1. Chapter 5 Applications of Exponential and Natural Logarithm Functions

2. We will rely significantly on the following fact throughout chapter 5: The function y = Ce kt satisfies the differential equation y’ = ky. Conversely, if y = f(t) satisfies the differential equation y’ = ky, then y = Ce kt for some constant C.

3. It is important to note that if f(t) = Ce kt, then by setting t = 0, we have f(0) = Ce 0 = C So, C is the value of f(t) at t = 0.

4. 5.1 Exponential Growth and Decay

5. In many areas, including biology, chemistry and economics, it is necessary to study the behavior of a quantity that increases as time passes. Many of the functions we will encounter are functions of time t. If, at every instant, the rate of increase of a quantity is proportional to the quantity at that instant, then we say that the quantity is growing exponentially or exhibiting exponential growth.

6. Consider the growth of a bacteria culture. Under ideal lab conditions, a bacteria culture grows at a rate proportional to the number of bacteria present. Bacteria grow through division. The more bacteria that is present at a given instant, the greater the rate of growth will be.

7. Let P(t) denote the number of bacteria in a certain culture at time t. The rate of growth of the culture at time t is P’(t). We assume that the rate of growth is proportional to the size of the culture at time t so P’(t) = kP(t) k is called the positive constant of proportionality.

8. If we let y = P(t), then we can rewrite this equation as y’ = ky So, y = P(t) = P 0 e kt P 0 is the number of bacteria present at time t = 0 k is called the growth constant

9. Problem: Suppose that a certain bacteria culture grows at a rate proportional to its size. At time t = 0, approximately 20,000 bacteria are present. In 5 hours there are 400,000 bacteria. Determine a function that expresses the size of the culture as a function of time, measured in hours.

10. Let P(t) be the number of bacteria present at time t. We are given that the culture “grows at a rate proportional to its size” in the problem. We can then assume that P(t) satisfies a differential equation of the form y’ = ky. So, y = P(t) = P 0 e kt P 0 and k must be determined.

11. If we have data about the bacteria at two different times, then we can discover P 0 and k. The problems gives us the bacteria population size at times t= 0 and t = 5. So, P(0) = 20,000 P(5) = 400,000 From P(0) = 20,000, we directly get P 0 as 20,000. Now we have P(t) = 20,000e kt

12. By using the second condition P(5) = 400,000, we have P(5) = 20,000e k(5) = 400,000 so 20,000e 5k = 400,000 After dividing both sides by 20,000, we have e 5k = 20 then ln e 5k = ln 20 5k = ln 20 k = (ln 20) / 5 or approximately.60

13. The graph of P(t) = 20,000e.6t appears as

14. Exponential Decay

15. It is known that, at any instant, the rate at which a radioactive substance is decaying is proportional to the amount of the substance that has not yet disintegrated. Let P(t) be the quantity of the substance present at time t. Then, P’(t) is the rate of decay. P’(t) must be negative since P(t) is decreasing, and P’(t) = kP(t) for some negative constant k.

16. To emphasize that k is negative, k is often replaced by –λ, where λ is a positive constant. Now, P(t) satisfies the differential equation P’(t) = –λP(t) and P(t) has the form P(t) = P 0 e -λt P(t) is called an exponential decay function λ is called the decay constant

17. Problem: The decay constant for strontium 90 is λ =.0244, where time is measured in years. How long will it take for the quantity P 0 of strontium 90 to decay to half its original mass?

18. We have P(t) = P 0 e -.0244t Now, we want to know when P(t) =.5 P 0 We substitute for P(t) P 0 e -.0244t =.5 P 0 So, e -.0244t =.5 ln e -.0244t = ln.5 -.0244t = ln.5 t = (ln.5) / -.0244 which is approximately 28 years

19. From our problem, the half-life of strontium 90 is 28 years. We can infer that it will take an additional 28 years for the original mass to decay to 1/4th of its original mass (56 years total) and still another 28 years to decay to 1/8th of its original mass (74 total years).