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

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8.8 – Exponential Growth & Decay
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Transcript of 8.8 – Exponential Growth & Decay. Decay: 1. Fixed rate.

  • Slide 1
  • 8.8 Exponential Growth & Decay
  • Slide 2
  • Decay:
  • Slide 3
  • 1. Fixed rate
  • Slide 4
  • Decay: 1. Fixed rate: y = a(1 r) t
  • Slide 5
  • Decay: 1. Fixed rate: y = a(1 r) t where a = original amount
  • Slide 6
  • Decay: 1. Fixed rate: y = a(1 r) t where a = original amount r = rate of decrease
  • Slide 7
  • Decay: 1. Fixed rate: y = a(1 r) t where a = original amount r = rate of decrease t = time
  • Slide 8
  • Decay: 1. Fixed rate: y = a(1 r) t where a = original amount r = rate of decrease t = time y = new amount
  • Slide 9
  • 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 persons body?
  • Slide 10
  • 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 persons body? 11% indicates that it is fixed-rate decay.
  • Slide 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 persons body? 11% indicates that it is fixed-rate decay. y = a(1 r) t
  • Slide 12
  • 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 persons body? 11% indicates that it is fixed-rate decay. y = a(1 r) t a = 130
  • Slide 13
  • 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 persons body? 11% indicates that it is fixed-rate decay. y = a(1 r) t a = 130 r = 0.11
  • Slide 14
  • 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 persons body? 11% indicates that it is fixed-rate decay. y = a(1 r) t a = 130 r = 0.11 y =
  • Slide 15
  • 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 persons body? 11% indicates that it is fixed-rate decay. y = a(1 r) t a = 130 r = 0.11 y = 65
  • Slide 16
  • 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 persons body? 11% indicates that it is fixed-rate decay. y = a(1 r) t a = 130 r = 0.11 y = 65 t = ???
  • Slide 17
  • 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 persons 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 = ???
  • Slide 18
  • 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 persons 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 = ???
  • Slide 19
  • 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 persons 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 = ???
  • Slide 20
  • 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 persons 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
  • Slide 21
  • 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 persons 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
  • Slide 22
  • 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 persons 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)
  • Slide 23
  • 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 persons 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
  • Slide 24
  • 2. Natural rate:
  • Slide 25
  • 2. Natural rate: y = ae -kt
  • Slide 26
  • a = original amount k = constant of variation t = time y = new amount
  • Slide 27
  • 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 its constant of variation is 0.00012.
  • Slide 28
  • 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 its constant of variation is 0.00012. *No rate given so must be Natural.
  • Slide 29
  • 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 its constant of variation is 0.00012. *No rate given so must be Natural. y = ae -kt
  • Slide 30
  • 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 its constant of variation is 0.00012. *No rate given so must be Natural. y = ae -kt a = 1
  • Slide 31
  • 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 its constant of variation is 0.00012. *No rate given so must be Natural. y = ae -kt a = 1 y = 0.5
  • Slide 32
  • 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 its constant of variation is 0.00012. *No rate given so must be Natural. y = ae -kt a = 1 y = 0.5 k = 0.00012
  • Slide 33
  • 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 its 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 = ???
  • Slide 34
  • 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 its 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 = ???
  • Slide 35
  • 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 its 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 = ???
  • Slide 36
  • 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 its 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 = ???
  • Slide 37
  • 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 its 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
  • Slide 38
  • 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 its 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
  • Slide 39
  • 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 its 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
  • Slide 40
  • 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 its 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 its original amount.
  • Slide 41
  • Growth:
  • Slide 42
  • 1. Fixed Rate:
  • Slide 43
  • Growth: 1. Fixed Rate: y = a(1 + r) t
  • Slide 44
  • 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?
  • Slide 45
  • 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
  • Slide 46
  • 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
  • Slide 47
  • 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
  • Slide 48
  • 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
  • Slide 49
  • 2. Natural Rate:
  • Slide 50
  • 2. Natural Rate: y = ae kt
  • Slide 51
  • Ex. 4 The population of Indianapolis, IN was 781,870 in 2000. It then rose to 784,118 by 2005.
  • Slide 52
  • 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.
  • Slide 53
  • 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
  • Slide 54
  • 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
  • Slide 55
  • 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
  • Slide 56
  • 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
  • Slide 57
  • 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
  • Slide 58
  • 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
  • Slide 59
  • 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
  • Slide 60
  • 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
  • Slide 61
  • b. Use your equation to predict the population of Indianapolis in 2010.
  • Slide 62
  • y = ae 0.000579t
  • Slide 63
  • 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)
  • Slide 64
  • 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
  • Slide 65
  • 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