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? (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 Value Principal (P) – initial amount invested/deposited or Present Value Interest (I) – percentage earned for t period I = PrtA = 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 earn 8% for the next year. How much will Annie have in her account at the end of the 2 nd 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)= after 3 years? $1000(1.08)(1.08) (1.08) = ( ) (1080) end of 1 st year end of 2 nd year Compound Interest – interest that is paid on both the original principal and the accumulated interest. end of 2 nd year end of 3 rd year A = P(1 + rt) Simple Interest A = 1000( ·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 a 1, a 2, a 3, a 4,..... a n,... 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.

Aim: Compound Interest Course: Math Literacy The n th 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 n th Term of a Geometric Sequence a 1 a 2 a 3 a a n.... a 1 a 1 r a 1 r 2 a 1 r a n = a 1 r n – 1, a 1 r n

Aim: Compound Interest Course: Math Literacy Compound Interest &Geometric Sequence $1000(1.08) = $1000(1.08)(1.08) = $1000(1.08)(1.08)(1.08) end of 1 st year end of 2 nd year end of 3 rd year = 1080 a n = a 1 r n – = 1000(1.08) 4 – = 1000(1.08) 3 end of 3 rd year $1000 initial deposit = $1000 a1a1 a2a2 a3a3 a4a4 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 a 1, a 1 r 2, a 1 r 3, a 1 r 4,.... a 1 r n with common ratio r  1 is given by

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) = After 3 years, Annie had $ (1.08) 3 = Post growth A, Pre-growth P rate r, time t y = a b x A = P(1 + r) t Exponential function recall: In general terms A = P(1 + rt) Simple Interest

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 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 be 260,000,000( ) The growth factor is %  = 261,820,000 c. y = 260,000,000( ) 8 = 274,921,758

Aim: Compound Interest Course: Math Literacy Exponential Growth and Decay Exponential decay in general terms Post decay y, Pre-decay P rate r of decay, time t y = a b x Exponential function Post growth y, Pre-growth P rate r of growth, time t Exponential growth in 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 car depreciates by 11% a year. What is the car’s value after one year? = $19,135 Value after 1 yeardepreciation depreciation rate 21,500 Original value y = P(1 - r) t Exponential decay in general terms (0.08) = $1080 principal End of year balance interest earned or 1000(1.08) Recall interest problem - a growth problem: 21,500( ) or 21,500(0.89)= $19,135 Post decay y, Pre-decay P rate 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 car Depreciates by 11% a year. What is the car’s Value after two years? y = P(1 - r) t Exponential decay in general terms Post decay y, Pre-decay P rate r of decay, time t y = 21,500( ) 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 car depreciates by 14% a year. What is the car’s value after four years? y = P(1 - r) t Exponential decay in general terms Post decay y, Pre-decay P rate r of decay, time t y = 32,950( ) 4 y = 32,950(0.86) 4 y = $18,023.92

Aim: Compound Interest Course: Math Literacy More Money in the Bank (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 = (1.02)(1.02) = End of 2nd quarter 1000(1.02)(1.02)(1.02) = End of 3rd quarter 1000(1.02)(1.02)(1.02)(1.02) = End 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) = End of 2nd quarter (1.02) = End of 3rd quarter (1.02) = End of 4th quarter 1000(1.02) = 1020 End of 1st quarter Principal + r/n) 4 = Ending Balance or A x (1 If Annie decides to keep her money in the bank for two years how much would she then have? 1000( ) 4 2 = Balance A, Principal P, rate r, # of compoundings n, time t

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 = $2, = $2, = $ = $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.

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 1 st year Additions Value at end of Quarter Jan 1 April 1 July 1 Oct 1 Account balance at end of year $ The sum represents a finite geometric series where a 1 = 25.13, r = and n = 4

Aim: Compound Interest Course: Math Literacy Present Value 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?

Aim: Compound Interest Course: Math Literacy Model Problem 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?