8.8 – Exponential Growth & Decay. Decay: 1. Fixed rate.

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Transcript of 8.8 – Exponential Growth & Decay. Decay: 1. Fixed rate.
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
Decay:
1. Fixed rate: y = a(1 – r)t
where a = original amount
r = rate of decrease
t = time
Decay:
1. Fixed rate: y = a(1 – r)t
where a = original amount
r = rate of decrease
t = time
y = new amount
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?
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 fixedrate 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 fixedrate decay.
y = a(1 – r)t
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 fixedrate decay.
y = a(1 – r)t
a = 130
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 fixedrate decay.
y = a(1 – r)t
a = 130r = 0.11
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 fixedrate decay.
y = a(1 – r)t
a = 130r = 0.11y =
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 fixedrate decay.
y = a(1 – r)t
a = 130r = 0.11y = 65
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 fixedrate decay.
y = a(1 – r)t
a = 130 r = 0.11y = 65t = ???
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 fixedrate decay.
y = a(1 – r)t
a = 130 65 = 130(1 – 0.11)t
r = 0.11y = 65t = ???
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 fixedrate decay.
y = a(1 – r)t
a = 130 65 = 130(1 – 0.11)t
r = 0.11 65 = 130(0.89)t
y = 65t = ???
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 fixedrate 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 = ???
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 fixedrate 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
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 fixedrate 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
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 fixedrate 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)
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 fixedrate 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
2. Natural rate:
2. Natural rate: y = aekt
2. Natural rate: y = aekt
a = original amount
k = constant of variation
t = time
y = new amount
2. Natural rate: y = aekt
a = original amount
k = constant of variation
t = time
y = new amount
Ex. 2 Determine the halflife of Carbon14 if it’s constant of variation is 0.00012.
2. Natural rate: y = aekt
a = original amount
k = constant of variation
t = time
y = new amount
Ex. 2 Determine the halflife of Carbon14 if it’s constant of variation is 0.00012.
*No rate given so must be ‘Natural.’
2. Natural rate: y = aekt
a = original amount
k = constant of variation
t = time
y = new amount
Ex. 2 Determine the halflife of Carbon14 if it’s constant of variation is 0.00012.
*No rate given so must be ‘Natural.’
y = aekt
2. Natural rate: y = aekt
a = original amount
k = constant of variation
t = time
y = new amount
Ex. 2 Determine the halflife of Carbon14 if it’s constant of variation is 0.00012.
*No rate given so must be ‘Natural.’
y = aekt
a = 1
2. Natural rate: y = aekt
a = original amount
k = constant of variation
t = time
y = new amount
Ex. 2 Determine the halflife of Carbon14 if it’s constant of variation is 0.00012.
*No rate given so must be ‘Natural.’
y = aekt
a = 1
y = 0.5
2. Natural rate: y = aekt
a = original amountk = constant of variation
t = time y = new amount
Ex. 2 Determine the halflife of Carbon14 if it’s constant of variation is 0.00012.
*No rate given so must be ‘Natural.’y = aekt
a = 1y = 0.5k = 0.00012
2. Natural rate: y = aekt
a = original amountk = constant of variation
t = time y = new amount
Ex. 2 Determine the halflife of Carbon14 if it’s constant of variation is 0.00012.
*No rate given so must be ‘Natural.’y = aekt
a = 1y = 0.5k = 0.00012t = ???
2. Natural rate: y = aekt
a = original amountk = constant of variation
t = time y = new amount
Ex. 2 Determine the halflife of Carbon14 if it’s constant of variation is 0.00012.
*No rate given so must be ‘Natural.’y = aekt
a = 1 0.5 = 1e0.00012t
y = 0.5k = 0.00012t = ???
2. Natural rate: y = aekt
a = original amountk = constant of variation
t = time y = new amount
Ex. 2 Determine the halflife of Carbon14 if it’s constant of variation is 0.00012.
*No rate given so must be ‘Natural.’y = aekt
a = 1 0.5 = 1e0.00012t
y = 0.5 0.5 = e0.00012t
k = 0.00012t = ???
2. Natural rate: y = aekt
a = original amountk = constant of variation
t = time y = new amount
Ex. 2 Determine the halflife of Carbon14 if it’s constant of variation is 0.00012.
*No rate given so must be ‘Natural.’y = aekt
a = 1 0.5 = 1e0.00012t
y = 0.5 0.5 = e0.00012t
k = 0.00012 ln(0.5) = ln e0.00012t
t = ???
2. Natural rate: y = aekt
a = original amountk = constant of variation
t = time y = new amount
Ex. 2 Determine the halflife of Carbon14 if it’s constant of variation is 0.00012.
*No rate given so must be ‘Natural.’y = aekt
a = 1 0.5 = 1e0.00012t
y = 0.5 0.5 = e0.00012t
k = 0.00012 ln(0.5) = ln e0.00012t
t = ??? ln(0.5) = 0.00012t
2. Natural rate: y = aekt
a = original amountk = constant of variation
t = time y = new amount
Ex. 2 Determine the halflife of Carbon14 if it’s constant of variation is 0.00012.
*No rate given so must be ‘Natural.’y = aekt
a = 1 0.5 = 1e0.00012t
y = 0.5 0.5 = e0.00012t
k = 0.00012 ln(0.5) = ln e0.00012t
t = ??? 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 amount
Ex. 2 Determine the halflife of Carbon14 if it’s constant of variation is 0.00012.
*No rate given so must be ‘Natural.’y = aekt
a = 1 0.5 = 1e0.00012t
y = 0.5 0.5 = e0.00012t
k = 0.00012 ln(0.5) = ln e0.00012t
t = ??? 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 amount
Ex. 2 Determine the halflife of Carbon14 if it’s constant of variation is 0.00012.
*No rate given so must be ‘Natural.’y = aekt
a = 1 0.5 = 1e0.00012t
y = 0.5 0.5 = e0.00012t
k = 0.00012 ln(0.5) = ln e0.00012t
t = ??? 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 it’s 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)10
y = 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)10
y = 100,000(1.04)10
y = $148,024.43
2. Natural Rate:
2. Natural Rate: y = aekt
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.
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.
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
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
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
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
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
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
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
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
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 = ae0.000579t
y = 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 = ae0.000579t
y = 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 = ae0.000579t
y = 781,870e0.000579(10)
y ≈ 786,410
Info obtained from http://www.idcide.com/citydata/in/indianapolis.htm