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Aim: Compound Interest Course: Math Literacy Aim: How does the exponential model fit into our lives? Do Now:

Aim: Compound Interest Course: Math Literacy Aim: How does the exponential model fit into our lives? Do Now:

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Page 1: Aim: Compound Interest Course: Math Literacy Aim: How does the exponential model fit into our lives? Do Now:

Aim: Compound Interest Course: Math Literacy

Aim: How does the exponential model fit into our lives?

Do Now:

Page 2: Aim: Compound Interest Course: Math Literacy Aim: How does the exponential model fit into our lives? Do Now:

Aim: Compound Interest Course: Math Literacy

Simple Interest – Money in the Bank

Annie deposits $1000 in a local bank at 8% interest for 1 year. How much does Annie have in her account after 1 year?

1000 + 1000(0.08)(1) = $1080

P = A - End of

year balance

+ Prt

= 1000(1.08)

Accumulated Amount (A) – total amount yielded after an amount of time t usually a year, also called Future ValuePrincipal (P) – initial amount invested/deposited or Present ValueInterest (I) – percentage earned for t period

I = Prt A = P + I = P + Prt = P(1 + rt)= P(1 + rt)

Page 3: Aim: Compound Interest Course: Math Literacy Aim: How does the exponential model fit into our lives? Do Now:

Aim: Compound Interest Course: Math Literacy

Compound Interest – More Money!

Annie deposits $1000 in a local bank at 8% interest for 1 year. At the end of the year Annie leaves her accumulated Amount in the bank. The accumulated amount will now earn8% for the next year. How much will Anniehave in her account at the end of the 2nd year?

Compound Interest – interest that is paid on both the original principal and the accumulated interest.

Simple interest - paid only on the initial principal.

Page 4: Aim: Compound Interest Course: Math Literacy Aim: How does the exponential model fit into our lives? Do Now:

Aim: Compound Interest Course: Math Literacy

Money in the Bank 1,2,3 Years

Annie deposits $1000 in a local bank at 8%. Interest is compounded annually. How much will Annie have in her account at the end of the 2nd year?

after 2 years?

$1000(1.08) (1.08) = 1166.40

after 3 years?

$1000(1.08)(1.08) (1.08) = 1259.712(1166.40)

(1080)end of 1st year

end of 2nd year

Compound Interest – interest that is paid on both the original principal and the accumulated interest.

end of 2nd year

end of 3rd year

A = P(1 + rt)Simple Interest

A = 1000(1 + 0.08·2) = 1160

Page 5: Aim: Compound Interest Course: Math Literacy Aim: How does the exponential model fit into our lives? Do Now:

Aim: Compound Interest Course: Math Literacy

Definition of Geometric Sequence

A sequence is geometric if the ratios of consecutive terms are the same.Sequence

a1, a2, a3, a4, . . . . . an, . . .

is geometric if there is a number r, r 0, such that

and so on. The number r is the common ratio of the geometric sequence.

,,,3

4

2

3

1

2 ra

ar

a

ar

a

a

Page 6: Aim: Compound Interest Course: Math Literacy Aim: How does the exponential model fit into our lives? Do Now:

Aim: Compound Interest Course: Math Literacy

The nth term of an geometric sequence has the form

where r is the common ratio between consecutive terms of the sequence. Thus ever geometric sequence can be written in the following form

The nth Term of a Geometric Sequence

a1 a2 a3 a4 . . . . . an . . . .

a1 a1r a1r2 a1r3 . . . .

an = a1rn – 1,

a1rn - 1 . . . .

Page 7: Aim: Compound Interest Course: Math Literacy Aim: How does the exponential model fit into our lives? Do Now:

Aim: Compound Interest Course: Math Literacy

Compound Interest &Geometric Sequence

$1000(1.08)

= 1166.40$1000(1.08)(1.08)

= 1259.712$1000(1.08)(1.08)(1.08)

end of 1st year

end of 2nd year

end of 3rd year

= 1080

an = a1rn – 1

1259.712 = 1000(1.08)4 – 1

1259.712 = 1000(1.08)3end of 3rd year

$1000initial deposit = $1000a1

a2

a3

a4

r = 1.08

Page 8: Aim: Compound Interest Course: Math Literacy Aim: How does the exponential model fit into our lives? Do Now:

Aim: Compound Interest Course: Math Literacy

Find the sum of the first eight terms of the geometric sequence 1, 3, 9, 27, . . .

previous problem

The Sum of a Finite Geometric Sequence

The sum of the finite geometric sequence a1, a1r2, a1r3, a1r4, . . . . a1rn - 1 . . . . with common ratio r 1 is given by

r

raS

n

1

11

31

311

8

S 32802

6560

Page 9: Aim: Compound Interest Course: Math Literacy Aim: How does the exponential model fit into our lives? Do Now:

Aim: Compound Interest Course: Math Literacy

Compound Interest & Exponential Growth

Principal(1 + interest rate)number of years = Ending balance for 3 years

$1000(1.08)(1.08)(1.08) = 1259.712After 3 years, Annie had $1259.71

1000(1.08)3 = 1259.71

Post growth A, Pre-

growth Prate r, time t

y = a • bx

A = P(1 + r)t

Exponential function

recall:

In general terms

A = P(1 + rt)Simple Interest

r

raS

n

1

11

Page 10: Aim: Compound Interest Course: Math Literacy Aim: How does the exponential model fit into our lives? Do Now:

Aim: Compound Interest Course: Math Literacy

Population Growth

The population of the United States in 1994 was 260 million, with an annual growth rate of 0.7%. a. What is the growth factor for the population? b. Suppose the rate of growth continues. Write an equation that models the future growth. c. Predict the population of the U.S. in the year 2002.

b. y = P(1 + r)t where y is the ending population, A is the starting population,

r is the growth factor and t is the number of years.

a. After 1 year the population would be260,000,000(1 + 0.007)

The growth factor is 1.007

0.7% 0.007

= 261,820,000

c. y = 260,000,000(1 + 0.007)8 = 274,921,758

Page 11: Aim: Compound Interest Course: Math Literacy Aim: How does the exponential model fit into our lives? Do Now:

Aim: Compound Interest Course: Math Literacy

Exponential Growth and Decay

Exponential decayin general terms

Post decay y, Pre-decay

Prate r of decay,

time t

y = a • bxExponential function

Post growth y, Pre-

growth Prate r of growth,

time t

Exponential growthin general terms

y = P(1 - r)t

y = P(1 + r)t

b > 1: growth a is initial amount

or value

b < 1: decay a is initial amount

or value

Page 12: Aim: Compound Interest Course: Math Literacy Aim: How does the exponential model fit into our lives? Do Now:

Aim: Compound Interest Course: Math Literacy

Depreciation/Decay

John buys a new car for $21,500. The cardepreciates by 11% a year. What is the car’svalue after one year?

= $19,135Value after 1 yeardepreciation

depreciation rate

21,500 Original value

y = P(1 - r)tExponential decayin general terms

1000 + 1000(0.08) = $1080principal End of year

balanceinterest earned

or 1000(1.08)

Recall interest problem - a growth problem:

21,500(1 - 0.11) or 21,500(0.89) = $19,135

Post decay y, Pre-decay

Prate r of decay,

time t

21,500(0.11)-

Page 13: Aim: Compound Interest Course: Math Literacy Aim: How does the exponential model fit into our lives? Do Now:

Aim: Compound Interest Course: Math Literacy

Depreciation

John buys a new car for $21,500. The carDepreciates by 11% a year. What is the car’sValue after two years?

y = P(1 - r)tExponential decayin general terms

Post decay y, Pre-decay

Prate r of decay,

time ty = 21,500(1 - 0.11)2

y = 21,500(0.89)2

y = 21,500(0.89)2

y = $17,030.15

Page 14: Aim: Compound Interest Course: Math Literacy Aim: How does the exponential model fit into our lives? Do Now:

Aim: Compound Interest Course: Math Literacy

Depreciation

Mary buys a new car for $32,950. The cardepreciates by 14% a year. What is the car’svalue after four years?

y = P(1 - r)tExponential decayin general terms

Post decay y, Pre-decay

Prate r of decay,

time ty = 32,950(1 - 0.14)4

y = 32,950(0.86)4

y = $18,023.92

Page 15: Aim: Compound Interest Course: Math Literacy Aim: How does the exponential model fit into our lives? Do Now:

Aim: Compound Interest Course: Math Literacy

More Money in the Bank

1000 + 1000(0.02) = 1020 End of 1st quarter

Compound interest - paid on the initial principal and previously earned interest.

Annie deposits $1000 in another bank at 8%. interest is compounded quarterly. How much does Annie earn after 1 year?

Since Annie accumulates interest 4 times a year, she earns 2% every 3 months. If r = rate and n is the number of compoundings per year, r/n is the interest earned after each compounding. 0.08/4 = 0.02

1000(1.02)(1.02) = 1040.40 End of 2nd quarter

1000(1.02)(1.02)(1.02) = 1061.21 End of 3rd quarter

1000(1.02)(1.02)(1.02)(1.02) = 1082.43End of 4th quarter

1000(1.02) = 1020 End of 1st quarter

Page 16: Aim: Compound Interest Course: Math Literacy Aim: How does the exponential model fit into our lives? Do Now:

Aim: Compound Interest Course: Math Literacy

Annie’s account

Compound Interest

1020(1.02) = 1040.40 End of 2nd quarter

1040.40(1.02) = 1061.21 End of 3rd quarter

1061.21(1.02) = 1082.43 End of 4th quarter

1000(1.02) = 1020 End of 1st quarter

Principal + r/n)4 = Ending Balance or Ax (1

If Annie decides to keep her money in the bank for two years how much would she then have?

1000(1 + 0.02)4 • 2 = 1171.66

Balance A, Principal P, rate r, # of compoundings n, time t

A P(1 r

n)nt

Page 17: Aim: Compound Interest Course: Math Literacy Aim: How does the exponential model fit into our lives? Do Now:

Aim: Compound Interest Course: Math Literacy

Model Problem

Find the future value of $1000 invested for 10 years at 8%

a) compounded annually

b) compounded semiannually

c) compounded quarterly

d) compounded daily

P = 1000, r = 0,08, t = 10

A P(1 r

n)nt

1 10.08

1000 11

A

= $2,158.92

2 10.08

1000 12

A

= $2,191.12

4 10.08

1000 14

A

= $2.208.04

360 10.08

1000 1360

A

= $2,225.34

Page 18: Aim: Compound Interest Course: Math Literacy Aim: How does the exponential model fit into our lives? Do Now:

Aim: Compound Interest Course: Math Literacy

Model Problem

A certificate of Deposit pays a fixed rate of interest for a term specified in advance, from a month to 10 years. If the best rate available on a 4 year CD is currently is 4.8% compounded monthly, how much is needed to set aside now in such a CD to have $12,000 in four years.

12n

A P(1 r

n)nt 0.048r 4t

12 40.04812000 (1 )

12P

48

120000.048

(1 )12

P

$9907.48

Page 19: Aim: Compound Interest Course: Math Literacy Aim: How does the exponential model fit into our lives? Do Now:

Aim: Compound Interest Course: Math Literacy

Interest Problems

Suppose at the beginning of each quarter you deposit $25 in a savings account that pays an APR of 2% compounded quarterly. Banks post interest at end of quarters. What would be the balance at year’s end?

Date of Deposit

1st year Additions Value at end of

Quarter

Jan 1

April 1

July 1

Oct 1

Account balance at end of year

nt

n

rPA

1

4005.125

3005.125

2005.125

1005.125

25.50

25.38

25.25

25.13

$101.26

The sum represents a finite geometric series where a1 = 25.13, r = 1.005 and n = 4

27.101005.11

005.1113.25

4

S

Page 20: Aim: Compound Interest Course: Math Literacy Aim: How does the exponential model fit into our lives? Do Now:

Aim: Compound Interest Course: Math Literacy

Present Value

(1 ) ntrP A

n

You wish to take a trip to Tahiti in 5 years and you decide that you will need $5000. To have that much money set aside in 5 years, how much money should you deposit now into a bank account paying 6% compounded quarterly?

A P(1 r

n)nt

4 50.065000 (1 )

4P

5000 (1.34685 )P 5000

(1.34685 )P

3712.37P

Page 21: Aim: Compound Interest Course: Math Literacy Aim: How does the exponential model fit into our lives? Do Now:

Aim: Compound Interest Course: Math Literacy

Model Problem

(1 ) ntrP A

n

An insurance agent wishes to sell you a policy that will pay you $100,000 in 30 years. What is the value of this policy in today’s dollars, if we assume a 9% annual inflation rate?

A P(1 r

n)nt

1 300.09100,000(1 )

1P

30100,000(1.09)P

$7,537.11P