171S5.6o Applications and Models: Growth and Decay; and Compound Interest
November 21, 2011
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
5.6 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.
Click globe on left to see TI Calculator tutorials at http://cfcc.edu/faculty/cmoore/TI83Modeling.htm
Population Growth
Population Growth ­ Graph
kt
The function P(t) = P0 e , 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.
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171S5.6o Applications and Models: Growth and Decay; and Compound Interest
November 21, 2011
Example
In 2006, the population of China was about 1.314 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 2010.
d) After how long will the population be double what it was in 2006?
Example (continued)
c) In 2010, t = 4. To find the population in 2010 we substitute 4 for t:
Example (continued)
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)
Example (continued)
d) We are looking for the doubling time; T such that P(T) = 2 • 1.314 = 2.628. Solve
P(4) = 1.314e0.006(4) = 1.314e0.024
≈ 1.346
The population will be approximately 1.346 billion in 2010. The graph also displays this value.
The population of China will be double what it was in 2006 about 115.5 years after 2006.
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171S5.6o Applications and Models: Growth and Decay; and Compound Interest
November 21, 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.
The population of China will be double that of 2006 about 115.5 years after 2006.
Example
Suppose that $2000 is invested at interest rate k, compounded continuously, and grows to $2504.65 after 5 years. 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:
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 interest rate is about 0.045 or 4.5%.
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171S5.6o Applications and Models: Growth and Decay; and Compound Interest
November 21, 2011
Example (continued)
Solution:
b. The exponential growth function is
Example (continued)
d. To find the doubling time T, we set
P(T) = 2 • P0= 2 • $2000 = $4000 and solve for T.
P(t) = 2000e0.045t . 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
Example
The growth rate k and doubling time T are related by
The population of the world is now doubling every 60.8 yr. What is the exponential growth rate?
kT = ln 2 or or Solution:
Note that the relationship between k and T does not depend on P0 .
The growth rate of the world population is about 1.14% per year.
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171S5.6o Applications and Models: Growth and Decay; and Compound Interest
November 21, 2011
Models of Limited Growth
Models of Limited Growth ­ Graph
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
Growth Graph
Decay Graph
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.
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171S5.6o Applications and Models: Growth and Decay; and Compound Interest
November 21, 2011
Decay Rate and Half­Life
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 (continued)
Solution:
First find k when the half­life T is 5750 yr:
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?
Example (continued)
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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171S5.6o Applications and Models: Growth and Decay; and Compound Interest
November 21, 2011
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
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.
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171S5.6o Applications and Models: Growth and Decay; and Compound Interest
November 21, 2011
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 . . . Example (continued)
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.
Example (continued)
455/3. Population Growth. Complete the following table.
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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171S5.6o Applications and Models: Growth and Decay; and Compound Interest
November 21, 2011
456/8. Interest Compounded Continuously. 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 1954. 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 R0 of the painting has grown exponentially: a) Find the value of k and determine the exponential growth function, assuming R0 = 900 and t is the number of years since 1960. b) Estimate the value of the painting in 2010. 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?
457/11. Radioactive Decay. Complete the following table.
460. 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) = ax2 + 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, 9
171S5.6o Applications and Models: Growth and Decay; and Compound Interest
November 21, 2011
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
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