171S5.6o Applications and Models: Growth and Decay; and Compound Interest November 21, 2011

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1 MAT 171 Precalculus Algebra Dr. Claude Moore Cape Fear Community College CHAPTER 5: Exponential and Logarithmic Functions 5.1 Inverse Functions 5.2 Exponential Functions and Graphs 5.3 Logarithmic Functions and Graphs 5.4 Properties of Logarithmic Functions 5.5 Solving Exponential and Logarithmic Equations 5.6 Applications and Models: Growth and Decay; and Compound Interest Click globe on left to see TI Calculator tutorials at Applications and Models: Growth and Decay; and Compound Interest 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) = P 0 e kt, k > 0 can model many kinds of population growths. Population Growth Graph In this function: P 0 = 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. 1

2 Example In 2006, the population of China was about billon, and the exponential growth rate was 0.6% per year. a) Find the exponential growth function. b) Graph the exponential growth function. c) Estimate the population in d) After how long will the population be double what it was in 2006? a) At t = 0 (2006), the population was about billion. We substitute for P 0 and for k to obtain the exponential growth function. b) P(t) = 1.314e 0.006t c) In 2010, t = 4. To find the population in 2010 we substitute 4 for t: P(4) = 1.314e 0.006(4) = 1.314e The population will be approximately billion in The graph also displays this value. d) We are looking for the doubling time; T such that P(T) = = Solve The population of China will be double what it was in 2006 about years after

3 d) Using the Intersect method we graph and find the first coordinate of their point of intersection. The population of China will be double that of 2006 about years after Interest Compound Continuously The function P(t) = P 0 e kt can be used to calculate interest that is compounded continuously. In this function: P 0 = 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 $ 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? a. At t = 0, P(0) = P 0 = $2000. Thus the exponential growth function is P(t) = 2000e kt. We know that P(5) = $ Substitute and solve for k: The interest rate is about or 4.5%. 3

4 b. The exponential growth function is d. To find the doubling time T, we set P(T) = 2 P 0 = 2 $2000 = $4000 and solve for T. P(t) = 2000e 0.045t. c. The balance after 10 years is Thus the orginal 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 Example The population of the world is now doubling every 60.8 yr. What is the exponential growth rate? Note that the relationship between k and T does not depend on P 0. The growth rate of the world population is about 1.14% per year. 4

5 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 Models of Limited Growth Graph 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. Exponential Decay Decay, or decline, of a population is represented by the function P(t) = P 0 e kt, k > 0. In this function: P 0 = 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. Growth Graph Decay Graph 5

6 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? First find k when the half life T is 5750 yr: If the linen wrapping lost 22.3% of its carbon 14 from the initial amount P 0, 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. 6

7 Exponential Curve Fitting We have added several new functions that can be considered when we fit curves to data. Logarithmic Curve Fitting Logarithmic Logistic Example The number of U.S. communities using surveillance cameras at intersections has greatly increased in recent years, as show in the table. 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

8 a. Fit an equation of the type y = a b x, where x is the number of years since Enter the data... b. Here s the graph of the function with the scatter plot. The correlation coefficient is close to 1, indicating the exponential function fits the data well. c. Using the VALUE feature in the CALC menu, we evaluate the function for x = 11 ( = 11), and estimate the number of communities using surveillance cameras at intersections in 2010 to be about /3. Population Growth. Complete the following table. 8

9 456/8. Interest Compounded Continuously. Complete the following table. 457/11. Radioactive Decay. Complete the following table. 458/15. Norman Rockwell Painting. Breaking Home Ties, painted by Norman Rockwell, appeared on the cover of the Saturday Evening Post in In 1960, the original sold for only $ 900; in 2006, this painting was sold at Sothebys auction house for $ 15.4 million ( Source: Associated Press, Indianapolis Star, December 2, 2006, p. B5). Assuming the value R 0 of the painting has grown exponentially: a) Find the value of k and determine the exponential growth function, assuming R 0 = 900 and t is the number of years since b) Estimate the value of the painting in c) What is the doubling time for the value of the painting? d) After how long will the value of the painting be $ 25 million, assuming there is no change in the growth rate? 460. In Exercises 23 28, determine which, if any, of these functions might be used as a model for the data in the scatterplot. a) Quadratic, f(x) = ax 2 + bx + c b) Polynomial, not quadratic c) Exponential, f(x) = ab x, or P oe kx, k > 0 d) Exponential, f(x) = ab x, or P oe kx, k > 0 e) Logarithmic, f(x) = a + b ln x f) Logistic, 9

10 460/30. Forgetting. In an art class, students were given a final exam at the end of the course. Then they were retested with an equivalent test at subsequent time intervals. Their scores after time x, in months, are given in the following table. a) Use a graphing calculator to fit a logarithmic function to the data. b) Use the function to predict test scores after 8, 10, 24, and 36 months. c) After how long will the test scores fall below 82%? f(x) = a + b ln x 10