C. Applications Involving Annuities and Amortization
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Section 5.6 Applications from Business, Finance, and Science
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To develop an annuity formula, we multiply the annuity equation by 1.08, then
subtract the original equation. This leaves only the first and last terms, since the other
(interior) terms add to zero:
WORTHY OF NOTE
It is often assumed that the first
payment into an annuity is made at
the end of a compounding period,
and hence earns no interest. This is
why the first $100 deposit is not
multiplied by the interest factor.
These terms are actually the terms
of a geometric sequence, which
we will study later in Section 9.3.
1.08A ϭ 10011.082 ϩ 10011.082 2 ϩ 10011.082 3 ϩ 10011.082 4
2
›
›
›
ϪA ϭ Ϫ 3100 ϩ 10011.082 ϩ 10011.082 ϩ 10011.082 4
1
3
1.08A Ϫ A ϭ 10011.082 Ϫ 100
4
0.08A ϭ 100 3 11.082 4 Ϫ 1 4
multiply by 1.08
original equation
subtract (“interior
terms” sum to zero)
factor out 100
100 3 11.082 Ϫ 1 4
4
Aϭ
solve for A
0.08
This result can be generalized for any periodic payment P, interest rate r, number
of compounding periods n, and number of years t. This would give
r nt
P c a1 ϩ b Ϫ 1 d
n
Aϭ
r
n
The formula can be made less formidable using R ϭ nr , where R is the interest rate
per compounding period.
Accumulated Value of an Annuity
If a periodic payment P is deposited n times per year at an annual interest rate r with
interest compounded n times per year for t years, the accumulated value is given by
Aϭ
P
r
3 11 ϩ R2 nt Ϫ 1 4, where R ϭ
n
R
This is also referred to as the future value of the account.
EXAMPLE 5
ᮣ
Solving an Application of Annuities
Since he was a young child, Fitisemanu’s parents have been depositing $50 each
month into an annuity that pays 6% annually and is compounded monthly. If the
account is now worth $9875, how long has it been open?
Solution
ᮣ
Algebraic Solution
ᮣ
In this case P ϭ 50, r ϭ 0.06, n ϭ 12, R ϭ 0.005, and A ϭ 9875.
P
3 11 ϩ R2 nt Ϫ 1 4
R
50
9875 ϭ
3 11.0052 11221t2 Ϫ 1 4
0.005
1.9875 ϭ 1.00512t
ln11.98752 ϭ 12t1ln 1.0052
ln11.98752
ϭt
12 ln11.0052
11.5 Ϸ t
Aϭ
future value formula
substitute given values
simplify and isolate variable term
apply base-e logarithms; power property
solve for t (exact form)
approximate form
The account has been open approximately 11.5 yr.
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Graphical Solution
ᮣ
Here we’ll use the intersection-of-graphs method. Entering
Y1 ϭ a
50
b 11.00512X Ϫ 12 and
0.005
20
Y2 ϭ 9875, we must next determine an
appropriate window size. Since the goal is
0
15,000
$9875, we’ll use [0, 15,000] for y, leaving
a large frame around the window. If no
interest were paid, it would take
9875
0
Ϸ 16.5 yr to save 9875, so [0, 20]
501122
will also give a window size with plenty of room. The result is shown in the figure,
and indicates the account has been open for about 11.5 yr.
Now try Exercises 41 through 44
ᮣ
The periodic payment required to meet a future goal or obligation can be comAR
puted by solving for P in the future value formula: P ϭ
. In this form,
3 11 ϩ R2 nt Ϫ 1 4
P is referred to as a sinking fund.
EXAMPLE 6
ᮣ
Solving an Application of Sinking Funds
Sheila is determined to stay out of debt and decides to save $20,000 to pay cash for a
new car in 4 yr. The best investment vehicle she can find pays 9% compounded
monthly. If $300 is the most she can invest each month, can she meet her “4-yr” goal?
Solution
C. You’ve just seen how
we can solve applications of
annuities and amortization
ᮣ
Here we have P ϭ 300, A ϭ 20,000, r ϭ 0.09, n ϭ 12, and R ϭ 0.0075. The
sinking fund formula gives
AR
sinking fund
Pϭ
3 11 ϩ R2 nt Ϫ 1 4
120,000210.00752
300 ϭ
substitute 300 for P, 20,000 for A, 0.0075
11.00752 12t Ϫ 1 for R, and 12 for n
multiply in numerator, clear denominators
30011.007512t Ϫ 12 ϭ 150
isolate variable term
1.007512t ϭ 1.5
apply base-e logarithms; power property
12t ln 11.00752 ϭ ln1.5
ln11.52
solve for t (exact form)
tϭ
12 ln11.00752
approximate form
Ϸ 4.5
No. She is close, but misses her original 4-yr goal.
Now try Exercises 45 and 46
For Example 6, we could have substituted 4 for
t while leaving P and A unknown, to see if a payment
of $300 per month would be sufficient. Using
x10.00752
Y1 ϭ
and the TABLE feature of a
1.007548 Ϫ 1
calculator shows that just over $17,000 would be saved
for monthly deposits of $300 (Figure 5.52), and that
deposits of $347.70 would be required to save $20,000.
Figure 5.52
ᮣ
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Section 5.6 Applications from Business, Finance, and Science
For additional practice with the formulas for interest earned or paid, the Working
with Formulas portion of this Exercise Set has been expanded. See Exercises 47
through 54.
WORTHY OF NOTE
D. Applications Involving Exponential Growth and Decay
Notice the formula for exponential
growth is virtually identical to the
formula for interest compounded
continuously. In fact, both are
based on the same principles. If we
let A(t) represent the amount in an
account after t years and A0
represent the initial deposit (instead
of P), we have: A1t2 ϭ A0ert versus
Q1t2 ϭ Q0ert and the two cannot be
distinguished.
Closely related to interest compounded continuously are applications of exponential
growth and exponential decay. If Q (quantity) and t (time) are variables, then Q grows
exponentially as a function of t if Q1t2 ϭ Q0ert for positive constants Q0 and r. Careful
studies have shown that population growth, whether it be humans, bats, or bacteria, can
be modeled by these “base-e” exponential growth functions. If Q1t2 ϭ Q0eϪrt, then we
say Q decreases or decays exponentially over time. The constant r determines how rapidly a quantity grows or decays and is known as the growth rate or decay rate constant.
EXAMPLE 7
ᮣ
Solving an Application of Exponential Growth
Because fruit flies multiply very quickly, they are often used in studies of genetics.
Given the necessary space and food supply, a certain population of fruit flies is
known to double every 12 days. If there were 100 flies to begin, find (a) the growth
rate r and (b) the number of days until the population reaches 2000 flies.
ᮢ
Algebraic Solution
a. Using the formula for exponential growth with
Q0 ϭ 100, t ϭ 12, and Q1t2 ϭ 200, we can
solve for the growth rate r.
exponential growth function
Q1t2 ϭ Q0ert
12r
substitute 200 for Q(t ) 100
200 ϭ 100e
for Q0, and 12 for t
12r
isolate variable term
2ϭe
base-e logarithms;
ln 2 ϭ 12r ln e apply
power property
ln 2
ϭr
solve for r (exact form)
12
approximate form
0.05776 Ϸ r
b. To find the number of days until the fly population
reaches 2000, we substitute 0.05776 for r in the
exponential growth function.
Q1t2 ϭ Q0ert
exponential growth function
0.05776t
substitute 2000 for Q(t), 100
2000 ϭ 100e
for Q0, and 0.05776 for r
0.05776t
isolate variable term
20 ϭ e
ln 20 ϭ 0.05776t ln e apply base-e logarithms;
power property
ln 20
ϭt
solve for t (exact form)
0.05776
approximate form
51.87 Ϸ t
The growth rate is approximately 5.78%.
ᮢ
The fruit fly population will reach 2000 on day 51.
Graphical Solution
Figure 5.53
a. After substituting the values given, we input Y1 ϭ 100e and Y2 ϭ 200
to use the intersection-of-graphs method. While growth rates vary widely
for animal populations, we might expect the growth rate to be a decimal
between and 0 and 0.2 (0% to 20%), but can adjust the window afterward
if needed. Since a population of 200 is the target, we can use [0, 300]
for y and [0, 0.2] for x. As seen in Figure 5.53, the graphs intersect at
x Ϸ 0.05776.
300
12x
0
0.2
0
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b. In part (a) we solved for the growth rate r. Here we’ll use this growth
rate and the intersection-of-graphs method to find the time t required
for an initial population of 100 flies to grow to 2000. Begin by entering
100e0.05776X as Y1 and 2000 as Y2. Setting the window for y is no problem,
as we have a target population of 2000. For the x-values, consider the
approximation 100e0.06t, and note that t ϭ 10 gives too small a value
1100e0.6 Ϸ 1822 , while t ϭ 100 gives too large a value 1100e6.0 Ϸ 40,3432 .
Using the window x ʦ 3 0, 100 4 and y ʦ 3 0, 30004 , we find the graphs
intersect at x Ϸ 51.87, and the population of flies will reach 2000 in just
less than 52 days. See Figure 5.54.
Figure 5.54
3000
0
100
0
Now try Exercises 55 and 56
ᮣ
Perhaps the best-known examples of exponential decay involve radioactivity. Ever
since the end of World War II, common citizens have been aware of the existence of
radioactive elements and the power of atomic energy. Today, hundreds of additional
applications have been found for these materials, from areas as diverse as biological
research, radiology, medicine, and archeology. Radioactive elements decay of their
own accord by emitting radiation. The rate of decay is measured using the half-life of
the substance, which is the time required for a mass of radioactive material to decay
until only one-half of its original mass remains. This half-life is used to find the rate of
decay r, first mentioned in Section 5.5. In general, if h represents the half-life of the
substance, one-half the initial amount remains when t ϭ h.
Q1t2 ϭ Q0eϪrt
1
Q0 ϭ Q0eϪrh
2
1
1
ϭ rh
2
e
2 ϭ erh
ln 2 ϭ rh ln e
ln 2
ϭr
h
exponential decay function
substitute 12 Q0 for Q (t ), h for t
divide by Q0; rewrite expression
property of ratios
apply base-e logarithms; power property
solve for r (ln e ϭ 1)
Radioactive Rate of Decay
If h represents the half-life of a radioactive substance per unit time, the nominal rate
of decay per a like unit of time is given by
rϭ
ln 2
h
The rate of decay for known radioactive elements varies greatly. For example, the
element carbon-14 has a half-life of about 5730 yr, while the element lead-211 has a
half-life of only about 3.5 min. Radioactive elements can be detected in extremely
small amounts. If a drug is “labeled” (mixed with) a radioactive element and injected
into a living organism, its passage through the organism can be traced and information
on the health of internal organs can be obtained.