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C. Applications Involving Annuities and Amortization

# C. Applications Involving Annuities and Amortization

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Page 543

College Algebra G&M—

5–65

Section 5.6 Applications from Business, Finance, and Science

543

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

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

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

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

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

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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CHAPTER 5 Exponential and Logarithmic Functions

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

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)

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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College Algebra G&M—

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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

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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CHAPTER 5 Exponential and Logarithmic Functions

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

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)

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

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.

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C. Applications Involving Annuities and Amortization

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