8.8 – Exponential Growth & Decay

Decay:

Decay:

1. Fixed rate

Decay:

1. Fixed rate: y = a(1 – r)t

Decay:

1. Fixed rate: y = a(1 – r)t

where a = original amount

Decay:

1. Fixed rate: y = a(1 – r)t

where a = original amount

r = rate of decrease

Decay:

1. Fixed rate: y = a(1 – r)t

where a = original amount

r = rate of decrease

t = time

Decay:

1. Fixed rate: y = a(1 – r)t

where a = original amount

r = rate of decrease

t = time

y = new amount

Decay:1. Fixed rate: y = a(1 – r)t

where a = original amount r = rate of decrease t = time y = new amount

Ex. 1 A cup of coffee contains 130mg. of caffeine. If caffeine is eliminated from the body at a rate of 11% per hour, how long will it take for half of this caffeine to be eliminated from a person’s body?

Decay:

1. Fixed rate: y = a(1 – r)t

where a = original amount

r = rate of decrease

t = time

y = new amount

Ex. 1 A cup of coffee contains 130mg. of caffeine. If caffeine is eliminated from the body at a rate of 11% per hour, how long will it take for half of this caffeine to be eliminated from a person’s body?

11% indicates that it is fixed-rate decay.

Decay:1. Fixed rate: y = a(1 – r)t

where a = original amount r = rate of decrease t = time y = new amount

Ex. 1 A cup of coffee contains 130mg. of caffeine. If caffeine is eliminated from the body at a rate of 11% per hour, how long will it take for half of this caffeine to be eliminated from a person’s body? 11% indicates that it is fixed-rate decay.

y = a(1 – r)t

Decay:1. Fixed rate: y = a(1 – r)t

where a = original amount r = rate of decrease t = time y = new amount

Ex. 1 A cup of coffee contains 130mg. of caffeine. If caffeine is eliminated from the body at a rate of 11% per hour, how long will it take for half of this caffeine to be eliminated from a person’s body? 11% indicates that it is fixed-rate decay.

y = a(1 – r)t

a = 130

Decay:1. Fixed rate: y = a(1 – r)t

where a = original amount r = rate of decrease t = time y = new amount

Ex. 1 A cup of coffee contains 130mg. of caffeine. If caffeine is eliminated from the body at a rate of 11% per hour, how long will it take for half of this caffeine to be eliminated from a person’s body? 11% indicates that it is fixed-rate decay.

y = a(1 – r)t

a = 130r = 0.11

Decay:1. Fixed rate: y = a(1 – r)t

where a = original amount r = rate of decrease t = time y = new amount

Ex. 1 A cup of coffee contains 130mg. of caffeine. If caffeine is eliminated from the body at a rate of 11% per hour, how long will it take for half of this caffeine to be eliminated from a person’s body? 11% indicates that it is fixed-rate decay.

y = a(1 – r)t

a = 130r = 0.11y =

Decay:1. Fixed rate: y = a(1 – r)t

where a = original amount r = rate of decrease t = time y = new amount

Ex. 1 A cup of coffee contains 130mg. of caffeine. If caffeine is eliminated from the body at a rate of 11% per hour, how long will it take for half of this caffeine to be eliminated from a person’s body? 11% indicates that it is fixed-rate decay.

y = a(1 – r)t

a = 130r = 0.11y = 65

Decay:1. Fixed rate: y = a(1 – r)t

where a = original amount r = rate of decrease t = time y = new amount

Ex. 1 A cup of coffee contains 130mg. of caffeine. If caffeine is eliminated from the body at a rate of 11% per hour, how long will it take for half of this caffeine to be eliminated from a person’s body? 11% indicates that it is fixed-rate decay.

y = a(1 – r)t

a = 130 r = 0.11y = 65t = ???

Decay:1. Fixed rate: y = a(1 – r)t

where a = original amount r = rate of decrease t = time y = new amount

Ex. 1 A cup of coffee contains 130mg. of caffeine. If caffeine is eliminated from the body at a rate of 11% per hour, how long will it take for half of this caffeine to be eliminated from a person’s body? 11% indicates that it is fixed-rate decay.

y = a(1 – r)t

a = 130 65 = 130(1 – 0.11)t

r = 0.11y = 65t = ???

Decay:1. Fixed rate: y = a(1 – r)t

where a = original amount r = rate of decrease t = time y = new amount

Ex. 1 A cup of coffee contains 130mg. of caffeine. If caffeine is eliminated from the body at a rate of 11% per hour, how long will it take for half of this caffeine to be eliminated from a person’s body? 11% indicates that it is fixed-rate decay.

y = a(1 – r)t

a = 130 65 = 130(1 – 0.11)t

r = 0.11 65 = 130(0.89)t

y = 65t = ???

Decay:1. Fixed rate: y = a(1 – r)t

where a = original amount r = rate of decrease t = time y = new amount

Ex. 1 A cup of coffee contains 130mg. of caffeine. If caffeine is eliminated from the body at a rate of 11% per hour, how long will it take for half of this caffeine to be eliminated from a person’s body? 11% indicates that it is fixed-rate decay.

y = a(1 – r)t

a = 130 65 = 130(1 – 0.11)t

r = 0.11 65 = 130(0.89)t

y = 65 0.5 = (0.89)t

t = ???

Decay:1. Fixed rate: y = a(1 – r)t

where a = original amount r = rate of decrease t = time y = new amount

Ex. 1 A cup of coffee contains 130mg. of caffeine. If caffeine is eliminated from the body at a rate of 11% per hour, how long will it take for half of this caffeine to be eliminated from a person’s body? 11% indicates that it is fixed-rate decay.

y = a(1 – r)t

a = 130 65 = 130(1 – 0.11)t

r = 0.11 65 = 130(0.89)t

y = 65 0.5 = (0.89)t

t = ??? log(0.5) = log(0.89)t

Decay:1. Fixed rate: y = a(1 – r)t

where a = original amount r = rate of decrease t = time y = new amount

Ex. 1 A cup of coffee contains 130mg. of caffeine. If caffeine is eliminated from the body at a rate of 11% per hour, how long will it take for half of this caffeine to be eliminated from a person’s body? 11% indicates that it is fixed-rate decay.

y = a(1 – r)t

a = 130 65 = 130(1 – 0.11)t

r = 0.11 65 = 130(0.89)t

y = 65 0.5 = (0.89)t

t = ??? log(0.5) = log(0.89)t

log(0.5) = tlog(0.89) Power Property

Decay:1. Fixed rate: y = a(1 – r)t

where a = original amount r = rate of decrease t = time y = new amount

Ex. 1 A cup of coffee contains 130mg. of caffeine. If caffeine is eliminated from the body at a rate of 11% per hour, how long will it take for half of this caffeine to be eliminated from a person’s body? 11% indicates that it is fixed-rate decay.

y = a(1 – r)t

a = 130 65 = 130(1 – 0.11)t

r = 0.11 65 = 130(0.89)t

y = 65 0.5 = (0.89)t

t = ??? log(0.5) = log(0.89)t

log(0.5) = tlog(0.89) Power Property log(0.5) = tlog(0.89)

Decay:1. Fixed rate: y = a(1 – r)t

where a = original amount r = rate of decrease t = time y = new amount

Ex. 1 A cup of coffee contains 130mg. of caffeine. If caffeine is eliminated from the body at a rate of 11% per hour, how long will it take for half of this caffeine to be eliminated from a person’s body? 11% indicates that it is fixed-rate decay.

y = a(1 – r)t

a = 130 65 = 130(1 – 0.11)t

r = 0.11 65 = 130(0.89)t

y = 65 0.5 = (0.89)t

t = ??? log(0.5) = log(0.89)t

log(0.5) = tlog(0.89) Power Property log(0.5) = tlog(0.89) 5.9480 ≈ t

2. Natural rate:

2. Natural rate: y = ae-kt

2. Natural rate: y = ae-kt

a = original amount

k = constant of variation

t = time

y = new amount

2. Natural rate: y = ae-kt

a = original amount

k = constant of variation

t = time

y = new amount

Ex. 2 Determine the half-life of Carbon-14 if it’s constant of variation is 0.00012.

2. Natural rate: y = ae-kt

a = original amount

k = constant of variation

t = time

y = new amount

Ex. 2 Determine the half-life of Carbon-14 if it’s constant of variation is 0.00012.

*No rate given so must be ‘Natural.’

2. Natural rate: y = ae-kt

a = original amount

k = constant of variation

t = time

y = new amount

Ex. 2 Determine the half-life of Carbon-14 if it’s constant of variation is 0.00012.

*No rate given so must be ‘Natural.’

y = ae-kt

2. Natural rate: y = ae-kt

a = original amount

k = constant of variation

t = time

y = new amount

Ex. 2 Determine the half-life of Carbon-14 if it’s constant of variation is 0.00012.

*No rate given so must be ‘Natural.’

y = ae-kt

a = 1

2. Natural rate: y = ae-kt

a = original amount

k = constant of variation

t = time

y = new amount

Ex. 2 Determine the half-life of Carbon-14 if it’s constant of variation is 0.00012.

*No rate given so must be ‘Natural.’

y = ae-kt

a = 1

y = 0.5

2. Natural rate: y = ae-kt

a = original amountk = constant of variation

t = time y = new amount

Ex. 2 Determine the half-life of Carbon-14 if it’s constant of variation is 0.00012.

*No rate given so must be ‘Natural.’y = ae-kt

a = 1y = 0.5k = 0.00012

2. Natural rate: y = ae-kt

a = original amountk = constant of variation

t = time y = new amount

Ex. 2 Determine the half-life of Carbon-14 if it’s constant of variation is 0.00012.

*No rate given so must be ‘Natural.’y = ae-kt

a = 1y = 0.5k = 0.00012t = ???

2. Natural rate: y = ae-kt

a = original amountk = constant of variation

t = time y = new amount

Ex. 2 Determine the half-life of Carbon-14 if it’s constant of variation is 0.00012.

*No rate given so must be ‘Natural.’y = ae-kt

a = 1 0.5 = 1e-0.00012t

y = 0.5k = 0.00012t = ???

2. Natural rate: y = ae-kt

a = original amountk = constant of variation

t = time y = new amount

Ex. 2 Determine the half-life of Carbon-14 if it’s constant of variation is 0.00012.

*No rate given so must be ‘Natural.’y = ae-kt

a = 1 0.5 = 1e-0.00012t

y = 0.5 0.5 = e-0.00012t

k = 0.00012t = ???

2. Natural rate: y = ae-kt

a = original amountk = constant of variation

t = time y = new amount

Ex. 2 Determine the half-life of Carbon-14 if it’s constant of variation is 0.00012.

*No rate given so must be ‘Natural.’y = ae-kt

a = 1 0.5 = 1e-0.00012t

y = 0.5 0.5 = e-0.00012t

k = 0.00012 ln(0.5) = ln e-0.00012t

t = ???

2. Natural rate: y = ae-kt

a = original amountk = constant of variation

t = time y = new amount

Ex. 2 Determine the half-life of Carbon-14 if it’s constant of variation is 0.00012.

*No rate given so must be ‘Natural.’y = ae-kt

a = 1 0.5 = 1e-0.00012t

y = 0.5 0.5 = e-0.00012t

k = 0.00012 ln(0.5) = ln e-0.00012t

t = ??? ln(0.5) = -0.00012t

2. Natural rate: y = ae-kt

a = original amountk = constant of variation

t = time y = new amount

Ex. 2 Determine the half-life of Carbon-14 if it’s constant of variation is 0.00012.

*No rate given so must be ‘Natural.’y = ae-kt

a = 1 0.5 = 1e-0.00012t

y = 0.5 0.5 = e-0.00012t

k = 0.00012 ln(0.5) = ln e-0.00012t

t = ??? ln(0.5) = -0.00012t ln(0.5) = t-0.00012

2. Natural rate: y = ae-kt

a = original amountk = constant of variation

t = time y = new amount

Ex. 2 Determine the half-life of Carbon-14 if it’s constant of variation is 0.00012.

*No rate given so must be ‘Natural.’y = ae-kt

a = 1 0.5 = 1e-0.00012t

y = 0.5 0.5 = e-0.00012t

k = 0.00012 ln(0.5) = ln e-0.00012t

t = ??? ln(0.5) = -0.00012t ln(0.5) = t-0.000125,776 ≈ t

2. Natural rate: y = ae-kt

a = original amountk = constant of variation

t = time y = new amount

Ex. 2 Determine the half-life of Carbon-14 if it’s constant of variation is 0.00012.

*No rate given so must be ‘Natural.’y = ae-kt

a = 1 0.5 = 1e-0.00012t

y = 0.5 0.5 = e-0.00012t

k = 0.00012 ln(0.5) = ln e-0.00012t

t = ??? ln(0.5) = -0.00012t ln(0.5) = t-0.000125,776 ≈ t

*It takes about 5,776 years for Carbon-14 to decay to half of it’s original amount.

Growth:

Growth:

1. Fixed Rate:

Growth:

1. Fixed Rate: y = a(1 + r)t

Growth:

1. Fixed Rate: y = a(1 + r)t

Ex. 3 Suppose you buy a house for $100,000. If the house appreciates at most 4% a year, how much will the house be worth in 10 years?

Growth:

1. Fixed Rate: y = a(1 + r)t

Ex. 3 Suppose you buy a house for $100,000. If the house appreciates at most 4% a year, how much will the house be worth in 10 years?

y = a(1 + r)t

Growth:

1. Fixed Rate: y = a(1 + r)t

Ex. 3 Suppose you buy a house for $100,000. If the house appreciates at most 4% a year, how much will the house be worth in 10 years?

y = a(1 + r)t

y = 100,000(1 + 0.04)10

Growth:

1. Fixed Rate: y = a(1 + r)t

Ex. 3 Suppose you buy a house for $100,000. If the house appreciates at most 4% a year, how much will the house be worth in 10 years?

y = a(1 + r)t

y = 100,000(1 + 0.04)10

y = 100,000(1.04)10

Growth:

1. Fixed Rate: y = a(1 + r)t

Ex. 3 Suppose you buy a house for $100,000. If the house appreciates at most 4% a year, how much will the house be worth in 10 years?

y = a(1 + r)t

y = 100,000(1 + 0.04)10

y = 100,000(1.04)10

y = $148,024.43

2. Natural Rate:

2. Natural Rate: y = aekt

2. Natural Rate: y = aekt

Ex. 4 The population of Indianapolis, IN was 781,870 in 2000. It then rose to 784,118 by 2005.

2. Natural Rate: y = aekt

Ex. 4 The population of Indianapolis, IN was 781,870 in 2000. It then rose to 784,118 by 2005.

a. Write an exponential growth equation for the data where t is the number of years since 2000.

2. Natural Rate: y = aekt

Ex. 4 The population of Indianapolis, IN was 781,870 in 2000. It then rose to 784,118 by 2005.

a. Write an exponential growth equation for the data where t is the number of years since 2000.

y = aekt

2. Natural Rate: y = aekt

Ex. 4 The population of Indianapolis, IN was 781,870 in 2000. It then rose to 784,118 by 2005.

a. Write an exponential growth equation for the data where t is the number of years since 2000.

y = aekt

784,118 = 781,870e5k

2. Natural Rate: y = aekt

Ex. 4 The population of Indianapolis, IN was 781,870 in 2000. It then rose to 784,118 by 2005.

a. Write an exponential growth equation for the data where t is the number of years since 2000.

y = aekt

784,118 = 781,870e5k

1.0029 = e5k

2. Natural Rate: y = aekt

Ex. 4 The population of Indianapolis, IN was 781,870 in 2000. It then rose to 784,118 by 2005.

a. Write an exponential growth equation for the data where t is the number of years since 2000.

y = aekt

784,118 = 781,870e5k

1.0029 = e5k

ln(1.0029) = ln e5k

2. Natural Rate: y = aekt

Ex. 4 The population of Indianapolis, IN was 781,870 in 2000. It then rose to 784,118 by 2005.

a. Write an exponential growth equation for the data where t is the number of years since 2000.

y = aekt

784,118 = 781,870e5k

1.0029 = e5k

ln(1.0029) = ln e5k

ln(1.0029) = 5k

2. Natural Rate: y = aekt

Ex. 4 The population of Indianapolis, IN was 781,870 in 2000. It then rose to 784,118 by 2005.

a. Write an exponential growth equation for the data where t is the number of years since 2000.

y = aekt

784,118 = 781,870e5k

1.0029 = e5k

ln(1.0029) = ln e5k

ln(1.0029) = 5kln(1.0029) = k 5

2. Natural Rate: y = aekt

Ex. 4 The population of Indianapolis, IN was 781,870 in 2000. It then rose to 784,118 by 2005.

a. Write an exponential growth equation for the data where t is the number of years since 2000.

y = aekt

784,118 = 781,870e5k

1.0029 = e5k

ln(1.0029) = ln e5k

ln(1.0029) = 5kln(1.0029) = k 5 0.000579 = k

2. Natural Rate: y = aekt

Ex. 4 The population of Indianapolis, IN was 781,870 in 2000. It then rose to 784,118 by 2005.

a. Write an exponential growth equation for the data where t is the number of years since 2000.

y = aekt

784,118 = 781,870e5k

1.0029 = e5k

ln(1.0029) = ln e5k

ln(1.0029) = 5kln(1.0029) = k 5 0.000579 = k y =

ae0.000579t

b. Use your equation to predict the population of Indianapolis in 2010.

b. Use your equation to predict the population of Indianapolis in 2010.

y = ae0.000579t

Ex. 4 The population of Indianapolis, IN was 781,870 in 2000. It then rose to 784,118 by 2005.

b. Use your equation to predict the population of Indianapolis in 2010.

y = ae0.000579t

y = 781,870e0.000579(10)

Ex. 4 The population of Indianapolis, IN was 781,870 in 2000. It then rose to 784,118 by 2005.

b. Use your equation to predict the population of Indianapolis in 2010.

y = ae0.000579t

y = 781,870e0.000579(10)

y ≈ 786,410

Ex. 4 The population of Indianapolis, IN was 781,870 in 2000. It then rose to 784,118 by 2005.

b. Use your equation to predict the population of Indianapolis in 2010.

y = ae0.000579t

y = 781,870e0.000579(10)

y ≈ 786,410

Info obtained from http://www.idcide.com/citydata/in/indianapolis.htm