Exponential Growth & Decay Modeling Data

1. Exponential Growth & DecayModeling Data • Objectives • Model exponential growth & decay • Model data with exponential & logarithmic functions.

2. Exponential Growth & Decay Models • A0 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

3. Graphs A0 A0

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

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?

6. Solution • At t = 0 (1990), the population was about 249 million. We substitute 249 for A0 and 0.08 for k to obtain the exponential growth function. A(t) = 249e0.08t • In2020, 3 decades later, t = 3. To find the population in 2020 we substitute 3 for t: A(3) = 249e0.08(3) = 249e0.24 317. The population will be approximately 317 million in 2020.

7. Solution continued • We are looking for the doubling time T. 498 = 249e0.08T 2 = e0.08T ln 2 = ln e0.08T ln 2 = 0.08T (ln ex = 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.

8. Interest Compound Continuously • The function A(t) = A0ekt can be used to calculate interest that is compounded continuously. In this function: A0 = amount of money invested, A = balance of the account, t = years, k = interest rate compounded continuously.

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?

10. Solution • At t = 0, A(0) = A0 = $2000. Thus the exponential growth function is A(t) = 2000ekt. We know that A(12) = $5889.36. We then substitute and solve for k: $5889.36 = 2000e12k The interest rate is approximately 9.5%

11. Solution continued • The exponential growth function is A(t) = 2000e0.09t. • The balance after 5 years is A(5) = 2000e0.09(5) = 2000e0.45  $3136.62

12. Solution continued • To find the doubling time T, set A(T) = 2  A0 = $4000 and solve for T. 4000 = 2000e0.09T 2 = e0.09T ln 2 = ln e0.09T ln 2 = 0.09T = T 7.7  T The original investment of $2000 doubled in about 7.7 years.

13. Growth Rate and Doubling Time The growth ratek and the doubling timeT are related by • kT = ln 2 or or * The relationship between k and T does not depend on A0.

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

15. Exponential Decay • Decay, or decline, is represented by the function A(t) = A0ekt, k < 0. In this function: A0 = 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.

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) = A0e-0.00012t. If the charcoal has lost 7.3% of its carbon-14 from its initial amount A0, then 92.7%A0 is the amount present.

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