Download exponential growth

Quiz 3-1a
1. Write the equation that
models the data under the
column labeled g(x).
g ( x)  ab x
g ( x)  4(3) x
2. Write the equation that models the data under the column
labeled f(x) above. f ( x)  ab x
f ( x)  8(0.25) x
3. Without using your calculator, determine if the following
x
function growth or decay?
f ( x)  2
4. Without using your calculator, determine if the following
function growth or decay?
0.5 x
f ( x)  3e
3.1B
Applications of Exponential
Functions
Exponential Function
f ( x)  ab
Input variable
x
Growth factor:
Initial value
What does ‘b’ equal
In order for it to be “growth”?
What does ‘b’ equal
In order for it to be “decay”?
What is the value of ‘y’ where
the graph crosses the y-axis?
Your turn:
Graph the functions:
1. Where does it cross the y-axis?
f ( x)  3(4)
x
2. What is the “intial value of f(t) ?
f (t )  2(5)t (where ' t' is time)
Population Growth
P0  initial population
If population grows at a constant percentage rate over a
year time frame, (the final population is the initial population
plus a percentage of the orginial population) then the
population at the end of the first year would be:
P1  P0  rP0
P1  P0 (1  r )
Percent rate of change
(in decimal form)
At the end of the second year the population would be:
P2  P1  rP1
P2  P1 (1  r )
Population Growth
P0  initial population
P1  population after 1 year
P1  P0 (1  r )
P2  population after 2 years
P2  P1  rP1
P2  P0 (1  r )  P0 (1  r )r
P2  P0  2rP0  P0 r
P2  P0 (1  r )
2
2
Quadratic equation!
Population Growth
P0  initial population
P1  population after 1 year
P2  population after 2 years
P1  P0 (1  r )
P2  P1 (1  r )
P2  ( P0  rP0 )  ( P0  rP0 )r
P2  P0  2rP0  P0 r
P2  P0 (1  r )
2
2
Quadratic equation!
Population Growth
P0  initial population
P1  population after 1 year
P2  population after 2 years
P3  P2  P2 r
P1  P0 (1  r )
P2  P0 (1  r )
2
P3  P0 (1  r )  ( P0 (1  r )r )r
2
2
3
P3  P0 (1  2r  r  r  2r  r )
2
2
P3  P0 (r  3r  3r  1)
3
2
P3  P0 (1  r )
3
P4  ?
Special cubic!
Pn  ?
Population Growth
Percent rate of change
(in decimal form)
P(t )  P0 (1  r )
Population (as a
function of time)
Initial
population
f ( x)  ab
t
time
Growth
rate
x
Growth factor:
Initial value
Word problems P (t )  P0 (1  r )
t
There are 4 quantities in the equation:
1. Population “t” years/min/sec in the future
2. Initial population
3. Growth rate
4. time
The words in the problem will give you three of the four
quantities. You just have to “plug them in” to the equation
and solve for the unknown quantity.
Population Growth
Percent rate of change
(in decimal form)
P(t )  P0 (1  r )
t
Population (at time
“t”) in the future
time
Initial
Growth
population
rate
The initial population of a colony of bacteria is 1000. The
population increases by 50% every hour. What is the
population after 5 hours?
P(5)  1000(1  0.50)
5
Unknown value
P(5)  1000(1.5)5
P(5)  7593
Simple Interest (savings account)
A(t )  P(1  r )
Amount (as a
Initial amount
function of time)
(“principle”)
t
time
Interest
rate
A bank account pays 3.5% interest per year. If you initially
invest $200, how much money will you have after 5 years?
A(5)  $200(1.035)
Unknown value
5
A(5)  $237.54
Your turn:
A(t )  P(1  r )
t
3. A bank account pays 14% interest per year. If you initially
invest $2500, how much money will you have after 7 years?
4. The population of a small town was 1500 in 1990. The
population increases by 3% every year. What is the
population in 2009?
Solve by
graphing
San Jose, CA
Year
1990
2000
Population
782,248
895,193
Assuming exponential growth, when will
the population equal 1 million?
Let ‘t’ = years since 1990
P(t )  P0bt
We must find the growth factor ‘b’ P(10)  782,248b10  895,193
895,193
10
895,193
10 10
b 
b  10
782,248
782,248
‘b’ = 1.0136
P(t )  782,248(1.0136)t
1,000,000  782,248(1.0136)t
Unknown value
Example
P(t )  782,248(1.0136)
P(? years )  1,000,000
t
1,000,000
‘t’ = approximately 18
18 years AFTER 1990  2008
Later in the chapter we will learn how to solve for the
unknown exponent algebraically.
Your Turn:
The population of “Smallville” in the year 1890 was 6250.
Assume the population increased at a rate of 2.75% per year.
5. When did the population reach 50,000 ?
Your turn:
Year
1990
2009
Population USA
248,709,873
307,006,550
6. Assuming exponential growth, when will
the population exceed 400 million?
We must find the growth factor ‘b’
P(19)  248,709,873b19  307,006,550
19
19
b

19
307,006,550
248,709,873
b19 
P(t )  P0bt
307,006,550
248,709,873
‘b’ = 1.0111
P(t )  248,709,873(1.0111)t
400,000,000  248,709,873(1.0111)t
43 yrs after
t = 0 (1990)
2033
Year
1900
2000
Your turn:
Population USA
76.21 million
248.71 million
7. Assuming exponential growth, when will
the population exceed 400 million?
We must find the growth factor ‘b’
100
P(100)  76.21b
100
100
b100  100
248.71
76.21
 248.71
b
P(t )  P0bt
248.71

76.21
‘b’ = 1.0119
P(t )  76.21(1.0119)t
400,000,000  76.21(1.0119)t
140.2 yrs after
t = 0 (1900)
2040.2
Finding an Exponential Function
$500 was deposited into an account that pays “simple
interest” (interest paid at the end of the year).
25 years later, the account contained $1250. What was
the percentage rate of change?
A(t )  P0 (1  r )
P0  500
t
1250  500(1  r )
1250
 (1  r ) 25
500
1250
25
 (1  r )
500
A(25)  1250
25
Unknown value
1.037  (1  r )
r  0.037
r  3.7 %
Your Turn:
8. The population of “Smallville” in the year 1890 was 6250.
Assume the population increased at a rate of 2.75% per year.
What is the population in 1915 ?
9. The population of “Bigville” in the year 1900 was 25,200. In
1955 the population was 37,200. What was the percentage
rate of change?
10. The population of “Ghost-town” in the year 1900 was 3500.
In 1935 the population was 200. What was the percentage
rate of change?
Finding Growth and Decay Rates
Is the following population model an
exponential growth or decay function?
P(t )  782,248(1.0136) t
Find the constant percentage growth (decay) rate.
P(t )  P0 (1  r )
t
P(t )  782,248(1  0.0136)
‘r’ > 0, therefore this is exponential growth.
‘r’ = 0.0136
or 1.36%
t
Your turn:
P(t )  50(1.5)
t
11. Is it growth or decay?
f ( x)  ab x
b = 1.5
b>0
Growth!
12. Find the constant percentage growth (decay) rate.
P(t )  P0 (1  r )
‘r’ = 0.5
t
P(t )  50(1  0.5)
t
or 50%  % rate of growth is 50%
‘r’ > 0, therefore this is exponential growth.
Finding an Exponential Function
Determine the exponential function with initial value = 10,
increasing at a rate of 5% per year.
P(t )  P0 (1  r )
t
P0  10
P(t )  10(1.05)t or
‘r’ = 0.05
f ( x)  10(1.05) x
Modeling Bacteria Growth
P(t )  P0 (1  r )
t
Suppose a culture of 100 bacteria cells are put into a petri dish
and the culture doubles every hour.
Predict when the number of bacteria will be 350,000.
What is the growth factor?
P(t )  P0 (1  r ) t
P(0) = 100
P(t )  P0 (2)t
P(t) = 350000
350000  100(2)t
5
3
.
5

10
3
2t 

3
.
5

10
 3500
2
110
2t  3500
Modeling Bacteria Growth
Suppose a culture of 100 bacteria cells are put into a petri dish
and the culture doubles every hour.
Predict when the number of bacteria will be 350,000.
2t  3500
x
y1  2
y2  3500
Where do the two
graphs cross?
x  11.77
t = 11 hours + 0.77hrs
t = 11 hours + 0.77hrs * 60 min
hr
t = 11 hours + 46 min
Your turn:
P(t )  P0 (1  r ) t
f ( x)  ab x
13. A family of 10 rabbits doubles every 2 years. When
will the family have 225 members?
225  10(2)t
b=2
t = 7.8 years
t = 7 years 6 months
Modeling U.S. Population Using
Exponential Regression
Use the 1900-2000 data and
exponential regression to predict
the U.S. population for 2003.
(Don’t enter the 2003 value).
Let P(t) = population,
“t” years after 1900.
Enter the data into your
calculator and use
exponential regression
to determine the model (equation).
Exponential Regression
Stat p/b  gives lists
Enter the data:
Let L1 be years since initial value
Let L2 be population
Stat p/b  calc p/b
scroll down to exponential regression
“ExpReg” displayed:
enter the lists: “L1,L2”
f ( x)  ab
x
The calculator will display the
values for ‘a’ and ‘b’.
Your turn:
14. What is your equation?
15. What is your predicted population in 2003 ?
16. Why isn’t your predicted value the same as the
actual value of 290.8 million?
Find the amout of material after ‘20’ days if the
initial mass is 5 grams and it doubles every 4 days:
f (t )  ab
t
The issue is units !!!
Initial value ‘a’  units of grams
The mass (# of grams) at some time “t” in the future is
the initial mass (# of grams) times some number b t .
Can the exponent have any units? NO !!!
This doubles every 4 days. How many times does it
double in 20 days?
5
f ( x)  5 grams (2)
Units of the exponent
A(t )  8(3)t
The input value is time (with units of seconds, minutes, hrs, etc.).
How can the input value have units and the exponent not have
any units (since that is where the input value is inserted into the
equation)?
IF the input value has the units of time in seconds , then the exponent
really has the units of “# of times the base is used as a factor / day”
to make the units work out.
Since the base is a 3, then this could be shortened to
“# of triples/ day”
This could be shortened to “per day” which in math is “1/day”
A(7 days)  8(3)
7
Find the amout of material after 20 days if the
initial mass is 5 grams and it doubles every 4 days:
A(t )  ab
t
(amount (grams) as
a function of time)
Initial mass = 5 grams
mass doubles every 4 days
A(" t" days)  5 grams(2)
"t" days*(1
)
4 days
20 days*(1
)
4 days
A(16 days)  5 grams(2)
5
A(16 days)  5 grams(2)
A(16 days )  160 grams
No units remain in the
exponent.
Find the amout of material after 20 days if the
initial mass is 5 grams and it doubles every 4 days:
A(t )  ab
t
(amount (grams) as
a function of time)
Initial mass = 5 grams
mass doubles every 4 days
So you could just write it as:
A(" t" days)  5 grams(2)
"t" days*(1
)
4 days
A(16 days)  5 grams(2)
20 days*(1
)
4 days
A(16 days)  5 grams(2)
5
A(16 days )  160 grams
Your turn:
17. The crowd in front of the Tunisian parlament
building increased by a factor of 4 every 3 hours.
If the initial crowd had 500 people in it, how many
people would there be after 12 hours?
18. The amount of radioactive Rubidium 88 decreases
by a factor of 2 every 8 minutes. If there was 5
grams of the material at the start, how much would
there be after 30 minutes?
HOMEWORK