Transcript 5.3 Applications of Exponential Functions

5.3 Applications of
Exponential Functions
Objective:
•Create and use exponential models for a
variety of exponential growth and decay
application problems
Compound Interest

When interest is paid on a balance that
includes interest accumulated from the
previous time periods it is called compound
interest.

Example 1:
– If you invest $9000 at 4% interest,
compounded annually, how much is in the
account at the end of 5 years?
Example 1: Solution

After one year, the account balance is
– 9000 + .04(9000)
Principal + Interest
– 9000(1+0.04)
Factor out 9000
– 9000(1.04)
Simplify (104% of Principal)
– $9360
Evaluate

Note: The account balance changed by a factor of
1.04. If this amount is left in the account, that
balance will change by a factor of 1.04 after the
second year.
– 9360(1.04) OR…
– 9000(1.04)(1.04)= 9000(1.04)2
Example 1: Solution

Continuing with this pattern shows that
the account balance at the end of t
years can be modeled by the function
B(t)=9000(1.04)t.

Therefore, after 5 years, an investment
of $9000 at 4% interest will be:
– B(5)=9000(1.04)5=$10,949.88
Compound Interest
Formula

If P dollars is invested at interest rate r
(expressed as a decimal) per time
period t, compounded n times per
period, then A is the amount after t
periods.
A  P1  r 
t
**NOTE: You are expected to know this formula!**
Example 2:
Different Compounding Periods

Determine the amount that a $5000 investment over ten
years at an annual interest rate of 4.8% is worth for each
compounding period.
– NOTE: Interest rate per period and the number of periods
may be changing!

A. annually

B. quarterly

C. monthly

D. daily
Example 2: Solution

Determine the amount that a $5000
investment over ten years at an annual
interest rate of 4.8% is worth for each
compounding period.

A. annually  A = 5000(1+.048)10=$7990.66

B. quarterly  A = 5000(1+.048/4)10(4)=$8057.32

C. monthly  A = 5000(1+.048/12)10(12)=$8072.64

D. daily  A = 5000(1+.048/365)10(365)=$8080.12
Example 3: Solving for the
Time Period

If $7000 is invested at 5% annual
interest, compounded monthly, when
will the investment be worth $8500?
Example 3: Solution

If $7000 is invested at 5% annual interest,
compounded monthly, when will the
investment be worth $8500?

8500=7000(1+.05/12)t

Graphing each side of the equation allows
us to use the Intersection Method to
determine when they are equal.

After about 47 months, or 3.9 years, the
investment will be worth $8500.
Continuous Compounding
and the Number e

As the previous examples have shown, the
more often interest is compounded, the larger
the final amount will be. However, there is a
limit that is reached.

Consider the following example:
– Example 4: Suppose you invest $1 for one
year at 100% annual interest, compounded n
times per year. Find the maximum value of
the investment in one year.
Continuous Compounding
and the Number e

The annual interest rate is 1, so the
interest rate period is 1/n, and the
number of periods is n.
– A = (1+1/n)n

Now observe what happens to the final
amount as n grows larger and larger…
Continuous Compounding
and the Number e
Compounding Period
n
Annually
1
Semiannually
2
Quarterly
4
Monthly
12
Daily
365
Hourly
8760
Every Minute
525,600
Every Second
31,536,000
(1+1/n)n
Continuous Compounding
and the Number e
Compounding Period
n
(1+1/n)n
Annually
1
=2
Semiannually
2
=2.25
Quarterly
4
≈2.4414
Monthly
12
≈2.6130
Daily
365
≈2.71457
Hourly
8760
≈2.718127
Every Minute
525,600
≈2.7182792
Every Second
31,536,000
≈2.7182825
The maximum amount of the $1 investment after one
year is approximately $2.72, no matter how large n is.
Continuous Compounding
and the Number e

When the number of compounding periods increases without
bound, the process is called continuous compounding.
(This suggests that n, the compounding period, approaches
infinity.) Note that the last entry in the preceding table is the
same as the number e to five decimal places. This example is
the case where P=1, r=100%, and t=1. A similar result occurs
in the general case and leads to the following formula:
A=Pert
**NOTE: You are expected to know this formula!**
Example 5: Continuous
Compounding

If you invest $3500 at 3% annual
interest compounded continuously,
how much is in the account at the end
of 4 years?
Example 5: Solution

If you invest $3500 at 3% annual
interest compounded continuously,
how much is in the account at the end
of 4 years?

A=3500e(.03)(4)=$3946.24
Exponential Growth and Decay

Exponential growth or decay can be described
by a function of the form f(x)=Pax where f(x) is
the quantity at time x, P is the initial quantity,
and a is the factor by which the quantity
changes (grows or decays) when x increases
by 1.

If the quantity f(x) is changing at a rate r per
time period, then a=1+r or a=1-r (depending on
the type of change) and f(x)=Pax can be written
as
Example 6: Population
Growth

The population of Tokyo, Japan, in the
year 2000 was about 26.4 million and
is projected to increase at a rate of
approximately 0.19% per year. Write
the function that gives the population
of Tokyo in year x, where x=0
corresponds to 2000.
Example 6: Solution

The population of Tokyo, Japan, in the
year 2000 was about 26.4 million and
is projected to increase at a rate of
approximately 0.19% per year. Write
the function that gives the population
of Tokyo in year x, where x=0
corresponds to 2000.
Example 7: Population
Growth

A newly formed lake is stocked with
900 fish. After 6 months, biologists
estimate there are 1710 fish in the
lake. Assuming the fish population
grows exponentially, how many fish
will there be after 24 months?
Example 7: Solution

Using the formula f(x)=Pax, we have f(x)=900ax,
which leaves us to determine a. Use the other
given data about the population growth to
determine a.

In 6 months, there were 1710 fish:
– f(6)=1710  1710= 900a6  a=1.91/6

f(x)=900(1.91/6)x

f(24)= 900(1.91/6)24 = 11,728.89

There will be about 11,729 fish in the lake after 24
months.
Example 8: Chlorine
Evaporation
Each day, 15% of the chlorine in a
swimming pool evaporates. After how
many days will 60% of the chlorine
have evaporated?
Example 8: Solution

Since 15% of the chlorine evaporates
each day, there is 85% remaining. This is
the rate, or a value.
 f(x)=(1-0.15)x=0.85x
• We want to know when 60% has evaporated;
or, when the evaporation has left 40%.
 f(x)=0.85x
 0.40=0.85x
Example 8: Solution
• Solve for x by finding the point of
intersection.
• x ≈ 5.638; After about 6 days, 60% of
the chlorine has evaporated