# 8.8 – Exponential Growth & Decay. Decay: 1. Fixed rate.

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8.8 – Exponential Growth & 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 = 130 r = 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 = 130 r = 0.11 y =

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 = 0.11 y = 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 = 0.11 y = 65 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 y = 65 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 = 0.1165 = 130(0.89) t y = 65 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 = 0.1165 = 130(0.89) t y = 650.5 = (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 = 0.1165 = 130(0.89) t y = 650.5 = (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) = t log(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) = t log(0.89) 5.9480 ≈ t

2. Natural rate:

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 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 k = 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.’ y = ae -kt a = 1 y = 0.5 k = 0.00012 t = ???

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 = 10.5 = 1e -0.00012t y = 0.5 k = 0.00012 t = ???

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 = 10.5 = 1e -0.00012t y = 0.50.5 = e -0.00012t k = 0.00012 t = ???

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 = 10.5 = 1e -0.00012t y = 0.50.5 = e -0.00012t k = 0.00012 ln(0.5) = ln e -0.00012t t = ???

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 = 10.5 = 1e -0.00012t y = 0.50.5 = 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 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 = 10.5 = 1e -0.00012t y = 0.50.5 = 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 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 = 10.5 = 1e -0.00012t y = 0.50.5 = 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 5,776 ≈ t

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 = 10.5 = 1e -0.00012t y = 0.50.5 = 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 5,776 ≈ t *It takes about 5,776 years for Carbon-14 to decay to half of it’s original amount.

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 = ae kt

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

2.Natural Rate: y = ae kt 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 = ae kt 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 = ae kt

2.Natural Rate: y = ae kt 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 = ae kt 784,118 = 781,870e 5k

2.Natural Rate: y = ae kt 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 = ae kt 784,118 = 781,870e 5k 1.0029 = e 5k

2.Natural Rate: y = ae kt 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 = ae kt 784,118 = 781,870e 5k 1.0029 = e 5k ln(1.0029) = ln e 5k

2.Natural Rate: y = ae kt 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 = ae kt 784,118 = 781,870e 5k 1.0029 = e 5k ln(1.0029) = ln e 5k ln(1.0029) = 5k

2.Natural Rate: y = ae kt 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 = ae kt 784,118 = 781,870e 5k 1.0029 = e 5k ln(1.0029) = ln e 5k ln(1.0029) = 5k ln(1.0029) = k 5

2.Natural Rate: y = ae kt 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 = ae kt 784,118 = 781,870e 5k 1.0029 = e 5k ln(1.0029) = ln e 5k ln(1.0029) = 5k ln(1.0029) = k 5 0.000579 = k

2.Natural Rate: y = ae kt 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 = ae kt 784,118 = 781,870e 5k 1.0029 = e 5k ln(1.0029) = ln e 5k ln(1.0029) = 5k ln(1.0029) = k 5 0.000579 = k y = ae 0.000579t

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

y = ae 0.000579t

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 = ae 0.000579t y = 781,870e 0.000579(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 = ae 0.000579t y = 781,870e 0.000579(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 = ae 0.000579t y = 781,870e 0.000579(10) y ≈ 786,410 Info obtained from http://www.idcide.com/citydata/in/india napolis.htm

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