1. Lesson 9-4 Exponential Growth and Decay

2. Generally these take on the form Where p 0 is the initial condition at time t= 0 population shrinking  decay (negative exponent) population expanding  growth (positive exponent) k is the exponential rate of change Although these equations can be derived like is demonstrated in your notes, most students will use the equation above. P = p 0 e ±kt

3. Example 1 - Growth The number of bacteria in a rapidly growing culture was estimated to be 10,000 at noon and 40000 after 2 hours. How many bacteria will there be at 5 pm? P = p 0 e ±kt General Equation: growth k > 0; p 0 = 10000 P(2)= 40000 = 10000e 2k 4 = e 2k ln 4 = ln e 2k = 2k ln 2² / 2 = k ln 2 = k Equation: Use to solve for k P(5) = 10000e 5ln2 = 320,000 bacteria

4. Example 2 - Decay Carbon 14 is radioactive and decays at a rate proportional to the amount present. Its half life is 5730 years. If 10 grams were present originally, how much will be left after 2000 years? P = p 0 e ±kt General Equation: decay k < 0; p 0 = 10 grams P(5730)= 5 = 10e 5730k 0.5 = e 5730k ln 0.5 = ln e 5730k = 5730k ln 2 -1 / 5730 = k -(ln 2)/5730 = k = -0.000121 Equation: Use to solve for k P(2000) = 10e 2000(-0.000121) = 7.851 grams

5. Compounded Interest Compounded Interest Formula: where A 0 is the initial amount r is the interest rate n is the number of times compounded per year t is the time period in years Compounded Continuously (n→∞): A(t) = A 0 (1 + r/n) nt A(t) = A 0 e rt

6. Example 3a Suppose Joe put $500 in the bank at 4% interest. How much will it be worth after 5 years if it is compounded annually? Annually yields time increments of years, so n = 1 A(5) = 500 (1 + 0.04/1) (1)5 A(5) = 500(1.04) 5 A(5) = $608.33 A(t) = A 0 (1 + r/n) nt

7. Example 3b Suppose Joe put $500 in the bank at 4% interest. How much will it be worth after 5 years if it is compounded monthly? Monthly yields time increments of months, so n = 12 A(5) = 500 (1 + 0.04/12) (12)5 A(5) = 500(1.003333) 60 A(5) = $610.50 A(t) = A 0 (1 + r/n) nt

8. Example 3c Suppose Joe put $500 in the bank at 4% interest. How much will it be worth after 5 years if it is compounded weekly? Weekly yields time increments of weeks, so n = 52 A(5) = 500 (1 + 0.04/52) (52)5 A(5) = 500(1.000769231) 260 A(5) = $610.65 A(t) = A 0 (1 + r/n) nt

9. Example 3d Suppose Joe put $500 in the bank at 4% interest. How much will it be worth after 5 years if it is compounded daily? Daily yields time increments of days, so n = 365 A(5) = 500 (1 + 0.04/365) (365)5 A(5) = 500(1.000109589) 1825 A(5) = $610.69 A(t) = A 0 (1 + r/n) nt

10. Example 3e Suppose Joe put $500 in the bank at 4% interest. How much will it be worth after 5 years if it is compounded continuously? Continuously yields infinite time increments, so n   A(5) = 500 (2.718281828) (0.04)5 A(5) = 500(2.718281828) 0.2 A(5) = $610.70 A(t) = A 0 e rt 2.718281828

11. So why should we save? Time is the key to savings growth and 5 years is just too short a time period! What if we put away $5000 the day our child was born with just 4% interest, how much would they have at age 65?: If we got 10% interest (average SP500 growth): A(65) = 5000 (1 + 0.04/365) (365)65 A(65) = 5000(1.000109589) 23725 A(65) = $67,309.10 A(65) = 5000 (1 + 0.1/365) (365)65 A(65) = 5000(1.000273973) 23725 A(65) = $3,322,748.59

12. Summary & Homework Summary: –Exponential Growth and Decay is a common real- world problem that can be solved using differential equations –Interest earning accounts need long periods of time to earn significant amounts of money Homework: –pg 620 – 621: Day One: 3, 4, 8, 9, Day Two: 13, 19. 20