Transcript Slide 1

Section 9C
Exponential Modeling
(pages 557 – 572)
Exponential Growth (Decay) occurs when a
quantity increases (decreases) by the same relative
amount—that is, by the same percentage—in each
unit of time. (Powertown: -- 5% each year, investments)
A Exponential Function grows (or decays) by
the same relative amount per unit of time.
Independent variable: t
Dependent variable: Q
decimal growth rate: r
Initial Value: Q0
(dependent) = initial value x (1 + r)independent
or
Q = Q0 x (1 + r)t
Comments
Q = Q0 x (1 + r)t
•Units for r and for t must be the same.
•If Q0 is the initial value at time t0, then t must be
measured in units since t0.
•In this formula, r is the decimal form of the
percentage growth/decay rate
•If r > 0, then quantity grows exponentially.
If r < 0, then quantity decays exponentially.
Ex1/559 The 2000 census found a US population of about 281
million. a) Write an equation for the US population that assumes
a exponential growth at 0.7% per year.
b) Use the equation to predict the US population in 2100.
initial value (in 2000): 281 million
growth rate: .007 per year
independent variable: years since 2000
dependent variable: population
Q = 281million x (1+.007)t
year 2100 is 100 years since 2000, so t = 100
Q = 281million x (1.007)100 = 564 million
Ex2/560 China’s one-child policy was originally implemented with the
goal of reducing China’s population to 700 million by 2050. China’s 2000
population was about 1.2 billion. Suppose China’s population declines at
a rate of 0.5% per year.
a) write an equation for the exponential decay of the population
b) will this rate of decline be sufficient to meet the original goal?
initial value (in 2000): 1.2 billion
rate: -.005 per year
independent variable: years since 2000
dependent variable: population
Q = 1.2billion x (1-.005)t
year 2050 is 50 years since 2000, so t = 50
Q = 1.2billion x (.995)50 = .934billion
With this model, the predicted population in 2050 is
934,000,000 and so the goal of 700,000,000 will not be met.
What do graphs look like?
33/570 Your starting salary at a new job is $2000 per month and you get
annual raises of 5% per year.
a) create an exponential function.
b) create a table showing Q values for the first 15 units of time.
c) make a graph of the exponential function.
Q
$24,000.00
$25,200.00
$26,460.00
$27,783.00
$29,172.15
$30,630.76
$32,162.30
$33,770.41
$35,458.93
$37,231.88
$39,093.47
$41,048.14
$43,100.55
$45,255.58
$47,518.36
$49,894.28
Q = 24000 x (1+.05)t
$55,000.00
$50,000.00
$45,000.00
annual salary
t
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
$40,000.00
$35,000.00
$30,000.00
$25,000.00
$20,000.00
0
curvy!
5
10
years on the job
15
20
Using the graph
33/570 Your starting salary at a new job is $2000 per month and you get
annual raises of 5% per year.Using the graph determine:
a) your salary after 12 years.
b) when your salary will be $30000.
$55,000.00
When t = 12, Q = _______.
$50,000.00
annual salary
$45,000.00
Q = $30,000 when t = _____
$40,000.00
$35,000.00
$30,000.00
$25,000.00
$20,000.00
0
5
10
years on the job
15
20
Can we solve exactly using the equation?
Use Properties of Logarithms
log10 (A  B) = log10 (A) + log10 (B)
A
log10 ( ) = log10 (A) - log10 (B)
B
log10 (A ) = B  log10 (A)
B
Can we solve exactly using the equation?
Q = 24000 x
(1+.05)t
30000 = 24000 x (1+.05)t
30000
t
 (1.05)
24000
1.25 = (1.05)t
log101.25 = log10 (1.05)t
log101.25 = t  log10 (1.05)
log10 (1.25)
= t
log10 (1.05)
0.09691
= t
.021189
4.57360 = t
The salary will be
$30,000 in 4.6 years
What do graphs look like?
29/570 A privately owned forest that had 1 million acres of old growth is
being clear cut at a rate of 7% per year.
a) create an exponential function.
b) create a table showing Q values for the first 15 units of time.
c) make a graph of the exponential function.
Q
1000000
930000
864900
804357
748052
695688
646990
601701
559582
520411
483982
450104
418596
389295
362044
336701
curvy!
Q = 1million x (1-.07)t = 1mill.x (.93)t
1220000
1020000
820000
acres
t
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
620000
420000
220000
20000
0
5
10
years
15
20
When will the acreage be reduced by half?
Q = 1000000 x (0.93)t
500000 = 1000000 x (0.93)t
-0.30103
= t
-0.03152
1
t
 (0.93)
2
log10 0.5 = log10 (0.93)
log10 (0.5)
= t
log10 (0.93)
t
log10 0.5 = t  log10 (0.93)
9.55134 = t
The acreage will be reduced
by half in 9.55 years.
Other forms for Exponential Functions
Independent variable: t
Dependent variable: Q
doubling time: Td
Initial Value: Q0
Q = Q0  2
t
Td
Independent variable: t
Dependent variable: Q
half-life: TH
Initial Value: Q0
Q = Q0  (0.5)
t
TH
Units for t and Td (and TH) must be the same.
41/571 The drug Valium is eliminated from the bloodstream exponentially
with a half-life of 36 hours. Suppose that a patient receives an initial
dose of 20 milligrams of Valium at midnight.
a) [equation] How much Valium is in the patient’s blood at noon the next
day?
b) [graph] Estimate when the Valium concentration will reach 10% of its
initial level.
initial value (at midnight): 20 milligrams
half-life: 36 hours
independent variable: hours since midnight
dependent variable: milligrams of Valium
Q = 20  (0.5)
t
36
noon the next day is 12 hours past midnight
12
36
Q = 20  (0.5)  15.9 mg
Q = 20  (0.5)
t
0
10
20
30
40
50
100
150
Q
20.00
16.50
13.61
11.22
9.26
7.64
2.92
1.11
milligrams of Valium
41/571 The drug Valium is eliminated from the bloodstream exponentially
with a half-life of 36 hours. Suppose that a patient receives an initial
dose of 20 milligrams of Valium at midnight.
a) [equation] How much Valium is in the patient’s blood at noon the next
day?
b) [graph] Estimate when the Valium concentration will reach 10% of its
initial level.
t
25.00
36
20.00
15.00
10.00
5.00
0.00
0
50
100
hours since midnight
10% of its initial value is 10% of 20 mg or .10x20 = 2 mg
Q is 2 mg when t is about _________
150
200
Can we solve exactly using the equation?
Q = 20  (0.5)
2 = 20  (0.5)
0.1 = (0.5)
t
36
log10 (0.1)
t
=( )
log10 (0.5)
36
t
36
log10 (0.1)
36 
= t
log10 (0.5)
t
36
log10 0.1 = log10 (0.5)
t
36
t
log10 0.1 = ( ) log10 (0.5)
36
-1
36 
= t
-.30103
119.59 hours = t
43/571 Uranium-238 has a half-life of 4.5 billion years.You find a rock
containing a mixture of uranium-238 and lead. You determine that 85%
of the original uranium-238 remains; the other 15% decayed into lead.
How old is the rock?
Q = 100  (0.5)
85 = 100  (0.5)
.85 = (0.5)
t
4.5
log10 .85
t
=
log10 (0.5)
4.5
t
4.5
log10 .85
4.5 
= t
log10 (0.5)
t
4.5
log10 .85 = log10 (0.5)
t
log10 .85 = ( ) log10 (0.5)
4.5
t
4.5
- .07058
4.5 
= t
.30103
1.05508 billion years = t
Homework
Pages 569-572
#28,#30,#32,#42,#44a