Download Exponential Growth and Decay Models Exercise

Exponential Growth and Decay Models
1. Modeling Bacteria Growth: Suppose a population of bacteria is growing 50% every 4
hours and the current population is 60 bacteria. Let t = the number of hours in the future and
N = number of bacteria after t hours.
a) Fill in the chart with the missing values of N.
t
N
0
60.0
4
90.0
8
135.0
12
202.5
16
303.8
20
455.6
Key
1
Growth factor = (100% + 50%) = 1.5 and the period of exponential grow = 4 hours.
b) Write a formula that expresses N as a function of t.
N(t) = 60(1.5t/4 )
c) Find the population of the bacteria after 10 hours.
N(10) = 60(1.510/4 ) = 165.3
d) Find the population 45 minutes ago.
N(-0.75) = 44.3
e) What formula would describe the relationship between
t and N if the bacteria population was increasing 40%
every 30 minutes? (Per iod of exponential growth = ½ hour.)
N(t) = 60(1.4t/0.5) = 60(1.42t )
2. Modeling Population Growth: Suppose the current population of a town is 12,000 and the
town’s population is expected to grow by 4% of the previous year's population per year for the
next 20 years. Let t = future time in years and P = the town’s population after t years.
t
P
0
12,000
1
12,480
2
12,979
3
13,498
10
17,763
20
26,293
a) Fill in the chart with the missing values of P for t = 0 , 1, 2, and 3.
Growth factor = (100% + 4%) = 1.04 and the period of exponential growth = 1 year.
b) Write a formula that expresses P as a function of t.
P(t) = 12,000(1.04t )
c) Fill in the chart with the missing values of P for t = 10 & 20 years.
d) What was the population of the town one year ago? N(-1) = 11,538
e) What equation would describe the relationship between t and P
if the town's population was increasing 15% every 5 years?
P(t) = 12,000(1.15t/5)
3. Modeling Depreciation on Equipment: A farmer purchased a new tractor for $320,000.
The value of the tractor is expected to decrease by 12% of previous year's value each year for the next
10 years. Let t = number of years in the future and V = the value of the tractor after t years.
t
V
a) Fill in the chart with the missing values of V for t = 0, 1, 2, and 3.
0
320,000.00
Decay factor = (100% - 12%) = 0.88 and the period of exponential decay = 1 year.
1
281,600.00
2
247,808.00
c) Fill in the chart with the missing values of V for t = 5 and 10 years.
3
218,071.04
d) What was the value of the tractor 6 months after purchase?
5
168,874.21
10
89,120.30
b) Write a formula that expresses V as a function of t.
V(t) = 320,000(0.88t)
V(0.5) = $300,186.61
V(3.75) = $198,134.45
e) What was the value of the tractor 3 years and 9 months after purchase?
4. Modeling Radioactive Decay: Suppose the half-life of a certain radioactive substance is
20 days. Let t = number of days in the future and N = the number of grams of the substance
remaining after t days. Suppose there is currently 160 grams of the substance.
t
N
0
160.00
20
80.00
40
40.00
60
20.00
80
10.00
100
5.00
2
a) Fill in the chart with the missing values of N for t = 0, 20, 40, 60, 80
and 100 days.
Decay factor = 0.5 and the period of exponential decay equals 20 days.
b) Write a formula that expresses N as a function of t.
N(t) = 160(0.5t/20 )
c) How many grams of radio active material remained after 215 days?
N(215) = 160(0.5215/20) = 0.0929 g
d) How many grams of radio active material remained 60 hours ago?
N(-2.5) = 905.10 g
e) What formula describes the relationship between N and t if the halflife is 6 hours instead of 20 days? (t still equals number of days.)
N(t) = 160(0.5t/0.25 ) = 160(0.54t)
(Period of decay = ¼ day.)
5. Modeling Compound Interest: Suppose a person invests $10,000 in a CD that will earn interest at
8%/year and interest is compounded monthly. Let t = the number of years in the future and V = the
value of the investment after t years. Use standard banking convention of 360 days per year.
t
V
0
$10,000.00
1
$10,830.00
2
$11,728.88
3
$12,702.37
4
$13,756.66
5
$14,898.46
a) Fill in chart with the missing values of V for t = 0 . . . 5 years.
Grawth factor = (100% + 8%/12) = 1 + .08/12 and the period of exponential growth = 1/12 year.
b) Write a formula that expresses V as a function of t .
V(t) = 10,000(1 + .08/12)t/(1/12) = 10,000(1 + .08/12)12t
c) Find the value of the investment after 20 years.
V(20) = $49,268.03
d) What was the value of the investment after 9 months?
V(.75) = $10,616.25
e) What formula describes the relationship between t and V if interest
is compounded hourly? V = 10,000(1 + 0.08/8,640)8,640 t
6. Modeling Inflation: Suppose the rate of inflation averages 3%/year for the next 20 years and
Elmo needs $60,000/year to maintain his current life style. Let t = the number of years in the future
and C = the cost to maintain Elmo’s life style t years in the future.
t
C
0
$60,000.00
1
$61,800.00
2
$63,654.00
c) Use your formula to find C when t = 10 and 20 years.
3
$65,563.00
10
$80,634.98
d) If Elmo now pays $2.00 for a loaf of bread, how much can he
expect pay for a loaf of bread 20 years from now?
20
$108,366.67
a) Fill in chart with the missing values of C for t = 0 . . . 3 years.
Growth factor = (100% + 3%) = 1.03 and the exponential growth period = 1 year.
b) Write a formula that expresses cost C as a function of t years .
C(t) = 60,000(1.03t)
Cost of bread = 2(1.0320) = $3.61 C(t) = 60,000(1.05t/2)
e) Find a formula that expresses the relationship between C and t if
the inflation rate changes to 5% every two years.
Remarks,
I give each student this handout when I first present the concept of exponential growth and decay. In
the first few examples students need a lot of coaching, but much less coaching in the last few
examples. If students carefully read the problem, I have found that it is easy to guide them in filling in
the first few rows of the table. By filling in the first few rows of the table, students get a better
understanding and feel of how exponential growth and decay works and the teacher has a much easier
job of explaining why the equation works. Teachers should never derive the equation until the first
few rows of the table are filled in.
As the lesson progresses, teaches should be asking questions. Why is the table set up the way it is?
What is the periodic growth/decay factor for this problem? Is this problem about exponent growth or
exponential decay and how do you know? What is the exponential growth/decay pariod?
Students should be using their graphing calculators to calculate function values and graph the
exponential functions in the lesson. Of course teachers can use the program Basic Trig Functions to
show the graphs. Almost every student will immediately see the difference between an exponential
growth graph and an exponential decay graph. Teachers will need to coach students on how to set up
the graphing window.
Discuss the general exponential growth/decay equation A(t) = A0*bt/k where t is some unit of time.
What does A0 mean? From a graphing point of view, that does A0 tell us?
What does the b parameter tell us and what are the possible values of b?
What does the k parameter tell us?
Why are the equations A(t) = A0*bt/k and A(t) = A0*(1/b)-t/k equivalent?
Have students graph the following equations:(Xmin = 0, Xmax = 50, Ymin = 0, Ymax = 900)
y = 800(0.5x/5 ) and y = 800(2-x/5)
Notice that none of examples in this handout mention the irrational number e. When exponential
growth/ decay relationships are expressed using the approach presented in this handout, it is easy to
determine the important features of the model by inspection. When exponential growth/decay
relationships are expressed in terms of base e, only God can determine the important features of the
model by simple inspection. Of course it is true that e is very important in many exponential growth/
decay models, however, a base other than e makes it easy to find an exponential grow/decay equation
that models the problem situation. After filling in the initial table values, the equation is almost
automatic.
Now consider the exponential growth equation for compound interest A(t) = P(1 + r/k) kt . t = time in
years, P = the principal or initial amount invested, r = the annual interest rate expressed as a decimal, and
k = the number of times per year interest is compounded. The limit as k approaches infinity gives us the
continuous compound interest formula A(t) = Pert which follows from the definition of the irrational
number e. No matter how many times per year we compound interest, the future value of the account
will never exceed Pert. Pert is the upper limit of our greed!
I have found the following activity a helpful way to teach continuous compounding of interest.
Set P = $20,000, r = 6%/year = 0.06, and t = 30 years. Then have students calculate the future value of
the account for k = 1, 12, 360, and 8,640. For the grand final, have students calculate the future value of
the account if interest is compounded continuously. Sometimes a teacher will hear a student gasp.