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

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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