5.3 Applications of Exponential Functions

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5.3 Applications of Exponential Functions

• Objective
• Create and use exponential models for a variety
  of exponential growth and decay application
  problems

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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?

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Example 1 Solution

• After one year, the account balance is
• 9000 .04(9000) Principal Interest
• 9000(10.04) 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

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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)510,949.88

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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.
• NOTE You are expected to know this formula!

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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

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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)107990.66
• 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

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Example 3 Solving for the Time Period

• If 7000 is invested at 5 annual interest,
  compounded monthly, when will the investment be
  worth 8500?

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Example 3 Solution

• If 7000 is invested at 5 annual interest,
  compounded monthly, when will the investment be
  worth 8500?
• 85007000(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.

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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.

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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 (11/n)n
• Now observe what happens to the final amount as n
  grows larger and larger

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Continuous Compounding and the Number e
Compounding Period n (11/n)n
Annually 1
Semiannually 2
Quarterly 4
Monthly 12
Daily 365
Hourly 8760
Every Minute 525,600
Every Second 31,536,000
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Continuous Compounding and the Number e
Compounding Period n (11/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.
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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 P1, r100, and t1. A
  similar result occurs in the general case and
  leads to the following formula
• APert
• NOTE You are expected to know this formula!

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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?

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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?
• A3500e(.03)(4)3946.24

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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 a1r or a1-r (depending on
  the type of change) and f(x)Pax can be written
  as
• f(x)P(1r) x or f(x)P(1-r) x
• NOTE You are expected to know this formula!

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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 x0
  corresponds to 2000.

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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 x0
  corresponds to 2000.
• f(x)26.4(1.0019)x

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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?

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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 ? a1.91/6
• 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.

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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?

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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)x0.85x

• We want to know when 60 has evaporated or,
  when the evaporation has left 40.
• f(x)0.85x
• 0.400.85x

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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