EXPONENTIAL RELATIONSMBF3C

Lesson #4: Modelling Exponential Growth and Decay

ITEM VALUE The value of an item increases by 2% each year. The

graph shows the value of this item over time.

a) Suzanne says that the relation could be exponential, but it could also be quadratic. Do you agree? Explain.

b) Referring to the graph, after how many years was the value of the item $200?

ITEM VALUEc) The item's value can be modelled by the relation C = 100(1.02)t, where C represents the value of the item in dollars and t represents the time in years. Use the relation. After how many years was the value of the item $200?

C = 100(1.02)t

C = 100(1.02)50

=269.16

C = 100(1.02)t

C = 100(1.02)48

=258.70

C = 100(1.02)t

C = 100(1.02)40

=220.80

C = 100(1.02)t

C = 100(1.02)35 =200

ITEM VALUEd) Determine how long it took for the value of the item to reach $300. Use both the graph and the relation. How do the answers compare?

USING THE GRAPH, WE CAN SEE THAT

WHEN THE VALUE OF THE ITEM REACHES $300, approximately

56 YEARS HAVE PASSED

USING THE EQUATION:

C = 100(1.02)t

C = 100(1.02)56 =303.17

C = 100(1.02)t

C = 100(1.02)55 =297.17

EXPONENTIAL GROWTH AND DECAY WORD PROBLEMS The general formula that can be used for all exponential growth and

decay problems is:

where A represents the initial value

b represents the growth or decay rate

x represents the length of time

y=A(b)x

EXPONENTIAL GROWTH: TWO TYPES OF PROBLEMS

eg. 1 Jason decides to grow bacteria for a living. The bacteria doubles in size every 3 minutes. If there are 1000 bacteria initially present, how many will there be in 15 minutes? The question involves doubling period.

Doubling Time and Half-LifeExponential growth leads to

repeated doubling. The time required for the quantity to double is called the

doubling time.

Exponential decay leads to repeated halving. The time required for the quantity to

diminish by ½ is called the half-life.

Exponential growth leads to repeated doubling. The time required for the quantity to double is called the doubling time.

Doubling Time

Examples• Doubling time for bacteria was 1 minute.• Doubles each week

KENDRA WON A CONTEST. SHE WILL BE PAID A SUM EACH WEEK FOR 26 WEEKS. THE FIRST WEEK SHE WILL BE PAID IS 1 CENT. THE AMOUNT DOUBLES EACH WEEK.

A. How much will she be paid in the eighth week?

B. How much will she be paid in the 20th week?

C. How much will she be paid in the last week (week 26)?

KENDRA WON A CONTEST. SHE WILL BE PAID A SUM EACH WEEK FOR 26 WEEKS. THE FIRST WEEK SHE WILL BE PAID IS 1 CENT. THE AMOUNT DOUBLES EACH WEEK.

A. How much will she be paid in the eighth week?

B. How much will she be paid in the 20th week?

C. How much will she be paid in the last week (week 26)?

WHAT IS THE DOUBLING PERIOD?

Week Pay Out

After 1 week the payout has doubled once. 0.01×21 = $0.02

After 2 weeks the payout has doubled twice. 0.01×22 = $0.04

After 3 weeks the payout has doubled three times. 0.01×23 = $0.08

After 4 weeks the payout has doubled four times.

0.01×24 = $0.16

After 20 weeks the payout has doubled twenty times.

0.01×220 = $10 485.76

After 26 weeks the payout has doubled twenty-six times.

0.01×226 = $671,088.64

Week Pay Out

After 1 week the payout has doubled once. 0.01×21 = $0.02

After 2 weeks the payout has doubled twice. 0.01×22 = $0.04

After 3 weeks the payout has doubled three times. 0.01×23 = $0.08

After 4 weeks the payout has doubled four times.

0.01×24 = $0.16

After 20 weeks the payout has doubled twenty times.

0.01×220 = $10 485.76

After 26 weeks the payout has doubled twenty-six times.

0.01×226 = $671,088.64

Exponent

1/1 = 1

2/1 = 2

3/1 = 3

4/1 = 4

20/1 = 20

26/1 = 26

DOUBLING TIME FORMULA

New value

Amount before increase

Rate of increase

Timey=A(2)

𝑡𝑑

Doubling period

𝑡hP=I() P=Po (2)

𝑡𝒅

Terms indicating doubling time Doubles, Doubling time, Growing by a factor of 2

When the base of an exponential relation is 2, the relation is describing doubling time.

M=Mo ()• M final quantity• Mo initial quantity• t represents time• h represents the doubling

time• represents doubling

If we know doubling time, can we determine new values?

Years Population Exponent

After 10 years the population has doubled once. 10,000×21 = 20000 1/10 = 1

After 20 years the population has doubled twice. 10,000×22 = 40000 20/10 = 2

After 30 years the population has doubled three times. 10,000×23 = 80000 30/10 = 3

After t years the population has doubled t / 10 times. 10,000×2 t /10

Consider an initial population of 10,000 with a doubling time of 10 years.

t

10

Doubling time of population is approximately 14 years.

EXPONENTIAL GROWTH: TWO TYPES OF PROBLEMS

eg. 2 Mrs. James is accumulating money in her Swiss bank account from her oil stocks investment. The stocks give her an interest rate of 30% per year. She's looking to retire in 22 years from now. If she invests 1000 dollars today how much money will she make when she retires? This question involves growth rate.

Exponential Growth FORMULAExponential Growth: is growth when the quantity increases by the same percent in each unit of time.

tbAy )1( Amount after increase

Amount before increase

Rate of increase

Time

This growth function is used to find growth in population, interest, or any quantity that grows exponentially.

P=I(1+b)

t Terms indicating growth • Increases• Appreciates• Grows• Climbs• Matures• Rises• Soar• triples, doubles, quadruples, etc.• More???

Example: You deposit $500 in a bank account that pays 6% interest compounded yearly. Find the value of the account after 1 year, 2 years, 3 years, 20 years.

Use trCy )1( C = 500 r = .06 t = 1,2,3,20

1)06.1(500 y1)06.1(500y

)06.1(500y530$y

1 year 2 years2)06.1(500 y

2)06.1(500y)1236.1(500y

80.561$y

3 years3)06.1(500y

)191016.1(500yy = $595.508

20 years20)06.1(500y

y = 500(3.20713)

y = $1603.57

The longer you invest the more interest you earn!

Step 1: Determine if it’s an example of exponential growth or decay and write the formula.Step 2: Create an equation from the problem.Step 3: Substitute values with information given (make sure the rate % is converted to a decimal)Step 4: Simplify and solve (if doing by hand, make sure you use BEDMAS!)

EXPONENTIAL DECAY: TWO TYPES OF PROBLEMS

eg. 3

When Earnest Rutherford discovered radioactive decay of radium, he noticed it has a half life of 100 days. This means whatever mass of radium he has initially it will be half as much in 100 days. He starts with 320 mg of radium and stores it for 40 days. How much will be left after that time? This question involves half-life.

HALF TIME TIME FORMULA

New value

Amount before increase

Rate of decrease

Timey=A()

𝑡h

Half-life

The half life is the time it takes for a

quantity to decay to half of its original size

P=I()𝑡h P=Po ()

𝑡𝒉 M=Mo ()

𝑡𝒉

• M final quantity• Mo initial quantity• t represents time• h represents the half-life• represents half-life

Terms indicating half-time half-life, half-time

When the base of an exponential relation is 1/2, the relation is

describing half-time.

THE WORLD POPULATION DOUBLES EVERY 51.8 YEARS. AT THE BEGINNING OF THE TWENTIETH CENTURY, THE WORLD POPULATION WAS 1.6 BILLION. WHAT WAS THE POPULATION AT THE END OF THE TWENTIETH CENTURY?

Time (in years) Population

0 1.6 billion

FRAN DATA

Date Customers without Power

Sept. 6 1,159,000Sept. 7 804,000Sept. 8 515,000Sept. 9 340,500Sept. 10 195,200Sept. 11 136,300Sept. 12 77,000Sept. 13 37,600

EXPONENTIAL DECAY: TWO TYPES OF PROBLEMS

eg. 4 Carly buys a new car for $20,000. It is a nice one, but the value depreciates at a rate of 6% per year. If she wants to sell her car in 8 years from now how much will it be worth? This question involves depreciation.

Terms indicating decay Loses, falls, depreciates, drops. declines, decays, regresses

Exponential Decay: is decrease of a quantity by the same percent over a period of time.

Things that have exponential decay: medication, radioactive waste, bodies, car value(depreciation).

tbAy )1(

Amount after decrease Initial

amountbefore decrease

Rate of decrease

Time

A popular medication loses 23% of its effectiveness each hour. Find how much of the 200 mg of each pill is working after 5 hours.

Example:

5)23.1(200 y5)77(.200y

y = 200(.2706)y = 54.13mg

EXPONENTIAL DECAY FORMULA

*Remember, sometimes Io r Po is used to

denote this.

*Remember, sometimes r is used to denote this.

*Remember, sometimes P, M or C is used to denote this.

*Remember, sometimes n is used to number of years.

Step 1: Determine if it’s an example of exponential growth or decay and write the formula.Step 2: Create an equation from the problem.Step 3: Substitute values with information given (make sure the rate % is converted to a decimal)Step 4: Simplify and solve (if doing by hand, make sure you use BEDMAS!)

tbAy )1(

Example: The average new car loses 17% of its value each year. Find the value of a $20,000 car after 3 years.

3)17.1(20000 y3)83(.20000y

y = 20000(.5717)

y =$11,435.74

Depreciation

Step 1: Determine if it’s an example of exponential

growth or decay and write the formula.

Step 2: Create an equation from the problem.

Step 3: Substitute values with information given (make sure the rate % is converted

to a decimal)

Step 4: Simplify and solve (if doing by hand, make sure

you use BEDMAS!)

PROBLEMS RE: POPULATION DECAY OR GROWTH

LESSON 4 SUCCESS CRITERIA UP TO NOW: I can explain how the equation of an exponential function relates to

exponential growth and decay

I can identify formulas for: Doubling time Exponential growth Half-life Exponential decay

I can use the equation that represents exponential growth and decay to solve problems.

I can convert a growth or decay rate from a percentage to a decimal.

I can tell the difference between an equation that is an example of exponential growth and exponential decay.

ONE MORE FOR THIS LESSON:

I can write an equation for an

exponential growth or decay application

EXAMPLE 1:

The number of caterpillars in a nest increases by 7.6% every 2 months. There are 750 caterpillars now. Write an equation for the caterpillar population P after m months.

P=750

tbAy )1( Amount after increase

Amount before increase

Rate of increase

Time

(1+7.6)m

P=750(8.6) m

EXAMPLE 2:

A bee hive population triples every 4 weeks. Initially there are 300 bees. Write an equation for the population P of bees after w weeks.

P=300(3)w/4

EXAMPLE 3:

A population of 1500 mice decreases by 20% every day. Write an equation for the population of mice P after d days.

P=1500(0.80)d

EXAMPLE 4:

The mass of a plant goes up 18% every 3 months. The current mass of the plant is 500 grams. Write an equation for the mass of the plant M after m months.

M=500(1.18)m/3

EXAMPLE 5:

The population of Cambridge goes up 2% every year. In 2010 the population was 130 thousand. Write an equation that gives the population Cambridge in thousands, T, after y years since 2010.

T=130(1.02)y

EXAMPLE 6:WHICH OF THESE FUNCTIONS DESCRIBE EXPONENTIAL DECAY? EXPLAIN

EXAMPLE 7: IN 1990, A SUM OF $100 IS INVESTED AT A RATE OF 6% PER YEAR FOR 15 YEARS.

(a) What is the growth rate (i.e. the value of "b")

(b) What is the initial amount?

(c) Write an equation that models the growth of the investment, and use it to determine the value of the investment after 15 years?

EXAMPLE 8:IN EACH CASE, WRITE AN EQUATION THAT MODELS THE SITUATION DESCRIBED. EXPLAIN WHAT EACH PART OF THE EQUATION REPRESENTS

a. The percent colour remaining if your blue jeans lose 1% of their original colour every time they are washed

b. The population if a town had 2500 residents in 1990 and grew at a rate of 0.5% each year after that forty years.

c. The population of a colony if a single bacterium doubles every day, what is the population P after t days.

EXAMPLE 9 The Petersons purchased their house for $ 30 000 in 1970. Since then, the

value of their house has increased 5 % per year. State the value of the house in the year 2001.

A car purchased for $24,500 will depreciate at a rate of 18% per year for the first 6 years. Write anequation and graph over the first 6 years.

x y

012345

024500(1 .18)124500(1 .18)224500(1 .18)324500(1 .18)424500(1 .18)524500(1 .18)

24500200901647413509110779083

x

y

6 624500(1 .18) 7448

$30,0

00

$20,0

00

$10,0

00

1 2 3 4 5 6 7 8

xy 24500(1 .18)

Compare Growth and Decay Models

Exponential Growth Exponential Decay

$100 growth at 10% for 5 years.

$100 depreciate at 10% for 5 years.

xy 100(1 .10) xy 100(1 .10) x y012345

100110121133146161

160140120100 80 60 40 20

0 1 2 3 4 5

x y012345

1009081736659

160140120100 80 60 40 20

0 1 2 3 4 5

Classify each as exponential growth orexponential decay.

1)

2)

3)

4)

5)

6)

xy 10(1.04)

xy 0.8(1 0.3)

ty 7(2)

ty 12(0.97)

xy 15(1 0.6)

ty 0.5(4)

Exponential Growth

Exponential Growth

Exponential Growth

Exponential Decay

Exponential Decay

Exponential Growth

Find the value of a downtown office building thatcost 12 million dollars to build 20 years ago and depreciated at 9% per year.

20y 12(1 .09)

y 1.82 Million Dollars

20y 12(.91)

tbAy )1(