Aim: How do we model growth and decay using the exponential function?

1. Aim: How do we model growth and decay using the exponential function? Do Now: Annie deposits $1000 in a local bank at 8%. Interest is compounded annually. How Much does Annie earn after 1 year?

2. interest rate Money in the Bank 1,2,3 Years Annie deposits $1000 in a local bank at 8%. Interest is compounded annually. How Much does Annie earn after 1 year? Simple interest - paid only on the initial principal = $1080 1000 + 1000(0.08) or 1000(1.08) + = principal End of year balance interest earned How much does Annie earn after 2 years? $1000(1.08) = 1166.40 (1.08) 1080 How much does Annie earn after 3 years? $1000(1.08)(1.08) (1.08) = 1259.712 1166.40

3. Principal(1 + interest rate)number of years = Ending balance for 3 years recall: Exponential function Exponential Growth After 3 years, Annie had $1259.71 $1000(1.08) (1.08)(1.08) = 1259.712 1000(1.08)3 = 1259.71 In general terms Post growth y, Pre-growth A rate r, time t y = A(1 + r)t y = a • bx

4. 1259.71 t = 3 Exponential Growth - Graph y = a • bx y = A(1 + r)t y = 1000(1 + 0.08)t

5. 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. 0.7%  0.007 After 1 year the population would be 260,000,000(1 + 0.007) The growth factor is 1.007 = 261,820,000 b. y = A(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. c. y = 260,000,000(1 + 0.007)8 = 274,921,758

6. 1000 + 1000(0.02) More Money in the Bank Annie deposits $1000 in another bank at 8%. interest is compounded quarterly. How much does Annie earn after 1 year? Compound interest - paid on the initial principal and previously earned interest. Since Annie earns 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 = 1020 End of 1st quarter = 1020 1000(1.02) End of 1st quarter = 1040.40 1000(1.02(1.02) End of 2nd quarter = 1061.21 1000(1.02)(1.02)(1.02) End of 3rd quarter = 1082.43 1000(1.02)(1.02)(1.02)(1.02) End of 4th quarter

7. Annie’s account Principal If Annie decides to keep her money in the bank for two years how much would she then have? Balance A, Principal P, rate r, # of compoundings n, time t Compound Interest = 1020 1000(1.02) End of 1st quarter = 1040.40 1020(1.02) End of 2nd quarter = 1061.21 1040.40(1.02) End of 3rd quarter = 1082.43 1061.21(1.02) End of 4th quarter = Ending Balance or A x (1 + r/n) 4 1000(1 + 0.02)4 = 1171.66 • 2

8. depreciation rate 21,500 Original value = $1080 1000 + 1000(0.08) or 1000(1.08) Recall interest problem - a growth problem: principal interest earned End of year balance y = A(1 - r)t 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 21,500(0.11) depreciation Value after 1 year 21,500(1 - 0.11) or 21,500(0.89) = $19,135 Post decay y, Pre-decay A rate r of decay, time t Exponential decay in general terms

9. Post decay y, Pre-decay A rate r of decay, time t Exponential decay in general terms y = A(1 - r)t 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 = 21,500(1 - 0.11)2 y = 21,500(0.89)2 y = 21,500(0.89)2 y = $17,030.15

10. Post decay y, Pre-decay A rate r of decay, time t Exponential decay in general terms y = A(1 - r)t 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 = 32,950(1 - 0.14)4 y = 32,950(0.86)4 y = $18,023.92

11. Exponential Growth and Decay Exponential function y = a • bx Post growth y, Pre-growth A rate r of growth, time t Exponential growth in general terms y = A(1 + r)t b > 1: growth a is initial amount or value Post decay y, Pre-decay A rate r of decay, time t Exponential decay in general terms y = A(1 - r)t b < 1: decay a is initial amount or value

12. The b Affect y = a • bx y = (1/2)x y = 2x b > 1 0 < b < 1 y = 1 (0,1) If b > 1, the graph is decreasing - Growth If 0 < b < 1, the graph is decreasing - Decay If b is a positive number other than 1, the graphs of y = bx and y = (1/b)x are reflections through the y-axis of each other

13. x = 1 x = 10000 x = 100 Where’d e Come From? Graph y  2.7183 y  2.7183 is asymptotic to f(x).

14. Exponential function y = a • bx Exponential growth in general terms y = A(1 + r)t Exponential growth Compound Interest Where’d e Come From? or e Leonard Euler

15. e n  Exponential growth Continuous compounding Continuous growth/decay k is a constant (±) Continuous Compounding Exponential function y = a • bx Exponential growth in general terms y = A(1 + r)t Exponential growth Compound Interest

16. Regents Question The formula for continuously compounded interest is A =Pert, where A is the amount of money in the account, P is the initial investment, r is the interest rate, and t is the time in years. Using the formula, determine, to the nearest dollar, the amount in the account after 8 years if $750 is invested at an annual rate of 3%.

17. Regents Question • Akeem invests $25,000 in an account that pays 4.75% annual interest compounded continuously. Using the formula A = Pert, where A = the amount in the account after t years, P = principal invested, and r = the annual interest rate, how many years, to the nearest tenth, will it take for Akeem’s investment to triple? • 10.0 3) 23.1 • 14.6 4) 24.0

18. Regents Question The temperature, T, of a given cup of hot chocolate after it has been cooling for t minutes can best be modeled by the function below, where T0 is the temperature of the room and k is a constant. ln(T − T0 ) = −kt + 4.718 A cup of hot chocolate is placed in a room that has a temperature of 68°. After 3 minutes, the temperature of the hot chocolate is 150°. Compute the value of k to the nearest thousandth. [Only an algebraic solution can receive full credit.]Using this value of k, find the temperature, T, of this cup of hot chocolate if it has been sitting in this room for a total of 10 minutes. Express your answer to the nearest degree. [Only an algebraic solution can receive full credit.]

19. Regents Question A population of rabbits doubles every 60 days according to the computations. P = 10(2) t/60 formula, where P is the population of rabbits on day t. What is the value of t when the population is 320? (1) 240 (3) 660 (2) 300 (4) 960

20. Exponential growth Compound Interest Exponential growth Continuous compounding Application Compare the balance after 25 years of a $10,000 investment earning 6.75% interest compounded continuously to the same investment compounded semi-annually. One earns $1484.49 more when compounded continuously

21. Application DDT, a pesticide, was used for many years before the EPA banned its use in 1973. Over time DDT degrades into harmless materials. In 1973 there was still 1.0 x 109 kilograms in the environment. The k for DDT is k = -0.0211. Write a function to model the amount of DDT remaining in the environment. Find the amount of DDT that will be in the environment in 2005. a. Continuous growth/decay k is a constant (±) N = (1 x 109)e-0.0211t b. 2005 – 1973 = 32 t = 32 N = (1 x 109)e-0.0211(32) N = 5.1 x 108 kilograms

22. Application According to Newton, a beaker of liquid cools exponentially when removed from a source of heat. Assume the initial temperature Ti, is 90ºF and that k = -0.275. Write a function to model the rate at which the liquid cools. Find the temperature T of the liquid after 4 minutes. a. Continuous growth/decay N = N0ekt T = Tiekt b. t = 4 T= (90)e-0.275(4) T = 29.95839753º

23. Model Problem 1 During the 19th Century, rabbits were brought to Australia. Since the rabbits had no natural enemies on that continent, their population increased rapidly. Suppose there were 65,000 rabbits in Australia in 1865 and 2,500,000 rabbits in 1867. Write an exponential equation that could be used to model the rabbit population in Australia. Write the equation in terms of the number of years elapsed since 1865. Estimate the Australian rabbit population in 1873.

24. Model Problem 2 The diameter of the base of a tree trunk in centimeters varies directly with the 3/2 power of its height in meters. A young sequoia tree is 6 meters tall, and the diameter of its base is 19.1 centimeters. Use this information to write an equation for the diameter d of the base of a sequoia tree if its height is h meters high. One of the oldest living things on Earth is the General Sherman Tree in Sequoia National Park in California. This sequoia is between 2200 and 2500 years old. If it is about 83.8 meters high, find the diameter at its base.

25. Model Problem 1 Sociologists have found that information spreads among a population at an exponential rate. Suppose that the function y = 525(1 – e-0.038t) models the number people in a town of 525 people who have heard news within t hours of its distribution. How many people will have heard about the opening of a new grocery store within 24 hours of the announcement? Graph the function on a graphing calculator. When will 90% of the people have heard about the grocery store opening?