171S5.6q Applications and Models: Growth and Decay; ?and Compound Interest
April 24, 2013
MAT 171 Precalculus Algebra
5.6 Applications and Models: Growth and Decay; and Compound Interest
Dr. Claude Moore
Cape Fear Community College
CHAPTER 5: Exponential and Logarithmic Functions
exponential growth • Solve applied problems involving and decay.
compound interest.
• Solve applied problems involving 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
• 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.
Click globe on left to see TI Calculator tutorials at http://cfcc.edu/faculty/cmoore/TI83Modeling.htm
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.
Mathematica Interactive Figures
are available through Tools for Success, Activities and Projects in CourseCompass. You may access these through CourseCompass or from the Important Links webpage. You must Login to MML to use this link.
There are two Interactive Figures for this section. (1) Exponential Growth Models (2) Graphs of Logistic Functions
Use this GeoGebra program to calculate Exponential Growth and Exponential Decay. http://cfcc.edu/mathlab/geogebra/growth_decay.html
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Example (continued)
Population Growth ­ Graph
Solution:
a) At t = 0 (2006), the population was about 1.314 billion. We substitute 1.314 for P0 and 0.006 for k to obtain the exponential growth function.
P(t) = 1.314e0.006t
b)
c) In 2010, t = 4. To find the population in 2010 we substitute 4 for t:
Example
In 2006, the population of China was about 1.314 billon, and the exponential growth rate was 0.6% per year.
P(4) = 1.314e0.006(4) = 1.314e0.024
a) Find the exponential growth function.
b) Graph the exponential growth function.
c) Estimate the population in 2010.
d) After how long will the population be double what it was in 2006?
≈ 1.346
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The population will be approximately 1.346 billion in 2010. The graph also displays this value.
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171S5.6q Applications and Models: Growth and Decay; ?and Compound Interest
April 24, 2013
Example (continued)
d) We are looking for the doubling time; T such that P(T) = 2 • 1.314 = 2.628. Solve
Interest Compound Continuously
The function P(t) = P0ekt can be used to calculate interest that is compounded continuously.
The population of China will be double what it was in 2006 about 115.5 years after 2006.
d) Using the Intersect method we graph
and find the first coordinate of their point of intersection.
In this function:
P0 = amount of money invested,
P(t) = balance of the account after t years,
t = years,
k = interest rate compounded continuously.
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?
The population of China will be double that of 2006 about 115.5 years after 2006.
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Example (continued)
Solution:
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:
Growth Rate and Doubling Time
The growth rate k and doubling time T are related by
The interest rate is about 0.045 or 4.5%.
b. The exponential growth function is
P(t) = 2000 e 0.045
t kT = ln 2 or or Note that the relationship between k and T does not depend on P0 .
. c. The balance after 10 years is
Example
The population of the world is now doubling every 60.8 yr. What is the exponential growth rate?
d. To find the doubling time T, we set
P(T) = 2 • P0= 2 • $2000 = $4000 and solve for T.
Solution:
The growth rate of the world population is about 1.14% per year.
Thus the orginal investment of $2000 will double in about 15.4 yr.
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171S5.6q Applications and Models: Growth and Decay; ?and Compound Interest
April 24, 2013
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.
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 half­life is the amount of time it takes for a substance to decay to half of the original amount.
Models of Limited Growth ­ Graph
Growth Graph
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Decay Graph
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Example (continued)
Decay Rate and Half­Life
Solution:
First find k when the half­life T is 5750 yr:
The decay rate k and the half­life T are related by
kT = ln 2 or or Note that the relationship between decay rate and half­life is the same as that between growth rate and doubling time.
Example
Carbon Dating . The radioactive element carbon­14 has a half­life of 5750 years. The percentage of carbon­14 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 carbon­14 at the time it was found. How old was the linen wrapping?
If the linen wrapping lost 22.3% of its carbon­14 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.
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171S5.6q Applications and Models: Growth and Decay; ?and Compound Interest
April 24, 2013
Exponential Curve Fitting
We have added several new functions that can be considered when we fit curves to data.
Example
The number of U.S. communities using surveillance cameras at intersections has greatly increased in recent years, as show in the table.
Logarithmic Curve Fitting
Logarithmic
Logistic
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.
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Example (continued)
Solution:
a. Fit an equation of the type y = a • bx, where x is the number of years since 1999. Enter the data . . . 463/3a, c. Population Growth
following table.
. Complete the The correlation coefficient is close to 1, indicating the exponential function fits the data well.
b. Here’s the graph of the function with the scatter plot.
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.
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171S5.6q Applications and Models: Growth and Decay; ?and Compound Interest
April 24, 2013
463/3b, d. Population Growth
following table.
. Complete the Nov 22­1:27 PM
463/3h, j. Population Growth
following table.
463/3e, f. Population Growth
following table.
. Complete the Nov 22­1:27 PM
. Complete the 464/8a. Interest Compounded Continuously
Complete the following table.
Nov 22­1:27 PM
. Nov 22­1:27 PM
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171S5.6q Applications and Models: Growth and Decay; ?and Compound Interest
April 24, 2013
464/8b. Interest Compounded Continuously
Complete the following table.
. Nov 22­1:27 PM
464/8f. Interest Compounded Continuously
Complete the following table.
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464/8c. Interest Compounded Continuously
Complete the following table.
. Nov 22­1:27 PM
. 465/11a. Radioactive Decay
following table.
. Complete the Nov 22­1:27 PM
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171S5.6q Applications and Models: Growth and Decay; ?and Compound Interest
April 24, 2013
465/11c. Radioactive Decay
following table.
. Complete the 465/11d. Radioactive Decay
following table.
Nov 22­1:27 PM
465/11e. Radioactive Decay
following table.
. Complete the Nov 22­1:27 PM
. Complete the 467. In Exercises 23­28 , 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 Poe kx , k > 0
d) Exponential, f(x) = ab ­x , or Poe ­kx , k > 0
e) Logarithmic, f(x) = a + b ln x
f) Logistic, 23.
24.
25.
26.
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171S5.6q Applications and Models: Growth and Decay; ?and Compound Interest
April 24, 2013
468/30. Forgetting . In an economics 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 listed in the table below.
468/31. Unoccupied Homes in a Census. After door­to­door canvassing, census workers determine the number of unoccupied homes. The table below lists the number of homes, in millions, that were unoccupied in four census years.
a) Use a graphing calculator to fit a logarithmic function y = a + b ln x 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%?
a) Use a graphing calculator to fit an exponential function to the data, where x is the number of years after 1980. b) Graph the function found in part (a). c) Use the function found in part (a) to project the number of unoccupied homes in the 2020 Census.
f(x) = a + b ln x
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468/32. Older Americans in Debt.
The debt among older Americans has been exponentially increasing in recent years. The table below lists, for selected years, the average debt per household, where the primary head is age 55 or older.
a) Use a graphing calculator to fit the data with an exponential function, where x is the number of years since 2000. b) Graph the function found in part (a). c) Use the function in part (a) to estimate the debt in 2014. d) In what year is average debt among older Americans expected to exceed $140,000?
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469/34. Effect of Advertising.
A company introduced a new software product on a trial run in a city. They advertised the product on television and found the following data regarding the percent P of people who bought the product after x ads were run.
a) Use a graphing calculator to fit a logistic function to the data.
b) What percent of people bought the product when 55 ads were run? 100 ads? c) Find the horizontal asymptote for the graph. Interpret the asymptote in terms of the advertising situation. Nov 22­1:57 PM
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