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8.8 – Exponential Growth & Decay

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

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8.8 – Exponential Growth & Decay

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  1. 8.8 – Exponential Growth & Decay

  2. Decay:

  3. Decay: 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 = 13065 = 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 = 13065 = 130(1 – 0.11)t r = 0.11 65 = 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 = 13065 = 130(1 – 0.11)t r = 0.11 65 = 130(0.89)t y = 65 0.5 = (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 = 13065 = 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

  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. 2. Natural rate: y = ae-kt 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 = 0.5 0.5 = 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 = 0.5 0.5 = 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 = 1 0.5 = 1e-0.00012t y = 0.5 0.5 = 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 = 1 0.5 = 1e-0.00012t y = 0.5 0.5 = 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 = 1 0.5 = 1e-0.00012t y = 0.5 0.5 = 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 = 1 0.5 = 1e-0.00012t y = 0.5 0.5 = 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. Growth: 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 = aekt

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