Chapter 5 Applications of the Exponential and Natural Logarithm Functions

1. Chapter 5Applications of the Exponential and Natural Logarithm Functions

2. Chapter Outline • Exponential Growth and Decay • Compound Interest • Applications of the Natural Logarithm Function to Economics • Further Exponential Models

3. §5.1 Exponential Growth and Decay

4. Section Outline • Exponential Growth • The Exponential Growth and Decay Model • Exponential Growth in Application • Exponential Decay • Exponential Decay in Application

5. Exponential Growth

6. Exponential Growth & Decay Model

7. Exponential Growth in Application EXAMPLE (World’s Population) The world’s population was 5.51 billion on January 1, 1993 and 5.88 billion on January 1, 1998. Assume that at any time the population grows at a rate proportional to the population at that time. In what year will the world’s population reach 7 billion? SOLUTION Since the “oldest” information we’re given corresponds to 1993, that will serve as our initial time. Therefore the year 1993 will be the year t = 0 and the population at time t = 0 is 5.51 (measured in billions). Therefore, the year 1998 will be year t = 5 and the population at time t = 5 is 5.88 (measured in billions). Since the population grows at a rate proportional to the size of the population, we can use the exponential growth model P(t) = P0ekt to describe the population of the world. Since P0 is the initial quantity, P0 = 5.51. Therefore, our formula becomes

8. Exponential Growth in Application CONTINUED Now we use the other given information (5.88 billion in 1998) to determine k. This is our function so far. When t = 5, the population is 5.88 billion people. Divide. Rewrite in logarithmic form. Solve for k. Therefore, our formula to model this situation is Now we can determine when the world’s population will be 7 billion.

9. Exponential Growth in Application CONTINUED This the derived function. Replace P(t) with 7. Divide. Rewrite in logarithmic form. Solve for t. Therefore, the world’s population will be 7 billion people about 18.36 years after our initial year, 1993. This would be the year 1993 + 18.36 = 2011.36. That is, around the year 2011. The graph is given below.

10. Exponential Growth in Application CONTINUED

11. Exponential Decay

12. Exponential Decay in Application EXAMPLE (Radioactive Decay) Radium-226 is used in cancer radiotherapy, as a neutron source for some research purposes, and as a constituent of luminescent paints. Let P(t) be the number of grams of radium-226 in a sample remaining after t years, and suppose that P(t) satisfies the differential equation (a) Find the formula for P(t). (b) What was the initial amount? (c) What is the decay constant? (d) Approximately how much of the radium will remain after 943 years? (e) How fast is the sample disintegrating when just one gram remains? Use the differential equation.

13. Exponential Decay in Application CONTINUED (f) What is the weight of the sample when it is disintegrating at the rate of 0.004 grams per year? (g) The radioactive material has a half-life of about 1612 years. How much will remain after 1612 years? 3224 years? SOLUTION (a) Since the function y = Cekt satisfies the differential equation y΄ = ky, the function P(t) = Cekt = Ce-0.00043t (where k = -0.00043). Since for the function y = Cekt, C is always the initial quantity (at time t = 0), C = 12 (since P(0) = 12). Therefore, our function is

14. Exponential Decay in Application CONTINUED (b) We were given P(0) = 12. Therefore the initial amount is 12 grams. (c) Since our exponential decay function is , the decay constant, being the coefficient of t, is -0.00043. (d) To determine approximately how much of the radium will remain after 943 years, we will evaluate the function at t = 943. This is the decay function. Evaluate the function at t = 943. Simplify. Therefore, after 943 years, there will be approximately 8 grams remaining.

15. Exponential Decay in Application CONTINUED (e) To determine how fast the sample is disintegrating when just one gram remains, we must first recognize that this is a situation concerning the rate of change of a quantity, namely the rate at which the radium is disintegrating. This of course involves the derivative function. This function was given to us and is . Now we will determine the value of the derivative function at P(t) = 1 (when one gram remains). This is the derivative function. Replace P(t) with 1. So, when there is just one gram remaining, the radium is disintegrating at a rate of 0.00043 grams/year. (f) To determine the weight of the sample when it is disintegrating at the rate of 0.004 grams per year, we must determine P(t) when P΄(t) = -0.004.

16. Exponential Decay in Application CONTINUED This is the derivative function. Replace P΄(t) with -0.004. Solve for P(t). So, the weight of the sample when it is disintegrating at the rate of 0.004 grams per year, is 9.3 grams. (g) To determine how much of the radium will remain after 1612 years, that is one half-life, we will simply recognize that after one half-life, half of the original amount of radium will be disintegrated. That is, 12/2 = 6 grams will be disintegrated and therefore 6 grams will remain. After 3224 years, two half-lives, half of what was remaining at the end of the first 1612 years (6 grams) will remain. That is, 6/2 = 3 grams. These results can be verified using the formula for P(t).