 SOLVE APPLIED PROBLEMS INVOLVING EXPONENTIAL GROWTH AND DECAY.  SOLVE APPLIED PROBLEMS INVOLVING COMPOUND INTEREST. Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley 5.6 Applications and Models: Growth and Decay; and Compound Interest

Population Growth Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley The function P(t) = P 0 e kt, k > 0 can model many kinds of population growths. 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.

Population Growth - Graph Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley

Example Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley In 2009, the population of Mexico was about million, and the exponential growth rate was 1.13% 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 2009?

Interest Compound Continuously Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley 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 Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley 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?

Growth Rate and Doubling Time Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley 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 P 0.

Example Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley The population of Kenya is now doubling every 25.8 years. What is the exponential growth rate?

Models of Limited Growth Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley 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 Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley

Exponential Decay Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley 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.

Graphs Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley

Decay Rate and Half-Life Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley 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 Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley 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? Now we have the function

Example (continued) Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley 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: