1. Exponential Growth & Decay Modeling Data Objectives –Model exponential growth & decay –Model data with exponential & logarithmic functions. 1

2. Exponential Growth & Decay Models A 0 is the amount you start with, t is the time, and k=constant of growth (or decay) f k>0, the amount is GROWING (getting larger), as in the money in a savings account that is having interest compounded over time If k<0, the amount is SHRINKING (getting smaller), as in the amount of radioactive substance remaining after the substance decays over time 2

3. Graphs A0A0 A0A0 3

4. Could the following graph model exponential growth or decay? 1) Growth model. 2) Decay model. 4

5. Example Population Growth of the United States. In 1990 the population in the United States was about 249 million and the exponential growth rate was 8% per decade. (Source: U.S. Census Bureau) –Find the exponential growth function. –What will the population be in 2020? –After how long will the population be double what it was in 1990? 5

6. Solution At t = 0 (1990), the population was about 249 million. We substitute 249 for A 0 and 0.08 for k to obtain the exponential growth function. A(t) = 249e 0.08t In 2020, 3 decades later, t = 3. To find the population in 2020 we substitute 3 for t: A(3) = 249e 0.08(3) = 249e 0.24  317. The population will be approximately 317 million in 2020. 6

7. Solution continued We are looking for the doubling time T. 498 = 249e 0.08T 2 = e 0.08T ln 2 = ln e 0.08T ln 2 = 0.08T (ln e x = x) = T 8.7  T The population of the U.S. will double in about 8.7 decades or 87 years. This will be approximately in 2077. 7

8. Interest Compound Continuously The function A(t) = A 0 e kt can be used to calculate interest that is compounded continuously. In this function: A 0 = amount of money invested, A = balance of the account, t = years, k = interest rate compounded continuously. 8

9. Example Suppose that $2000 is deposited into an IRA at an interest rate k, and grows to $5889.36 after 12 years. –What is the interest rate? –Find the exponential growth function. –What will the balance be after the first 5 years? –How long did it take the $2000 to double? 9

10. Solution At t = 0, A(0) = A 0 = $2000. Thus the exponential growth function is A(t) = 2000e kt. We know that A(12) = $5889.36. We then substitute and solve for k: $5889.36 = 2000e 12k The interest rate is approximately 9.5% 10

11. Solution continued The exponential growth function is A(t) = 2000e 0.09t. The balance after 5 years is A(5) = 2000e 0.09(5) = 2000e 0.45  $3136.62 11

12. Solution continued To find the doubling time T, set A(T) = 2  A 0 = $4000 and solve for T. 4000 = 2000e 0.09T 2 = e 0.09T ln 2 = ln e 0.09T ln 2 = 0.09T = T 7.7  T The original investment of $2000 doubled in about 7.7 years. 12

13. Growth Rate and Doubling Time The growth rate k and the doubling time T are related by – kT = ln 2 or – or – * The relationship between k and T does not depend on A 0. 13

14. Example A certain town’s population is doubling every 37.4 years. What is the exponential growth rate? Solution: 14

15. Exponential Decay Decay, or decline, is represented by the function A(t) = A 0 e kt, k < 0. In this function: A 0 = initial amount of the substance, A = amount of the substance left after time, t = time, k = decay rate. The half-life is the amount of time it takes for half of an amount of substance to decay. 15

16. Example Carbon Dating. The radioactive element carbon-14 has a half-life of 5715 years. If a piece of charcoal that had lost 7.3% of its original amount of carbon, was discovered from an ancient campsite, how could the age of the charcoal be determined? Solution: The function for carbon dating is A(t) = A 0 e -0.00012t. If the charcoal has lost 7.3% of its carbon-14 from its initial amount A 0, then 92.7%A 0 is the amount present. 16

17. Example continued To find the age of the charcoal, we solve the equation for t : The charcoal was about 632 years old. 17