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Exponential Applications

EQ: How do we graph and use exponential functions?M2 Unit 5a: Day 7

Real-World Applications:•Internet traffic growth•The number of microorganisms growing in a culture•The spread of a virus (SARS, West Nile, small pox, etc)•Human population•High profits for a few initial investors in Pyramid schemes or Ponzi schemes•Example on the right: # of cell phone users from 1986-1995•Compound interest

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The Exponential Growth Model When a real-life quantity is increases by a fixed

percent ,r, each year (or other time period), the amount, y, of the quantity after t years can be modeled in this equation:

a = initial amount r = % increase (1 + r) = growth factor

= + 20500(1 .08)

(1 )ty a r= +

=$2330.48(1 )ty a r= +

Example 1:A diamond ring was purchased 20 years ago for $500.The value of the ring increased by 8% each year.What is the value of the ring today?

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Example 2: In 1985, there were 285 cell phone subscribers in the small

town of Centerville.  The number of subscribers increased by 75% per year after 1985.  How many cell phone subscribers were in Centerville in 1994 if it can be found using the formula a is the initial amount, r is the growth rate, and x is the number of years since 1985?

ya(1 r )x

ya(1 r )x

285(1 .75)9

43,871

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The Exponential Decay Model The exponential decay model has the

form where y is the quantity after t years, a is the initial amount, r is the percent decrease expressed as a decimal, and the quantity 1 – r is called the decay factor.

= -(1 )ty a r

8.8 gramsC=100010(0.99987)C=

Example 3:Ten grams of Carbon 14 is stored in a container. The amount C (in grams of Carbon 14 present after t years can be modeled by . How much carbon 14 is present after 1000 years?

10(0.99987)tC=

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Example 4: A man purchased a brand new

Outlander 800 ATV for $13,000. It depreciates at a rate of 15% per year. What is the value of the Outlander after 5 years?

= -(1 )ty a r= - 513,000(1 .15)y=$5,768.17y

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Compound interest:

AP 1 rn

nt

Example 5:

The amount of money, A, accrued at the end of n years when a certain

amount P, is invested at a compound annual rate, r, is given by

. If a person invests $550 in an account that pays 7%

interest compounded annually, find the balance after 5 years.

AP 1 rn

nt

AP 1 rn

nt

550 1 .071

15

$771.40

A = amount of moneyP = principle or amount initially investedr = compound annual raten = number of times compounded annually t = time (in years)

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Compound interest:

AP 1 rn

nt

A = amount of moneyP = principle or amount initially investedr = compound annual raten = number of times compounded annually t = time (in years)

Example 6:You deposit $1000 in an account that pays 8% annual interest. Find the balance after 1 year if compounded with the given frequency:(a) Annually (b) Quarterly (c) Daily11.081000 1 1A ×æ ö÷ç ÷= +ç ÷ç ÷çè ø$1080A=

41.081000 1 4A ×æ ö÷ç ÷= +ç ÷ç ÷çè ø$1082.43A=

3651.081000 1 365A ×æ ö÷ç ÷= +ç ÷ç ÷çè ø$1083.28A=

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Homework

Exponential Handout