Transcript Section 4.6 - Shelton State Community College

Section 4.6
Applications and Models:
Growth and Decay
Population Growth
• The function P (t ) = P0e kt, k > 0 can model many
kinds of population growths.
In this function:
P0 = population at time 0,
P = population after time,
t = amount of time,
k = exponential growth rate.
The growth rate unit must be the same as the
time unit.
Example
• Population Growth of the United States.
In 1990 the population in the United States
was about 249 million and the exponential
growth rate was 8% per decade. (Source:
U.S. Census Bureau)
– Find the exponential growth function.
– What will the population be in 2020?
– After how long will the population be double
what it was in 1990?
Solution
•
At t = 0 (1990), the population was about 249 million.
We substitute 249 for P0 and 0.08 for k to obtain the
exponential growth function.
•
In 2020, 3 decades later, t = 3. To find the population in
2020 we substitute 3 for t:
The population will be approximately
million in 2020.
Solution continued
•
We are looking for the doubling time T.
Interest Compound Continuously
• The function P (t ) = P0e kt can be used to
calculate interest that is compounded
continuously.
In this function:
P0 = amount of money invested,
P = balance of the account,
t = years,
k = interest rate compounded continuously.
Example
• Suppose that $2000 is deposited into an
IRA at an interest rate k, and grows to
$5889.36 after 12 years.
–
–
–
–
What is the interest rate?
Find the exponential growth function.
What will the balance be after the first 5 years?
How long did it take the $2000 to double?
Solution
•
At t = 0, P (0) = P0 = $2000.
Thus the exponential growth function is
Solution continued
Solution continued
•
The exponential growth function is
•
The balance after 5 years is
Solution continued
Growth Rate and Doubling Time
The growth rate k and the doubling time
T are related by
kT = ln 2
ln 2
– k
T
or
or
– T  ln 2
k
* The relationship between k and T does not
depend on P0.
Example
• A certain town’s population is doubling
every 37.4 years.
What is the exponential growth rate?
Solution:
ln 2 ln 2
k

 1.9%
T
37.4
Models of Limited Growth
In previous examples, we have modeled population
growth. However, in some populations, there can be
factors that prevent a population from exceeding some
limiting value.
a
P(t ) 
 kt
One model of such growth is
1  be
which is called a logistic function.
This function increases toward a limiting value a
as t approaches infinity.
Thus, y = a is the horizontal asymptote of the graph.
Exponential Decay
• Decay, or decline, of a population is represented by
the function P (t ) = P0e kt, k > 0.
In this function:
P0 = initial amount of the substance,
P = amount of the substance left after time,
t = time,
k = decay rate.
• The half-life is the amount of time it takes for half
of an amount of substance to decay.
Graphs
Example
Carbon Dating. The radioactive element carbon-14
has a half-life of 5750 years. If a piece of charcoal
that had lost 7.3% of its original amount of carbon,
was discovered from an ancient campsite, how
could the age of the charcoal be determined?
Solution: The function for carbon dating is
P (t ) = P0e -0.00012t.
If the charcoal has lost 7.3% of its carbon-14 from
its initial amount P0, then 92.7%P0 is the amount
present.
Example continued
To find the age of the charcoal, we solve the equation
for t :