicted population for the year 2010?year 2010?

Example 6Example 6a)a) We know We know

QQ = 1,260,000,000. = 1,260,000,000.• Since there was an increase of Since there was an increase of

132,200,000, the population in 1990 132,200,000, the population in 1990 must have been must have been PP = 1,260,000,000 - = 1,260,000,000 - 132,200,000 = 1,127,800,000.132,200,000 = 1,127,800,000.

•

1101260000000 1 1.11%

1127800000r ⎛ ⎞= − ≈⎜ ⎟⎝ ⎠

Example 6Example 6b)b) Solution: Predict the population of Solution: Predict the population of

China in 2010.China in 2010.• We have We have PP = 1,127,800,000, = 1,127,800,000,

QQ = 1,260,000,000, = 1,260,000,000, nn = 20, and = 20, and mm = 10. = 10.

•

2010

20126000000011278000001127800000

1, 407,696,400

P ⎛ ⎞= ⎜ ⎟⎝ ⎠≈

Ponzi SchemesPonzi Schemes

• A confidence man named Charles A confidence man named Charles Ponzi developed a fraudulent Ponzi developed a fraudulent investment scheme which became investment scheme which became known as the known as the Ponzi schemePonzi scheme..• Instead of offering a true investment, Instead of offering a true investment,

Ponzi just paid old investors with the Ponzi just paid old investors with the money received from new investors.money received from new investors.

Ponzi SchemesPonzi Schemes• Suppose a Ponzi scheme offers an interest Suppose a Ponzi scheme offers an interest

rate of 40% in 90 days (a quarter year).rate of 40% in 90 days (a quarter year).• Let Let SSmm be the number of investors in the be the number of investors in the mmth th

quarter.quarter.• Ponzi must have enough investors in the Ponzi must have enough investors in the

next quarter to pay the interest to these next quarter to pay the interest to these current investors.current investors.• The number of investors in the The number of investors in the mm + 1 quarter, + 1 quarter,

SSm+1m+1 , must be at least 1.4( , must be at least 1.4(SSmm).).

Example 7Example 7

• Suppose the number of investors in Suppose the number of investors in the first quarter of a Ponzi scheme the first quarter of a Ponzi scheme is 18.is 18.

• What is the minimum number of What is the minimum number of investors necessary after 12 years, investors necessary after 12 years, or 48 quarters?or 48 quarters?

Example 7Example 7

• Solution: We know that Solution: We know that SS11 = 18 and = 18 and we need to find we need to find SS4949..• SinceSince and and and and

so on, we find thatso on, we find that2 11.4S S≥ 3 21.4S S≥

( )1 11.4 mmS S+ ≥

Example 7Example 7• We can use this formula to find We can use this formula to find SS4949..

•

• The scheme started with 18 investors, but The scheme started with 18 investors, but after 12 years he must have 185,959,435 after 12 years he must have 185,959,435 investors to keep the scheme going.investors to keep the scheme going.

( ) ( )4849 1.4 18 185,959, 435S ≥ ≈

Chain LettersChain Letters• A chain letter arrives with a list of A chain letter arrives with a list of mm names. names.

• The number of names on the list is called the The number of names on the list is called the number of number of levelslevels,,

• The recipient mails something valuable to The recipient mails something valuable to the person at the top of the list.the person at the top of the list.• The recipient then removes that name from the The recipient then removes that name from the

top of the list and adds his or her name to the top of the list and adds his or her name to the bottom.bottom.

• He or she sends the letter to He or she sends the letter to nn new people. new people.

Chain LettersChain Letters• For a chain letter with For a chain letter with mm levels, levels, nn new new

participants for each letter, and a price to participants for each letter, and a price to participate of participate of PP, the payoff in rising from the , the payoff in rising from the bottom to the top of the letter is bottom to the top of the letter is

• The total number of people who must join for The total number of people who must join for an individual to make it to the top of the list is an individual to make it to the top of the list is

( )mP n

1 11

mnn

+ −−

Example 8Example 8• Suppose a chain letter has 7 levels, Suppose a chain letter has 7 levels,

asks you to send $10 to the person at asks you to send $10 to the person at the top of the list, and requires you to the top of the list, and requires you to send out 5 new letters.send out 5 new letters.

a)a) How much is the payoff in going from How much is the payoff in going from the bottom to the top of the list?the bottom to the top of the list?

b)b) How many people must participate?How many people must participate?

Example 8Example 8

a)a) Solution: We know that Solution: We know that mm = 7 levels, = 7 levels,

n n = 5 new participants, and = 5 new participants, and PP = $10. = $10.•

• The payoff in going from the bottom to The payoff in going from the bottom to the top of the list is $781,250.the top of the list is $781,250.

( )710 5 781, 250=

Example 8Example 8

b)b) Solution: Since we know that Solution: Since we know that mm = 7 = 7 and and n n = 5. Use:= 5. Use:

•

• A total of 97,656 people must participate A total of 97,656 people must participate for the payoff to be achieved. for the payoff to be achieved.

85 1 97,6565 1

−=

−

1 11

mnn

+ −−

Section 12.2Section 12.2Population Decrease,Population Decrease,

Radioactive DecayRadioactive Decay• GoalsGoals

• Study radioactive decayStudy radioactive decay• Study half-lifeStudy half-life• Study carbon-14 datingStudy carbon-14 dating

12.2 Initial Problem12.2 Initial Problem• The tranquilizer LibriumThe tranquilizer Librium®® has a half-life has a half-life

of between 24 and 48 hours.of between 24 and 48 hours.• What is the hourly rate at which What is the hourly rate at which

LibriumLibrium®® leaves the bloodstream as the leaves the bloodstream as the drug is metabolized?drug is metabolized?• The solution will be given at the end of the The solution will be given at the end of the

section.section.

Example 1Example 1• The population of Lake County, The population of Lake County,

Oregon was 7532 in 1980 and 7186 Oregon was 7532 in 1980 and 7186 in 1990. Assume a Malthusian in 1990. Assume a Malthusian model.model.

a)a) Determine the annual rate of decline Determine the annual rate of decline between 1980 and 1990.between 1980 and 1990.

b)b) Estimate the population in 2000.Estimate the population in 2000.

Example 1Example 1a)a) Solution: We have Solution: We have PP = 7532, = 7532,

QQ = 7186, and = 7186, and mm = 10. = 10.

•

• If the growth rate remained constant, the If the growth rate remained constant, the population decreased by 0.469% each population decreased by 0.469% each year between 1980 and 1990.year between 1980 and 1990.

1 11071861 1 0.00469

7532mQr

P⎛ ⎞ ⎛ ⎞= − = − ≈−⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

Example 1Example 1b)b) Solution: We have Solution: We have PP00 = 7532, = 7532,

rr = -0.00469, and = -0.00469, and mm = 20. = 20.•

• If the population continued to decrease If the population continued to decrease by 0.469% each year, the population in by 0.469% each year, the population in 2000 would have been 6856 people.2000 would have been 6856 people.

( )( )

20 0

20

1

1 .00469 7532 6856

mP r P= + ⋅

= − ⋅ ≈

Example 1Example 1

• The population of Lake County, The population of Lake County, Oregon in 2000 was actually 7422.Oregon in 2000 was actually 7422.

• This indicates that the population did not This indicates that the population did not continue to decrease at a constant rate.continue to decrease at a constant rate.

• Many factors affect population growth or Many factors affect population growth or decrease making reliable predictions decrease making reliable predictions difficult.difficult.

Radioactive DecayRadioactive Decay

• If a radioactive substance has an annual If a radioactive substance has an annual rate of decay of rate of decay of dd and there are initially and there are initially AA00 units of the substance present, then the units of the substance present, then the amount of the radioactive substance amount of the radioactive substance present after present after mm years will be: (This is the years will be: (This is the same formula as before except same formula as before except dd is used in is used in place of place of r r and and AA is used in place of is used in place of PP))

( ) 01 mmA d A= − ⋅

Example 2Example 2

• Suppose a radioactive substance has Suppose a radioactive substance has an annual decay rate of 1%.an annual decay rate of 1%.

• If a sample contains 100 grams of the If a sample contains 100 grams of the substance, how much will remain after substance, how much will remain after 25 years?25 years?

Example 2Example 2

• Solution: The decay rate is Solution: The decay rate is dd = 0.01, = 0.01, the initial amount is the initial amount is AA00 = 100, and the = 100, and the length of time is length of time is mm = 25. = 25.•

• After 25 years, 77.8 grams will remain.After 25 years, 77.8 grams will remain.

( )2525 1 0.01 100 77.8A = − ⋅ ≈

Half-LifeHalf-Life• The The half-lifehalf-life of a radioactive substance is of a radioactive substance is

the time at which exactly half of the initial the time at which exactly half of the initial amount remains.amount remains.

Decay Rate FormulaDecay Rate Formula• If If dd is the annual decay rate of a is the annual decay rate of a

substance and substance and hh is the half-life of the is the half-life of the substance, then:substance, then:

1

112

hd ⎛ ⎞= −⎜ ⎟⎝ ⎠

Example 3Example 3

• Suppose the half-life of a particular Suppose the half-life of a particular radioactive substance is 25 years.radioactive substance is 25 years.

• What is the annual decay rate?What is the annual decay rate?

Example 3Example 3

• Solution: We have Solution: We have hh = 25. = 25.

•

• The decay rate is about 2.73%.The decay rate is about 2.73%.

1 1251 11 1 0.0273

2 2h

d ⎛ ⎞ ⎛ ⎞= − = − ≈⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

Example 4Example 4

• Suppose in 1995 you had 128 grams Suppose in 1995 you had 128 grams of the substance from the previous of the substance from the previous example. example.

• How much will remain in 2095?How much will remain in 2095?

Example 4Example 4

• Solution: We know Solution: We know hh = 25 and = 25 and

dd = 0.0273. = 0.0273.• The initial amount was The initial amount was AA00 = 128 and the = 128 and the

length of time is length of time is mm = 100. = 100.•

• You would have about 8 grams left in 2095.You would have about 8 grams left in 2095.

( )100100 1 0.0273 128 8.04A = − ⋅ ≈

Radioactive Decay FormulaRadioactive Decay Formula• If a radioactive substance has a half-If a radioactive substance has a half-

life of life of hh and and AA00 units of the substance units of the substance are initially present, then the units of are initially present, then the units of the substance present after the substance present after mm years years will be: will be:

012

mh

mA A⎛ ⎞= ⋅⎜ ⎟⎝ ⎠

Example 5Example 5

• Suppose a radioactive substance has Suppose a radioactive substance has a half-life of 300 years.a half-life of 300 years.

• If a sample contains 150 grams of the If a sample contains 150 grams of the substance, how much will remain after substance, how much will remain after 1000 years?1000 years?

Example 5Example 5

• Solution: We have Solution: We have hh = 300 years, = 300 years,

mm = 1000 years, and = 1000 years, and AA00 = 150 grams. = 150 grams.

•

• After 1000 years, just under 15 grams will After 1000 years, just under 15 grams will remain.remain.

1000300

10001 150 14.92

A ⎛ ⎞= ⋅ ≈⎜ ⎟⎝ ⎠

Half-LifeHalf-Life

Carbon-14 DatingCarbon-14 Dating• Carbon-14 dating uses the known half-Carbon-14 dating uses the known half-

life of the radioactive isotope of carbon life of the radioactive isotope of carbon to estimate the time since the death of to estimate the time since the death of living organisms.living organisms.• Scientists assume the ratio of normal Scientists assume the ratio of normal

carbon, C-12, to radioactive carbon, carbon, C-12, to radioactive carbon, C-14, has been constant for millions of C-14, has been constant for millions of years.years.

Carbon-14 DatingCarbon-14 Dating• All living things have All living things have

the same ratio of C-12 the same ratio of C-12 to C-14.to C-14.

• As soon as an As soon as an organism dies, its organism dies, its amount of C-14 begins amount of C-14 begins to decay.to decay.

• Percentages of C-14 Percentages of C-14 present are given in the present are given in the table.table.

12.2 Initial Problem Solution12.2 Initial Problem Solution

• The tranquilizer LibriumThe tranquilizer Librium®® has a half-life has a half-life of between 24 and 48 hours. of between 24 and 48 hours. What is What is the hourly rate at which Libriumthe hourly rate at which Librium®® leaves the bloodstream as the drug is leaves the bloodstream as the drug is metabolized?metabolized?

Initial Problem Solution, cont’dInitial Problem Solution, cont’d• Use the decay rate formula, with the Use the decay rate formula, with the

time unit of hours.time unit of hours.• For For hh = 24 hours: = 24 hours:

12411 0.028

2d ⎛ ⎞= − ≈⎜ ⎟⎝ ⎠

Initial Problem Solution, cont’dInitial Problem Solution, cont’d• For For hh = 48 hours: = 48 hours:

• The drug leaves the bloodstream at a The drug leaves the bloodstream at a rate of 1.4% to 2.8% per hour.rate of 1.4% to 2.8% per hour.

14811 0.014

2d ⎛ ⎞= − ≈⎜ ⎟⎝ ⎠

Section 12.3Section 12.3Logistic Population ModelsLogistic Population Models

• GoalsGoals• Study logistic growth modelsStudy logistic growth models

• Use the logistic growth lawUse the logistic growth law• Study steady-state populationsStudy steady-state populations• Study steady-state population fractionsStudy steady-state population fractions

Logistic Growth ModelLogistic Growth Model

• For a population with limited resources, For a population with limited resources, there may be a maximum population there may be a maximum population size, called the size, called the carrying capacitycarrying capacity..

• A model for population growth that A model for population growth that takes into account the carrying capacity takes into account the carrying capacity is called the is called the logistic population modellogistic population model..

Logistic Growth ModelLogistic Growth Model• Graphs comparing the Malthusian model and Graphs comparing the Malthusian model and

logistic model are shown below.logistic model are shown below.

Logistic Growth ModelLogistic Growth Model

• The growth rate in a logistic model The growth rate in a logistic model varies over time.varies over time.

• The The natural growth ratenatural growth rate, , rr, is the rate at , is the rate at which the population would grow if there which the population would grow if there were no limitations. were no limitations.

Logistic Growth LawLogistic Growth Law• If a population has a natural growth rate of If a population has a natural growth rate of rr

from 0 to 3, and the environment has a from 0 to 3, and the environment has a carrying capacity of carrying capacity of cc, then after , then after m m + 1 + 1 breeding seasons, the population will be: breeding seasons, the population will be: (You will NOT use this formula in a problem. (You will NOT use this formula in a problem. I just want you to have seen it.)I just want you to have seen it.)

( ) 21

11m m mrP r P Pc++⎛ ⎞= + ⋅ − ⋅⎜ ⎟⎝ ⎠

Chapter 12 AssignmentDue 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.