aim: compound interest course: math literacy aim: how does the exponential model fit into our lives?...
TRANSCRIPT
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)
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.
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
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
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 . . . .
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
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
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
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
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
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)-
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
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
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
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
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
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
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
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
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