MBF3C
Lesson #4: Modelling Exponential Growth and Decay

 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?

c) 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

d) 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

 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



leads to
. The time required for the quantity to double is called the
.
leads to
.
The time required for the quantity to diminish by ½ is called the
.

Exponential growth leads to repeated doubling. The time required for the quantity to double is called the doubling time .
Examples
• Doubling time for bacteria was 1 minute.
• Doubles each week

A.
B.
C.
How much will she be paid in the eighth week?
How much will she be paid in the 20 th week?
How much will she be paid in the last week (week 26)?

A.
B.
C.
How much will she be paid in the eighth week?
How much will she be paid in the 20 th week?
How much will she be paid in the last week (week 26)?

Week Pay Out
After 1 week the payout has doubled once .
0.01
×2 1 = $0.02
After 2 weeks the payout has doubled twice .
0.01
×2 2 = $0.04
After 3 weeks the payout has doubled three times.
After 4 weeks the payout has doubled four times.
0.01
×2 3 = $0.08
0.01
×2 4 = $0.16
After 20 weeks the payout has doubled twenty times.
0.01
×2 20 = $10 485.76
After 26 weeks the payout has doubled twenty-six times.
0.01
×2 26 = $671,088.64

Week Pay Out Exponent
After 1 week the payout has doubled once .
0.01
×2 1 = $0.02
After 2 weeks the payout has doubled twice .
0.01
×2 2 = $0.04
After 3 weeks the payout has doubled three times.
After 4 weeks the payout has doubled four times.
0.01
×2 3 = $0.08
0.01
×2 4 = $0.16
After 20 weeks the payout has doubled twenty times.
0.01
×2 20 = $10 485.76
After 26 weeks the payout has doubled twenty-six times.
0.01
×2 26 = $671,088.64
1/
2/
3/
4/
1
1
1
1
= 1
= 2
= 3
= 4
20/ 1 = 20
26/ 1 = 26

New value
Amount before increase
•
• M final quantity
M o initial quantity
• t represents time
• h represents the doubling time represents doubling
P=I( 𝑏 ) ℎ 𝑡 y=A(2) 𝑑 𝑡
Time
Doubling period
Rate of increase
When the base of an exponential relation is 2, the relation is describing doubling time
.
Terms indicating doubling time  Doubles, Doubling time, Growing by a factor of 2

If we know doubling time, can we determine new values?
Consider an initial population of 10,000 with a doubling time of 10 years.
Years Population Exponent
After 10 years the population has doubled once .
After 20 years the population has doubled twice .
After 30 years the population has doubled three times.
After t years the population has doubled t / 10 times.
10,000
10,000
×2
×2
1
2
= 20000
= 40000
1/
20/
10
10
= 1
= 2
10,000 ×2 3 = 80000 30/ 10 = 3
10,000 ×2 t
10 t /10

Doubling time of population is approximately 14 years.

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: is growth when the quantity increases by the same percent in each unit of time.
Amount after increase y

A ( 1
 b ) t
Time
Rate of increase
Amount before increase
This growth function is used to find growth in population, interest, or any quantity that grows exponentially.
Terms indicating growth 
• Increases
• Appreciates
• Grows
• Climbs
• Matures
• Rises
• Soar
• triples, doubles, quadruples, etc.
• More???

Example:
Use
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.
y

C ( 1
 r ) t
1 year y y y y



500 ( 1

500 ( 1 .
06 )
1
500

$
( 1
.
06 )
1
.
06 )
530
C = 500 r = .06 t = 1,2,3,20 y y y y




2 years
500 ( 1

.
06 )
2
500 ( 1 .
06 )
2
500 ( 1 .
1236 )
$ 561 .
80
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!)
3 years y

500 ( 1 .
06 )
3 y

500 ( 1 .
191016
y = $595.508
) y

20 years
500 ( 1 .
06 )
20
y = 500(3.20713)
y = $1603.57
The longer you invest the more interest you earn!

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.

New value
When the base of an exponential relation is 1/2, the relation is describing half- time .
P=I( 𝑏 ) ℎ 𝑡 y=A(
1
2
) 𝑡 ℎ
Rate of decrease
Time
Half-life
Amount before increase
The half life is the time it takes for a quantity to decay to half of i t s original size
Terms indicating half-time  half-life, half-time
• M final quantity
• M o initial quantity
• t represents time
• h represents the half-life represents half-life

Time (in years)
0
Population
1.6 billion

Date
Sept. 6
Sept. 7
Sept. 8
Sept. 9
Sept. 10
Sept. 11
Sept. 12
Sept. 13
Customers without Power
1,159,000
804,000
515,000
340,500
195,200
136,300
77,000
37,600

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
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).
Time y

A ( 1
 b ) t
Amount after decrease Initial amount before decrease
Rate of decrease
Example:
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.
y y


200 ( 1

200 (.
.
23 )
77
y = 200(.2706)
)
5
5
y = 54.13mg
y

A ( 1
 b ) t
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!)

Example:
Depreciation
Step 1: Determine if it’s an example of exponential growth or decay and write the formula.
The average new car loses 17% of its value each year. Find the value of a $20,000 car after
3 years. y y


20000 ( 1

20000 (.
y = 20000(.5717)
.
17
83 )
)
3
3
Step 2: Create an equation from the problem.
y =$11,435.74
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!)

 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.



 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.
Time
Amount after increase y

A ( 1
 b ) t
Rate of increase
Amount before increase
P=750 (1+7.6) m
P=750(8.6) m

 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

 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

 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

 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

 (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?

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.

 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 an equation and graph over the first 6 years. x y y
  x
$30,00 y
0
1
2
3
4
5
6
24500(1 .18)
24500(1 .18)
1
2
0 


 5
3 
4 

6 
24500
20090
16474
13509
11077
9083
7448
0
$20,00
0
1 2 3 4 5 6 7 8 x

Compare Growth and Decay Models
Exponential Growth
0
1 x
2
3
4
5
$100 growth at 10% for 5 years.
y
  y
100
110
121
133
146
160
140
120
100
80
60
40
20
161 x
0 1 2 3 4 5
Exponential Decay
0
1
$100 depreciate at 10% for 5 years.
x
2
3
4 y
  y
100
90
81
73
66
160
140
120
100
80
60
40
20 x
5 59 0 1 2 3 4 5

Classify each as exponential growth or exponential decay.
1) y

10(1.04)
Exponential
Growth x
4) y

12(0.97) t
Exponential
Decay
2) y
 
Exponential
Growth
3) y

7(2)
Exponential
Growth t x
5) y

15(1 0.6) x
Exponential
Decay
6) y

0.5(4)
Exponential
Growth t

Find the value of a downtown office building that cost 12 million dollars to build 20 years ago and depreciated at 9% per year. y

A ( 1
 b ) y

12(1 .09)
20 t y

12(.91)
20 y

1.82 Million Dollars