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8.8 – Exponential Growth & Decay

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Decay:

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1. Fixed rate

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Decay: 1. Fixed rate: y = a(1 – r) t

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Decay: 1. Fixed rate: y = a(1 – r) t where a = original amount

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Decay: 1. Fixed rate: y = a(1 – r) t where a = original amount r = rate of decrease

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Decay: 1. Fixed rate: y = a(1 – r) t where a = original amount r = rate of decrease t = time

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Decay: 1. Fixed rate: y = a(1 – r) t where a = original amount r = rate of decrease t = time y = new amount

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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?

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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.

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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

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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

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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

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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 =

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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

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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 = ???

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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(1 – 0.11) t r = 0.11 y = 65 t = ???

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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(1 – 0.11) t r = = 130(0.89) t y = 65 t = ???

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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(1 – 0.11) t r = = 130(0.89) t y = = (0.89) t t = ???

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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(1 – 0.11) t r = = 130(0.89) t y = = (0.89) t t = ??? log(0.5) = log(0.89) t

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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(1 – 0.11) t r = = 130(0.89) t y = = (0.89) t t = ??? log(0.5) = log(0.89) t log(0.5) = tlog(0.89) Power Property

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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(1 – 0.11) t r = = 130(0.89) t y = = (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)

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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(1 – 0.11) t r = = 130(0.89) t y = = (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) ≈ t

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2. Natural rate:

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2. Natural rate: y = ae -kt

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a = original amount k = constant of variation t = time y = new amount

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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

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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 *No rate given so must be ‘Natural.’

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

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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 *No rate given so must be ‘Natural.’ y = ae -kt a = 1

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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 *No rate given so must be ‘Natural.’ y = ae -kt a = 1 y = 0.5

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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 *No rate given so must be ‘Natural.’ y = ae -kt a = 1 y = 0.5 k =

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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 *No rate given so must be ‘Natural.’ y = ae -kt a = 1 y = 0.5 k = t = ???

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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 *No rate given so must be ‘Natural.’ y = ae -kt a = 10.5 = 1e t y = 0.5 k = t = ???

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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 *No rate given so must be ‘Natural.’ y = ae -kt a = 10.5 = 1e t y = = e t k = t = ???

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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 *No rate given so must be ‘Natural.’ y = ae -kt a = 10.5 = 1e t y = = e t k = ln(0.5) = ln e t t = ???

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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 *No rate given so must be ‘Natural.’ y = ae -kt a = 10.5 = 1e t y = = e t k = ln(0.5) = ln e t t = ??? ln(0.5) = t

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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 *No rate given so must be ‘Natural.’ y = ae -kt a = 10.5 = 1e t y = = e t k = ln(0.5) = ln e t t = ??? ln(0.5) = t ln(0.5) = t

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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 *No rate given so must be ‘Natural.’ y = ae -kt a = 10.5 = 1e t y = = e t k = ln(0.5) = ln e t t = ??? ln(0.5) = t ln(0.5) = t ,776 ≈ t

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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 *No rate given so must be ‘Natural.’ y = ae -kt a = 10.5 = 1e t y = = e t k = ln(0.5) = ln e t t = ??? ln(0.5) = t ln(0.5) = t ,776 ≈ t *It takes about 5,776 years for Carbon-14 to decay to half of it’s original amount.

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Growth:

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1. Fixed Rate:

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

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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?

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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

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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( ) 10

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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( ) 10 y = 100,000(1.04) 10

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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( ) 10 y = 100,000(1.04) 10 y = $148,024.43

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2. Natural Rate:

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2. Natural Rate: y = ae kt

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

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2.Natural Rate: y = ae kt Ex. 4 The population of Indianapolis, IN was 781,870 in It then rose to 784,118 by a. Write an exponential growth equation for the data where t is the number of years since 2000.

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2.Natural Rate: y = ae kt Ex. 4 The population of Indianapolis, IN was 781,870 in It then rose to 784,118 by a. Write an exponential growth equation for the data where t is the number of years since y = ae kt

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2.Natural Rate: y = ae kt Ex. 4 The population of Indianapolis, IN was 781,870 in It then rose to 784,118 by a. Write an exponential growth equation for the data where t is the number of years since y = ae kt 784,118 = 781,870e 5k

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2.Natural Rate: y = ae kt Ex. 4 The population of Indianapolis, IN was 781,870 in It then rose to 784,118 by a. Write an exponential growth equation for the data where t is the number of years since y = ae kt 784,118 = 781,870e 5k = e 5k

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2.Natural Rate: y = ae kt Ex. 4 The population of Indianapolis, IN was 781,870 in It then rose to 784,118 by a. Write an exponential growth equation for the data where t is the number of years since y = ae kt 784,118 = 781,870e 5k = e 5k ln(1.0029) = ln e 5k

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2.Natural Rate: y = ae kt Ex. 4 The population of Indianapolis, IN was 781,870 in It then rose to 784,118 by a. Write an exponential growth equation for the data where t is the number of years since y = ae kt 784,118 = 781,870e 5k = e 5k ln(1.0029) = ln e 5k ln(1.0029) = 5k

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2.Natural Rate: y = ae kt Ex. 4 The population of Indianapolis, IN was 781,870 in It then rose to 784,118 by a. Write an exponential growth equation for the data where t is the number of years since y = ae kt 784,118 = 781,870e 5k = e 5k ln(1.0029) = ln e 5k ln(1.0029) = 5k ln(1.0029) = k 5

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2.Natural Rate: y = ae kt Ex. 4 The population of Indianapolis, IN was 781,870 in It then rose to 784,118 by a. Write an exponential growth equation for the data where t is the number of years since y = ae kt 784,118 = 781,870e 5k = e 5k ln(1.0029) = ln e 5k ln(1.0029) = 5k ln(1.0029) = k = k

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2.Natural Rate: y = ae kt Ex. 4 The population of Indianapolis, IN was 781,870 in It then rose to 784,118 by a. Write an exponential growth equation for the data where t is the number of years since y = ae kt 784,118 = 781,870e 5k = e 5k ln(1.0029) = ln e 5k ln(1.0029) = 5k ln(1.0029) = k = k y = ae t

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

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y = ae t

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Ex. 4 The population of Indianapolis, IN was 781,870 in It then rose to 784,118 by b. Use your equation to predict the population of Indianapolis in y = ae t y = 781,870e (10)

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Ex. 4 The population of Indianapolis, IN was 781,870 in It then rose to 784,118 by b. Use your equation to predict the population of Indianapolis in y = ae t y = 781,870e (10) y ≈ 786,410

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Ex. 4 The population of Indianapolis, IN was 781,870 in It then rose to 784,118 by b. Use your equation to predict the population of Indianapolis in y = ae t y = 781,870e (10) y ≈ 786,410 Info obtained from napolis.htm

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