Section 9CExponential 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: Qdecimal 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 yearindependent variable: years since 2000dependent variable: population

Q = 281million x (1+.007)t

Q = 281million x (1.007)100 = 564 million

year 2100 is 100 years since 2000, so t = 100

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 populationb) will this rate of decline be sufficient to meet the original goal? initial value (in 2000): 1.2 billion

rate: -.005 per yearindependent variable: years since 2000dependent 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 = .934billionWith 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.

t Q0 $24,000.001 $25,200.002 $26,460.003 $27,783.004 $29,172.155 $30,630.766 $32,162.307 $33,770.418 $35,458.939 $37,231.8810 $39,093.4711 $41,048.1412 $43,100.5513 $45,255.5814 $47,518.3615 $49,894.28

Q = 24000 x (1+.05)t

$20,000.00

$25,000.00

$30,000.00

$35,000.00

$40,000.00

$45,000.00

$50,000.00

$55,000.00

0 5 10 15 20

years on the job

ann

ual

sal

ary

curvy!

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.

$20,000.00

$25,000.00

$30,000.00

$35,000.00

$40,000.00

$45,000.00

$50,000.00

$55,000.00

0 5 10 15 20

years on the job

ann

ual

sal

ary

When t = 12, Q = _______.

Q = $30,000 when t = _____

Use Properties of Logarithms

10 10 10log (A B) = log (A) + log (B)

10 10 10

Alog ( ) = log (A) - log (B)

B

B10 10log (A ) = B log (A)

Can we solve exactly using the equation?

Q = 24000 x (1+.05)t

30000 = 24000 x (1+.05)t

t30000(1.05)

24000

1.25 = (1.05)t

t10 10log 1.25 = log (1.05)

10 10log 1.25 = t log (1.05)

10

10

log (1.25) = t

log (1.05)

0.09691 = t

.021189

4.57360 = t

The salary will be $30,000 in 4.6 years

Can we solve exactly using the equation?

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 = 1million x (1-.07)t = 1mill.x (.93)t

t Q0 10000001 9300002 8649003 8043574 7480525 6956886 6469907 6017018 5595829 52041110 48398211 45010412 41859613 38929514 36204415 336701

20000

220000

420000

620000

820000

1020000

1220000

0 5 10 15 20

years

acre

s

curvy!

Q = 1000000 x (0.93)t

500000 = 1000000 x (0.93)t

t1(0.93)

2

t10 10log 0.5 = log (0.93)

10 10log 0.5 = t log (0.93)

10

10

log (0.5) = t

log (0.93)

-0.30103 = t

-0.03152

9.55134 = t

The acreage will be reduced by half in 9.55

years.

When will the acreage be reduced by half?

Other forms for Exponential FunctionsIndependent variable: tDependent variable: Q

doubling time: Td

Initial Value: Q0

d

t

T0Q = Q 2

Independent variable: tDependent variable: Q

half-life: TH

Initial Value: Q0

H

t

T0Q = Q (0.5)

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.

t

36Q = 20 (0.5)

initial value (at midnight): 20 milligramshalf-life: 36 hoursindependent variable: hours since midnightdependent variable: milligrams of Valium

noon the next day is 12 hours past midnight

12

36Q = 20 (0.5) 15.9mg

t Q0 20.0010 16.5020 13.6130 11.2240 9.2650 7.64

100 2.92150 1.11

0.00

5.00

10.00

15.00

20.00

25.00

0 50 100 150 200

hours since midnight

mill

igra

ms o

f V

aliu

m

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

36Q = 20 (0.5)

10% of its initial value is 10% of 20 mg or .10x20 = 2 mg

Q is 2 mg when t is about _________

Can we solve exactly using the equation?t

36Q = 20 (0.5)t

362 = 20 (0.5)t

360.1 = (0.5)

t

3610 10log 0.1 = log (0.5)

10 10

tlog 0.1 = ( ) log (0.5)

36

10

10

log (0.1) t = ( )

log (0.5) 36

10

10

log (0.1)36 = t

log (0.5)

-136 = 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?

t

4.5Q = 100 (0.5)

t

4.585 = 100 (0.5)

t

4.5.85 = (0.5)

t

4.510 10log .85 = log (0.5)

10 10

tlog .85 = ( ) log (0.5)

4.5

10

10

log .85 t =

log (0.5) 4.5

10

10

log .854.5 = t

log (0.5)

- .070584.5 = t

.30103

1.05508 billion years = t

Homework

Pages 569-572

#28,#30,#32,#42,#44a