1. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 1 of 47 Chapter 5 Applications of the Exponential and Natural Logarithm Functions

2. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 2 of 47  Exponential Growth and Decay  Compound Interest  Applications of the Natural Logarithm Function to Economics  Further Exponential Models Chapter Outline

3. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 3 of 47 § 5.1 Exponential Growth and Decay

4. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 4 of 47  Exponential Growth  The Exponential Growth and Decay Model  Exponential Growth in Application  Exponential Decay  Exponential Decay in Application Section Outline

5. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 5 of 47 Exponential Growth DefinitionExample Exponential Growth: A quantity, such that, at every instant the rate of increase of the quantity is proportional to the amount of the quantity at that instant

6. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 6 of 47 Exponential Growth & Decay Model

7. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 7 of 47 Exponential Growth in ApplicationEXAMPLE SOLUTION (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? 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) = P 0 e kt to describe the population of the world. Since P 0 is the initial quantity, P 0 = 5.51. Therefore, our formula becomes

8. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 8 of 47 Exponential Growth in Application Now we use the other given information (5.88 billion in 1998) to determine k. This is our function so far. CONTINUED 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. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 9 of 47 Exponential Growth in ApplicationCONTINUED This the derived function. 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. Replace P(t) with 7. Divide. Rewrite in logarithmic form. Solve for t.

10. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 10 of 47 Exponential Growth in ApplicationCONTINUED

11. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 11 of 47 Exponential Decay DefinitionExample Exponential Decay: A quantity, such that, at every instant the rate of decrease of the quantity is proportional to the amount of the quantity at that instant

12. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 12 of 47 Exponential Decay in ApplicationEXAMPLE (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. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 13 of 47 Exponential Decay in Application (a) Since the function y = Ce kt satisfies the differential equation y΄ = ky, the function P(t) = Ce kt = Ce -0.00043t (where k = -0.00043). Since for the function y = Ce kt, C is always the initial quantity (at time t = 0), C = 12 (since P(0) = 12). Therefore, our function is 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

14. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 14 of 47 Exponential Decay in Application (c) Since our exponential decay function is, the decay constant, being the coefficient of t, is -0.00043. (b) We were given P(0) = 12. Therefore the initial amount is 12 grams. CONTINUED (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. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 15 of 47 Exponential Decay in ApplicationCONTINUED This is the derivative function. (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). 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. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 16 of 47 Exponential Decay in ApplicationCONTINUED 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).