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 65 = 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 65 = 130(1 – 0.11) t r = 0.1165 = 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 65 = 130(1 – 0.11) t r = 0.1165 = 130(0.89) t y = 650.5 = (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 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

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 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

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 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)

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 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

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 0.00012.

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 0.00012. *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 0.00012. *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 0.00012. *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 0.00012. *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 0.00012. *No rate given so must be ‘Natural.’ y = ae -kt a = 1 y = 0.5 k = 0.00012

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 0.00012. *No rate given so must be ‘Natural.’ y = ae -kt a = 1 y = 0.5 k = 0.00012 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 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 = ???

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 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 = ???

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 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 = ???

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 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

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 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

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 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

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 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.

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(1 + 0.04) 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(1 + 0.04) 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(1 + 0.04) 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 2000. 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 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.

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

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

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

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

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

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

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

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

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

62. y = ae 0.000579t

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)

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

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