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8.8 – Exponential Growth & Decay
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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. - PowerPoint PPT Presentation

### Transcript of 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)twhere a = original amount

• Decay:1. Fixed rate: y = a(1 r)twhere a = original amount r = rate of decrease

• Decay:1. Fixed rate: y = a(1 r)twhere a = original amount r = rate of decrease t = time

• Decay:1. Fixed rate: y = a(1 r)twhere a = original amount r = rate of decrease t = time y = new amount

• Decay:1. Fixed rate: y = a(1 r)twhere a = original amount r = rate of decrease t = time y = new amountEx. 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?

• Decay:1. Fixed rate: y = a(1 r)twhere a = original amount r = rate of decrease t = time y = new amountEx. 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.

• Decay:1. Fixed rate: y = a(1 r)twhere a = original amount r = rate of decrease t = time y = new amountEx. 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

• Decay:1. Fixed rate: y = a(1 r)twhere a = original amount r = rate of decrease t = time y = new amountEx. 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)ta = 130

• Decay:1. Fixed rate: y = a(1 r)twhere a = original amount r = rate of decrease t = time y = new amountEx. 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)ta = 130r = 0.11

• Decay:1. Fixed rate: y = a(1 r)twhere a = original amount r = rate of decrease t = time y = new amountEx. 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)ta = 130r = 0.11y =

• Decay:1. Fixed rate: y = a(1 r)twhere a = original amount r = rate of decrease t = time y = new amountEx. 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)ta = 130r = 0.11y = 65

• Decay:1. Fixed rate: y = a(1 r)twhere a = original amount r = rate of decrease t = time y = new amountEx. 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)ta = 130 r = 0.11y = 65t = ???

• Decay:1. Fixed rate: y = a(1 r)twhere a = original amount r = rate of decrease t = time y = new amountEx. 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)ta = 130 65 = 130(1 0.11)tr = 0.11y = 65t = ???

• Decay:1. Fixed rate: y = a(1 r)twhere a = original amount r = rate of decrease t = time y = new amountEx. 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)ta = 130 65 = 130(1 0.11)tr = 0.1165 = 130(0.89)ty = 65t = ???

• Decay:1. Fixed rate: y = a(1 r)twhere a = original amount r = rate of decrease t = time y = new amountEx. 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)ta = 130 65 = 130(1 0.11)tr = 0.1165 = 130(0.89)ty = 650.5 = (0.89)tt = ???

• Decay:1. Fixed rate: y = a(1 r)twhere a = original amount r = rate of decrease t = time y = new amountEx. 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)ta = 130 65 = 130(1 0.11)tr = 0.1165 = 130(0.89)ty = 650.5 = (0.89)tt = ??? log(0.5) = log(0.89)t

• Decay:1. Fixed rate: y = a(1 r)twhere a = original amount r = rate of decrease t = time y = new amountEx. 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)ta = 130 65 = 130(1 0.11)tr = 0.11 65 = 130(0.89)ty = 65 0.5 = (0.89)tt = ??? log(0.5) = log(0.89)t log(0.5) = tlog(0.89) Power Property

• Decay:1. Fixed rate: y = a(1 r)twhere a = original amount r = rate of decrease t = time y = new amountEx. 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)ta = 130 65 = 130(1 0.11)tr = 0.11 65 = 130(0.89)ty = 65 0.5 = (0.89)tt = ??? log(0.5) = log(0.89)t log(0.5) = tlog(0.89) Power Property log(0.5) = tlog(0.89)

• Decay:1. Fixed rate: y = a(1 r)twhere a = original amount r = rate of decrease t = time y = new amountEx. 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)ta = 130 65 = 130(1 0.11)tr = 0.11 65 = 130(0.89)ty = 65 0.5 = (0.89)tt = ??? log(0.5) = log(0.89)t log(0.5) = tlog(0.89) Power Property log(0.5) = tlog(0.89) 5.9480 t

• 2. Natural rate:

• 2. Natural rate: y = ae-kt

• 2. Natural rate: y = ae-kta = original amountk = constant of variation t = time y = new amount

• 2. Natural rate: y = ae-kt a = original amountk = constant of variation t = time y = new amountEx. 2 Determine the half-life of Carbon-14 if its constant of variation is 0.00012.

• 2. Natural rate: y = ae-kt a = original amountk = constant of variation t = time y = new amountEx. 2 Determine the half-life of Carbon-14 if its constant of variation is 0.00012.*No rate given so must be Natural.

• 2. Natural rate: y = ae-kt a = original amountk = constant of variation t = time y = new amountEx. 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

• 2. Natural rate: y = ae-kt a = original amountk = constant of variation t = time y = new amountEx. 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-kta = 1

• 2. Natural rate: y = ae-kt a = original amountk = constant of variation t = time y = new amountEx. 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-kta = 1y = 0.5

• 2. Natural rate: y = ae-kt a = original amountk = constant of variation t = time y = new amountEx. 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-kta = 1y = 0.5k = 0.00012

• 2. Natural rate: y = ae-kt a = original amountk = constant of variation t = time y = new amountEx. 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-kta = 1y = 0.5k = 0.00012t = ???

• 2. Natural rate: y = ae-kt a = original amountk = constant of variation t = time y = new amountEx. 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-kta = 10.5 = 1e-0.00012ty = 0.5k = 0.00012t = ???

• 2. Natural rate: y = ae-kt a = original amountk = constant of variation t = time y = new amountEx. 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-kta = 10.5 = 1e-0.00012ty = 0.50.5 = e-0.00012tk = 0.00012t = ???

• 2. Natural rate: y = ae-kt a = original amountk = constant of variation t = time y = new amountEx. 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-kta = 10.5 = 1e-0.00012ty = 0.50.5 = e-0.00012tk = 0.00012 ln(0.5) = ln e-0.00012tt = ???

• 2. Natural rate: y = ae-kt a = original amountk = constant of variation t = time y = new amountEx. 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-kta = 10.5 = 1e-0.00012ty = 0.50.5 = e-0.00012tk = 0.00012 ln(0.5) = ln e-0.00012tt = ??? ln(0.5) = -0.00012t

• 2. Natural rate: y = ae-kt a = original amountk = constant of variation t = time y = new amountEx. 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-kta = 10.5 = 1e-0.00012ty = 0.50.5 = e-0.00012tk = 0.00012 ln(0.5) = ln e-0.00012tt = ??? ln(0.5) = -0.00012t ln(0.5) = t-0.00012

• 2. Natural rate: y = ae-kt a = original amountk = constant of variation t = time y = new amountEx. 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-kta = 10.5 = 1e-0.00012ty = 0.50.5 = e-0.00012tk = 0.00012 ln(0.5) = ln e-0.00012tt = ??? ln(0.5) = -0.00012t ln(0.5) = t-0.000125,776 t

• 2. Natural rate: y = ae-kt a = original amountk = constant of variation t = time y = new amountEx. 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-kta = 10.5 = 1e-0.00012ty = 0.50.5 = e-0.00012tk = 0.00012 ln(0.5) = ln e-0.00012tt = ??? ln(0.5) = -0.00012t ln(0.5) = t-0.000125,776 t*It takes about 5,776 years for Carbon-14 to decay to half of its original amount.

• Growth:

• Growth:1. Fixed Rate:

• Growth:1. Fixed Rate: y = a(1 + r)t

• 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?

• 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

• 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

• 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)10y = 100,000(1.04)10

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

• 2. Natural Rate:

• 2. Natural Rate: y = aekt

• Natural Rate: y = aektEx. 4 The population of Indianapolis, IN was 781,870 in 2000. It then rose to 784,118 by 2005.

• Natural Rate: y = aektEx. 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.

• Natural Rate: y = aektEx. 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 = aekt

• Natural Rate: y = aektEx. 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 = aekt784,118 = 781,870e5k

• Natural Rate: y = aektEx. 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 = aekt784,118 = 781,870e5k1.0029 = e5k

• Natural Rate: y = aektEx. 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 = aekt784,118 = 781,870e5k1.0029 = e5kln(1.0029) = ln e5k

• Natural Rate: y = aektEx. 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 = aekt784,118 = 781,870e5k1.0029 = e5kln(1.0029) = ln e5k ln(1.0029) = 5k

• Natural Rate: y = aektEx. 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 = aekt784,118 = 781,870e5k1.0029 = e5kln(1.0029) = ln e5k ln(1.0029) = 5kln(1.0029) = k 5

• Natural Rate: y = aektEx. 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 = aekt784,118 = 781,870e5k1.0029 = e5kln(1.0029) = ln e5k ln(1.0029) = 5kln(1.0029) = k 5 0.000579 = k

• Natural Rate: y = aektEx. 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 = aekt784,118 = 781,870e5k1.0029 = e5kln(1.0029) = ln e5k ln(1.0029) = 5kln(1.0029) = k 5 0.000579 = k y = ae0.000579t

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

• b. Use your equation to predict the population of Indianapolis in 2010. y = ae0.000579t

• 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 = ae0.000579ty = 781,870e0.000579(10)

• 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 = ae0.000579ty = 781,870e0.000579(10)y 786,410

• 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 = ae0.000579ty = 781,870e0.000579(10)y 786,410

Info obtained from http://www.idcide.com/citydata/in/indianapolis.htm