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8.8 – Exponential Growth & Decay

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

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Page 1: 8.8 – Exponential Growth & Decay

8.8 – Exponential Growth & Decay

Page 2: 8.8 – Exponential Growth & Decay

Decay:

Page 3: 8.8 – Exponential Growth & Decay

Decay:

1. Fixed rate

Page 4: 8.8 – Exponential Growth & Decay

Decay:

1. Fixed rate: y = a(1 – r)t

Page 5: 8.8 – Exponential Growth & Decay

Decay:

1. Fixed rate: y = a(1 – r)t

where a = original amount

Page 6: 8.8 – Exponential Growth & Decay

Decay:

1. Fixed rate: y = a(1 – r)t

where a = original amount

r = rate of decrease

Page 7: 8.8 – Exponential Growth & Decay

Decay:

1. Fixed rate: y = a(1 – r)t

where a = original amount

r = rate of decrease

t = time

Page 8: 8.8 – Exponential Growth & Decay

Decay:

1. Fixed rate: y = a(1 – r)t

where a = original amount

r = rate of decrease

t = time

y = new amount

Page 9: 8.8 – Exponential Growth & Decay

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?

Page 10: 8.8 – Exponential Growth & Decay

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.

Page 11: 8.8 – Exponential Growth & Decay

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

Page 12: 8.8 – Exponential Growth & Decay

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

Page 13: 8.8 – Exponential Growth & Decay

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 = 130r = 0.11

Page 14: 8.8 – Exponential Growth & Decay

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 = 130r = 0.11y =

Page 15: 8.8 – Exponential Growth & Decay

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 = 130r = 0.11y = 65

Page 16: 8.8 – Exponential Growth & Decay

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 = 0.11y = 65t = ???

Page 17: 8.8 – Exponential Growth & Decay

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 = 0.11y = 65t = ???

Page 18: 8.8 – Exponential Growth & Decay

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 = 65t = ???

Page 19: 8.8 – Exponential Growth & Decay

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 = ???

Page 20: 8.8 – Exponential Growth & Decay

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

Page 21: 8.8 – Exponential Growth & Decay

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

Page 22: 8.8 – Exponential Growth & Decay

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) = tlog(0.89)

Page 23: 8.8 – Exponential Growth & Decay

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) = tlog(0.89) 5.9480 ≈ t

Page 24: 8.8 – Exponential Growth & Decay

2. Natural rate:

Page 25: 8.8 – Exponential Growth & Decay

2. Natural rate: y = ae-kt

Page 26: 8.8 – Exponential Growth & Decay

2. Natural rate: y = ae-kt

a = original amount

k = constant of variation

t = time

y = new amount

Page 27: 8.8 – Exponential Growth & Decay

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.

Page 28: 8.8 – Exponential Growth & Decay

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.’

Page 29: 8.8 – Exponential Growth & Decay

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

Page 30: 8.8 – Exponential Growth & Decay

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

Page 31: 8.8 – Exponential Growth & Decay

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

Page 32: 8.8 – Exponential Growth & Decay

2. Natural rate: y = ae-kt

a = original amountk = 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 = 1y = 0.5k = 0.00012

Page 33: 8.8 – Exponential Growth & Decay

2. Natural rate: y = ae-kt

a = original amountk = 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 = 1y = 0.5k = 0.00012t = ???

Page 34: 8.8 – Exponential Growth & Decay

2. Natural rate: y = ae-kt

a = original amountk = 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 = 0.5k = 0.00012t = ???

Page 35: 8.8 – Exponential Growth & Decay

2. Natural rate: y = ae-kt

a = original amountk = 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 = 0.00012t = ???

Page 36: 8.8 – Exponential Growth & Decay

2. Natural rate: y = ae-kt

a = original amountk = 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 = ???

Page 37: 8.8 – Exponential Growth & Decay

2. Natural rate: y = ae-kt

a = original amountk = 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

Page 38: 8.8 – Exponential Growth & Decay

2. Natural rate: y = ae-kt

a = original amountk = 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

Page 39: 8.8 – Exponential Growth & Decay

2. Natural rate: y = ae-kt

a = original amountk = 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-0.000125,776 ≈ t

Page 40: 8.8 – Exponential Growth & Decay

2. Natural rate: y = ae-kt

a = original amountk = 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-0.000125,776 ≈ t

*It takes about 5,776 years for Carbon-14 to decay to half of it’s original amount.

Page 41: 8.8 – Exponential Growth & Decay

Growth:

Page 42: 8.8 – Exponential Growth & Decay

Growth:

1. Fixed Rate:

Page 43: 8.8 – Exponential Growth & Decay

Growth:

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

Page 44: 8.8 – Exponential Growth & Decay

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?

Page 45: 8.8 – Exponential Growth & Decay

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

Page 46: 8.8 – Exponential Growth & Decay

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

Page 47: 8.8 – Exponential Growth & Decay

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

Page 48: 8.8 – Exponential Growth & Decay

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

Page 49: 8.8 – Exponential Growth & Decay

2. Natural Rate:

Page 50: 8.8 – Exponential Growth & Decay

2. Natural Rate: y = aekt

Page 51: 8.8 – Exponential Growth & Decay

2. Natural Rate: y = aekt

Ex. 4 The population of Indianapolis, IN was 781,870 in 2000. It then rose to 784,118 by 2005.

Page 52: 8.8 – Exponential Growth & Decay

2. Natural Rate: y = aekt

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.

Page 53: 8.8 – Exponential Growth & Decay

2. Natural Rate: y = aekt

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 = aekt

Page 54: 8.8 – Exponential Growth & Decay

2. Natural Rate: y = aekt

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 = aekt

784,118 = 781,870e5k

Page 55: 8.8 – Exponential Growth & Decay

2. Natural Rate: y = aekt

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 = aekt

784,118 = 781,870e5k

1.0029 = e5k

Page 56: 8.8 – Exponential Growth & Decay

2. Natural Rate: y = aekt

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 = aekt

784,118 = 781,870e5k

1.0029 = e5k

ln(1.0029) = ln e5k

Page 57: 8.8 – Exponential Growth & Decay

2. Natural Rate: y = aekt

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 = aekt

784,118 = 781,870e5k

1.0029 = e5k

ln(1.0029) = ln e5k

ln(1.0029) = 5k

Page 58: 8.8 – Exponential Growth & Decay

2. Natural Rate: y = aekt

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 = aekt

784,118 = 781,870e5k

1.0029 = e5k

ln(1.0029) = ln e5k

ln(1.0029) = 5kln(1.0029) = k 5

Page 59: 8.8 – Exponential Growth & Decay

2. Natural Rate: y = aekt

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 = aekt

784,118 = 781,870e5k

1.0029 = e5k

ln(1.0029) = ln e5k

ln(1.0029) = 5kln(1.0029) = k 5 0.000579 = k

Page 60: 8.8 – Exponential Growth & Decay

2. Natural Rate: y = aekt

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 = aekt

784,118 = 781,870e5k

1.0029 = e5k

ln(1.0029) = ln e5k

ln(1.0029) = 5kln(1.0029) = k 5 0.000579 = k y =

ae0.000579t

Page 61: 8.8 – Exponential Growth & Decay

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

Page 62: 8.8 – Exponential Growth & Decay

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

y = ae0.000579t

Page 63: 8.8 – Exponential Growth & Decay

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 = ae0.000579t

y = 781,870e0.000579(10)

Page 64: 8.8 – Exponential Growth & Decay

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 = ae0.000579t

y = 781,870e0.000579(10)

y ≈ 786,410

Page 65: 8.8 – Exponential Growth & Decay

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 = ae0.000579t

y = 781,870e0.000579(10)

y ≈ 786,410

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