Applications and Models: Growth and Decay; and Compound Interest Section 5.6 Applications and Models: Growth and Decay; and Compound Interest Copyright ©2013, 2009, 2006, 2001 Pearson Education, Inc.

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

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.

Population Growth - Graph

Example In 2011, the population of Mexico was about 113.7 million, and the exponential growth rate was 1.1% per year. a) Find the exponential growth function. b) Graph the exponential growth function. c) Estimate the population in 2015. d) After how long will the population be double what it was in 2011?

Example (continued) a) At t = 0 (2011), the population was about 113.7 million. We substitute 113.7 for P0 and 1.1% or 0.011 for k to obtain the exponential growth function. b) P(t) = 113.7e0.011t

Example (continued) c) In 2015, t = 4; that is 4 years have passed since 2011. To find the population in 2015, we substitute 4 for t. The population will be about 118.8 million in 2015.

Example (continued) d) We are looking for the doubling time; T such that P(T) = 2 • 113.7 or 227.4. The population of Mexico will be double what it was in 2011 about 63.0 years after 2011.

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

Interest Compound Continuously The function P(t) = P0ekt can be used to calculate interest that is compounded continuously. In this function: P0 = amount of money invested, P(t) = balance of the account after t years, t = years, k = interest rate compounded continuously.

Example Suppose that $2000 is invested at interest rate k, compounded continuously, and grows to $2504.65 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?

Example (continued) a. At t = 0, P(0) = P0 = $2000. Thus the exponential growth function is P(t) = 2000ekt. We know that P(5) = $2504.65. Substitute and solve for k: The interest rate is about 0.045 or 4.5%.

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

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 yr.

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 .

Example The population of Kenya is now doubling every 28.2 years. What is the exponential growth rate? The growth rate of the population of Kenya is about 2.46% per year.

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 which 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.

Models of Limited Growth - Graph

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.

Graphs

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.

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?

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

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.

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

Logarithmic Curve Fitting Logistic

Example The number of U.S. communities using surveillance cameras at intersections has greatly increased in recent years, as show in the table.

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. Estimate the number of U.S. communities using surveillance cameras at intersections in 2010.

Example (continued) a. Fit an equation of the type y = a • bx, where x is the number of years since 1999. Enter the data . . . The equation is The correlation coefficient is close to 1, indicating the exponential function fits the data well.

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

Example (continued) c. Using the VALUE feature in the CALC menu, we evaluate the function for x = 11 (2010 – 1999 = 11), and estimate the number of communities using surveillance cameras at intersections in 2010 to be about 651.