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)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 fixedrate 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 fixedrate 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 fixedrate 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 fixedrate 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 fixedrate 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 fixedrate 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 fixedrate 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 fixedrate 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 fixedrate 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 fixedrate 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 fixedrate 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 fixedrate 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 fixedrate 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 fixedrate 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 = aekt

2. Natural rate: y = aekta = original amountk = constant of variation t = time y = new amount

2. Natural rate: y = aekt a = original amountk = constant of variation t = time y = new amountEx. 2 Determine the halflife of Carbon14 if its constant of variation is 0.00012.

2. Natural rate: y = aekt a = original amountk = constant of variation t = time y = new amountEx. 2 Determine the halflife of Carbon14 if its constant of variation is 0.00012.*No rate given so must be Natural.

2. Natural rate: y = aekt a = original amountk = constant of variation t = time y = new amountEx. 2 Determine the halflife of Carbon14 if its constant of variation is 0.00012.*No rate given so must be Natural.y = aekt

2. Natural rate: y = aekt a = original amountk = constant of variation t = time y = new amountEx. 2 Determine the halflife of Carbon14 if its constant of variation is 0.00012.*No rate given so must be Natural.y = aekta = 1

2. Natural rate: y = aekt a = original amountk = constant of variation t = time y = new amountEx. 2 Determine the halflife of Carbon14 if its constant of variation is 0.00012.*No rate given so must be Natural.y = aekta = 1y = 0.5

2. Natural rate: y = aekt a = original amountk = constant of variation t = time y = new amountEx. 2 Determine the halflife of Carbon14 if its constant of variation is 0.00012.*No rate given so must be Natural.y = aekta = 1y = 0.5k = 0.00012

2. Natural rate: y = aekt a = original amountk = constant of variation t = time y = new amountEx. 2 Determine the halflife of Carbon14 if its constant of variation is 0.00012.*No rate given so must be Natural.y = aekta = 1y = 0.5k = 0.00012t = ???

2. Natural rate: y = aekt a = original amountk = constant of variation t = time y = new amountEx. 2 Determine the halflife of Carbon14 if its constant of variation is 0.00012.*No rate given so must be Natural.y = aekta = 10.5 = 1e0.00012ty = 0.5k = 0.00012t = ???

2. Natural rate: y = aekt a = original amountk = constant of variation t = time y = new amountEx. 2 Determine the halflife of Carbon14 if its constant of variation is 0.00012.*No rate given so must be Natural.y = aekta = 10.5 = 1e0.00012ty = 0.50.5 = e0.00012tk = 0.00012t = ???

2. Natural rate: y = aekt a = original amountk = constant of variation t = time y = new amountEx. 2 Determine the halflife of Carbon14 if its constant of variation is 0.00012.*No rate given so must be Natural.y = aekta = 10.5 = 1e0.00012ty = 0.50.5 = e0.00012tk = 0.00012 ln(0.5) = ln e0.00012tt = ???

2. Natural rate: y = aekt a = original amountk = constant of variation t = time y = new amountEx. 2 Determine the halflife of Carbon14 if its constant of variation is 0.00012.*No rate given so must be Natural.y = aekta = 10.5 = 1e0.00012ty = 0.50.5 = e0.00012tk = 0.00012 ln(0.5) = ln e0.00012tt = ??? ln(0.5) = 0.00012t

2. Natural rate: y = aekt a = original amountk = constant of variation t = time y = new amountEx. 2 Determine the halflife of Carbon14 if its constant of variation is 0.00012.*No rate given so must be Natural.y = aekta = 10.5 = 1e0.00012ty = 0.50.5 = e0.00012tk = 0.00012 ln(0.5) = ln e0.00012tt = ??? ln(0.5) = 0.00012t ln(0.5) = t0.00012

2. Natural rate: y = aekt a = original amountk = constant of variation t = time y = new amountEx. 2 Determine the halflife of Carbon14 if its constant of variation is 0.00012.*No rate given so must be Natural.y = aekta = 10.5 = 1e0.00012ty = 0.50.5 = e0.00012tk = 0.00012 ln(0.5) = ln e0.00012tt = ??? ln(0.5) = 0.00012t ln(0.5) = t0.000125,776 t

2. Natural rate: y = aekt a = original amountk = constant of variation t = time y = new amountEx. 2 Determine the halflife of Carbon14 if its constant of variation is 0.00012.*No rate given so must be Natural.y = aekta = 10.5 = 1e0.00012ty = 0.50.5 = e0.00012tk = 0.00012 ln(0.5) = ln e0.00012tt = ??? ln(0.5) = 0.00012t ln(0.5) = t0.000125,776 t*It takes about 5,776 years for Carbon14 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