1. Applications and Models: Growth and Decay; and Compound Interest
Section 5.6
Applications and Models: Growth and Decay; and Compound Interest

2. Objectives
Solve applied problems involving exponential growth and exponential decay.
Solve applied problems involving compound interest.
Find models involving exponential functions and logarithmic functions.

3. Population Growth
The function P(t) = P0 ekt, k > 0 can model many kinds of population growths. In this function: P0 = population at time 0, P(t) = population after time t, t = amount of time, k = exponential growth rate. The growth rate unit must be the same as the time unit.

4. Population Growth - Graph

5. Example
Population Growth of Ghana. In 2014, the population of Ghana, located on the west coast of Africa, was about 25.8 million, and the exponential growth rate was 2.19% per year (Source: CIA World Factbook, 2015).
a) Find the exponential growth function.
b) Graph the exponential growth function.
c) Estimate the population in 2018.
d) At this growth rate, when will the population be 40 million?

6. Example (continued)
a) At t = 0 (2014), the population was about 25.8 million, and the exponential growth rate 2as 2.19% per year. We substitute 25.8 for P0 and 2.19% or for k to obtain the exponential growth function b)

7. Example (continued)
c) In 2018, t = 4; that is 4 years have passed since To find the population in 2018, we substitute 4 for t. The population will be about 28.2 million in 2018.

8. Example (continued)
d) We are looking for the time t for which P(t) = 40. To find t, we solve the equation
The population of Ghana will be 40 million about 20 years after 2014.

9. Example (continued)
d) Using the Intersect method we graph the equations and find the first coordinate of their point of intersection.

10. Interest Compound Continuously
When interest is paid on interest, we call it compound interest. Suppose that an amount P0 is invested in a savings account at interest rate k compounded continuously. The amount P(t) in the account after t years is given by the exponential function P(t) = P0ekt In this function: P0 = amount of money invested, P(t) = balance of the account after t years, t = years, k = interest rate compounded continuously.

11. Example
Suppose that $2000 is invested at interest rate k, compounded continuously, and grows to $ after 5 years. a. What is the interest rate? b. Find the exponential growth function. c. What will the balance be after 10 years? d. After how long will the $2000 have doubled?

12. Example (continued)
a. At t = 0, P(0) = P0 = $2000. Thus the exponential growth function is P(t) = 2000ekt. We know that P(5) = $ Substitute and solve for k:
The interest rate is about or 4.5%.

13. Example (continued)
b. The exponential growth function is P(t) = 2000e0.045t .
c. The balance after 10 years is

14. Example (continued)
d. To find the doubling time T, we set P(T) = 2 • P0= 2 • $2000 = $4000 and solve for T.
Thus the original investment of $2000 will double in about 15.4 years

15. Growth Rate and Doubling Time
The growth rate k and doubling time T are related by kT = ln 2 or or Note that the relationship between k and T does not depend on P0 .

16. Example
The population of the Philippines is now doubling every 37.7 years. What is the exponential growth rate?
The growth rate of the population of the Philippines is about 1.84% per year.

17. Models of Limited Growth
In previous examples, we have modeled population growth. However, in some populations, there can be factors that prevent a population from exceeding some limiting value. One model of such growth is This is called a logistic function. This function increases toward a limiting value a as t approaches infinity. Thus, y = a is the horizontal asymptote of the graph.

18. Exponential Decay
Decay, or decline, of a population is represented by the function P(t) = P0ekt, k > 0. In this function: P0 = initial amount of the substance (at time t = 0), P(t) = amount of the substance left after time, t = time, k = decay rate. The half-life is the amount of time it takes for a substance to decay to half of the original amount.

19. Graphs

20. Decay Rate and Half-Life
The decay rate k and the half-life T are related by kT = ln 2 or or Note that the relationship between decay rate and half-life is the same as that between growth rate and doubling time.

21. Example
Carbon Dating. The radioactive element carbon-14 has a half-life of 5750 years. The percentage of carbon-14 present in the remains of organic matter can be used to determine the age of that organic matter. Archaeologists discovered that the linen wrapping from one of the Dead Sea Scrolls had lost 22.3% of its carbon-14 at the time it was found. How old was the linen wrapping?

22. Example (continued) First find k when the half-life T is 5750 yr:
Now we have the function

23. Example (continued)
If the linen wrapping lost 22.3% of its carbon-14 from the initial amount P0, then 77.7% is the amount present. To find the age t of the wrapping, solve for t:
The linen wrapping on the Dead Sea Scrolls was about 2103 years old when it was found.

24. Exponential Curve Fitting
We have added several new functions that can be considered when we fit curves to data.

25. Logarithmic Curve Fitting

26. Example
Health Expenditures. Health costs have increased exponentially over the past few decades. The following lists national health expenditures in selected years from 1960 to 2012.

27. Example (continued)
a. Use a graphing calculator to fit an exponential function to the data. b. Graph the function with the scatter plot of the data. c. Use the function to estimate national health expenditures in 1985, in 2003, and in Then use the function to estimate health expenditures in 2016.

28. Example (continued)
a. Fit an equation of the type y = a • bx, where x is the number of years after Enter the data . . .
The equation is
The correlation coefficient is close to 1, indicating the exponential function fits the data well.

29. Example (continued)
b. Here’s the graph of the function with the scatter plot.

30. Example (continued)
c. Using the VALUE feature in the CALC menu, we evaluate the function for x = 25 (1985 – 1960 = 25), and estimate national health expenditures in 1985 to be about $320 billion.