Transcript Growth and Decay

4
Inverse,
Exponential,
and
Logarithmic
Functions
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4.6
Applications and Models of
Exponential Growth and Decay
• The Exponential Growth or Decay Function
• Growth Function Models
• Decay Function Models
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Exponential Growth or Decay Function
In many situations that occur in ecology,
biology, economics, and the social
sciences, a quantity changes at a rate
proportional to the amount present. In
such cases the amount present at time t
is a special function of t called an
exponential growth or decay function.
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Exponential Growth or Decay
Function
Let y0 be the amount or number
present at time t = 0. Then, under
certain conditions, the amount present
at any time t is modeled by
y  y 0ekt ,
where k is a constant.
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Exponential Growth or Decay Function
When k > 0, the function describes
growth. In Section 4.2, we saw examples
of exponential growth: compound interest
and atmospheric carbon dioxide, for
example.
When k < 0, the function describes
decay; one example of exponential decay
is radioactivity.
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Example 1
DETERMINING AN EXPONENTIAL
FUNCTION TO MODEL THE INCREASE
OF CARBON DIOXIDE
In Example 11, Section
4.2, we discussed the
growth of atmospheric
carbon dioxide over time.
A function based on the
data from the table was
given in that example.
Now we can see how to
determine such a function
from the data.
Year
1990
2000
2075
2175
2275
Carbon
Dioxide (ppm)
353
375
590
1090
2000
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Example 1
DETERMINING AN EXPONENTIAL
FUNCTION TO MODEL THE INCREASE
OF CARBON DIOXIDE
(a) Find an exponential function that gives the amount
of carbon dioxide y in year x.
Solution
Recall that the graph of the data points showed
exponential growth, so the equation will take
the form y = y0ekt. We must find the values of y0
and k. The data began with the year 1990, so to
simplify our work we let 1990 correspond to
x = 0, 1991 correspond to x = 1, and so on.
Since y0 is the initial amount, y0 = 353 in 1990
when x = 0.
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Example 1
DETERMINING AN EXPONENTIAL
FUNCTION TO MODEL THE INCREASE
OF CARBON DIOXIDE
(a) Find an exponential function that gives the amount
of carbon dioxide y in year x.
Solution
Thus the equation is
y  353e .
kx
From the last pair of values in the table, we know
that in 2275 the carbon dioxide level is expected
to be 2000 ppm. The year 2275 corresponds to
2275 – 1990 = 285. Substitute 2000 for y and
285 for x and solve for k.
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Example 1
DETERMINING AN EXPONENTIAL
FUNCTION TO MODEL THE INCREASE
OF CARBON DIOXIDE
Solution y  353ekx
2000  353e
k (285)
2000
 e285 k
353
 2000 
285 k
In 

In
e



 353 
 2000 
In 
  285k
 353 
Divide by 353.
Take the logarithm on
each side.
In ex = x, for all x.
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Example 1
DETERMINING AN EXPONENTIAL
FUNCTION TO MODEL THE INCREASE
OF CARBON DIOXIDE
Solution
1
 2000
k
In
 353 
285
k  0.00609
1
Multiply by
and
285
rewrite.
Use a calculator.
A function that models the data is
y  353e0.00609 x .
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Example 1
DETERMINING AN EXPONENTIAL
FUNCTION TO MODEL THE INCREASE
OF CARBON DIOXIDE
(b) Estimate the year when future levels of carbon
dioxide will be double the preindustrial level of
280 ppm.
Solution
Let y = 2(280) = 560 in y = 353e0.00609x , and find x.
560  353e
560
0.00609 x
e
353
0.00609 x
 560
0.00609 x
In
 In e
 353
Divide by 353.
Take the logarithm on
each side.
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DETERMINING AN EXPONENTIAL
FUNCTION TO MODEL THE INCREASE
OF CARBON DIOXIDE
Example 1
Solution
 560
In
 0.00609 x
 353
1
 560
x
In
 353 
0.00609
x  75.8
In ex = x, for all x.
1
Multiply by 0.00609 and
rewrite.
Use a calculator.
Since x = 0 corresponds to 1990, the preindustrial
carbon dioxide level will double in the 75th year
after 1990, or during 2065, according to this model.
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FINDING DOUBLING TIME FOR
MONEY
Example 2
How long will it take for the money in an account
that accrues interest at a rate of 3%, compounded
continuously, to double?
Solution
A  P er t
2P  P e0.03 t
2e
0.03 t
In 2  In e
0.03 t
Continuous compounding
formula
Let A = 2P and r = 0.03.
Divide by P.
Take the logarithm on
each side.
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FINDING DOUBLING TIME FOR
MONEY
Example 2
How long will it take for the money in an account
that accrues interest at a rate of 3%, compounded
continuously, to double?
Solution
In 2  0.03t
In ex = x
In 2
t
0.03
Divide by 0.03
23.10  t
Use a calculator.
It will take about 23 yr for the amount to double.
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Example 3
DETERMINING AN EXPONENTIAL FUNCTION
TO MODEL POPULATION GROWTH
According to the U.S. Census Bureau, the
world population reached 6 billion people
during 1999 and was growing
exponentially. By the end of 2010, the
population had grown to 6.947 billion. The
projected world population (in billions of
people) t years after 2010, is given by the
function
f (t )  6.947e0.00745t .
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Example 3
DETERMINING AN EXPONENTIAL FUNCTION
TO MODEL POPULATION GROWTH
(a) Based on this model, what will the world population
be in 2020?
Solution
Since t = 0 represents the year 2010, in 2020, t
would be 2020 – 2010 = 10 yr. We must find (t)
when t is 10.
f (t )  6.947e0.00745 t
f (10)  6.947e
0.00745(10 )
Given formula
Let t = 10.
f (10)  7.484
The population will be 7.484 billion at the end of
2020.
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Example 3
DETERMINING AN EXPONENTIAL FUNCTION
TO MODEL POPULATION GROWTH
(b) In what year will the world population reach
9 billion?
Solution
0.00745 t
f (t )  6.947e
9  6.947e
0.00745 t
9
0.00745 t
e
6.947
9
0.00745 t
In
 In e
6.947
Let f(t) = 9.
Divide by 6.947.
Take the logarithm on
each side.
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Example 3
DETERMINING AN EXPONENTIAL FUNCTION
TO MODEL POPULATION GROWTH
(b) In what year will the world population reach
9 billion?
Solution
9
In
 0.00745t
6.947
9
In
t  6.947
0.00745
t  34.8
In ex = x, for all x.
Divide by 0.00745
and rewrite.
Use a calculator.
Thus, 34.8 yr after 2010, during the year 2044,
world population will reach 9 billion.
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Example 4
DETERMINING AN EXPONENTIAL
FUNCTION TO MODEL RADIOACTIVE
DECAY
Suppose 600 g of a radioactive substance are
present initially and 3 yr later only 300 g remain.
(a) Determine the exponential equation that
models this decay.
Solution
To express the situation as an exponential
equation y = y0ekt, we use the given values first to
find y0 and then to find k.
y  y 0ekt
600  y 0ek ( 0 )
600  y 0
Let y = 600 and t = 0.
e0 = 1
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Example 4
DETERMINING AN EXPONENTIAL
FUNCTION TO MODEL RADIOACTIVE
DECAY
Solution
Thus, y0 = 600, which gives y = 600ekt. Because
the initial amount (600 g) decays to half that
amount (300 g) in 3 yr, its half-life is 3 yr. Now we
solve this exponential equation for k.
y  600e
300  600e
kt
3k
0.5  e3 k
In 0.5  In e
Let y = 300 and t = 3.
Divide by 600.
3k
Take logarithms on
each side.
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Example 4
DETERMINING AN EXPONENTIAL
FUNCTION TO MODEL RADIOACTIVE
DECAY
Solution
In 0.5  3k
In ex = x, for all x.
In 0.5
k
3
Divide by 3.
k  0.231
Use a calculator.
Thus, the exponential decay equation is
y = 600e –0.231t .
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Example 4
DETERMINING AN EXPONENTIAL
FUNCTION TO MODEL RADIOACTIVE
DECAY
(b) How much of the substance will be present after
6 yr?
Solution
To find the amount present after 6 yr, let t = 6.
y  600e
0.231( 6 )
 600e
1.386
 150
After 6 yr, 150 g of the substance will remain.
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Example 5
SOLVING A CARBON DATING
PROBLEM
Carbon-14, also known as radiocarbon, is a
radioactive form of carbon that is found in all living
plants and animals. After a plant or animal dies,
the radiocarbon disintegrates. Scientists can
determine the age of the remains by comparing
the amount of radiocarbon with the amount
present in living plants and animals. This
technique is called carbon dating. The amount of
radiocarbon present after t years is given by
y  y 0e
0.0001216 t
,
where y0 is the amount present in living plants
and animals.
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Example 5
SOLVING A CARBON DATING
PROBLEM
(a) Find the half-life of carbon-14.
Solution
If y0 is the amount of radiocarbon present in a
1
living thing, then 2 y0 is half this initial amount.
Thus, we substitute and solve the given
equation for t.
y  y 0e
0.0001216 t
1
y 0  y 0e 0.0001216t
2
Given equation
1
Let y = y0 .
2
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Example 5
SOLVING A CARBON DATING
PROBLEM
(a) Find the half-life of carbon-14.
Solution
1
Divide by y0.
 e 0.0001216t
2
Take the logarithm
1
0.0001216 t
In  In e
on each side.
2
1
In  0.0001216t
2
In ex = x, for all x.
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Example 5
SOLVING A CARBON DATING
PROBLEM
(a) Find the half-life of carbon-14.
Solution
1
In
2
t
0.0001216
5700  t
Divide by –0.0001216.
Use a calculator.
The half-life is about 5700 yr.
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Example 5
SOLVING A CARBON DATING
PROBLEM
(b) Charcoal from an ancient fire pit on Java
contained ¼ the carbon-14 of a living sample of
the same size. Estimate the age of the charcoal.
Solution
Solve again for t, this time letting the amount
1
y = y 0.
4
y  y 0e
0.0001216 t
1
y 0  y 0e 0.0001216t
4
Given equation
1
Let y = y0 .
4
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Example 5
Solution
SOLVING A CARBON DATING
PROBLEM
1
 e 0.0001216t
4
1
0.0001216 t
In  In e
4
1
In
4
t
0.0001216
t  11,400
Divide by y0 .
Take the logarithm
on each side.
In ex = x; Divide by
−0.0001216.
Use a calculator.
The charcoal is about 11,400 yr old.
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Example 6
MODELING NEWTON’S LAW OF
COOLING
Newton’s law of cooling says that the rate at
which a body cools is proportional to the
difference in temperature between the body and
the environment around it. The temperature (t)
of the body at time t in appropriate units after
being introduced into an environment having
constant temperature T0 is
f (t )  T0  Ce kt ,
where C and k are constants.
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Example 6
MODELING NEWTON’S LAW OF
COOLING
A pot of coffee with a temperature of 100°C is set
down in a room with a temperature of 20°C. The
coffee cools to 60°C after 1 hr.
(a) Write an equation to model the data.
Solution
We must find values of C and k in the given formula.
From the information, when t = 0, T0 = 20, and the
temperature of the coffee is f(0) = 100. Also, when
t = 1, f(1) = 60. Substitute the first pair of values into
the function along with T0 = 20.
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MODELING NEWTON’S LAW OF
COOLING
Example 6
(a) Write an equation to model the data.
Solution
 kt
Given formula
f (t )  T0  Ce
100  20  Ce
0 k
100  20  C
Let t = 0, (0) = 100,
and T0 = 20.
e0 = 1
80  C
Subtract 20.
The function that models the data can now be given.
f (t )  20  80e
 kt
Let T0 = 20 and C = 80.
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Example 6
MODELING NEWTON’S LAW OF
COOLING
Solution
Now use the remaining pair of values in this
function to find k.
f (t )  20  80e  kt
60  20  80e 1k
40  80e  k
1 k
e
2
1
k
In  In e
2
Formula with T0 = 20
and C = 80.
Let t = 1 and (1) = 60.
Subtract 20.
Divide by 80.
Take the logarithm
on each side.
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MODELING NEWTON’S LAW OF
COOLING
Example 6
Solution
1
In  k
2
1
k  In
2
k  0.693
In ex = x for all x.
Multiply by –1 and rewrite.
Use a calculator.
Thus, the model is f (t )  20  80e 0.693 t .
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Example 6
MODELING NEWTON’S LAW OF
COOLING
(b) Find the temperature after half an hour.
Solution
To find the temperature after ½ hr, let t = ½ in the
model from part (a).
f (t )  20  80e
0.693 t
Model from part (a)
 1
f
 20  80e( 0.693)(1/ 2 )
 2
Let t = ½ .
 1
f
 76.6C
 2
Use a calculator.
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Example 6
MODELING NEWTON’S LAW OF
COOLING
(c) How long will it take for the coffee to cool to 50°C?
Solution
To find how long it will take for the coffee to cool to
50°C, let (t) = 50.
50  20  80e 0.693 t
30  80e
0.693 t
3
0.693 t
e
8
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Let (t) = 50.
Subtract 20.
Divide by 80.
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Example 6
MODELING NEWTON’S LAW OF
COOLING
(c) How long will it take for the coffee to cool to 50°C?
Solution
3
0.693 t
Take the logarithm on
In  In e
each side.
8
3
In ex = x, for all x.
In  0.693t
8
3
In
8  1.415 hr, or about 1 hr, 25 min
t
0.693
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