Growth & Decay
Using Exponential Functions
Growth and Decay Word Problems This equation is used to represent a variety of growth and decay word problems—depending on the type of problem b will change y = ending amount a = initial amount t = time (time number of time cycles) b = base growth (1 + r) (where r = rate) decay (1 – r) half-life (.5) doubles, triples, etc. (2, 3,etc.)
Examples The population of Pasadena in the year 2000 was 142,000 people. Assume that the population is increasing at a rate of 2.25% per year. What will the population be in the year 2030? in 2050?
Examples The population of Leonardville in 1890 was 89,000. Assume the population is decreasing at a rate of 2.75% per year. What was the population in 1920? in 1930?
Examples The population of Bernadetteton in 1985 was 560,000. If the population is increasing at a rate of 3% each year, when will the population reach a million?
Examples The half-life of Howardium-234 is 25 days. If you start with 70 grams initially, how much is left after 100 days? after 400 days?
Examples The half-life of Sheldonium-14 is 5730 years. If you start with 600 grams initially, how much do you have left after 34,380 years?
Examples Amy Farrah Fowler purchased a house in 1990 for $100,000. The annual rate of appreciation is 3.25%. How much will the house be worth in 2010?
Compound Interest p=initial investment r=interest rate n=# of times compounded in a year Annually n=1 Monthly n=12 Quarterly n=4 Weekly n=52 Bimonthly n=24 t=time in years
Compound Interest Example Raj invests $500 into an account earning 7% annual interest compounded monthly. When will his investment be doubled?
Compound Interest Examples Stuart invests $3500 into an account earning 12% annual interest compounded bimonthly. How much will he have after 7 years?
Compound Interest Examples Mrs. Wolowitz invested $10,000 into an account earning 3.6% annual interest compounded weekly. How long did it take for her account to reach $12,000?
Continuously Compounded Interest p=principal (initial investment) e=ex 2.71… r=interest rate t=time
Continuous Compounding Example Penny invests $23,000 into an account earning 6.2% annual interest compounded continuously. How much will she have after 6 years?