Transcript 10.6 – 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
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 fixed-rate 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
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
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
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=
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
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 = ???
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
y = 65
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 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
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 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 = ???
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
Decay:
1. Fixed rate:
where
y = a(1 – r)t
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
Decay:
1. Fixed rate:
where
y = a(1 – r)t
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)
Decay:
1. Fixed rate:
where
y = a(1 – r)t
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
2. Natural rate:
2. Natural rate: y = ae-kt
2. Natural rate: y = ae-kt
a = original amount
k = constant of variation
t = time
y = new amount
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.
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.’
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
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
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
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
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 = ???
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
k = 0.00012
t = ???
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
t = ???
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 = ???
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
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
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
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.
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) = 5k
ln(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) = 5k
ln(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) = 5k
ln(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/india
napolis.htm