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

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Presentation on theme: "8.8 – Exponential Growth & Decay. Decay: 1. Fixed rate."— Presentation transcript:

1 8.8 – Exponential Growth & Decay

2 Decay:

3 1. Fixed rate

4 Decay: 1. Fixed rate: y = a(1 – r) t

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

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

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

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

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 person’s body?

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 person’s body? 11% indicates that it is fixed-rate decay.

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

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 person’s body? 11% indicates that it is fixed-rate decay. y = a(1 – r) t a = 130

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 person’s body? 11% indicates that it is fixed-rate decay. y = a(1 – r) t a = 130 r = 0.11

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 person’s body? 11% indicates that it is fixed-rate decay. y = a(1 – r) t a = 130 r = 0.11 y =

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 person’s body? 11% indicates that it is fixed-rate decay. y = a(1 – r) t a = 130 r = 0.11 y = 65

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 person’s body? 11% indicates that it is fixed-rate decay. y = a(1 – r) t a = 130 r = 0.11 y = 65 t = ???

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 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 = ???

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 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 = ???

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 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 = ???

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 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

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 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

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 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)

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 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

24 2. Natural rate:

25 2. Natural rate: y = ae -kt

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

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 it’s constant of variation is

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 it’s constant of variation is *No rate given so must be ‘Natural.’

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 it’s constant of variation is *No rate given so must be ‘Natural.’ y = ae -kt

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 it’s constant of variation is *No rate given so must be ‘Natural.’ y = ae -kt a = 1

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 it’s constant of variation is *No rate given so must be ‘Natural.’ y = ae -kt a = 1 y = 0.5

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 it’s constant of variation is *No rate given so must be ‘Natural.’ y = ae -kt a = 1 y = 0.5 k =

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 it’s constant of variation is *No rate given so must be ‘Natural.’ y = ae -kt a = 1 y = 0.5 k = t = ???

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 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 = ???

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 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 = ???

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 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 = ???

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 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

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 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

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 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

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 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.

41 Growth:

42 1. Fixed Rate:

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

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?

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

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( ) 10

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

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

49 2. Natural Rate:

50 2. Natural Rate: y = ae kt

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

52 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.

53 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

54 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

55 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

56 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

57 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

58 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

59 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

60 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

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

62 y = ae t

63 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)

64 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

65 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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