A Graphical Approach to Precalculus with Limits: A Unit Circle Approach illustrates how the graph of a function can be u
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English Pages ii, 934 pages: illustrations (colour; 28 cm [939] Year 2013;2014
Table of contents :
Cover......Page 1
Table of Contents......Page 4
Chapter 1 Linear Functions, Equations, and Inequalities......Page 6
Chapter 2 Analysis of Graphs of Functions......Page 86
Chapter 3 Polynomial Functions......Page 164
Chapter 4 Rational, Power, and Root Functions......Page 260
Chapter 5 Inverse, Exponential, and Logarithmic Functions......Page 324
Chapter 6 Analytic Geometry......Page 396
Chapter 7 Systems of Equations and Inequalities Matrices......Page 441
Chapter 8 The Unit Circle and the Functions of Trigonometry......Page 532
Chapter 9 Trigonometric Identities and Equations......Page 635
Chapter 10 Applications of Trigonometry and Vectors......Page 696
Chapter 11 Further Topics in Algebra......Page 769
Chapter 12 Limits, Derivatives, and Definite Integrals......Page 834
Chapter R Reference: Basic Algebraic Concepts......Page 878
Appendices......Page 912
Photo Credits......Page 927
C......Page 928
D......Page 929
F......Page 930
H......Page 931
L......Page 932
N......Page 933
P......Page 934
R......Page 935
S......Page 936
V......Page 937
Z......Page 938
9 781292 040806
A Graphical Approach to Precalculus with Limits Hornsby et al. 5e
ISBN 9781292040806
A Graphical Approach to Precalculus with Limits Hornsby Lial Rockswold Fifth Edition
Pearson New International Edition A Graphical Approach to Precalculus with Limits Hornsby Lial Rockswold Fifth Edition
Pearson Education Limited Edinburgh Gate Harlow Essex CM20 2JE England and Associated Companies throughout the world Visit us on the World Wide Web at: www.pearsoned.co.uk © Pearson Education Limited 2014
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ISBN 10: 1292040807 ISBN 10: 1269374508 ISBN 13: 9781292040806 ISBN 13: 9781269374507
British Library CataloguinginPublication Data A catalogue record for this book is available from the British Library
Printed in the United States of America
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Table of Contents Chapter 1 Linear Functions, Equations, and Inequalities John Hornsby/Margaret Lial/Gary K. Rockswold
1
Chapter 2 Analysis of Graphs of Functions John Hornsby/Margaret Lial/Gary K. Rockswold
81
Chapter 3 Polynomial Functions John Hornsby/Margaret Lial/Gary K. Rockswold
159
Chapter 4 Rational, Power, and Root Functions John Hornsby/Margaret Lial/Gary K. Rockswold
255
Chapter 5 Inverse, Exponential, and Logarithmic Functions John Hornsby/Margaret Lial/Gary K. Rockswold
319
Chapter 6 Analytic Geometry John Hornsby/Margaret Lial/Gary K. Rockswold
391
Chapter 7 Systems of Equations and Inequalities; Matrices John Hornsby/Margaret Lial/Gary K. Rockswold
436
Chapter 8 The Unit Circle and the Functions of Trigonometry John Hornsby/Margaret Lial/Gary K. Rockswold
527
Chapter 9 Trigonometric Identities and Equations John Hornsby/Margaret Lial/Gary K. Rockswold
630
Chapter 10 Applications of Trigonometry and Vectors John Hornsby/Margaret Lial/Gary K. Rockswold
691
Chapter 11 Further Topics in Algebra John Hornsby/Margaret Lial/Gary K. Rockswold
764
Chapter 12 Limits, Derivatives, and Definite Integrals John Hornsby/Margaret Lial/Gary K. Rockswold
829
Chapter R Reference: Basic Algebraic Concepts John Hornsby/Margaret Lial/Gary K. Rockswold
873
I
II
Appendices John Hornsby/Margaret Lial/Gary K. Rockswold
907
Photo Credits John Hornsby/Margaret Lial/Gary K. Rockswold
922
Index
923
1 Linear Functions, Equations, and Inequalities DURING RECENT YEARS, temperatures have risen over much of the Arctic and Antarctic regions. A sustained warming of 4°F to 11°F above average twentiethcentury temperatures could cause the Greenland ice sheet to melt. Sea levels may rise as much as 25 feet. (See Exercise 57 in Section 1.6.) This melting phenomenon is also occurring in the Antarctic, where the Wilkins Ice Shelf is melting at an alarming rate. During the past 50 years, the rate of warming at Antarctica has been about 0.9°F per decade. If the entire Antarctic ice cap were to melt, sea levels could rise over 200 feet. To bring attention to global warming, British adventurer and endurance swimmer Lewis Gordon Pugh dove into a region of the Antarctic that used to be covered by ice. What will the future bring? Linear functions can help us make predictions. In this chapter, we learn how to model many different phenomena with linear functions and equations. Source: National Snow and Ice Data Center, March 25, 2008.
CHAPTER OUTLINE 1.1 Real Numbers and the Rectangular Coordinate System 1.2 Introduction to Relations and Functions 1.3 Linear Functions 1.4 Equations of Lines and Linear Models 1.5 Linear Equations and Inequalities 1.6 Applications of Linear Functions
From Chapter 1 of A Graphical Approach to Precalculus with Limits: A Unit Circle Approach, Fifth Edition. John Hornsby, Margaret Lial, Gark K. Rockswold. Copyright © 2011 by Pearson Education, Inc. All rights reserved.
2
CHAPTER 1
Linear Functions, Equations, and Inequalities
Real Numbers and the Rectangular Coordinate System
1.1
Sets of Real Numbers • The Rectangular Coordinate System • Viewing Windows • Approximations of Real Numbers • Distance and Midpoint Formulas
Sets of Real Numbers There are several important sets of numbers that are used in mathematics. Some of these sets are listed in the following table. Sets of Numbers Set Natural Numbers Whole Numbers Integers Rational Numbers Irrational Numbers Real Numbers
Origin
–5 – 4 –3 –2 –1 0
1 2
3 4
5
Graph of {–3, –1, 0, 1, 3, 5} FIGURE 1
– 5 – 4 –3 – 2 – 1 0 1 2 3 4 5
Graph of the set of real numbers FIGURE 2
–
–1
2 3
√2 √5 0
1
Graph of
–
2 2 3,
3
0, √2, √5, , 4
FIGURE 3
4
Description 51, 2, 3, 4, Á 6 50, 1, 2, 3, 4, Á 6 5 Á ,  2,  1, 0, 1, 2, Á 6
E q  p and q are integers, q Z 0 F p
5x  x is not rational6
5x  x is a decimal number6
Examples 1, 45, 127, 103 0, 86, 345, 23 0,  5,  10 2, 99 0,  56,  2, 22 7 , 0.5 22, p,  2 37  26, p, 23, 245, 0.41
Whole numbers include the natural numbers; integers include the whole numbers and the natural numbers. The result of dividing two integers (with a nonzero divisor) is a rational number, or fraction. Rational numbers include the natural numbers, whole numbers, and integers. For example, the integer  3 is a rational number because it can be written as 13. Every rational number can be written as a repeating or terminating decimal. For example, 0.6 = 0.66666 Á represents the rational number 23. Numbers that can be written as decimal numbers are real numbers. Real numbers include rational numbers and can be shown pictorially—that is, graphed—on a number line. See FIGURE 1. The point corresponding to 0 is called the origin. Every real number corresponds to one and only one point on the number line, and each point corresponds to one and only one real number. This correspondence is called a coordinate system. The number associated with a given point is called the coordinate of the point. The set of all real numbers is graphed in FIGURE 2. Some real numbers cannot be represented by quotients of integers or by repeating or terminating decimals. These numbers are called irrational numbers. Examples of 3 10, and 2 5 20, but not 21, 24, 29, Á , irrational numbers include 23, 25, 2 which equal 1, 2, 3, Á , and hence are rational numbers. If a is a natural number but 2a is not a natural number, then 2a is an irrational number. Another irrational number is p, which is approximately equal to 3.14159. In FIGURE 3 the numbers in the set E  23, 0, 22, 25, p, 4 F are located on a number line. Note that 22 is approximately equal to 1.41, so it is located between 1 and 2, slightly closer to 1.
The Rectangular Coordinate System If we place two number lines at right angles, intersecting at their origins, we obtain a twodimensional rectangular coordinate system. The number lines intersect at the
1.1
yaxis
Quadrant Quadrant II I b P(a, b) a
xaxis
0
Origin
Quadrant Quadrant III IV Rectangular coordinate system FIGURE 4 y
A(3, 4)
B(–5, 6)
Real Numbers and the Rectangular Coordinate System
3
origin of the system, designated 0. The horizontal line is called the xaxis, and the vertical line is called the yaxis. On the xaxis, positive numbers are located to the right of the origin, and negative numbers are located to the left. On the yaxis, positive numbers are located above the origin, negative numbers below. This rectangular coordinate system is also called the Cartesian coordinate system, named after René Descartes (1596–1650). The plane into which the coordinate system is introduced is the coordinate plane, or xyplane. The xaxis and yaxis divide the plane into four regions, or quadrants, as shown in FIGURE 4. The points on the xaxis or yaxis belong to no quadrant. Each point P in the xyplane corresponds to a unique ordered pair (a, b) of real numbers. We call a the xcoordinate and b the ycoordinate of point P. The point P corresponding to the ordered pair (a, b) is often written as P(a, b), as in FIGURE 4, and referred to as “the point (a, b).” FIGURE 5 illustrates how to plot the point A(3, 4). Additional points are labeled B–E. The coordinates of the origin are (0, 0).
4 E(–3, 0)
0
C(–2, –4)
x
3
Viewing Windows D(4, – 3)
The rectangular (Cartesian) coordinate system extends indefinitely in all directions. We can show only a portion of such a system in a text figure. Similar limitations occur with the viewing “window” on a calculator screen. FIGURE 6 shows a calculator screen that has been set to have a minimum xvalue of  10, a maximum xvalue of 10, a minimum yvalue of  10, and a maximum yvalue of 10. The tick marks on the axes have been set to be 1 unit apart. Thus, there are 10 tick marks on the positive xaxis. This window is called the standard viewing window. To convey information about a viewing window, we use the following abbreviations:
Plotting points in the xyplane FIGURE 5
10
–10
10
Xmin: minimum value of x Xmax: maximum value of x Xscl: scale (distance between tick marks) on the xaxis
–10 Standard viewing window FIGURE 6
GCM
Ymin: minimum value of y Ymax: maximum value of y Yscl: scale (distance between tick marks) on the yaxis.
To further condense this information, we use the following symbolism, which gives viewing information for the window in FIGURE 6.
TECHNOLOGY NOTE You should consult the graphing calculator manual that accompanies this text or your owner’s manual to see how to set the viewing window on your screen. Remember that different settings will result in different views of graphs.
Xmax Ymin
Xmin
3 10, 104 by 3 10, 104. Xscl = 1 Yscl = 1
Ymax
FIGURE 7 shows several other viewing windows. Notice that FIGURES 7(b) and 7(c) look exactly alike, and unless we are told what the settings are, we have no way of distinguishing between them. What are Xscl and Yscl in each figure?
3.1
–4.7
100
10
4.7
–10
10
–100
100
–3.1
–10
–100
[–4.7, 4.7] by [–3.1, 3.1]
[–10, 10] by [–10, 10]
[–100, 100] by [–100, 100]
(a)
(b)
(c)
FIGURE 7
4
CHAPTER 1
Linear Functions, Equations, and Inequalities
WHAT WENT WRONG? A student learning how to use a graphing calculator could not understand why the axes on the graph were so “thick,” as seen in FIGURE A , while those on a friend’s calculator were not, as seen in FIGURE B. 40
–60
40
60
–60
60
–40
–40
Xscl = 1, Yscl = 1
Xscl = 10, Yscl = 10
FIGURE A
FIGURE B
What Went Wrong? How can the student correct the problem in FIGURE A so that the axes look like those in FIGURE B?
Approximations of Real Numbers Although calculators have the capability to express numbers like 22, 2 3 5, and p to many decimal places, we often ask that answers be rounded. The following table reviews rounding numbers to the nearest tenth, hundredth, or thousandth. TI84 Plus FIGURE 8
Rounding Numbers Number 1.3782 201.6666 0.0819
FIGURE 9
Nearest Tenth
Nearest Hundredth
Nearest Thousandth
1.4 201.7 0.1
1.38 201.67 0.08
1.378 201.667 0.082
In FIGURE 8, the TI84 Plus graphing calculator is set to round values to the nearest hundredth (two decimal places). In FIGURE 9, the numbers from the preceding table are rounded to the nearest hundredth. The symbol L indicates that two expressions are approximately equal. For example, p L 3.14, but p Z 3.14, since p = 3.141592654. Á When using p in calculations, be sure to use the builtin key for p rather than 3.14. See FIGURE 10.
FIGURE 10
Answer to What Went Wrong? Since Xscl = 1 and Yscl = 1 in FIGURE A, there are 120 tick marks along the xaxis and 80 tick marks along the yaxis. The resolution of the graphing calculator screen is not high enough to show all these tick marks, so the axes appear as heavy black lines instead. The values for Xscl and Yscl need to be larger, as in FIGURE B.
1.1
GCM
Real Numbers and the Rectangular Coordinate System
5
EXAMPLE 1
Finding Roots on a Calculator Approximate each root to the nearest thousandth. (Note: You can use the fact that n 2a = a 1/n to find roots.) (a) 223
3 87 (b) 2
4 12 (c) 2
Solution (a) The screen in FIGURE 11(a) shows that an approximation for 223, to the nearest thousandth, is 4.796. It is displayed twice, once for 223 and once for 231/2. 3 87 L 4.431. See FIGURE 11(b). (b) To the nearest thousandth, 2 (c)
FIGURE 11(c)
indicates 2 4 12 L 1.861 in three different ways.
TECHNOLOGY NOTE
In all the screens, note the inclusion of parentheses.
Many graphing calculators have builtin keys for calculating square roots and menus for calculating other types of roots.
(a)
(b)
(c)
FIGURE 11
EXAMPLE 2
Approximating Expressions with a Calculator Approximate each expression to the nearest hundredth. 3.8  1.4 (a) (b) 3p4  9 2 (c) 2(4  1) 2 + ( 3  2) 2 5.4 + 3.5
TECHNOLOGY NOTE Some graphing calculators display leading zeros in decimal numbers, whereas others do not. For example, 14 might be displayed as either 0.25 or .25. In this text, graphing calculator screens do not usually show leading zeros. See FIGURES 9 and 12(a).
Solution (a) See FIGURE 12(a). To the nearest hundredth, 3.8  1.4 L 0.27. 5.4 + 3.5 (b) Many calculators also have a special key to calculate the square of a number. To the nearest hundredth, 3p4  9 2 L 211.23. See FIGURE 12(b). (c) From FIGURE 12(c), 2(4  1) 2 + ( 3  2) 2 L 5.83. Do not confuse the negation and subtraction symbols.
Insert parentheses around both the numerator and the denominator.
(a)
(b) FIGURE 12
Use parentheses carefully.
(c)
6
CHAPTER 1
Linear Functions, Equations, and Inequalities
WHAT WENT WRONG? Two students were asked to compute the expression (2 + 9)  (8 + 13) on a TI83/84 Plus calculator. One student obtained the answer  10, as seen in FIGURE A , while the other obtained  231, as seen in FIGURE B.
FIGURE A
FIGURE B
What Went Wrong? Compute the expression by hand to determine which screen gives the correct answer. Why is the answer on the other screen incorrect?
Distance and Midpoint Formulas The Pythagorean theorem can be used to calculate the lengths of the sides of a right triangle.
Pythagorean Theorem In a right triangle, the sum of the squares of the lengths of the legs is equal to the square of the length of the hypotenuse.
a2 ⴙ b 2 ⴝ c 2
Leg a
Hypotenuse c Leg b
y
NOTE The converse of the Pythagorean theorem is also true. That is, if a, b, and c are lengths of the sides of a triangle and a 2 + b 2 = c 2, then the triangle is a right triangle with hypotenuse c. For example, if a triangle has sides with lengths 3, 4, and 5, then it is a right triangle with hypotenuse of length 5 because 32 + 42 = 52.
⏐x2 – x1⏐ P(x1 , y1)
Q(x2 , y1)
d
x
0
⏐y2 – y1⏐ R(x2 , y2 )
d(P, R) = √(x2 – x1)2 + ( y2 – y1)2 FIGURE 13
To derive a formula to find the distance between two points in the xyplane, let P(x1, y1) and R(x2, y2) be any two distinct points in the plane, as shown in FIGURE 13. Complete a right triangle by locating point Q with coordinates (x2, y1). The Pythagorean theorem gives the distance between P and R as d(P, R) = 2(x2  x1) 2 + ( y2  y1) 2.
Answer to What Went Wrong? The correct answer is  10, as shown in FIGURE A. FIGURE B gives an incorrect answer because the negation symbol is used, rather than the subtraction symbol. The calculator computed 2 + 9 = 11 and then multiplied by the negative of 8 + 13 (that is,  21), to obtain the incorrect answer,  231.
1.1
Real Numbers and the Rectangular Coordinate System
7
NOTE Absolute value bars are not necessary in this formula, since, for all real numbers a and b,  a  b  2 = 1a  b2 2.
Distance Formula Suppose that P(x1, y1) and R(x2, y2) are two points in a coordinate plane. Then the distance between P and R, written d(P, R), is given by the distance formula d(P, R) ⴝ 2(x2 ⴚ x1)2 ⴙ ( y2 ⴚ y1)2.
y
EXAMPLE 3
Using the Distance Formula Use the distance formula to find d(P, Q) in FIGURE 14. Solution
P(–8, 4) 4
d(P, Q) = 2(x2  x1) 2 + ( y2  y1) 2
2 x
= 233  ( 8)42 + ( 2  4) 2
Be careful when subtracting a = 2112 + ( 6) 2 negative number.
Q(3, –2)
FIGURE 14
(x2, y2)
M (x1, y1) Point M is the midpoint of the segment joining (x1, y1) and (x2, y2).
x1 =  8, y1 = 4, x2 = 3, y2 =  2
= 2121 + 36
Apply exponents.
= 2157
Leave in radical form.
The midpoint of a line segment is the point on the segment that lies the same distance from both endpoints. See FIGURE 15. The coordinates of the midpoint are found by calculating the average of the xcoordinates and the average of the ycoordinates of the endpoints of the segment.
y
0
Distance formula
x
Midpoint Formula The midpoint of the line segment with endpoints (x1, y1) and (x2, y2) has coordinates a
FIGURE 15
x1 ⴙ x2 y1 ⴙ y2 , b. 2 2
EXAMPLE 4
Using the Midpoint Formula Find the coordinates of the midpoint M of the segment with endpoints (8,  4) and ( 9, 6). Solution Let (x1, y1) = (8,  4) and (x2, y2) = ( 9, 6) in the midpoint formula. a
x1 + x2 y1 + y2 8 + ( 9)  4 + 6 , b = a , b 2 2 2 2 1 = a  , 1b 2
Substitute.
Simplify.
8
CHAPTER 1
Linear Functions, Equations, and Inequalities
EXAMPLE 5
Estimating Tuition and Fees In 2002, average tuition and fees at public colleges and universities were $4081, whereas they were $6585 in 2008. Use the midpoint formula to estimate tuition and fees in 2005. Compare your estimate with the actual value of $5491. (Source: The College Board.) Solution The year 2005 lies midway between 2002 and 2008. Therefore, we can find the midpoint of the line segment joining (2002, 4081) and (2008, 6585). a
2002 + 2008 4081 + 6585 , ≤ = (2005, 5333) 2 2
The midpoint formula estimates tuition and fees at public colleges and universities to be $5333 in 2005. This is within $158 of the actual value.
1.1
EXERCISES
For each set, list all elements that belong to the (a) natural numbers, (b) whole numbers, (c) integers, (d) rational numbers, (e) irrational numbers, and (f ) real numbers. 1. e  6, 
12 5 ,  ,  23, 0, 0.31, 0.3, 2p, 10, 217 f 4 8
3. e  2100, 
13 22 ,  1, 5.23, 9.14, 3.14, f 6 7
2. e  8, 
14 6 ,  0.245, 0, , 8, 281, 212 f 7 2
4. 5 249,  0.405,  0.3, 0.1, 3, 18, 6p, 566
Classify each number as one or more of the following: natural number, integer, rational number, or real number. 5. 10,600,000,000,000 (The federal debt in dollars in January 2009)
6. 700,000,000,000 (The federal 2008 bailout fund in dollars)
7.  17 (The percent change in CD sales from 2006 to 2007)
8.  3 (The annual percent change in the area of tropical rain forests)
(The fractional increase in the cost of living from 2001 to 2008)
10.  3.5 (The amount in billions of dollars that the Motion Picture Association of America estimates is lost annually due to piracy)
11. 5 22 (The length of the diagonal of a square measuring 5 units on each side)
12. p (The ratio of the circumference of a circle to its diameter)
9.
1 5
Concept Check For each measured quantity, state the set of numbers that is most appropriate to describe it. Choose from the natural numbers, integers, and rational numbers. 13. Populations of cities
14. Distances to nearby cities on road signs
15. Shoe sizes
16. Prices paid (in dollars and cents) for gasoline tank fillups
17. Daily low winter temperatures in U.S. cities
18. Golf scores relative to par
Graph each set of numbers on a number line. 19. 5 4,  3,  2,  1, 0, 16
20. 5 6,  5,  4,  3,  26
5 21. e  0.5, 0.75, , 3.5 f 3
9 13 22. e  0.6, , 2.5, f 8 4
1.1
9 23.
Real Numbers and the Rectangular Coordinate System
9
24. Concept Check Using her calculator, a student found the decimal 1.414213562 when she evaluated 22. Is this decimal the exact value of 22 or just an approximation of 22? Should she write 22 = 1.414213562 or 22 L 1.414213562?
Explain the distinction between a rational number and an irrational number.
Locate each point on a rectangular coordinate system. Identify the quadrant, if any, in which each point lies. 25. (2, 3)
26. ( 1, 2)
27. ( 3,  2)
28. (1,  4)
29. (0, 5)
30. ( 2,  4)
31. ( 2, 4)
32. (3, 0)
33. ( 2, 0)
34. (3,  3)
Name the possible quadrants in which the point (x, y) can lie if the given condition is true. x 35. xy 7 0 36. xy 6 0 37. 6 0 y 39. Concept Check If the xcoordinate of a point is 0, the point must lie on which axis?
38.
x 7 0 y
40. Concept Check If the ycoordinate of a point is 0, the point must lie on which axis?
Give the values of Xmin, Xmax, Ymin, and Ymax for each screen, given the values for Xscl and Yscl. Use the notation described in this section. 41.
42.
43.
Xscl = 10, Yscl = 50
Xscl = 5, Yscl = 1
Xscl = 1, Yscl = 5
44.
45.
Xscl = 50, Yscl = 10
46.
Xscl = 100, Yscl = 100
Xscl = 75, Yscl = 75
Set the viewing window of your calculator to the given specifications. Make a sketch of your window. 47. 3 10, 104 by 3 10, 104 Xscl = 1 Yscl = 1
50. 3 3.5, 3.54 by 3 4, 104 Xscl = 1 Yscl = 1
9 53.
48. 3 40, 404 by 3 30, 304 Xscl = 5 Yscl = 5
51. 3 100, 1004 by 3 50, 504 Xscl = 20 Yscl = 25
Set your viewing window to 3 10, 104 by 3 10, 104, and then set Xscl to 0 and Yscl to 0. What do you notice? Make a conjecture as to how to set the screen with no tick marks on the axes.
9 54.
49. 3 5, 104 by 3 5, 104 Xscl = 3 Yscl = 3
52. 3 4.7, 4.74 by 3 3.1, 3.14 Xscl = 1 Yscl = 1
Set your viewing window to 3 50, 504 by 3  50, 504, Xscl to 1, and Yscl to 1. Describe the appearance of the axes compared with those seen in the standard window. Why do you think they appear this way? How can you change your scale settings so that this “problem” is alleviated?
Find a decimal approximation of each root or power. Round answers to the nearest thousandth. 55. 258
56. 297
57. 2 3 33
58. 2 3 91
59. 2 4 86
60. 2 4 123
61. 19 1/2
62. 29 1/3
63. 461.5
64. 232.75
10
CHAPTER 1
Linear Functions, Equations, and Inequalities
Approximate each expression to the nearest hundredth. 65.
5.6  3.1 8.9 + 1.3
66.
34 + 25 23
67. 2p3 + 1
3 2.1  62 68. 2
69. 3(5.9) 2  2(5.9) + 6
70. 2p3  5p2  3
71. 2(4  6) 2 + (7 + 1) 2
72. 23 1  ( 3)42 + ( 5  3) 2
73.
74. 2 3 4.5 * 10 5 + 3.7 * 10 2
75.
2
2p  1 21 + p 4.5
76. 1 
1  2 35
3  22
Find the length of the unknown side of the right triangle. In each case, a and b represent the lengths of the legs and c represents the length of the hypotenuse. c a
b Typical labeling
77. a = 8, b = 15; find c
78. a = 7, b = 24; find c
79. a = 13, c = 85; find b
80. a = 14, c = 50; find b
81. a = 5, b = 8; find c
82. a = 9, b = 10; find c
83. b = 213, c = 229; find a
84. b = 27, c = 211; find a
Find (a) the distance between P and Q and (b) the coordinates of the midpoint of the segment joining P and Q. y
85.
y
86.
87.
y
88.
y
Q(1, 4)
P(–3, 4) P(–7, 4)
Q(2, 5) x
0
P(–4, 3) 0
x
x
0
Q(2, –1)
0
Q(6, –2) P(–3, –3)
89. P(5, 7), Q(2, 11)
90. P( 2, 5), Q(4,  3)
91. P( 8,  2), Q( 3,  5)
92. P( 6,  10), Q(6, 5)
93. P(9.2, 3.4), Q(6.2, 7.4)
94. P(8.9, 1.6), Q(3.9, 13.6)
95. P(13x,  23x), Q(6x, x), x 7 0
96. P(12y,  3y), Q(20y, 12y),
y 7 0
Suppose that P is an endpoint of a segment PQ and M is the midpoint of PQ. Find the coordinates of endpoint Q. 97. P(7,  4), M(8, 5) 99. P(5.64, 8.21), M( 4.04, 1.60)
98. P(13, 5), M( 2,  4) 100. P( 10.32, 8.55), M(1.55,  2.75)
x
1.1
Solve each problem. 101. (Modeling)Tuition and Fees From 2004–2008, the average annual cost (in dollars) of tuition and fees at private fouryear colleges rose in an approximately linear fashion. The graph depicts this growth with a line segment. Use the midpoint formula to approximate the cost during the year 2006.
Cost (in dollars)
y
(2008, 25,143)
25,000 20,000
(2004, 20,082) 15,000 10,000 x 2004
2006
2008
Year Source: The College Board.
102. (Modeling)TwoYear College Enrollment Estimated enrollments in twoyear colleges for 2005, 2007, and 2009 are shown in the table. Use the midpoint formula to estimate the enrollments to the nearest thousand for 2006 and 2008. Year
Enrollment (in thousands)
2005 2007 2009
6038 6146 6257
Real Numbers and the Rectangular Coordinate System
(b) An equilateral triangle has all sides of equal length. Determine whether the triangle with vertices ( 1,  1), (2, 3), and ( 4, 3) is equilateral. (c) Determine whether a triangle having vertices ( 1, 0), (1, 0) and (0, 23) is isosceles, equilateral, or neither. (d) Determine whether a triangle having vertices ( 3, 3), ( 2, 5) and ( 1, 3) is isosceles, equilateral, or neither. 105. Distance between Cars At 9:00 A.M., Car A is traveling north at 50 mph and is located 50 miles south of Car B. Car B is traveling west at 20 mph. (a) Let (0, 0) be the initial coordinates of Car B in the xyplane, where units are in miles. Plot the locations of each car at 9:00 A.M. and at 11:00 A.M. (b) Find the distance between the cars at 11:00 A.M. 106. Distance between Ships Two ships leave the same harbor at the same time. The first ship heads north at 20 mph and the second ship heads west at 15 mph. (a) Draw a sketch depicting their positions after t hours. (b) Write an expression that gives the distance between the ships after t hours. 107. One of the most popular proofs of the Pythagorean theorem uses the figure shown here. Determine the area of the figure in two ways. First, find the area of the large square, using the formula for the area of a square. Then, find its area as the sum of the areas of the smaller square and the four right triangles. Set these expressions equal to each other and simplify to obtain a 2 + b 2 = c 2.
Source: Statistical Abstract of the United States.
103. (Modeling)PovertyLevel Income Cutoffs The table lists povertylevel income cutoffs for a family of four for selected years. Use the midpoint formula to estimate the povertylevel cutoffs (rounded to the nearest dollar) in 2001 and 2005. Year
Income (in dollars)
1999 2003 2007
17,029 18,810 21,203
b a
a c c
b
c
b
c a
a b
108. Prove that the midpoint of the line segment joining endpoints P(x1, y1) and Q(x2, y2) is
Source: U.S. Census Bureau.
Ma 104. Geometry Triangles can be classified by their sides. (a) An isosceles triangle has at least two sides of equal length. Determine whether the triangle with vertices (0, 0), (3, 4), and (7, 1) is isosceles.
11
x1 + x2 y1 + y2 , b 2 2
by showing that the distance between P and M is equal to the distance between M and Q and that the sum of these distances is equal to the distance between P and Q.
12
CHAPTER 1
1.2
Linear Functions, Equations, and Inequalities
Introduction to Relations and Functions
SetBuilder Notation and Interval Notation • Relations, Domain, and Range • Functions • Tables and Graphing Calculators • Function Notation
SetBuilder Notation and Interval Notation Suppose we wish to symbolize the set of real numbers greater than  2. The following are two ways to do this.
1. In setbuilder notation, we write 5x  x 7  26, which is read “the set of all real numbers x such that x is greater than  2.” 2. In interval notation, we write ( 2, q). ■ In this case, a left parenthesis “(” indicates that the endpoint,  2, is not included. For 5x  x 6  26, we would use a right parenthesis and write ( q,  2). ■ A square bracket, either [ or ], would indicate that the endpoint is included. ■ The infinity symbol q does not represent a number; rather, it shows that the interval includes all numbers greater than  2. ■ A parenthesis is always used next to the infinity symbol.
–4
–2
0
2
5x  x 7  26, or ( 2, q)
4
A number line graph of 5x  x 7  26 is shown in FIGURE 16. The following chart summarizes setbuilder notation, interval notation, and graphs of intervals of real numbers. It is assumed that a 6 b.
FIGURE 16
Type of Interval
SetBuilder Notation
Corresponding Interval Notation
Open interval
5x  a 6 x 6 b6
(a, b)
Closed interval
5x  a … x … b6
3a, b4
Halfopen (or halfclosed) interval
Unbounded interval
All real numbers
⎧5x  a 6 x … b6 ⎪ ⎨ ⎪5x  a … x 6 b6 ⎩
⎧5x  x ⎪ ⎪ ⎪5x  x ⎪ ⎨ ⎪5x  x ⎪ ⎪ ⎪5x  x ⎩
Corresponding Graph
a
b
a
b
a
b
(a, b4 3a, b)
b
a
7 a6
(a, q)
Ú a6
3a, q)
6 a6
( q, a)
… a6
( q, a4
a
a
a
5x  x is real6
a
( q, q)
a
b
CAUTION Interval notation for the open interval ( a, b ) looks exactly like the notation for the ordered pair ( a, b ). When the need arises, we will distinguish between them by referring to “the interval ( a, b )” or “the point ( a, b ).”
1.2
Introduction to Relations and Functions
13
EXAMPLE 1
Writing Interval Notation Write each set of real numbers in interval notation. (a) 5x   3 6 x … 56 (b) 5x  x … 36 (c) 5x  3 6 x6 Solution (a) Use ( for 6 and 4 for … . Thus we write ( 3, 54. (b) This interval is unbounded, so we write ( q, 34. (c) Because 3 6 x is equivalent to x 7 3, we write (3, q).
Relations, Domain, and Range The table shows the atmospheric carbon dioxide (CO2) concentration in parts per million (ppm) for selected years. Atmospheric Carbon Dioxide Concentrations Year
1958
1975
1990
2005
CO2 (ppm)
315
335
355
380
Source: Mauna Loa Observatory.
Since each year in the table is paired with a CO2 concentration, we can depict this information as a set of ordered pairs in the form (year, concentration). The first component is the year, and the second component is the concentration in parts per million. 5(1958, 315), (1975, 335), (1990, 355), (2005, 380)6
Such a set of ordered pairs is called a relation.
Relation A relation is a set of ordered pairs.
If we denote the ordered pairs of a relation by (x, y), then the set of all xvalues is called the domain of the relation and the set of all yvalues is called the range of the relation. For the relation represented by the table, and
Domain = 51958, 1975, 1990, 20056 Range = 5315, 335, 355, 3806.
Here are three other examples of relations:
F = 5(1, 2), ( 2, 5), (3,  1)6, G = 5( 2, 1), ( 1, 0), (0, 1), (1, 2), (2, 2)6, and H = 5( 4, 1), ( 2, 1), ( 2, 0)6.
For the relations F, G, and H,
Domain of F = 51,  2, 36, Domain of G = 5  2,  1, 0, 1, 26, Domain of H = 5 4,  26,
Range of F = 52, 5,  16; Range of G = 51, 0, 26; Range of H = 51, 06.
Since a relation is a set of ordered pairs, it may be represented graphically in the rectangular coordinate system. See FIGURE 17 on the next page.
14
Linear Functions, Equations, and Inequalities
CHAPTER 1
F = 5(1, 2), ( 2, 5), (3,  1)6
G = 5( 2, 1), ( 1, 0), (0, 1), (1, 2), (2, 2)6
y
y
y
(–2, 5)
H = 5( 4, 1), ( 2, 1), ( 2, 0)6
(1, 2) (2, 2)
(1, 2)
(–2, 1)
1 0
(–4, 1) (–2, 1) 1
x 1
(–1, 0)
0
(0, 1)
1 x
1
(–2, 0)
x
0
1
(3, –1) Graph of F
Graph of G
Graph of H
FIGURE 17
Some relations contain infinitely many ordered pairs. For example, let M represent a relation consisting of all ordered pairs having the form (x, 2x), where x is a real number. Since there are infinitely many values for x, there are infinitely many ordered pairs in M. Five such ordered pairs are plotted in FIGURE 18(a) and suggest that the graph of M is a line. The graph of M includes all points (x, y), such that y = 2x, as shown in FIGURE 18(b).
x y –2 –4 –1 0 1 2
–2 0 2 4
y
y (2, 4) 4 3 2 (1, 2) 1 (0, 0) 4 3 2 1 1 1 2 3 4 (–1, –2) 2 3 4 (–2, –4)
x
4 3 2 1
y 2x
x
4 3 2 11 1 2 3 4 2 3 4
(a)
(b) FIGURE 18
–2
5
1
2
3
–1
Domain
Range Mapping of F FIGURE 19
A relation can be illustrated with a “mapping” diagram, as shown in FIGURE 19 for relation F. The arrow from 1 to 2 indicates that the ordered pair (1, 2) belongs to F. In summary, a relation can be represented by any of the following. 1. A graph, as illustrated in FIGURE 17 and FIGURE 18 NOTE A graph of a relation can consist of distinct points, a line, or a curve.
2. A table of xyvalues, as shown in FIGURE 18(a) 3. An equation, such as y = 2x in FIGURE 18(b) 4. A mapping or diagram, as illustrated in FIGURE 19 EXAMPLE 2
Determining Domains and Ranges from Graphs Give the domain and range of each relation from its graph. Domain (a) (b) (c) y y
y
6
4
x
Range
0
–4
0
x
0 –3
–6
2
x
1.2
Introduction to Relations and Functions
15
Solution (a) The xvalues of the points on the graph include all numbers between  4 and 4 inclusive. The yvalues include all numbers between  6 and 6 inclusive. Using interval notation, the domain is 3 4, 44, and the range is 3 6, 64. (b) The arrowheads indicate that the line extends indefinitely left and right, as well as upward and downward. Therefore, both the domain and range are the set of all real numbers, written ( q, q). (c) The arrowheads indicate that the graph extends indefinitely left and right, as well as upward. The domain is ( q, q). There is a least yvalue,  3, so the range includes all numbers greater than or equal to  3, written 3 3, q). 5
–5
EXAMPLE 3
5
–5 FIGURE 20
Finding Domain and Range from a Calculator Graph FIGURE 20 shows a graph on a screen with viewing window 3 5, 54 by 3 5, 54, Xscl = 1, Yscl = 1. Give the domain and range of this relation. Solution Since the scales on both axes are 1, the graph appears to have minimum xvalue  3, maximum xvalue 3, minimum yvalue  2, and maximum yvalue 2. Observation leads us to conclude that the domain is 3 3, 34 and the range is 3 2, 24.
Functions TECHNOLOGY NOTE In FIGURE 20, we see a calculator graph that is formed by a rather jagged curve. These representations are sometimes called jaggies and are typically found on lowresolution graphers, such as graphing calculators. In general, most curves in this book are smooth, and jaggies are just a part of the limitations of technology.
Suppose that a sales tax rate is 6%. Then a purchase of $200 results in a sales tax of 0.06 * $200 = $12. This calculation can be summarized by the ordered pair (200, 12). The ordered pair (50, 3) indicates that a purchase of $50 results in a sales tax of $3. For each purchase of x dollars, there is exactly one sales tax amount of y dollars. Calculating sales tax y on a purchase of x dollars results in a set of ordered pairs (x, y), where y = 0.06x. This set of ordered pairs represents a special type of relation called a function. In a function, each xvalue must correspond to exactly one yvalue.
Function A function is a relation in which each element in the domain corresponds to exactly one element in the range.* If x represents any element in the domain, x is called the independent variable. If y represents any element in the range, y is called the dependent variable, because the value of y depends on the value of x. For example, sales tax depends on purchase price. EXAMPLE 4
Deciding whether Relations are Functions Give the domain and range of each relation. Decide whether each relation is a function. (a) 5(1, 2), (3, 4), (5, 6), (7, 8), (9, 10)6 (b) 5(1, 1), (1, 2), (1, 3), (2, 4)6 (c)
D = 51, 3, 5, 7, 96 R = 52, 4, 6, 8, 106
x
4
3
2
1
0
1
y
2
2
2
2
2
2
(d) 5(x, y)  y = x  26
Solution (a) The domain D is 51, 3, 5, 7, 96, and the range R is 52, 4, 6, 8, 106. Since each element in the domain corresponds to exactly one element in the range, this set is a function. The correspondence is shown in the margin. (continued) *An alternative definition of function based on the idea of correspondence is given later in the section.
16
CHAPTER 1
Linear Functions, Equations, and Inequalities
(b) The domain D here is 51, 26, and the range R is 51, 2, 3, 46. As shown in the correspondence in the margin, one element in the domain, 1, has been assigned three different elements from the range, so this relation is not a function.
D = 51, 26 R = 51, 2, 3, 46
(c) In the table of ordered pairs, the domain is 5  4,  3,  2,  1, 0, 16 and the range is 526. Although every element in the domain corresponds to the same element in the range, this is a function because each element in the domain has exactly one range element assigned to it. (d) In 5(x, y)  y = x  26, y is always found by subtracting 2 from x. Each x corresponds to just one value of y, so this relation is a function. Any number can be used for x, and each x will give a number that is 2 less for y; thus, both the domain and range are the set of real numbers: ( q, q).
y
y3
y1
y2 x1
x2
x3
x
This is the graph of a function. (a) y
Functions are often defined by equations, such as y = x  2 in Example 4(d), where each (valid) xvalue determines a unique yvalue. Equations that define functions are usually solved for the independent variable y. There is a quick way to tell whether a given graph is the graph of a function. In the graph in FIGURE 21(a), each value of x corresponds to only one value of y, so this is the graph of a function. By contrast, the graph in FIGURE 21(b) is not the graph of a function. The vertical line through x1 intersects the graph at two points, showing that there are two values of y that correspond to this xvalue. This concept is known as the vertical line test for a function.
y1
Vertical Line Test x1
0
y2
This is not the graph of a function. (b) FIGURE 21
x
If every vertical line intersects a graph in no more than one point, then the graph is the graph of a function.
EXAMPLE 5
Using the Vertical Line Test
(a) Is the graph in FIGURE 22 the graph of a function? y
y 3
3
–3 0
–3 –1
FIGURE 22
0
x 3
x 3 –3
FIGURE 23
(b) Assuming that the graph in FIGURE 23 extends right indefinitely and upward and downward indefinitely, does it appear to be the graph of a function? Solution (a) Any vertical line will intersect the graph at most once. Therefore, the graph satisfies the vertical line test and is the graph of a function. (b) Any vertical line that is right of the origin intersects the graph twice. Therefore, the graph does not represent a function.
1.2
Introduction to Relations and Functions
17
While the concept of a function is crucial to the study of mathematics, its definition may vary from text to text. We now give an alternative definition.
Function (Alternative Definition) A function is a correspondence in which each element x from a set called the domain is paired with one and only one element y from a set called the range.
Tables and Graphing Calculators A convenient way to display ordered pairs in a function is by using a table. An equation such as y = 9x  5 describes a function. If we choose xvalues to be 0, 1, 2, Á , 6, then the corresponding yvalues are y = 9(0)  5 =  5
Use parentheses around substituted y = 9(2)  5 = 13 values to avoid y = 9(4)  5 = 31 errors.
y = 9(1)  5 = 4 y = 9(3)  5 = 22 y = 9(5)  5 = 40
y = 9(6)  5 = 49.
These ordered pairs, (0,  5), (1, 4), Á , (6, 49), can be organized in a table. A graphing calculator can efficiently generate this table, as shown in FIGURE 24. GCM
TECHNOLOGY NOTE
This screen indicates that the table will start with 0 and have an increment of 1. Both variables appear automatically. TblStart represents the initial value of the independent variable (0 in the table in FIGURE 24) and ⌬Tbl represents the difference, or increment, between successive values of the independent variable (1 in FIGURE 24 because 1 – 0 = 1, 2 – 1 = 1, 3 – 2 = 1, and so on).
x
y
0 1 2 3 4 5 6
5 4 13 22 31 40 49
FIGURE 24
Function Notation To say that y is a function of x means that for each value of x from the domain of the function, there is exactly one value of y. To emphasize that y is a function of x, or that y depends on x, it is common to write y ⴝ ƒ(x), with ƒ(x) read “ƒ of x.” This notation is called function notation. Function notation is used frequently when functions are defined by equations. For the function defined by y = 9x  5, we may name this function ƒ and write ƒ(x) = 9x  5. Note that ƒ(x) equals y. If x = 2, then we find y, or ƒ(2), by replacing x with 2. ƒ(2) = 9 # 2  5 = 13
The statement “if x = 2, then y = 13” represents the ordered pair (2, 13) and is abbreviated with function notation as ƒ(2) = 13.
Also, ƒ(0) = 9 # 0  5 =  5, and ƒ(  3) = 9( 3)  5 =  32.
18
CHAPTER 1
Linear Functions, Equations, and Inequalities
These ideas can be explained as follows: Name of the function
Defining expression
⎧ ⎨ ⎩
y =
ƒ(x)
Value of the function
= 9x  5. Name of the independent variable
CAUTION The symbol ƒ( x ) does not indicate “ƒ times x ,” but represents the yvalue for the indicated xvalue. For example, ƒ(2) is the yvalue that corresponds to the xvalue 2.
We can use variables other than ƒ, such as g and h, to represent functions.
Looking Ahead to Calculus
Functions are evaluated frequently in calculus. They can be evaluated in a variety of ways, as illustrated in Example 6.
EXAMPLE 6
Using Function Notation For each function, find ƒ(3). (a) ƒ(x) = 3x  7 (b) The function ƒ depicted in FIGURE 25 (c) The function ƒ graphed in FIGURE 26 y
f –2
6
4
3
5
2
10
12 0
Domain
y = f (x)
x 2
4
Range
FIGURE 25
FIGURE 26
(d) The function ƒ defined by the table x ƒ(x)
3
4
 15
 12
9
6
(b) In FIGURE 25, 3 in the domain is paired with 5 in the range, so ƒ(3) = 5.
y = f(x)
2 0
2
Solution (a) Replace x with 3 to get ƒ(3) = 3(3)  7 = 2.
y
4
1
x 2 3 4
FIGURE 27
(c) To evaluate ƒ(3), begin by finding 3 on the xaxis. See FIGURE 27. Then move upward until the graph of ƒ is reached. Moving horizontally to the yaxis gives 4 for the corresponding yvalue. Thus, ƒ(3) = 4. Note that the point (3, 4) lies on the graph of ƒ. (d) From the table, ƒ(3) =  9.
When x = 3, y =  9.
NOTE If ƒ ( a ) = b, then the point ( a , b ) lies on the graph of ƒ. Conversely, if the point ( a , b ) lies on the graph of ƒ, then ƒ ( a ) = b . Thus, each point on the graph of ƒ can be written in the form ( a , ƒ ( a )) .
1.2
1.2
Introduction to Relations and Functions
19
EXERCISES
Checking Analytic Skills Sketch the graph of ƒ by hand. Do not use a calculator. 1. ƒ(x) = x  3 1 x 2
5. ƒ(x) =
2. ƒ(x) = 1  2x
3. ƒ(x) = 3
4. ƒ(x) =  4
2 6. ƒ(x) =  x 3
7. ƒ(x) = x 2
8. ƒ(x) =  x 
Using interval notation, write each set. Then graph it on a number line. 9. 5x   1 6 x 6 46
10. 5x  x Ú  36
12. 5x  8 7 x 7 36
11. 5x  x 6 06
14. 5x   5 6 x …  46
13. 5x  1 … x 6 26
Using the variable x, write each interval using setbuilder notation. 16. 32, 7)
15. ( 4, 3)
17. ( q,  14
19.
20. –2
0
0
6
21.
8
22. –4
9 23.
18. (3, q)
0
0
3
24. Concept Check The threepart inequality a 6 x 6 b means “a is less than x, and x is less than b.” Which one of the following inequalities is not satisfied by some real number x? A.  3 6 x 6 5 B. 0 6 x 6 4 C.  3 6 x 6  2 D.  7 6 x 6  10
Explain how to determine whether a parenthesis or a square bracket is used when graphing an inequality on a number line.
Determine the domain D and range R of each relation, and tell whether the relation is a function. Assume that a calculator graph extends indefinitely and a table includes only the points shown. 25. 5(5, 1), (3, 2), (4, 9), (7, 6)6
26. 5(8, 0), (5, 4), (9, 3), (3, 8)6
27. 5(1, 6), (2, 6), (3, 6)6
28. 5( 10, 5), ( 20, 5), ( 30, 5)6
29. 5(4, 1), (3,  5), ( 2, 3), (3, 7)6
30. 5(0, 5), (1, 3), (0,  4)6
31. x
11
12
13
14
32. x
1
1
1
1
y
6
6
7
6
y
12
13
14
15
33. x
0
1
2
3
4
34. x
1
1 2
1 4
1 8
1 16
22
23
25
26
27
y
0
1
2
3
4
y
35.
36.
y
y
3
4
2 0
37.
y
2
–4
x
0 –3
0
x
x 4
–3
20
CHAPTER 1
Linear Functions, Equations, and Inequalities
y
38.
y
39.
y
40.
3 4
–2 0
3 –1 0
x
x 2
4
2
41.
42.
10
–10
4
43.
3.1
–4.7
10
x
0 –1
–3
4.7
–3.1
–10
44.
45.
46.
f 2 5 11 17 3
1 7 20
f 1 2
10 15
3 5
19 27
Find each function value, if defined. 47. ƒ( 2) if ƒ(x) = Y1 as defined in Exercise 43
48. ƒ(5) if ƒ(x) = Y1 as defined in Exercise 44
49. ƒ(11) for the function ƒ in Exercise 45
50. ƒ(5) for the function ƒ in Exercise 45
51. ƒ(1) for the function in Exercise 45
52. ƒ(10) for the function in Exercise 45
Find ƒ(x) at the indicated value of x. 53. ƒ(x) = 3x  4, x =  2
54. ƒ(x) = 5x + 6, x =  5
55. ƒ(x) = 2x 2  x + 3, x = 1
56. ƒ(x) = 3x 2 + 2x  5, x = 2
57. ƒ(x) =  x 2 + x + 2, x = 4
58. ƒ(x) =  x 2  x  6, x = 3
59. ƒ(x) = 5, x = 9
60. ƒ(x) =  4, x = 12
61. ƒ(x) = 2x 3 + 12, x =  2
62. ƒ(x) = 2 3 x 2  x + 6, x = 2
63. ƒ(x) =  5  2x  , x = 8
64. ƒ(x) =  6  12 x  , x = 20
Concept Check
Work each problem.
65. If ƒ( 2) = 3 , identify a point on the graph of ƒ.
66. If ƒ(3) =  9.7, identify a point on the graph of ƒ.
67. If the point (7, 8) lies on the graph of ƒ, then ƒ( ) = .
68. If the point ( 3, 2) lies on the graph of ƒ, then ƒ( ) = .
Use the graph of y = ƒ(x) to find each function value: (a) ƒ( 2), (b) ƒ(0), (c) ƒ(1), and (d) ƒ(4). y
69.
y
70.
6 4
6 4
2 0 –2 –2
y
71. 4 2 2
x 1 2 3 4
–2
0 –2
y
72.
1 2 3 4
x
–2
0 –4
4 2 x 1 2 3 4
–2
0 –2 –4
x 1 2 3 4
Reviewing Basic Concepts (Sections 1.1 and 1.2)
9 73.
9 76.
Explain each term in your own words. (a) Relation (b) Function (c) Domain of a function (d) Range of a function (e) Independent variable (f) Dependent variable
74. Radio Stations The function ƒ gives the number y in thousands of radio stations on the air during year x: ƒ = 5(1950, 2.8), (1975, 7.7), (1990, 10.8), (2000, 12.8), (2005, 13.5)6. (Source: M. Street Corporation.) (a) Use a mapping diagram (see FIGURE 19) to represent ƒ. 9 (b) Evaluate ƒ(2000) and explain what it means. (c) Identify the domain and range of ƒ. (Modeling) Solve each problem. 75. Distance to Lightning When a bolt of lightning strikes in the distance, there is often a delay between seeing the lightning and hearing the thunder. The function defined by ƒ(x) = 5x computes the approximate distance in miles between an observer and a bolt of lightning when the delay is x seconds. (Source: Weidner, R., and R. Sells, Elementary Classical Physics, Allyn and Bacon, Inc.) 9 (a) Find ƒ(15) and interpret the result. (b) Graph y = ƒ(x). Let the domain of ƒ be [0, 20].
SECTIONS 1.1 AND 1.2
21
Air Temperature When the relative humidity is 100%, air cools 5.8°F for every 1mile increase in altitude. If the temperature is 80°F on the ground, then ƒ(x) = 80  5.8x calculates the air temperature x miles above the ground. Find ƒ(3) and interpret the result. (Source: Battan, L., Weather in Your Life, W. H. Freeman.)
77. Sales Tax If the sales tax rate is 7.5%, write a function ƒ that calculates the sales tax on a purchase of x dollars. What is the sales tax on a purchase of $86? 78. Income and Level of Education Function ƒ gives the median 2006 individual income (in dollars) by educational attainment for people 25 years old and over. This function is defined by ƒ(N) = 21,788, ƒ(H) = 30,940, ƒ(B) = 50,024, and ƒ(M) = 59,280, where N denotes no high school diploma, H a high school diploma, B a bachelor’s degree, and M a master’s degree. (Source: U.S. Bureau of Labor Statistics.) (a) Write ƒ as a set of ordered pairs. (b) Give the domain and range of ƒ. 9 (c) Discuss the relationship between education and income. 79. Tuition and Fees If college tuition costs $192 per credit and fees are fixed at $275, write a formula for a function ƒ that calculates the tuition and fees for taking x credits. What is the total cost of taking 11 credits? 80. Converting Units of Measure Write a formula for a function ƒ that converts x gallons to quarts. How many quarts are there in 19 gallons?
REVIEWING BASIC CONCEPTS
1. Plot the points ( 3, 1), ( 2,  1), (2,  3), (1, 1), and (0, 2). Label each point.
7. Find ƒ( 5) if ƒ(x) = 3  4x. 8. Use the graph of y = ƒ(x) to find ƒ(2) and ƒ( 1).
2. Find the length and midpoint of the line segment that connects the points P( 4, 5) and Q(6,  2). 3. Use a calculator to approximate thousandth.
25 + p
2
4. The hypotenuse of a right triangle measures 61 inches, and one of its legs measures 11 inches. Find the length of the other leg. 5. Use interval notation to write the sets 5x   2 6 x … 56 and 5x  x Ú 46. y
0
–2
0
x 2
4
y = f(x)
9. Graph ƒ(x) = 12 x  1 by hand. 10. Find the distance between the points (12,  3) and ( 4, 27).
3
–2
4
to the nearest
2 33 + 1
6. Determine whether the relation shown in the graph is a function. What are its domain and range?
y
x 2
–3
22
CHAPTER 1
1.3
Linear Functions, Equations, and Inequalities
Linear Functions
Basic Concepts about Linear Functions • Slope of a Line • Slope–Intercept Form of the Equation of a Line
Basic Concepts about Linear Functions Suppose that by noon 2 inches of rain had fallen. Rain continued to fall at the rate of 1 2 inch per hour in the afternoon. Then the total rainfall x hours past noon is given by ƒ(x) =
1 x + 2. 2
Rate of rainfall
Amount of rainfall at noon
Hours of rainfall past noon
At 3:00 P.M., the total rainfall equaled ƒ(3) =
1 (3) + 2 = 3.5 inches. 2
This function ƒ satisfies the general form ƒ(x) = ax + b (where a = and is called a linear function.
1 2
and b = 2)
Linear Function A function ƒ defined by ƒ(x) ⴝ ax ⴙ b, where a and b are real numbers, is called a linear function.
TECHNOLOGY NOTE A graphing calculator can make a table of values for ƒ( x ) = 3x + 6 and graph ƒ, as shown in the figures below. Be sure to enter the formula for ƒ and then set an appropriate viewing rectangle.
The graph of ƒ(x) = ax + b is a line and can be found by graphing y = ax + b. For example, the graph of ƒ(x) = 3x + 6 is the line determined by y = 3x + 6. An equation such as y = 3x + 6 is called a linear equation in two variables. A solution is an ordered pair (x, y) that makes the equation true. Verify that ( 2, 0), ( 1, 3), (0, 6), and (1, 9) are all solutions of y = 3x + 6 and, therefore, lie on the graph of ƒ(x) = 3x + 6. Graphing linear equations by hand involves plotting points whose coordinates are solutions of the equation and then connecting them with a straight line. FIGURE 28(a) shows the ordered pairs just mentioned for the linear equation y = 3x + 6, accompanied by a table of values. Notice that the points appear to lie in a straight line; that is indeed the case. Since we may substitute any real number for x, we connect these points with a line to obtain the graph of ƒ(x) = 3x + 6, as shown in FIGURE 28(b). y
x y –2 0
(1, 9)
–1 3 0 6 1 9
10
–10
10
y (1, 9)
8
8
6 (0, 6)
6 (0, 6)
(–1, 3) 4
(–1, 3) 4
2
2
(–2, 0) 4 3 2 1
1 2 3 4
x
(a)
–10
(–2, 0) 4 3 1
1 2 3 4
(b) FIGURE 28
f (x) 3x
+6 x
1.3
Linear Functions
23
From geometry, we know that two distinct points determine a line. Therefore, if we know the coordinates of two points, we can graph the line. For the equation y = 3x + 6, suppose we let x = 0 and solve for y. Then suppose we let y = 0 and solve for x. y y y y
= = = =
3x + 6 3(0) + 6 0 + 6 6
Let x = 0. Multiply. Add.
y 0 6 x
= = = =
3x + 6 3x + 6 3x 2
Let y = 0. Subtract 6. Divide by 3; rewrite.
The points (0, 6) and ( 2, 0) lie on the graph of y = 3x + 6 and are sufficient for obtaining the graph in FIGURE 28(b) on the previous page. (Sometimes it is advisable to plot a third point. Why?) The numbers 6 and  2 are called the y and xintercepts of the line respectively.
y
f (c) = 0 c Zero of f
Locating x and yIntercepts
x
To find the xintercept of the graph of y = ax + b, let y = 0 and solve for x (assuming that a Z 0). To find the yintercept, let x = 0 and solve for y.
y = f (x)
(a)
The xintercept of the graph of a linear function is a value that makes ƒ(x) = 0 a true statement; that is, it causes the function value to equal zero. In general, such a number c is called a zero of the function. See FIGURE 29.
f(x) = –x – 2 3.1
–4.7
4.7
Zero of a Function Let ƒ be a function. Then any number c for which ƒ(c) = 0 is called a zero of the function ƒ.
–3.1 (b) FIGURE 29
NOTE If c is a real zero of ƒ, then ƒ(c) ⴝ 0 and c is an xintercept of the graph of ƒ. See
FIGURE 29(a).
A calculator can also be directed to find a zero, as in
FIGURE 29(b).
EXAMPLE 1
Graphing a Line Graph the function defined by ƒ(x) =  2x + 5. What is the zero of ƒ? Analytic Solution The graph of ƒ(x) =  2x + 5 and its intercepts are shown in FIGURE 30. The zero of ƒ is 2.5.
Graphing Calculator Solution A calculator graph is shown in FIGURE 31. The xintercept 2.5 is shown, so 2.5 is the zero of ƒ. f(x) = –2x + 5
y
f (x) = –2x + 5
x
y
0 2.5
5 0
5
3.1
Zero of f 3 x
–4.7
4.7
2
–3.1 FIGURE 30
FIGURE 31
24
CHAPTER 1
Linear Functions, Equations, and Inequalities
EXAMPLE 2
Finding a Formula for a Function A 100gallon tank full of water is being drained at a rate of 5 gallons per minute. (a) Write a formula for a linear function ƒ that models the number of gallons of water in the tank after x minutes. (b) How much water is in the tank after 4 minutes? (c) Use the x and yintercepts to graph ƒ. Interpret each intercept. Solution (a) The amount of water in the tank is decreasing at a constant rate of 5 gallons per minute, so the constant rate of change is  5. The initial amount of water is equal to 100 gallons, so
y
Water (in gallons)
125 100
ƒ(x) = (constant rate of change)x + (initial amount) ƒ(x) =  5x + 100.
75 50
(b) After 4 minutes, the tank held ƒ(4) =  5(4) + 100 = 80 gallons.
25 0
4
8
12 16 20 24
x
Time (in minutes) FIGURE 32
(c) The graph of y =  5x + 100 has yintercept 100 because y = 100 when x = 0. To find the xintercept, let y = 0 and solve the equation 0 =  5x + 100, obtaining x = 20. The graph of ƒ is shown in FIGURE 32. The xintercept corresponds to the time in minutes that it takes to empty the tank, and the yintercept corresponds to the number of gallons of water in the tank initially. Suppose that for a linear function defined by ƒ(x) = ax + b, we have a = 0. Then the function becomes ƒ(x) = b, where b is some real number.
y
Sketching the Graph of f ( x) ⴝ b Graph the function defined by ƒ(x) =  3. EXAMPLE 3
0
f(x) = –3
x
Horizontal line
Solution Since y always equals  3, the yintercept is  3. Since the value of y can never be 0, the graph has no xintercept and is a line parallel to the xaxis, as shown in FIGURE 33. Its graph is a horizontal line.
(0, –3) FIGURE 33
Constant Function A function defined by ƒ(x) ⴝ b, where b is a real number, is called a constant function. Its graph is a horizontal line with yintercept b. For b Z 0, it has no xintercept. (Every constant function is also linear.) Unless otherwise specified, the domain of a linear function is the set of all real numbers. The range of a nonconstant linear function is also the set of all real numbers. The range of a constant function defined by ƒ(x) ⴝ b is {b}. The choice of viewing window may give drastically different views of a calculator graph, as shown in FIGURE 34 for the graph of ƒ(x) = 3x + 6. f (x) = 3x + 6
5
–20
–10
10
–4 FIGURE 34
10
4
20
–5
f (x) = 3x + 6
f (x) = 3x + 6
–15
15
–10
1.3
Linear Functions
25
We usually want a window that shows the most important features of a particular graph. Such a graph is called a comprehensive graph.* The choice of window for a comprehensive graph is not unique—there are many acceptable ones. For a line, a comprehensive graph shows all intercepts of the line. EXAMPLE 4
Finding a Comprehensive Graph of a Line Find a comprehensive graph of g(x) =  0.75x + 12.5. Solution The window 3 10, 104 by 3 10, 104 of FIGURE 35(a) does not show either intercept, so it will not work. We can alter the window size to, for example, 3 10, 204 by 3 10, 204 to obtain a comprehensive graph. See FIGURE 35(b). g(x) = –0.75x + 12.5
g(x) = –0.75x + 12.5 10
20
–10
10 –10
20
–10 This is a comprehensive graph of the line, since both intercepts are visible.
–10 This is not a comprehensive graph of the line, since the intercepts are not visible. (a)
(b) FIGURE 35
WHAT WENT WRONG? A student learning to use a graphing calculator attempted to graph y = 12 x + 15. However, she obtained the blank screen shown here: 10
–10
10
–10
What Went Wrong?
How can she obtain a comprehensive graph of this function?
Looking Ahead to Calculus
Slope of a Line
The concept of the slope of a line is extended in calculus to general curves. The slope of a curve at a point is understood to mean the slope of the line tangent to the curve at that point.
In 1998, the average annual cost for tuition and fees at private fouryear colleges was $14,709. By 2008, this cost had increased to $25,143. The line (segment) graphed in FIGURE 36 on the next page is actually somewhat misleading, since it indicates that the increase in cost was the same from year to year. Answer to What Went Wrong? The yintercept is 15 and the xintercept is  30, so the window size must be increased to show the intercepts. For example, a window of 3  40, 404 by 3 30, 304 would work. *The term comprehensive graph was coined by Shoko Aogaichi Brant and Edward A. Zeidman in the text Intermediate Algebra: A Functional Approach (HarperCollins College Publishers, 1996), with the assistance of Professor Brant’s daughter Jennifer. The authors thank them for permission to use the terminology in this text.
26
CHAPTER 1
Linear Functions, Equations, and Inequalities
We can use the graph to determine the average yearly increase in cost. Over the 10year span, the cost increased $10,434. Therefore, the average yearly increase was $10,434 $25,143  $14,709 = L $1043. 2008  1998 10 This quotient is an illustration of the slope of the line joining (1998, 14,709) and (2008, 25,143). The concept of slope applies to any nonvertical line.
30,000
$25,143
25,000 20,000 15,000
$14,709 10,000 5,000 0 1998
2008
Source: The College Board.
FIGURE 36
A table of values and the graph of the line y = 3x + 1 are shown in FIGURE 37. For each unit increase in x, the yvalue increases by 3. The slope of the line is 3. y
xincrease is 1.
y
x
y
2 1 0 1 2
5 2 1 4 7
(2, 7)
y = 3x + 1
(1, 4) (0, 1) 0
yincrease is 3.
x
(–1, –2) (–2, –5)
Slope =
Δy Δx
(x 2, y2)
FIGURE 37
Δ y = y2 – y1 (x1, y1)
Δ x = x2 – x1
(x 2, y1)
x
0
FIGURE 38
Geometrically, slope is a numerical measure of the steepness of a line and may be interpreted as the ratio of rise to run. To calculate slope, start with the line through the two distinct points (x1, y1) and (x2, y2), where x1 Z x2. See FIGURE 38. The difference x2  x1 is called the change in x, denoted ≤x (read “delta x”), where ≤ is the Greek letter delta. In the same way, the change in y is denoted ≤y = y2  y1. The slope of a nonvertical line is defined as the quotient of the change in y and the change in x.
Slope The slope m of the line passing through the points (x1, y1) and (x2, y2) is mⴝ
y2 ⴚ y1 ≤y , where ¢x = x2  x1 Z 0. ⴝ x2 ⴚ x1 ≤x
1.3
Linear Functions
27
NOTE The preceding slope definition can be used to find an equation for a nonvertical line passing through the points with coordinates ( x1, y1) and ( x2, y2). If P ( x, y ) is any point y  y1 on this nonvertical line, then its equation can be written m = . See Section 1.4. x  x1 CAUTION When using the slope formula, it makes no difference which point is ( x1, y1) or ( x2, y2); however, be consistent. Start with the x and yvalues of either point, and subtract the corresponding values of the other point. EXAMPLE 5
x
 12
5
2
9
y
7
3
1
5
Using the Slope Formula to Find Slope A table of values for a linear function is shown in the margin. Determine the slope of the graph of the line. Sketch the graph. Solution Because the slope of a line is the same regardless of the two points chosen, we can choose any two points from the table. If we let (2,  1) = (x1, y1) and
y
(–5, 3)
–1 – 3 = –4
m=
=–
(2, –1)
4 7
y2  y1 x2  x1
=
3  ( 1) 4 =  . 5  2 7
Start with the x and yvalues of the same point.
x
0 –4 7
m =
then
2 – (–5) = 7
( 5, 3) = (x2, y2),
See FIGURE 39. By contrast, if we let
FIGURE 39
( 5, 3) = (x1, y1) and m =
then
EXAMPLE 6
y
(2,  1) = (x2, y2),
1  3 4 =  . 2  ( 5) 7
The slope is  47 no matter which point is considered first.
Using the Slope and a Point to Graph a Line
Graph the line that passes through (2, 1) and has slope  43. 2 0
x
Down 4 –4
Solution Start by locating the point (2, 1) on the graph. Find a second point on the line by using the definition of slope.
(2, 1) P(5, –3)
Right 3
(0, 4)
change in y 4 = change in x 3
Move down 4 units from (2, 1) and then 3 units to the right to obtain P(5,  3). Draw a line through this second point P and (2, 1), as shown in FIGURE 40.
FIGURE 40
y
slope =
The graph of a constant function is a horizontal line. Because there is no change in y, the slope of a horizontal line is 0. See FIGURE 41. f (x) = 4 (1, 4) Horizontal line– 0 slope x
0
m=
4–4 1–0
=0
Geometric Orientation Based on Slope For a line with slope m, 1. If m 7 0 (i.e., slope is positive), the line rises from left to right. 2. If m 6 0 (i.e., slope is negative), the line falls from left to right. 3. If m = 0 (i.e., slope is 0), the line is horizontal.
FIGURE 41
28
Linear Functions, Equations, and Inequalities
CHAPTER 1
In the slope formula, we have the condition ¢x = x2  x1 Z 0. This means that x2 Z x1. If we graph a line with two points having equal xvalues, we get a vertical line. See FIGURE 42. Notice that this is not the graph of a function, since 4 appears as the first number in more than one ordered pair. If we were to apply the slope formula, the denominator would be 0. As a result, the slope of a vertical line is undefined.
y
x=4 x
0
4
Vertical Line
Vertical line – undefined slope
A vertical line with xintercept a has an equation of the form x ⴝ a. Its slope is undefined.
FIGURE 42
Slope–Intercept Form of the Equation of a Line
y
In FIGURE 43, the slope m of the line y = 3x + 1 is 3 and the yintercept is 1. In general, if ƒ(x) = ax + b, then the slope of the graph of ƒ(x) is a and the yintercept is b. To verify this fact, notice that ƒ(0) = a(0) + b = b. Thus, the graph of ƒ passes through the point (0, b) and b is the yintercept. Since ƒ(1) = a(1) + b = a + b, the graph of ƒ also passes through the point (1, a + b). See FIGURE 44. The slope of the line that passes through the points (0, b) and (1, a + b) is
(2, 7)
y = 3x + 1
(1, 4) (0, 1) x
0
(–1, –2)
m =
(–2, –5)
a + b  b = a. 1  0
Because the slope of the graph of ƒ(x) = ax + b is a, it is often convenient to use m rather than a in the general form of the equation. Therefore, we can write either
FIGURE 43 y
ƒ(x) ⴝ mx ⴙ b or
to indicate a linear function. The slope is m, and the yintercept is b. This equation for ƒ(x) is generally called the slope–intercept form of the equation of a line.
(1, a + b)
f(x) = ax + b
y ⴝ mx ⴙ b
(0, b) x
0
Slope–Intercept Form The slope–intercept form of the equation of a line is y ⴝ mx ⴙ b, where m is the slope and b is the yintercept. FIGURE 44
EXAMPLE 7
Matching Graphs with Equations FIGURE 45 shows the graphs of four lines. Their equations are y = 2x + 3, y =  2x + 3, y = 2x  3, and y =  2x  3, but not necessarily in that order. Match each equation with its graph. y
A.
0
y
B.
x
0
C.
x
FIGURE 45
y
0
y
D.
x
0
x
1.3
Looking Ahead to Calculus
Slope represents a rate of change. In calculus, rates of change of nonlinear functions are studied extensively by using the derivative.
Linear Functions
29
Solution The sign of m determines whether the graph rises or falls from left to right. Also, if b 7 0, the yintercept is above the xaxis, and if b 6 0, the yintercept is below the xaxis. Therefore, y = 2x + 3
is shown in B, since the graph rises from left to right and the yintercept is positive;
y =  2x + 3 is shown in D, since the graph falls from left to right and the yintercept is positive; y = 2x  3
is shown in A, since the graph rises from left to right and the yintercept is negative;
y =  2x  3 is shown in C, since the graph falls from left to right and the yintercept is negative.
EXAMPLE 8
Interpreting Slope In 1980, passengers traveled a total of 4.5 billion miles on Amtrak, and in 2007 they traveled 5.8 billion miles. (Source: U.S. Department of Transportation.) (a) Find the slope m of the line passing through the points (1980, 4.5) and (2007, 5.8). (b) Interpret the slope. Solution (a) m =
5.8  4.5 1.3 13 = = L 0.05 2007  1980 27 270
(b) Because the slope is positive, the number of passengermiles traveled on Amtrak increased, on average, by about 0.05 billion, or 50 million miles per year, between 1980 and 2007. The number of passengermiles did not necessarily increase by 50 million each year. Thus, slope gives an average rate of change.
EXAMPLE 9
Interpreting Slope–Intercept Form In 2008, there were approximately 42 million people worldwide living with HIV/ AIDS. At that time, the infection rate was 5.1 million people per year. (Source: Joint United Nations Programme on HIV/AIDS (UNAIDS.org).) (a) Find values for m and b so that y = mx + b models the total number of people y in millions who lived with HIV/AIDS x years after 2008. (b) Find y for the year 2011. Interpret your result. Solution (a) Since there were 42 million people living with HIV/AIDS in 2008 and x = 0 corresponds to 2008, the line determined by y = mx + b must pass through the point (0, 42). Thus, the yintercept is b = 42. The infection rate is 5.1 million per year, so m = 5.1. The equation is y = 5.1x + 42. (b) The year 2011 corresponds to x = 3, so y = 5.1(3) + 42 = 57.3. Thus, this model estimates that 57.3 million people lived with HIV/AIDS from 2008 to 2011. (Note that this estimate includes people who may have died during this period.)
30
CHAPTER 1
Linear Functions, Equations, and Inequalities
y
EXAMPLE 10
Finding an Equation from a Graph Use the graph of the linear function ƒ in FIGURE 46 to complete the following. (a) Find the slope, yintercept, and xintercept.
1 –3
x
0
3
–1
(b) Write an equation defining ƒ. (c) Find any zeros of ƒ.
y = f(x)
Solution (a) The line falls 1 unit each time the xvalue increases by 3 units. Therefore, the slope is 31 =  13. The graph intersects the yaxis at the point (0,  1) and intersects the xaxis at the point ( 3, 0). Therefore, the yintercept is  1 and the xintercept is  3.
FIGURE 46
(b) Because the slope is  13 and the yintercept is  1, it follows that an equation defining ƒ is ƒ(x) = 
1 x  1. 3
(c) Zeros correspond to xintercepts, so the only zero of ƒ is  3.
1.3
EXERCISES
Checking Analytic Skills Graph each linear function. Give the (a) xintercept, (b) yintercept, (c) domain, (d) range, and (e) slope of the line. Do not use a calculator. 1. ƒ(x) = x  4 5. ƒ(x) = 
2 x + 2 5
2. ƒ(x) =  x + 4 6. ƒ(x) =
4 x  3 3
4. ƒ(x) =
7. ƒ(x) = 3x
8. ƒ(x) =  0.5x
Work each problem related to linear functions. (a) Evaluate ƒ( 2) and ƒ(4).
(b) Find the zero of ƒ.
(c) Graph ƒ. How can the graph of ƒ be used to determine the zero of ƒ? 9. ƒ(x) = x + 2 12. ƒ(x) =
1 1 x + 4 2
15. ƒ(x) = 0.4x + 0.15
18. ƒ(x) =
3  3x 6
10. ƒ(x) =  3x + 2 13. ƒ(x) =
1 x 3
11. ƒ(x) = 2 
1 x 2
14. ƒ(x) =  3x
16. ƒ(x) = x + 0.5
17. ƒ(x) =
2  x 4
3x + p 2
20. ƒ(x) =
4x + p 3
19. ƒ(x) =
2 x  2 3
3. ƒ(x) = 3x  6
1.3
Linear Functions
31
y
21. Concept Check Based on your graphs of the functions in Exercises 7 and 8, what conclusion can you make about one particular point that must lie on the graph of the line y = ax (where b = 0)?
(1, 4)
0
22. Concept Check Using the concept of slope and your answer in Exercise 21, give the equation of the line whose graph is shown at the right.
x
(0, 0)
Checking Analytic Skills Graph each line. Also, give the (a) xintercept (if any), (b) yintercept (if any), (c) domain, (d) range, and (e) slope of the line (if defined). Do not use a calculator. 23. ƒ(x) =  3 5 26. ƒ(x) = 4 29. Concept Check
24. ƒ(x) = 5
25. x =  1.5
27. x = 2
28. x =  3
What special name is given to the functions found in Exercises 23, 24, and 26?
Give the equation of the line illustrated. y
30. 3
y
31. (0, 3)
(4, 3)
32.
33.
y
y
(2, 3) (0, 2)
2 0
(4, 0)
x
0
x 2
(–2, 0)
34. Concept Check Answer each question. (a) What is the equation of the xaxis?
0
x
(2, 0)
0
x 2
(0, –2)
(b) What is the equation of the yaxis?
Graph each linear function on a graphing calculator, using the two different windows given. State which window gives a comprehensive graph. 35. ƒ(x) = 4x + 20 Window A: 3 10, 104 by 3 10, 104 Window B: 3 10, 104 by 3 5, 254
36. ƒ(x) =  5x + 30 Window A: 3 10, 104 by 3 10, 404 Window B: 3 5, 54 by 3 5, 404
37. ƒ(x) = 3x + 10 Window A: 3 3, 34 by 3 5, 54 Window B: 3 5, 54 by 3 10, 144
38. ƒ(x) =  6 Window A: 3 5, 54 by 3 5, 54 Window B: 3 10, 104 by 3 10, 104
Find the slope (if defined) of the line that passes through the given points. 39. ( 2, 1) and (3, 6)
40. ( 2, 3) and ( 1, 2)
41. (8, 4) and ( 1,  3)
42. ( 4,  3) and (5, 0)
43. ( 11, 3) and ( 11, 5)
44. ( 8, 2) and ( 8, 1)
2 1 45. a , 9 b and a , 9 b 3 2
46. (0.12, 0.36) and (0.18, 0.36)
3 1 1 2 47. a ,  b and a , b 2 3 4 6
9 48.
Given an equation having x and y as variables, explain how to determine the x and yintercepts.
32
Linear Functions, Equations, and Inequalities
CHAPTER 1
Match each equation with the graph that it most closely resembles.
Concept Check 49. y = 3x + 6
y
A.
y
B.
C.
y
y
D.
50. y =  3x + 6 51. y =  3x  6
x
0
x
0
x
0
x
0
52. y = 3x  6 53. y = 3x
y
E.
y
F.
G.
y
y
H.
54. y =  3x 55. y = 3
x
0
x
0
x
0
0
56. y =  3
The graph of a linear function ƒ is shown. (a) Identify the slope, yintercept, and xintercept. (b) Write a formula for ƒ. (c) Estimate the zero of ƒ. y
57.
y
58.
3
3
3
1
1
1
0
–3
y
59.
3
x
–3
0
1
–1
3
x
0
–3
y
y
61.
0
y
62.
300
2 –2
x
3
–3
–3
60.
1
2
4
150
100
x –3
–2
–1 0
50 1
3
x
–10
0
x
5 10
–300
–4
Concept Check A linear function ƒ has the ordered pairs listed in the table. Find the slope m of the graph of ƒ, use the table to find the yintercept of the line, and give an equation that defines ƒ. 63.
x
ƒ(x)
3 2 1 0 1
 10 6 2 2 6
64. x 2 1 0 1 2
ƒ(x)  11 8 5 2 1
65.
x  0.4  0.2 0 0.2 0.4
ƒ(x)  2.54  2.82  3.1  3.38  3.66
66.
x
ƒ(x)
 100  50 0 50 100
4 4 4 4 4
x
1.3
33
Linear Functions
Concept Check Match each equation in Exercises 67–70 with the line in choices A–D that would most closely resemble its graph, where k 7 0. 67. y = k A.
68. y =  k B.
y
x
0
69. x = k C.
y
x
0
70. x =  k D.
y
0
x
y
x
0
Sketch by hand the graph of the line passing through the given point and having the given slope. Label two points on the line. 71. Through ( 1, 3), m =
3 2
74. Through ( 2,  3), m = 77. Through (0,  4), m =
3 4
3 4
1 3
72. Through ( 2, 8), m =  1
73. Through (3,  4), m = 
75. Through ( 1, 4), m = 0
9 76. Through a , 2b, undefined slope 4
78. Through (0, 5), m =  2.5
79. Through ( 3, 0), undefined slope
80. Concept Check Refer to Exercises 77 and 78. (a) Give the equation of the line described in Exercise 77. (b) Give the equation of the line described in Exercise 78. (Modeling) Solve each problem. 81. Water Flow The graph gives the number of gallons of water in a small swimming pool after x hours. y
Water (in gallons)
8000
84. HIV/AIDS Infections In 2006, there were approximately 40 million people worldwide living with HIV/AIDS. At that time the infection rate was 4.3 million people per year. (a) Find values for m and b so that y = mx + b models the total number of people y in millions who lived with HIV/AIDS x years after 2006. (b) Find y for the year 2010. Interpret your result.
6000 4000
y = f(x)
2000 x 0
2
4
6
8
Time (in hours)
(a) Write a formula for ƒ(x).
9 (b) Interpret both the slope and the yintercept.
(c) Use the graph to estimate how much water was in the pool after 7 hours. Verify your answer by evaluating ƒ(x).
82. Fuel Consumption The table shows the distance y traveled in miles by a car burning x gallons of gasoline. x (gallons) y (miles)
83. Rainfall By noon, 3 inches of rain had fallen during a storm. Rain continued to fall at a rate of 14 inch per hour. (a) Find a formula for a linear function ƒ that models the amount of rainfall x hours past noon. (b) Find the total amount of rainfall by 2:30 P.M.
5
10
12
16
115
230
276
368
(a) Find values for a and b so that ƒ(x) = ax + b models the data exactly. That is, find values for a and b so that the graph of ƒ passes through the data points in the table. 9 (b) Interpret the slope of the graph of ƒ. (c) How many miles could be driven using 20 gallons of gasoline?
85. Birthrate In 1995, the number of births per 1000 people in the United States was 14.8 and was decreasing at 0.096 birth per 1000 people each year. (Source: U.S. Census Bureau.) (a) Write a formula for a linear function ƒ that models the birthrate in year x, where x = 0 corresponds to 1995, x = 1 to 1996, and so on. (b) Estimate the birthrate in 2003, and compare it with the actual value of 14.0. 86. Birthrate In 1998, the number of births per 1000 people in the United States was 14.6 and was decreasing at 0.16 birth per 1000 people each year. (Source: U.S. Census Bureau.) (a) Write a formula for a linear function ƒ that models the birthrate in year x, where x = 0 corresponds to 1998, x = 1 to 1999, and so on. (b) Estimate the birthrate in 2001, and compare it with the actual value of 14.1.
34
CHAPTER 1
1.4
Linear Functions, Equations, and Inequalities
Equations of Lines and Linear Models
Point–Slope Form of the Equation of a Line • Standard Form of the Equation of a Line • Parallel and Perpendicular Lines • Linear Models and Regression
Point–Slope Form of the Equation of a Line
y
Slope = m
FIGURE 47 shows a line passing through the fixed point (x1, y1) with slope m. Let (x, y) be any other point on the line. Then,
m =
(x1, y1)
(x, y)
x  x1 m(x  x1) = y  y1 y  y1 = m(x  x1).
x
0
FIGURE 47
y  y1
Slope formula Multiply each side by x  x1. Rewrite.
This result is called the point–slope form of the equation of a line.
Looking Ahead to Calculus
In calculus, it is often necessary to find the equation of a line, given its slope and a point on the line. The point–slope form is a valuable tool in these situations. y
The line with slope m passing through the point (x1, y1) has equation y ⴚ y1 ⴝ m(x ⴚ x1).
EXAMPLE 1
Using Point–Slope Form Find the slope–intercept form of the line passing through the two points shown in FIGURE 48.
7 5 3
Solution The points are (1, 7) and (3, 3). Find the slope of the line.
1 –1
Point–Slope Form
0
x
1
3
7
Start with the x and yvalues of the same point.
m =
4 7  3 = = 2 1  3 2
FIGURE 48
Now, use either point—say, (1, 7)—with m =  2 in the point–slope form. y  y1 = m(x  x1)
Be careful with signs.
Point–slope form
y  7 =  2(x  1)
y1 = 7, m =  2, x1 = 1
y  7 =  2x + 2
Distributive property
y =  2x + 9
Add 7 to obtain slope–intercept form.
EXAMPLE 2
x
y
2 3 4 5 6 7 8
3 5 7 9 11 13 15
Using Point–Slope Form The table in the margin lists points found on the line y = mx + b. Find the slope–intercept form of the equation of the line. Solution Choose any two points to find the slope of the line. m =
11  3 8 = = 2 6  2 4
Use (2, 3) and (6, 11).
Now, use a point on the line—say, (2, 3)—with m = 2 in the point–slope form.
1.4
y  y1 y  3 y  3 y
= = = =
m(x  x1) 2(x  2) 2x  4 2x  1
Equations of Lines and Linear Models
35
Point–slope form y1 = 3, m = 2, x1 = 2 Distributive property Add 3 to obtain slope–intercept form.
Standard Form of the Equation of a Line Another form of the equation of a line, standard form, can be used to represent any line, including vertical lines.
Standard Form A linear equation written in the form Ax ⴙ By ⴝ C, where A, B, and C are real numbers (A and B not both 0), is said to be in standard form.
NOTE When writing the standard form of a line, we will usually give A, B, and C as integers with greatest common factor 1 and A Ú 0. For example, the equation y =  12 x + 3 could be written in standard form as x + 2y = 6.
One advantage of standard form is that it allows the quick calculation of both intercepts. For example, given 3x + 2y = 6, we can find the xintercept by letting y = 0 and the yintercept by letting x = 0. yintercept: 3(0) + 2y = 6 2y = 6 y = 3
xintercept: 3x + 2(0) = 6 3x = 6 x = 2
This information is useful when sketching the graph of the line by hand. EXAMPLE 3
Graphing an Equation in Standard Form
Graph 3x + 2y = 6. Analytic Solution As just calculated, the points (2, 0) and (0, 3) correspond to the x and yintercepts respectively. Plot these two points and connect them with a straight line. See FIGURE 49. Sometimes it is advisable to plot a third point, such as A 1, 32 B , to check our work.
x y 0 3 2 0
3x + 2y = 6 2y =  3x + 6 y =  1.5x + 3
Given equation Subtract 3x. Divide by 2.
The desired graph is shown in FIGURE 50.
y 4 3 (0, 3) 2 1 (2, 0)
Graphing Calculator Solution Solve the equation for y so that it can be entered into a calculator.
3x + 2y = 6, or y = –1.5x + 3 10
x
4 3 2 11 1 2 3 4 2 3 3x + 2y 6 4
–10
10
–10 FIGURE 49
FIGURE 50
36
Linear Functions, Equations, and Inequalities
CHAPTER 1
NOTE Because of the usefulness of the slope–intercept form when graphing with a graphing calculator, we emphasize that form in much of our work.
FOR DISCUSSION
Parallel and Perpendicular Lines
In the standard viewing window of your calculator, graph the following four lines:
Two lines in a plane are parallel if they do not intersect. Although the exercise in the “For Discussion” box in the margin does not actually prove the result that follows, it provides visual support.
y1 = 2x  6, y2 = 2x  2, y3 = 2x,
Parallel Lines
y4 = 2x + 4.
Two distinct nonvertical lines are parallel if and only if they have the same slope.
What is the slope of each line? What geometric term seems to describe the set of lines?
EXAMPLE 4
Using the Slope Relationship for Parallel Lines Find the equation of the line that passes through the point (3, 5) and is parallel to the line with equation 2x + 5y = 4. Graph both lines in the standard viewing window. Solution Since the point (3, 5) is on the line, we need only find the slope by writing the given equation in slope–intercept form. (That is, solve for y.) 2x + 5y = 4 2 4 y =  x + 5 5
Given equation Subtract 2x; divide by 5.
The slope is  25. Since the lines are parallel,  25 is also the slope of the line whose equation we must find. Substitute into the point–slope form. y  y1 = m(x  x1) 2 y  5 =  (x  3) 5 5( y  5) =  2(x  3) 5y  25 =  2x + 6 Be careful with signs.
y2 = –
2 5
x+
31, 5
or
y2 = –0.4x + 6.2 10
–10
10
–10
y1 = –
2 5
x+
4, 5
y1 = –0.4 x + 0.8 FIGURE 51
or
5y =  2x + 31 2 31 y =  x + 5 5
Point–slope form y1 = 5, m =  25, x1 = 3 Multiply by 5 to clear fractions. Distributive property Add 25. Divide by 5 to obtain the slope–intercept form of the desired line.
An alternative method for finding this equation involves using the slope  25 and the point (3, 5) in the slope–intercept form to find b. y = mx + b 2 5 =  (3) + b 5 6 5 =  + b 5 31 b = 5
Slope–intercept form y = 5, m =  25, x = 3 Multiply. Add 65; rewrite.
Therefore, the equation is y =  25 x + 31 5 , which agrees with our earlier result. FIGURE 51 provides support, as the lines appear to be parallel.
1.4
y1 = –0.5x + 4 10
37
When using a graphing calculator, be aware that visual support (as seen in Example 4) does not necessarily prove the result. For example, FIGURE 52 shows the graphs of y1 =  0.5x + 4
10
–10
Equations of Lines and Linear Models
and
y2 =  0.5001x + 2.
Although they appear to be parallel by visual inspection, they are not parallel, because the slope of y1 is  0.5 and the slope of y2 is  0.5001. –10
y2 = –0.5001x + 2 FIGURE 52
A CAUTION AGAINST RELYING TOO HEAVILY ON TECHNOLOGY Analysis of a calculator graph is often not sufficient to draw correct conclusions. While this technology is incredibly powerful and pedagogically useful, we cannot rely on it alone in our work. We must understand the basic concepts of algebra as well.
FOR DISCUSSION
Using a “square” viewing window, such as 3 9.4, 9.44 by 3 6.2, 6.24, graph each pair of lines. Graph each group separately. I y1 = 4x + 1 y2 = 
1 x + 3 4
II 2 y1 =  x + 3 3 3 y2 = x  4 2
III y1 = 6x  3 y2 = 
1 x + 4 6
IV 13 y1 = x  3 7 7 y2 = x + 4 13
What geometric term applies to each pair of lines? What is the product of the slopes for each pair of lines?
As in the earlier “For Discussion” box, we have not proved the result that follows; rather, we have provided visual support for it. The proof for this statement is done in Exercise 59.
Perpendicular Lines Two lines, neither of which is vertical, are perpendicular if and only if their slopes have product  1.
NOTE If one line is vertical, then its slope is undefined and the statement does not apply.
TECHNOLOGY NOTE Many calculators can set a square viewing window automatically. Check the graphing calculator manual that accompanies this text or your owner’s manual, or look under the ZOOM menu.
For example, if the slope of a line is  34, the slope of any line perpendicular to it is 43, since  34 A 43 B =  1. We refer to numbers such as  34 and 43 as negative reciprocals. In a square viewing window, circles appear to be circular, squares appear to be square, and perpendicular lines appear to be perpendicular. On many calculators, a square viewing window requires that the distance along the yaxis be about twothirds the distance along the xaxis. Examples of square viewing windows on the TI83/84 Plus calculators are 3 4.7, 4.74 by 3 3.1, 3.14 and
FIGURE 53
3 9.4, 9.44 by 3 6.2, 6.24.
on the next page illustrates the importance of square viewing windows.
38
CHAPTER 1
Linear Functions, Equations, and Inequalities
y2 = 0.25x – 3
y1 = – 4x + 2
y2 = 0.25x – 3
10
–10
10
y1 = – 4x + 2 6.2
–9.4
9.4
–10 Although the graphs of these lines are perpendicular, they do not appear to be when graphed in a standard viewing window.
–6.2 Visual support for perpendicularity is more obvious using a square viewing window.
(a)
(b) FIGURE 53
GCM
y1 = –
2 5
4 5,
x+
EXAMPLE 5
Using the Slope Relationship for Perpendicular Lines Find the equation of the line that passes through the point (3, 5) and is perpendicular to the line with equation 2x + 5y = 4. Graph both lines in a square viewing window. Solution In Example 4, we found that the slope of the given line is  25, so the slope 5 of any line perpendicular to it is 2. We can use either method shown in Example 4 to find the equation of the line. The second method yields the following:
or
y = mx + b 5 5 = (3) + b 2 15 5 = + b 2 5 b =  . 2
y1 = –0.4 x + 0.8 6.2
–9.4
9.4
–6.2
y2 =
5 2
x–
5 2,
Slope–intercept form y = 5, m = 52, x = 3 Multiply. Subtract
15 2;
rewrite.
Thus, the equation is y = 52 x  52. Graphing both equations in slope–intercept form and using a square viewing window provides visual support for our answer. See FIGURE 54.
or
y2 = 2.5 x – 2.5 FIGURE 54
Linear Models and Regression When data points are plotted in the xyplane, the resulting graph is sometimes called a scatter diagram. Scatter diagrams are often helpful for analyzing trends in data.
x (year)
y (cost)
2002 2003 2004 2005 2006 2007
264 281 299 318 336 354
Source: U.S. Center for Medicare and Medicaid Services.
GCM
EXAMPLE 6 Modeling Medicare Costs with a Linear Function Estimates for Medicare costs (in billions of dollars) are shown in the table in the margin. (a) Make a scatter diagram of the data. Let x = 0 correspond to 2002, x = 1 to 2003, and so on. What type of function might model the data?
(b) Find a linear function ƒ that models the data. Graph ƒ and the data in the same viewing window. Interpret the slope m. (c) Use ƒ(x) to estimate Medicare costs in 2010. Solution (a) Since x = 0 corresponds to 2002, x = 1 corresponds to 2003, and so on, the data points can be expressed as the ordered pairs (0, 264),
(1, 281),
(2, 299),
(3, 318),
(4, 336),
and (5, 354).
1.4
Equations of Lines and Linear Models
39
Scatter diagrams are shown in FIGURE 55. The data appear to be approximately linear, so a linear function might be appropriate.
TECHNOLOGY NOTE To make a scatter diagram with a graphing calculator, you may need to use the list feature by entering the xvalues in list L1 and the yvalues in list L2. See FIGURE 56.
Medicare Costs (in billions of dollars)
y 375
375
350 325 300 275 250 x 0
1
2
3
4
5
Year (2002 ¶ 0) FIGURE 56
6
–1
6 200
FIGURE 55
(b) We start by choosing two data points that the line should pass through. For example, if we use (0, 264) and (3, 318), then the slope of the line is
f (x) = 18x + 264 375
Start with the x and yvalues of the same point.
m =
318  264 = 18. 3  0
The point (0, 264) indicates that the yintercept b is 264. Thus, –1
ƒ(x) = 18x + 264.
6
A graph of ƒ and the data are shown in FIGURE 57. The slope m = 18 indicates that Medicare costs increased, on average, by $18 billion per year.
200 FIGURE 57
(c) The value x = 8 corresponds to the year 2010. Since ƒ(8) = 18(8) + 264 = 408, this model estimates that Medicare costs reached $408 billion in 2010. NOTE The formula ƒ( x ) = 18x + 264 found in Example 6 is not unique. If two different points are chosen, a different formula for ƒ( x ) may result. However, all such formulas for ƒ( x ) should be in approximate agreement. TECHNOLOGY NOTE
The method for finding the model in Example 6 used algebraic concepts that do not always give a unique line. Graphing calculators are capable of finding the line of “best fit,” called the leastsquares regression line, by using a technique taught in statistics courses known as leastsquares regression.
To find the equation of a leastsquares regression line, refer to the graphing calculator manual that accompanies this text or your owner’s manual. GCM
EXAMPLE 7
Finding the LeastSquares Regression Line Use a graphing calculator to find the leastsquares regression line that models the Medicare costs presented in Example 6. Graph the data and the line in the same viewing window. Solution FIGURE 58 on the next page shows how a TI83/84 Plus graphing calculator finds the regression line for the data in Example 6. In FIGURE 58(a), the years are entered into list L 1, and Medicare costs are entered into list L 2. In FIGURE 58(b), the formula for the regression line is calculated to be y L 18.11x + 263.38. Notice that this equation is not exactly the same as the one found for ƒ(x) in Example 6. In FIGURE 58(c), both the data and the regression line are graphed. (continued)
40
CHAPTER 1
Linear Functions, Equations, and Inequalities
375
–1
6 200
(a)
(b)
(c)
FIGURE 58
TECHNOLOGY NOTE By choosing the DiagnosticOn option on the TI83/84 Plus, the correlation coefficient r and its square r 2 are displayed. See FIGURE 59, and compare with FIGURE 58(b).
Once an equation for the leastsquares regression line has been found, it is reasonable to ask, “Just how good is this line for predictive purposes?” If the line fits the observed data points, then future pairs of points might be expected to do so also. One common measure of the strength of the linear relationship in a data set is called the correlation coefficient, denoted r, where  1 … r … 1. Interpretations for various values of r are given in the following table. Correlation Coefficient r (where ⴚ1 ◊ r ◊ 1) Value of r rⴝ1
Comments There is an exact linear fit. The line passes through all data points and has a positive slope.
Sample Scatter Diagram y
x
r ⴝ ⴚ1
There is an exact linear fit. The line passes through all data points and has a negative slope.
y
FIGURE 59 x
00
x
0
Horizontal shift to the left FIGURE 21
EXAMPLE 2
Recognizing Horizontal Shifts Give the equation of each function graphed. y (a) (b) y
4 4 2 –2
2 0
x 4
8
–4 –2
0
x 2
Solution Both graphs are horizontal translations of the graph of y =  x . (a) The graph has been shifted 4 units to the right, so the equation is y =  x  4 . (b) The graph has been shifted 2 units to the left, so the equation is y =  x + 2 . (Be sure that you understand why the signs are as they appear here.)
Combinations of Vertical and Horizontal Shifts EXAMPLE 3
y
Applying Both Vertical and Horizontal Shifts Describe how the graph of each y2 would be obtained by translating the graph of y1 =  x . Sketch the graphs of y1 and y2 on the same xyplane by hand. (a) y2 =  x + 2   6 (b) y2 =  x  2  + 8
y1 x
8 4 8
4 2
(2, –6) 8
4
8
x
y2 x + 2 – 6
FIGURE 22
Solution (a) The function defined by y2 =  x + 2   6 is translated 2 units to the left (because of the  x + 2 ) and 6 units downward (because of  6) compared with the graph of y1 =  x . See FIGURE 22. Note that at the point ( 2,  6), the graph of y2 changes from decreasing to increasing. (continued)
98
CHAPTER 2
Analysis of Graphs of Functions
(b) The graph of y2 =  x  2  + 8 is obtained by translating the graph of y1 =  x  2 units to the right and 8 units upward. See FIGURE 23. Notice that the point (0, 0) has been translated to (2, 8). y
y2 x  2
12
(2, 8) y1 x
8 4 8
+8
4
4
8
x
4 FIGURE 23
Effects of Shifts on Domain and Range EXAMPLE 4
Determining Domains and Ranges of Shifted Graphs Give the domain and range of each function. (a) (b) (c) y y y
3 2 1 –1
0
3 2 1
y = (x – 1)2 + 2
–3 –2 –1 x
1 2
–3
y = √x – 2 – 3
1 0
x 1 2 3
y = (x + 1)3 – 2
–1 –3
0
x 2
4
(2, –3)
Solution (a) To obtain the graph of y = (x  1) 2 + 2, the graph of y = x 2 was translated 1 unit to the right and 2 units upward. The original domain ( q, q) is not affected. However, the range of this function is 32, q), because of the vertical translation. (b) The graph of y = (x + 1) 3  2 is obtained by vertical and horizontal shifts of the graph of y = x 3, a function that has both domain and range ( q, q). Neither is affected here, so the domain and range of y = (x + 1) 3  2 are also ( q, q). (c) The function y = 2x has domain 30, q). The graph of y = 2x  2  3 is obtained by shifting the basic graph 2 units to the right, so the new domain is 32, q). The original range, 30, q), has also been affected by the shift of the graph 3 units downward, so the new range is 3 3, q).
Horizontal Shifts Applied to Equations for Modeling In Examples 6 and 7 of Section 1.4, we examined data that showed estimates for Medicare costs (in billions of dollars) between the years 2002 and 2007. To simplify our work, x = 0 corresponded to the year 2002, x = 1 to 2003, and so on. We determined that the leastsquares regression line has equation y L 18.11x + 263.38.
2.2
Vertical and Horizontal Shifts of Graphs
99
The data and graph of this line are shown in FIGURE 24(a). To use the values of the years 2002 through 2007 directly in the regression equation, we must shift the graph of the line 2002 units to the right. The equation of this new line is
y = 18.11x + 263.38 400
y L 18.11(x  2002) + 263.38. Using a graphing calculator to plot the set of points –1
6 200 (a)
y = 18.11(x – 2002) + 263.38
5(2002, 264), (2003, 281), (2004, 299), (2005, 318), (2006, 336), (2007, 354)6
and graphing y = 18.11(x  2002) + 263.38 in the viewing window 32001, 20084 by 3200, 4004 gives the graph shown in FIGURE 24(b).
400
EXAMPLE 5
Applying a Horizontal Shift to an Equation Model The table lists U.S. sales of Toyota vehicles in millions.
2001 200
2008 (b) FIGURE 24
Year
Vehicles
2000 2001 2002 2003 2004
1.6 1.7 1.8 1.9 2.0
Source: Autodata.
These data can be modeled by the equation y = 0.1x + 1.6, where x = 0 corresponds to 2000, x = 1 to 2001, and so on. (a) Find the corresponding equation that allows direct input of the year. (b) Use the equation to estimate U.S. sales of Toyota vehicles in 2007. Solution (a) Because 2000 corresponds to 0, the graph of y = 0.1x + 1.6 would have to be shifted 2000 units to the right. Thus, the equation becomes y = 0.1(x  2000) + 1.6. (b) To estimate sales in 2007, substitute 2007 for x and evaluate the expression. y = 0.1(2007  2000) + 1.6 = 2.3 This model estimates U.S. sales of Toyota vehicles in 2007 to be about 2.3 million.
2.2
EXERCISES
Checking Analytic Skills Write the equation that results in the desired translation. Do not use a calculator. 1. The squaring function, shifted 3 units upward
2. The cubing function, shifted 2 units downward
3. The square root function, shifted 4 units downward
4. The cube root function, shifted 6 units upward
5. The absolute value function, shifted 4 units to the right
6. The absolute value function, shifted 3 units to the left
7. The cubing function, shifted 7 units to the left
8. The square root function, shifted 9 units to the right
9 9.
Explain how the graph of g(x) = ƒ(x) + 4 is obtained from the graph of y = ƒ(x).
9 10.
Explain how the graph of g(x) = ƒ(x + 4) is obtained from the graph of y = ƒ(x).
100
CHAPTER 2
Analysis of Graphs of Functions
Exercises 11–19 are grouped in threes: 11–13, 14–16, and 17–19. For each group, match the correct graph, A, B, or C, to the function. Confirm your answers with your calculator. 11. y = x 2  3 y A.
12. y = (x  3) 2 y B.
13. y = (x + 3) 2 y C.
4 2 –4
x
0
2
–2
14. y =  x  + 4 y A.
4
x 2 4 6
–2 –4 –6
17. y = (x  3) 3 y A.
4
2
2 0
6 4 x
0
x 2
4
4
16. y =  x  4   3 y C.
–6 –4 –2 –4 –6
2 4 6
18. y = (x  2) 3  4 y B.
4
x 2
–4
6 4 2 0
0
–4 –2
15. y =  x + 4   3 y B.
4 2
–4 –2
2
–4
–4
–6 –4 –2 –4 –6
x
0
–4 –2
4
4 2
–4 –2
0
x 4
19. y = (x + 2) 3  4 y C.
4 2 0
x 2
–4 –2
4
–4
0
x 4
–4
Concept Check The function Y2 is defined as Y1 + k for some real number k. Based on the table shown, what is the value of k? 20.
21.
22.
Concept Check The function Y2 is defined as Y1 + k for some real number k. Based on the two views of the graphs of Y1 and Y2 and the displays at the bottoms of the screens, what is the value of k? 23.
10 –10
–10
24.
10
10
10
10
10
–10
10 –10
10
–10
–10
–10
–10
(6, 2) lies on the graph of Y1. First view
(6, –1) lies on the graph of Y2. Second view
(– 4, 3) lies on the graph of Y1. First view
(– 4, 8) lies on the graph of Y2. Second view
2.2
101
Vertical and Horizontal Shifts of Graphs
Use the results of the specified exercises to determine (a) the domain and (b) the range of each function. 25. y = x 2  3 (Exercise 11)
26. y = (x  3) 2 (Exercise 12)
27. y =  x + 4   3 (Exercise 15)
28. y =  x  4   3 (Exercise 16)
29. y = (x  3) 3 (Exercise 17)
30. y = (x  2) 3  4 (Exercise 18)
Checking Analytic Skills Use translations of one of the basic functions defined by y = x 2, y = x 3, y = 2x, or y =  x  to sketch a graph of y = ƒ(x) by hand. Do not use a calculator. 31. y = (x  1) 2
32. y = 2x + 2
33. y = x 3 + 1
34. y =  x + 2 
35. y = (x  1) 3
36. y =  x   3
37. y = 2x  2  1
38. y = 2x + 3  4
39. y = (x + 2) 2 + 3
40. y = (x  4) 2  4
41. y =  x + 4   2
42. y = (x + 3) 3  1
Concept Check graph.
Suppose that h and k are both positive numbers. Match each equation with the correct
43. y = (x  h) 2  k
44. y = (x + h) 2  k
45. y = (x + h) 2 + k
46. y = (x  h) 2 + k
A.
B.
C.
D.
y
x
0
y
y
x
0
0
y
0
x
x
Concept Check Given the graph shown, sketch by hand the graph of each function described, indicating how the three points labeled on the original graph have been translated. 47. y = ƒ(x) + 2 48. y = ƒ(x)  2
y
(–1, 4)
y = f (x)
49. y = ƒ(x + 2)
(5, 0)
50. y = ƒ(x  2)
x
0
(–3, –2)
Each graph is a translation of the graph of one of the basic functions defined by y = x 2, y = x 3, y = 2x, or y =  x . Find the equation that defines each function. Then, using the concepts of increasing and decreasing functions discussed in Section 2.1, determine the interval of the domain over which the function is (a) increasing and (b) decreasing. 51.
52.
y
6 4 2 –6 –4 –2 –4 –6
53.
y
6 4 2
6 4 0
x 2 4 6
–6 –4 –2 –4 –6
y
0
x 2 4 6
–6 –4 –2 –4
0
x 4 6
102
CHAPTER 2
Analysis of Graphs of Functions
y
54.
y
55.
3
30
4 1
2
10 –30 –10
y
56.
0
x 10
30
–4 –2
0
x 2
0
–2
4
x 1
3
–3
–4
–30
RELATING CONCEPTS
For individual or group investigation (Exercises 57–60)
Use the xintercept method and the given graph to solve each equation or inequality. (Hint: Extend the concepts of Section 1.5 for graphical solution of linear equations and inequalities.) 57. (a) ƒ(x) = 0 (b) ƒ(x) 7 0 (c) ƒ(x) 6 0
58. (a) ƒ(x) = 0 (b) ƒ(x) 7 0 (c) ƒ(x) 6 0
y
f(x) = 4⏐x – 3.5⏐ – 2
0
f (x) = (x – √2)3
0
x
(3, 0)
y
(4, 0)
x
(√2, 0)
–2
59. (a) ƒ(x) = 0 (b) ƒ(x) Ú 0 (c) ƒ(x) … 0
y
(
f (x) = x – 1 2
60. (a) ƒ(x) = 0 (b) ƒ(x) Ú 0 (c) ƒ(x) … 0
)2 – 814
y
f (x) = √x – 1 + 2 (1, 2)
0 1
(–4, 0)
x
0
x
(5, 0)
–20
61. Concept Check The graph is a translation of y =  x . What are the values of h and k if the equation is of the form y =  x  h  + k?
62. Concept Check Suppose that the graph of y = x 2 is translated in such a way that its domain is ( q, q) and its range is 338, q). What values of h and k can be used if the new function is of the form y = (x  h) 2 + k ?
y
(– 3, 1) 0
x
(Modeling) Solve each problem. 63. Corn Production
The linear equation
64. Federal Debt
The linear equation
y = 1.311x + 68.28
y = 726.7x + 7421.6
gives an approximation of the number of acres of corn, in millions, harvested annually in the U.S, where x = 1 represents 2001, x = 2 represents 2002, and so on. (Source: Statistical Abstract of the United States.) (a) Write an equation that yields the same yvalues when the exact year is entered. (b) Approximate the number of acres of corn harvested in 2003.
gives an approximation of the total public debt y (in billions of dollars), where x = 0 corresponds to 2005, x = 1 corresponds to 2006, and so on. (Source: www. treasurydirect.gov) (a) Write an equation that yields the same yvalues when the exact year is entered. (b) Predict the total public debt in 2012.
2.2
65. Cost of Public College Education The table lists the average annual costs (in dollars) of tuition and fees at public fouryear colleges for selected years.
Year
Tuition and Fees (in dollars)
1993 1995 1997 1999 2001 2003 2005
2527 2860 3111 3356 3857 4632 5491
Source: The College Board.
Vertical and Horizontal Shifts of Graphs
103
66. Women in the Workforce The table shows how the percent of women in the civilian workforce has changed from 1970 to 2005. Year
Percent of Women in the Workforce
1970 1975 1980 1985 1990 1995 2000 2005
43.3 46.3 51.5 54.5 57.5 58.9 59.9 59.0
Source: U.S. Bureau of Labor Statistics.
(a) Use a calculator to find the leastsquares regression line for these data, where x = 0 corresponds to 1993, x = 2 corresponds to 1995, and so on. (b) Based on your result from part (a), write an equation that yields the same yvalues when the exact year is entered. (c) Estimate the cost of tuition and fees in 2009 to the nearest hundred dollars.
RELATING CONCEPTS
(a) Use a calculator to find the leastsquares regression line for these data, where x = 0 corresponds to 1970, x = 5 corresponds to 1975, and so on. (b) Based on your result from part (a), write an equation that yields the same yvalues when the exact year is entered. (c) Predict the percent of population of women in the civilian workforce in 2015.
For individual or group investigation (Exercises 67–74)
Recall from Chapter 1 that a unique line is determined by two distinct points on the line and that the values of m and b can then be determined for the general form of the linear function defined by ƒ(x) = mx + b.
68. Use the slope formula to find the slope of this line.
73. Graph both y1 and y2 in the standard viewing window of your calculator, and describe how the graph of y2 can be obtained by vertically translating the graph of y1. What is the value of the constant in this vertical translation? Where do you think it comes from?
69. Find the equation of the line, and write it in the form y1 = mx + b.
74. Fill in the blanks with the correct responses, based on your work in Exercises 67–73.
Work these exercises in order. 67. Sketch by hand the line that passes through the points (1,  2) and (3, 2).
70. Keeping the same two xvalues as given in Exercise 67, add 6 to each yvalue. What are the coordinates of the two new points? 71. Find the slope of the line passing through the points determined in Exercise 70. 72. Find the equation of this new line, and write it in the form y2 = mx + b.
If the points with coordinates (x1, y1) and (x2, y2) lie on a line, then when we add the positive constant c to each yvalue, we obtain the points with coordinates ) and (x2, y2 + ). The slope of (x1, y1 + the new line is the slope of (the same as/different from)
the original line. The graph of the new line can be obtained by shifting the graph of the original line units in the direction.
104
Analysis of Graphs of Functions
CHAPTER 2
2.3
Stretching, Shrinking, and Reflecting Graphs
Vertical Stretching • Vertical Shrinking • Horizontal Stretching and Shrinking • Reflecting across an Axis • Combining Transformations of Graphs
In the previous section, we saw how adding or subtracting a constant to ƒ(x) can cause a vertical or horizontal shift to the graph of y = ƒ(x). Now we will see how multiplying by a constant alters the graph of a function.
Vertical Stretching FOR DISCUSSION
In each group that follows, we give four related functions. Graph the functions in the first group (Group A), and then answer the questions. Repeat the process for Group B and Group C. Use the window specified for each group. TECHNOLOGY NOTE
By defining Y1 as directed in parts A, B, and C, defining Y2 , Y3 , and Y4 as shown here, you can minimize your keystrokes. (These graphs will not appear unless Y1 is defined.)
A 3 5, 54 by 3 5, 204
B 3 5, 154 by 3 5, 104
C 3 20, 204 by 3 10, 104
y1 = x 2
y1 = 2x
y1 = 2 3x
y2 = 2x 2
y2 = 2 2x
y2 = 2 2 3x
y3 = 3x 2
y3 = 3 2x
y3 = 3 2 3x
y4 =
y4 = 4 2x
3x y4 = 4 2
4x 2
1. How do the graphs of y2, y3, and y4 compare with the graph of y1? 2. If we choose c 7 4, how do you think the graph of y5 = c # y1 would compare with the graph of y4? Provide support by choosing such a value of c. In each group of functions in the preceding activity, we obtained a vertical stretch of the graph of the basic function with which we started. These observations can be generalized to any function.
y
y = cf (x), c > 1
x
0
Vertical Stretching of the Graph of a Function
y = f(x)
(x, y)
(x, cy) Vertical stretch
If a point (x, y) lies on the graph of y = ƒ(x), then the point (x, cy) lies on the graph of y = cƒ(x). If c 7 1, then the graph of y = cƒ(x) is a vertical stretching of the graph of y = ƒ(x) by applying a factor of c.
FIGURE 25 y
y2
y4
In FIGURE 25, we graphically interpret the preceding statement. Notice that the graphs have the same xintercepts.
y3
6
EXAMPLE 1
y1 = f (x) = ⏐x⏐ –4
0
x 4
FIGURE 26
Recognizing Vertical Stretches In FIGURE 26, the graph labeled y1 is that of the function defined by ƒ(x) =  x . The other three functions, y2, y3, and y4, are defined as follows, but not necessarily in the given order: 2 x , 3 x , and 4 x . Determine the correct equation for each graph.
2.3
Stretching, Shrinking, and Reflecting Graphs
105
Solution The given values of c are 2, 3, and 4. The vertical heights of the points with the same xcoordinates of the three graphs will correspond to the magnitudes of these c values. Thus, the graph just above y1 =  x  will be that of y = 2 x , the “highest” graph will be that of y = 4 x , and the graph of y = 3 x  will lie “between” the others. Therefore, y2 = 4 x , y3 = 2 x , and y4 = 3 x . If we graph each of these functions on the same screen of a graphing calculator, trace to any point on the graph of y1, and then move the tracing cursor to the other graphs one by one, the yvalues of the points are multiplied by the appropriate values of c. You may wish to experiment with your calculator in this way.
Vertical Shrinking FOR DISCUSSION
This discussion parallels the earlier one in this section. Follow the same general directions. (Note: The fractions 34, 12, and 14 may be entered as their decimal equivalents when plotting the graphs.) TECHNOLOGY NOTE
You can use a screen such as this to minimize your keystrokes in parts A, B, and C. Again, Y1 must be defined in order to obtain the other graphs.
A 3 5, 54 by 3 5, 204
B 3 5, 154 by 3 2, 54
C 3 10, 104 by 3 2, 104
y1 = x 2
y1 = 2x
y1 =  x 
y2 = 34 x 2
y2 = 34 2x
y2 = 34  x 
y3 = 12 x 2
y3 = 12 2x
y3 = 12  x 
y4 = 14 x 2
y4 = 14 2x
y4 = 14  x 
1. How do the graphs of y2, y3, and y4 compare with the graph of y1? 2. If we choose 0 6 c 6 14, how do you think the graph of y5 = c # y1 would compare with the graph of y4? Provide support by choosing such a value of c. In this “For Discussion” activity, we began with a basic function y1, and in each case the graph of y1 was vertically shrunk (or compressed). These observations can also be generalized to any function.
y
y = f (x) (x, cy)
x
0
(x, y) y = cf (x), 0 < c < 1 Vertical shrink
Vertical Shrinking of the Graph of a Function If a point (x, y) lies on the graph of y = ƒ(x), then the point (x, cy) lies on the graph of y = cƒ(x). If 0 6 c 6 1, then the graph of y = cƒ(x) is a vertical shrinking of the graph of y = ƒ(x) by applying a factor of c.
FIGURE 27 y
FIGURE 27
y3 y y 4 2
shows a graphical interpretation of vertical shrinking.
3
y1 = f (x) = x3 2
EXAMPLE 2
1 –3 –2
0 1
x 1
2
2 3
FIGURE 28
3
Recognizing Vertical Shrinks In FIGURE 28, the graph labeled y1 is that of the function defined by ƒ(x) = x 3. The other three functions, y2, y3, and y4, are defined as follows, but not necessarily in the given order: 0.5x 3, 0.3x 3, and 0.1x 3. Determine the correct equation for each graph. Solution The smaller the positive value of c, where 0 6 c 6 1, the more compressed toward the xaxis the graph will be. Since we have c = 0.5, 0.3, and 0.1, the function rules must be as follows: y2 = 0.1x 3, y3 = 0.5x 3, and y4 = 0.3x 3.
106
CHAPTER 2
Analysis of Graphs of Functions
Horizontal Stretching and Shrinking TECHNOLOGY NOTE
FOR DISCUSSION
This discussion parallels the two earlier ones in this section. Follow the same general directions. (Note: We give three related functions, defined by y1 = ƒ(x), y2 = ƒ(2x), and y3 = ƒ A 12 x B , in each group.) Use the standard viewing window to graph the equations. A To minimize keystrokes when graphing Y2 and Y3, you can use a screen such as this.
2 0
–2
4
y
y=f
( 12 x)
2
4
2 0
–4
y2 = 5  (2x) 2
y2 = (2x) 3  4(2x)
B
1 2 2x
y3 = A
y1 =  x   3
B  4A B
1 3 2x
1 2x
y2 =  2x   3 y3 =  12 x   3
x 2
FIGURE 29
stretch
y1 = x 3  4x
1. How do the graphs of y2 and y3 compare with the graph of y1? 2. How does the graph of y = ƒ(cx) compare with the graph of y = ƒ(x) when c 7 1 and when 0 6 c 6 1. Support your answer by graphing.
y = f (x)
–4
C
y1 = 5  x 2 y3 = 5  A
y
B
x
Horizontal stretch
In each group of functions in the preceding activity, we compared the graph of y1 = ƒ(x) with the graphs of functions of the form y = ƒ(cx). In each case, we obtained a graph that was either a horizontal stretch or shrink of y = ƒ(x). Consider the line graph in FIGURE 29. The graph can be stretched or shrunk horizontally. On one hand, if the line graph represents the graph of a function ƒ, then the graph of y = ƒ A 12 x B in FIGURE 30 is a horizontal stretching of the graph of y = ƒ(x). On the other hand, the graph of y = ƒ(2x) in FIGURE 31 represents a horizontal shrinking of the graph of y = ƒ(x). If one were to imagine the graph of y = ƒ(x) as a flexible wire, then a horizontal stretching would happen if the wire were pulled on each end, and a horizontal shrinking would happen if the wire were compressed. Horizontal stretching or shrinking does not change the height (maximum or minimum yvalues) of the graph, nor does it change the yintercept. These observations can be generalized to any function.
FIGURE 30 y
Horizontal Stretching or Shrinking of the Graph of a Function shrink
–4
0
–2
If a point (x, y) lies on the graph of y = ƒ(x), then the point A xc, y B lies on the graph of y = ƒ(cx).
y = f(2x)
x 2
4
–4
Horizontal shrink
(a) If 0 6 c 6 1, then the graph of y = ƒ(cx) is a horizontal stretching of the graph of y = ƒ(x). (b) If c 7 1, then the graph of y = ƒ(cx) is a horizontal shrinking of the graph of y = ƒ(x).
FIGURE 31
In FIGURE 32 on the next page, we interpret these statements graphically. For example, if the point (2, 0) is on the graph of y = ƒ(x), then the point A 22, 0 B = (1, 0) is on the 2 1 graph of y = ƒ(2x) and the point ¢ 1 , 0 ≤ = (4, 0) is on the graph of y = ƒ A 2 x B . Notice 2
in FIGURE 32, and also in the next example, that horizontal stretching or shrinking changes the xintercepts, but not the yintercept.
2.3
Stretching, Shrinking, and Reflecting Graphs
y
y = f (x)
0
107
y
y = f (cx) 01, then the graph of y ⴝ cƒ(x) is a vertical stretching of the graph of y = ƒ(x) by applying a factor of c. VERTICAL SHRINKING OF THE GRAPH OF A FUNCTION If a point (x, y) lies on the graph of y = ƒ(x), then the point (x, cy) lies on the graph of y = cƒ(x). If 03, where x is the weight of the bird in kilograms and S is the surface area of the wings in square meters. (Source: Pennycuick, C., Newton Rules Biology, Oxford University Press.) Approximate S(0.5) and interpret the result.
ƒ(x) = 0.117x 1.7, where x represents the man’s height in inches and ƒ(x) is his weight in pounds. What is the average weight of a 68inchtall man? 69. Animal Pulse Rate and Weight According to one model, the rate at which an animal’s heart beats varies with its weight. Smaller animals tend to have faster pulses, whereas larger animals tend to have slower pulses. The table in the next column lists average pulse rates in beats per minute (bpm) for animals with various weights in
9 72.
Explain why determining the domain of a function of the form n
ƒ(x) = 2ax + b requires two different considerations, depending upon the parity of n.
Checking Analytic Skills Determine the domain of each function. Do not use a calculator. 73. ƒ(x) = 25 + 4x
74. ƒ(x) = 29x + 18
75. ƒ(x) =  26  x
76. ƒ(x) =  22  0.5x
77. ƒ(x) = 2 3 8x  24
78. ƒ(x) = 2 5 x + 32
79. ƒ(x) = 249  x 2
80. ƒ(x) = 281  x 2
81. ƒ(x) = 2x 3  x
9 82.
Explain why the domain of ƒ(x) = 2x 2 + 1 is ( q, q).
4.5
Equations, Inequalities, and Applications Involving Root Functions
301
For Exercises 83–90, the domains were determined in Exercises 73–80. Use a graph to (a) find the range, (b) give the largest interval over which the function is increasing, (c) give the largest interval over which the function is decreasing, and (d) solve the equation ƒ(x) = 0 by observing the graph. 83. ƒ(x) = 25 + 4x
84. ƒ(x) = 29x + 18
85. ƒ(x) =  26  x
86. ƒ(x) =  22  0.5x
87. ƒ(x) = 2 3 8x  24
88. ƒ(x) = 2 5 x + 32
89. ƒ(x) = 249  x 2
90. ƒ(x) = 281  x 2
Concept Check Use transformations to explain how the graph of the given function can be obtained 3 x B. from the graph of the appropriate root function A y = 2x or y = 2 91. y = 29x + 27
92. y = 216x + 16
93. y = 24x + 16 + 4
94. y = 232  4x  3
95. y = 2 3 27x + 54  5
96. y = 2 3 8x  8
In Exercises 97–104, describe the graph of the equation as either a circle or a parabola with a horizontal axis of symmetry. Then, determine two functions, designated by y1 and y2, such that their union will give the graph of the given equation. Finally, graph the equation in the given viewing window. 97. x 2 + y 2 = 100; 3 15, 154 by 3 10, 104
98. x 2 + y 2 = 81; 3 15, 154 by 3 10, 104 100. (x + 3) 2 + y 2 = 16; 3 9.4, 9.44 by 3 6.2, 6.24
99. (x  2) 2 + y 2 = 9; 3 9.4, 9.44 by 3 6.2, 6.24 101. x = y 2 + 6y + 9; 3 10, 104 by 3 10, 104
102. x = y 2  8y + 16; 3 10, 104 by 3 10, 104
103. x = 2y 2 + 8y + 1; 3 10, 104 by 3 10, 104
104. x =  3y 2  6y + 2; 3 9.4, 9.44 by 3 6.2, 6.24
4.5
Equations, Inequalities, and Applications Involving Root Functions
Equations and Inequalities • An Application of Root Functions
Equations and Inequalities To solve an equation involving roots, such as 211  x  x = 1, in which the variable appears in a radicand, we use the following property.
Power Property If P and Q are algebraic expressions, then every solution of the equation P = Q is also a solution of the equation P n = Q n, for any positive integer n.
302
CHAPTER 4
Rational, Power, and Root Functions
CAUTION The power property does not say that the equations P ⴝ Q and P n ⴝ Q n are equivalent; it says only that each solution of the original equation P ⴝ Q is also a solution of the new equation P n ⴝ Q n.
When the power property is used to solve equations, the new equation may have more solutions than the original equation. We call these proposed solutions of the original equation. For example, the solution set of the equation x =  2 is obviously 5 26. If we square each side of the equation x =  2, we get the new equation x 2 = 4, which has solution set 5  2, 26. Since the solution sets are not equal, the equations are not equivalent. After applying the power property on equations that contain radicals or rational exponents, it is essential to check all proposed solutions in the original equation. To solve equations involving roots, follow these steps.
Solving an Equation Involving a Root Function Step 1 Isolate a term involving a root on one side of the equation. Step 2 Raise each side of the equation to a positive integer power that will eliminate the radical or rational exponent. Step 3 Solve the resulting equation. (If a root is still present after Step 2, repeat Steps 1 and 2.) Step 4 Check each proposed solution in the original equation.
EXAMPLE 1
Solving an Equation Involving a Square Root
Solve 211  x  x = 1. Analytic Solution 211  x  x = 1 211  x = 1 + x
Isolate the radical.
A 211  x B = (1 + x) 2 2
11  x = 2 x + 3x  10 = (x + 5)(x  2) = x + 5 = 0 or x =  5 or
Square each side.
1 + 2x + 0 0 x  2 = 0 x = 2
x2
( x + y)2 = x 2 + 2 x y + y 2 Standard form Factor. Zeroproduct property Solve each equation.
Check the proposed solutions,  5 and 2, in the original equation. 211  x  x = 1 Let x =  5. 211  ( 5)  ( 5) = 1 ? 216 + 5 = 1 ? 4 + 5 = 1 ? 9 = 1 False
Original equation
y1 = 211  x and
y2 = 1 + x
and observe the xcoordinate of the only point of intersection of the graphs, we see that it is 2, confirming the analytic solution. See FIGURE 58. Note that extraneous values do not occur in graphical solutions. y1 = √11 – x
Let x = 2.
y2 = 1 + x 10
211  2  2 = 1 ? 29  2 = 1 ? 3  2 = 1 ? 1 = 1 True
Squaring each side of the equation led to the extraneous value  5, as indicated by the false statement 9 = 1. Therefore, the only solution of the original equation is 2. The solution set is 526.
Graphing Calculator Solution The equation in the second line of the analytic solution has the same solution set as the original equation, because each side of the equation has not yet been squared. If we graph
–10
15
–10 FIGURE 58
4.5
Equations, Inequalities, and Applications Involving Root Functions
EXAMPLE 2
Solve (5 Analytic Solution (5  5x) 1/2 + x = 1 (5  5x) 1/2 = 1  x 3(5  5x) 1/242 5  5x 2 x + 3x  4 (x + 4)(x  1) x + 4 = 0 x = 4
5x) 1/2
Solving an Equation Involving a Rational Exponent + x = 1. Graphing Calculator Solution To use the xintercept method, we let y1 = (5  5x) 1/2 + x and
Isolate the term with the rational exponent.
= (1  x) 2 = 1  2x + x 2 = 0 = 0 or x  1 = 0 or x = 1
y2 = 1.
We graph y3 = y1  y2 to produce the curve shown in FIGURE 59. The displays at the bottom of the screens confirm the solutions,  4 and 1.
Square each side. ( x  y)2 = x 2  2 x y + y 2 Standard form Factor.
y3 = y1 – y2 = (5 – 5x)1/2 + x – 1
Zeroproduct property
6.2
Solve each equation.
Check the proposed solutions,  4 and 1, in the original equation. (5 
303
+ x = 1 Original equation Let x =  4. Let x 35  5(1)41/2 + (1) 35  5( 4)41/2 + ( 4) = 1 ? 0 1/2 + 1 251/2 + ( 4) = 1 ? 5 + ( 4) = 1 ? 0 + 1 1 = 1 True 1
–9.4
9.4
5x) 1/2
= = = = =
1. 1 1 1 1
Both proposed solutions check, so the solution set is 5  4, 16.
–6.2
?
y3 = y1 – y2 = (5 – 5x)1/2 + x – 1
?
6.2
? True
–9.4
9.4
–6.2 FIGURE 59
EXAMPLE 3
Solving an Equation Involving Cube Roots
Solve 2 3 x 2 + 3x = 2 3 5. Solution 2 3 x 2 + 3x = 2 35
3 x 2 + 3x B = A 2 3 5B A2 3
x2
+ 3x = 5
x 2 + 3x  5 = 0
3
Cube each side. Simplify. Standard form
Since this equation cannot be solved by usual factoring methods, we use the quadratic formula, with a = 1, b = 3, and c =  5. x =
 3 ⫾ 232  4(1)( 5)  3 ⫾ 229  b ⫾ 2b 2  4ac = = 2a 2(1) 2
While an analytic check is quite involved, it can be shown that both of these values are solutions. The solution set is e
 3 + 229  3  229 , f. 2 2
304
CHAPTER 4
Rational, Power, and Root Functions
EXAMPLE 4
Solving an Equation Involving Roots (Squaring Twice)
Solve 22x + 3  2x + 1 = 1. Solution Isolate the more complicated radical. 22x + 3  2x + 1 = 1 22x + 3 = 1 + 2x + 1
A 22x + 3 B = A 1 + 2x + 1 B 2
Remember the middle term.
(ab) m ⴝ ambm
Isolate 22 x + 3. 2
Square each side.
2x + 3 = 1 + 22x + 1 + x + 1
Simplify.
x + 1 = 2 2x + 1 (x +
1) 2
= A 2 2x + 1 B
Isolate the remaining radical. 2
Square each side again.
x 2 + 2x + 1 = 4(x + 1) x 2 + 2x + 1 = x 2  2x  3 = (x  3)(x + 1) = x  3 = 0 or x = 3 or
Simplify.
4x + 4 0 0 x + 1 = 0 x = 1
Distributive property Standard form Factor. Zeroproduct property Solve each equation.
Both proposed solutions check, giving 53,  16 as the solution set.
WHAT WENT WRONG? A student solved the following equation as shown. x + 4 x 2 + 8x + 16 x 2 + 7x + 10 (x + 5)(x + 2) x =  5 or x
= = = = =
2x + 6 x + 6 0 0 2
Square each side. Standard form Factor. Zeroproduct property
A check shows that  5 is extraneous, and the solution set is 5  26. To support his solution, he entered Y1 = X + 4 and Y2 = 2X + 6 into his calculator. Using the intersectionofgraphs method, he obtained the screen on the left below, and this assured him that he was correct. Next, he decided to use the xintercept method to show that  2 is a solution. He got the screen on the right below. Something was wrong. Y2 = √X + 6
Y1 = √X + 6 – X + 4
Y1 = X + 4 10
10 –10
–10
10
10
–10
What Went Wrong? How can he correct his work so that the screen on the right shows a zero of  2? Answer to What Went Wrong? He entered 2 (X + 6)  X + 4 rather than 2 (X + 6)  (X + 4). He should insert parentheses around X + 4 in Y1.
4.5
Equations, Inequalities, and Applications Involving Root Functions
EXAMPLE 5
305
Solving Inequalities Involving Roots
Solve each inequality. (b) 22x + 3  2x + 1 7 1
(a) 2 3 x 2 + 3x … 2 35
3
3
y = √x2 + 3x – √5
Solution (a) The associated equation was solved in Example 3. Its solutions are
10
 3 + 229 2 –10
and
 3  229 . 2
10
–10 FIGURE 60
We use the xintercept method to solve this inequality. As seen in FIGURE 60, the 3 5 lies on or below the xaxis in the interval 3 x 2 + 3x  2 graph of y = 2 between the two xintercepts, including the endpoints. Therefore, the solution set of the given inequality is c
 3  229  3 + 229 , d. 2 2
(b) This inequality is equivalent to 22x + 3  2x + 1  1 7 0. The associated equation was solved in Example 4; its solutions are 3 and  1. The graph of y3 = 22x + 3  2x + 1  1 lies above the xaxis for real numbers greater than 3, as we see in FIGURE 61. The solution set is (3, q). y3 = y1 – y2 = √2x + 3 – √x + 1 – 1 0.5
–6.2
12.6
–0.5 FIGURE 61
EXAMPLE 6
Solve
15x 2

19x 1
Solving an Equation Having Negative Exponents + 6 = 0.
Solution 15x 2  19x 1 + 6 = 0 x 2(15x 2  19x 1 + 6) = x 2(0) 15x 2x 2  19x 2x 1 + 6x 2 = 0 15  19x + 6x 2 = 0 6x 2
 19x + 15 = 0
(2x  3)(3x  5) = 0 x =
3 2
or x =
5 3
Multiply each side by x 2. Distributive property Properties of exponents Rewrite. Factor. Zeroproduct property
The solutions check. The solution set is E 32, 53 F .
306
CHAPTER 4
Rational, Power, and Root Functions
EXAMPLE 7
Solve
2x 2/3
+
5x 1/3
Solving an Equation Having Fractional Exponents  3 = 0.
Solution 2x 2/3 + 5x 1/3  3 = 2(x 1/3) 2 + 5(x 1/3)  3 = 2u 2 + 5u  3 = (2u  1) (u + 3) = 1 u = or u 2 1 or x 1/3 x 1/3 = 2 1 3 (x 1/3) 3 = a b or (x 1/3) 3 2 1 x = or x 8
0 0 0 0
Properties of exponents Let u = x 1/3 and substitute. Factor.
= 3
Zeroproduct property
= 3
Substitute x 1/3 for u.
= ( 3) 3
Cube each side.
=  27
Simplify.
The solutions check, so the solution set is E 18 ,  27 F .
An Application of Root Functions EXAMPLE 8
B Island k
6
Shoreline x
9–x
C
A FIGURE 62
Solving a Cable Installation Problem Involving a Root Function A company wishes to run a utility cable from point A on the shore (as shown in FIGURE 62) to an installation at point B on the island. The island is 6 miles from the shore. It costs $400 per mile to run the cable on land and $500 per mile under water. Assume that the cable starts at A, runs along the shoreline, and then angles and runs under water to the island. Let x represent the distance from C at which the underwater portion of the cable run begins, and let the distance between A and C be 9 miles. (a) What are the possible values of x in this problem? (b) Express the cost of laying the cable as a function of x. (c) Find the total cost if 3 miles of cable are on land. (d) Find the point at which the line should begin to angle to minimize the total cost. What is this total cost? Solve graphically. Solution (a) The value of x must be a real number greater than or equal to 0 and less than or equal to 9, meaning that x must be in the interval 30, 94. (b) The total cost is found by adding the cost of the cable on land to the cost of the cable under water. If we let k represent the length of the cable under water, then k 2 = 62 + x 2 k 2 = 36 + x 2 k = 236 + x 2.
Use the Pythagorean theorem.
Square root property; k 7 0
The cost of the cable on land is 400(9  x) dollars, while the cost of the cable under water is 500k, or 500236 + x 2, dollars. The total cost C(x) is given by C(x) = 400(9  x) + 500 236 + x 2.
Total cost in dollars
4.5
C(x) = 400 (9 – x) + 500√36 + x2 6500
Equations, Inequalities, and Applications Involving Root Functions
307
(c) According to FIGURE 62 on the preceding page, if 3 miles of cable are on land, then 3 = 9  x, giving x = 6. We must evaluate C(6). C(6) = 400(9  6) + 500 236 + 62 L 5442.64 dollars
–1.4
17.4
5200 FIGURE 63
(d) From FIGURE 63, the absolute minimum value of the function on the interval 30, 94 is found when x = 8, meaning that 9  8 = 1 mile should be along land and 236 + 82 = 10 miles should be under water. We must evaluate C(8).
xvalue
C(8) = 400(9  8) + 500236 + 82 = 5400 dollars
Minimum total cost
FIGURE 64 shows how the coordinates of the minimum point can be determined on the TI83/84 Plus without actually graphing the function.
yvalue FIGURE 64
4.5
EXERCISES
Checking Analytic Skills In Exercises 1–5, begin by drawing a rough sketch to determine the number of real solutions for the equation y1 = y2. Then solve this equation by hand. Give the solution set and any extraneous values that may occur. Do not use a calculator. 1. y1 = 2x y2 = 2x  1
9 6.
2. y1 = 2x y2 = x  6
3. y1 = 2x y2 =  x + 3
4. y1 = 2x y2 = 3x
5. y1 = 2 3x y2 = x 2
Use a handdrawn graph to explain why 2x =  x  5 has no real solutions.
Checking Analytic Skills Solve each equation by hand. Do not use a calculator. 7. x  4 = 23x  8
8. x  5 = 25x  1
9. 2x + 5 + 1 = x
10. 24  3x  8 = x
11. 22x + 3  2x + 1 = 1
12. 23x + 4  22x  4 = 2
13. 2 3 x + 1 = 3
14. 2 3x + 9 = 2
15. 2 3 3x 2 + 7 = 2 3 7  4x
16. 2 3 2x 2 + 1 = 2 31  x
17. 2 4x  2 + 4 = 6
18. 2 4 2x + 3 = 2x + 1
19. x 2/5 = 4
20. x 2/3 = 16
21. 2x 1/3  5 = 1
22. 4x 3/2 + 5 = 21
23. x 2 + 3x 1 + 2 = 0
24. 2x 2  x 1 = 3
25. 5x 2 + 13x 1 = 28
26. 3x 2  19x 1 + 20 = 0
27. x 2/3  x 1/3  6 = 0
28. x 2/3 + 9x 1/3 + 14 = 0
29. x 3/4  x 1/2  x 1/4 + 1 = 0
30. x 3/4  2x 1/2  4x 1/4 + 8 = 0
Use an analytic method to solve each equation in part (a). Support the solution with a graph. Then use the graph to solve the inequalities in parts (b) and (c). 31. (a) 23x + 7 = 2
32. (a) 22x + 13 = 3
33. (a) 24x + 13 = 2x  1
(b) 23x + 7 7 2
(b) 22x + 13 7 3
(b) 24x + 13 7 2x  1
(c) 23x + 7 6 2
(c) 22x + 13 6 3
(c) 24x + 13 6 2x  1
308
CHAPTER 4
Rational, Power, and Root Functions
34. (a) 23x + 7 = 3x + 5
35. (a) 25x + 1 + 2 = 2x
36. (a) 23x + 4 + x = 8
(b) 23x + 7 7 3x + 5
(b) 25x + 1 + 2 7 2x
(b) 23x + 4 + x 7 8
(c) 23x + 7 6 3x + 5
(c) 25x + 1 + 2 6 2x
(c) 23x + 4 + x 6 8
37. (a) 23x  6 + 2 = 25x  6
38. (a) 22x  4 + 2 = 23x + 4
39. (a) 2 3 x 2  2x = 2 3x
(b) 23x  6 + 2 7 25x  6
(b) 22x  4 + 2 7 23x + 4
(b) 2 3 x 2  2x 7 2 3x
(c) 23x  6 + 2 6 25x  6
(c) 22x  4 + 2 6 23x + 4
(c) 2 3 x 2  2x 6 2 3x
40. (a) 2 3 4x 2  4x + 1 = 2 3x (b)
2 3 4x 2
(c)
2 3 4x 2
41. (a) 2 4 3x + 1 = 1
42. (a) 2 4 x  15 = 2
 4x + 1 7 2 3x
(b) 2 4 3x + 1 7 1
(b) 2 4 x  15 7 2
 4x + 1 6 2 3x
(c) 2 4 3x + 1 6 1
(c) 2 4 x  15 6 2
43. (a) (2x  5) 1/2  2 = (x  2) 1/2
44. (a) (x + 5) 1/2  2 = (x  1) 1/2
(b) (2x  5) 1/2  2 Ú (x  2) 1/2
(b) (x + 5) 1/2  2 Ú (x  1) 1/2
(c) (2x  5) 1/2  2 … (x  2) 1/2
(c) (x + 5) 1/2  2 … (x  1) 1/2
45. (a) (x 2 + 6x) 1/4 = 2
46. (a) (x 2 + 2x) 1/4 = 31/4
(b) (x 2 + 6x) 1/4 7 2
(b) (x 2 + 2x) 1/4 7 31/4
(c) (x 2 + 6x) 1/4 6 2
(c) (x 2 + 2x) 1/4 6 31/4
47. (a) (2x  1) 2/3 = x 1/3
48. (a) (x  3) 2/5 = (4x) 1/5
(b) (2x  1) 2/3 7 x 1/3
(b) (x  3) 2/5 7 (4x) 1/5
(c) (2x  1) 2/3 6 x 1/3
(c) (x  3) 2/5 6 (4x) 1/5
RELATING CONCEPTS
For individual or group investigation (Exercises 49–60)
Exercises 49–60 incorporate many concepts from Chapter 3 with the method of solving equations involving roots. Work them in order. Consider the equation 2 3 4x  4 = 2x + 1. 49. Rewrite the equation, using rational exponents.
50. What is the least common denominator of the rational exponents found in Exercise 49?
51. Raise each side of the equation in Exercise 49 to the power indicated by your answer in Exercise 50.
52. Show that the equation in Exercise 51 is equivalent to x 3  13x 2 + 35x  15 = 0.
53. Graph the cubic function defined by the polynomial on the left side of the equation in Exercise 52 in the window 3 5, 104 by 3 100, 1004. How many real solutions does the equation have?
54. Use synthetic division to show that 3 is a zero of the polynomial
55. Use the result of Exercise 54 to factor P(x) so that one factor is linear and the other is quadratic.
56. Set the quadratic factor of P(x) from Exercise 55 equal to 0, and solve the equation, using the quadratic formula.
57. What are the three proposed solutions of the original equation,
58. Let y1 = 2 3 4x  4 and let y2 = 2x + 1. Graph y3 = y1  y2 in the window 3 2, 204 by 3 0.5, 0.54 to determine the number of real solutions of the original equation.
2 3 4x  4 = 2x + 1? 59. Use both an analytic method and your calculator to solve the original equation.
P(x) = x 3  13x 2 + 35x  15.
9 60. Write an explanation of how the solutions of the equa
tion in Exercise 52 relate to the solutions of the original equation. Discuss any extraneous solutions that may be involved.
4.5
Equations, Inequalities, and Applications Involving Root Functions
309
Solve each equation involving “nested” radicals for all real solutions analytically. Support your solutions with a graph. 61. 32x = x
62. 3 3 2 3x = x
64. 322x + 10 = 22x  2
3x + 6 65. 3232x = 2
67. Velocity of a Meteorite The velocity v of a meteorite approaching Earth is given by k 2d
68. Illumination The illumination I in footcandles produced by a light source is related to the distance d in feet from the light source by the equation
66. 32x + 63 = 2 3 2x + 6
72. Plant Species and Land Area A research biologist has shown that the number S of different plant species on a Galápagos Island is related to the area A of the island by S = 28.62 3 A.
,
measured in kilometers per second, where d is its distance from the center of Earth and k is a constant. If k = 350, what is the velocity of a meteorite that is 6000 kilometers away from the center of Earth? Round to the nearest tenth.
d =
3
3
(Modeling) Solve each problem.
v =
63. 3228x + 8 = 23x + 2
Find S for an island with each area. (a) 100 square miles (b) 1500 square miles 73. Wire between Two Poles Two vertical poles of lengths 12 feet and 16 feet are situated on level ground, 20 feet apart, as shown in the figure. A piece of wire is to be strung from the top of the 12foot pole to the top of the 16foot pole, attached to a stake in the ground at a point P on a line formed by the vertical poles. Let x represent the distance from P to D, the base of the 12foot pole.
k , BI
B
where k is a constant. If k = 400, how far from the source will the illumination be 14 footcandles? Round to the nearest hundredth of a foot.
A 16 ft
69. Period of a Pendulum The period P of a pendulum in seconds depends on its length L in feet and is given by P = 2p
12 ft D
L . B 32
If the length of a pendulum is 5 feet, what is its period? Round to the nearest tenth. 70. Visibility from an Airplane A formula for calculating the distance d one can see from an airplane to the horizon on a clear day is d = 1.22 2x, where x is the altitude of the plane in feet and d is given in miles. How far can one see to the horizon in a plane flying at each altitude? Give answers to the nearest mile. (a) 15,000 feet (b) 20,000 feet 71. Speed of a Car in an Accident To estimate the speed s at which a car was traveling at the time of an accident, a police officer drives a car like the one involved in the accident under conditions similar to those during which the accident took place and then skids to a stop. If the car is driven at 30 miles per hour, the speed s at the time of the accident is given by
P
C
(a) Express the distance from P to C in terms of x. (b) What are the restrictions on the value of x? (c) Use the Pythagorean theorem to express the lengths AP and BP in terms of x. (d) Give a function ƒ that expresses the total length of the wire used. 9 (e) Graph ƒ in the window 30, 204 by 30, 504. Use your calculator to find ƒ(4), and interpret your result. (f ) Find the value of x that will minimize the amount of wire used. What is this minimum? (g) 9 Write a short paragraph summarizing what this problem has examined and the results you have obtained.
9 74.
Wire between Two Poles Repeat Exercise 73 if the heights of the poles are 9 feet and 12 feet and the distance between the poles is 16 feet. Let P be x feet from the 9foot pole. In part (e), use the window 30, 164 by 30, 504. B
a , Ap
A
s = 30
where a is the length of the skid marks and p is the length of the marks in the police test. Find s if a = 900 feet and p = 97 feet.
x
12 ft
9 ft D
x
P
C
310
CHAPTER 4
Rational, Power, and Root Functions
75. Hunter Returns to His Cabin A hunter is at a point on a riverbank. He wants to get to his cabin, located 3 miles north and 8 miles west. He can travel 5 mph on the river but only 2 mph on this very rocky land. How far upriver, to the nearest hundredth of a mile, should he go in order to reach the cabin in a minimum amount of time? (Hint: distance = rate * time.)
77. Cruise Ship Travel At noon, the cruise ship Celebration is 60 miles due south of the cruise ship Inspiration and is sailing north at a rate of 30 mph. If the Inspiration is sailing west at a rate of 20 mph, find the time at which the distance d between the ships is a minimum. What is this distance to the nearest hundredth of a mile?
N
Inspiration
Cabin
E
W 3 x
8⫺x
Hunter d
76. Homing Pigeon Flight Homing pigeons avoid flying over large bodies of water, preferring to fly around them instead. (One possible explanation is the fact that extra energy is required to fly over water because air pressure drops over water in the daytime.) Assume that a pigeon released from a boat 1 mile from the shore of a lake (point B in the figure) flies first to point P on the shore and then along the straight edge of the lake to reach its home at L. If L is 2 miles from point A, the point on the shore closest to the boat, and if a pigeon needs 43 as much energy to fly over water as over land, find the location, to the nearest hundredth of a mile, of point P that minimizes the pigeon’s energy use.
Celebration S
78. Fisherman Returns to Camp “Mido” Simon is in his bass boat, the Lesley, 3 miles from the nearest point on the shore. He wishes to reach his camp at Maggie Point, 6 miles farther down the shoreline. If Mido’s motor is disabled, and he can row his boat at a rate of 4 mph and walk at a rate of 5 mph, find the least amount of time to the nearest hundredth of an hour that he will need to reach the camp.
6 mi
B
x 3 mi
Maggie Point 6–x
1 mi
P A
SECTIONS 4.4 AND 4.5
9 1.
L
REVIEWING BASIC CONCEPTS
Use a graphing calculator to graph y = x 0.7, y = x 1.2, and y = x 2.4 for x Ú 0 on the same screen. Describe what happens as the exponent n increases in value.
2. (Modeling) Wing Size Suppose that the relationship between the surface area of the wings of a species of bird and the weight of a bird is given by S(w) = 0.3w 3/4, where w is the weight of the bird in kilograms and S is the surface area of the wings in square meters. If a bird weighs 0.75 kilogram, approximate the surface area of the wings.
Lesley
2 mi
3. To graph the circle with equation x 2 + y 2 = 16 with a graphing calculator in function mode, what two expressions must be used for y1 and y2? Graph the circle in a decimal window. 4. To graph the horizontal parabola x = y 2 + 4y + 6 with a graphing calculator in function mode, what two expressions must be used for y1 and y2? Graph the parabola in a square window. 5. Solve the equation 23x + 4 = 8  x analytically. 6. Use a graph and the solution of the equation in Exercise 5 to solve the inequality 23x + 4 7 8  x.
CHAPTER 4
7. Use a graph and the solution of the equation in Exercise 5 to solve the inequality 23x + 4 6 8  x. 8. Solve the equation 23x + 4 + 25x + 6 = 2 analytically, and support the solution with a graph.
9 9.
Summary
311
In this formula, x is the length of the animal in inches and ƒ(x) is the approximate number of heartbeats per minute. (Source: Lancaster, H., Quantitative Methods in Biology and Medical Sciences, SpringerVerlag.)
Explain why the table shows error messages for all values of X less than 15.
10. (Modeling) Animal Pulse Rates An 18thcentury study found that the pulse rate of an animal could be approximated by ƒ(x) =
4
1607 2 4 x3
.
(a) Use ƒ to estimate the pulse rates of a 2foot cat and a 5.5foot person. (b) What length, to the nearest tenth, corresponds to a pulse rate of 400 beats per minute?
SUMMARY
KEY TERMS & SYMBOLS
KEY CONCEPTS
4.1 Rational Functions and Graphs
RATIONAL FUNCTION p A function ƒ of the form q defined by
rational expression rational function discontinuous reciprocal function vertical asymptote horizontal asymptote
ƒ(x) ⴝ
p(x) , q(x)
where p(x) and q(x) are polynomials, with q(x) Z 0, is called a rational function. 1 (the reciprocal function) decreases on the intervals ( q, 0) and (0, q). x It is discontinuous at x = 0. The yaxis is a vertical asymptote, and the xaxis is a horizontal asymptote. It is an odd function, and its graph is symmetric with respect to the origin.
ƒ(x) ⴝ ■ ■ ■
1 increases on the interval ( q, 0) and decreases on the interval (0, q). x2 It is discontinuous at x = 0. The yaxis is a vertical asymptote, and the xaxis is a horizontal asymptote. It is an even function, and its graph is symmetric with respect to the yaxis.
ƒ(x) ⴝ ■ ■ ■
(continued)
312
CHAPTER 4
Rational, Power, and Root Functions
KEY TERMS & SYMBOLS
KEY CONCEPTS
4.2 More on Rational Functions and Graphs
GRAPHING A RATIONAL FUNCTION
oblique (slant) asymptote pseudoasymptote
Let ƒ(x) = Step 1 Step 2 Step 3 Step 4
Step 6 Step 7
rational equation rational inequality inverse variation combined variation joint variation
define a function with the rational expression in lowest terms. To sketch
its graph, follow these steps:
Step 5
4.3 Rational Equations, Inequalities, Models, and Applications
p(x) q(x)
Find all vertical asymptotes. (See page 265.) Find all horizontal or oblique asymptotes. (See page 265.) Find the yintercept, if possible, by evaluating ƒ(0). Find the xintercepts, if any, by solving ƒ(x) = 0. (These will be the zeros of the numerator p(x).) Determine whether the graph will intersect its nonvertical asymptote by solving ƒ(x) = b, where b is the yvalue of the horizontal asymptote, or by solving ƒ(x) = mx + b, where y = mx + b is the equation of the oblique asymptote. Plot selected points as necessary. Choose an xvalue in each interval of the domain determined by the vertical asymptotes and xintercepts. Complete the sketch.
SOLVING A RATIONAL EQUATION Step 1 Step 2 Step 3 Step 4
Determine all values for which the rational equation has undefined expressions. Multiply each side of the equation by the LCD of all rational expressions in the equation. Solve the resulting equation. Reject any values found in Step 1.
SOLVING A RATIONAL INEQUALITY Step 1 Step 2
Step 3
Rewrite the inequality, if necessary, so that 0 is on one side and there is a single fraction on the other side. Determine the values that will cause either the numerator or the denominator of the rational expression to equal 0. These values determine the intervals on the number line to consider. Use a test value from each interval to determine which intervals form the solution set.
INVERSE VARIATION AS THE nTH POWER Let x and y denote two quantities and n be a positive number. Then y is inversely proportional to the nth power of x, or y varies inversely as the nth power of x, if there exists a nonzero number k such that yⴝ
k . xn
If y = kx, then y is inversely proportional to x, or y varies inversely as x. JOINT VARIATION Let m and n be real numbers. Then z varies jointly as the nth power of x and the mth power of y if a nonzero real number k exists such that z ⴝ k x n y m.
CHAPTER 4
KEY TERMS & SYMBOLS
KEY CONCEPTS
4.4 Functions Defined by Powers and Roots
POWER AND ROOT FUNCTIONS A function ƒ given by ƒ(x) ⴝ x b, where b is a constant, is a power function. If b = function given by
power function root function
ƒ(x) ⴝ x 1/n, ■ ■
4.5 Equations, Inequalities, and Applications Involving Root Functions proposed solution
1 n
for some integer n Ú 2, then ƒ is a root n
ƒ(x) ⴝ 2x.
or equivalently, n
For n even, the root function defined by ƒ(x) = 2x increases and is continuous on its entire domain, 30, q). n For n odd, the root function defined by ƒ(x) = 2x increases and is continuous on its entire domain, ( q, q).
SOLVING AN EQUATION INVOLVING A ROOT FUNCTION Step 1 Step 2 Step 3 Step 4
4
313
Review Exercises
Isolate a term involving a root on one side of the equation. Raise each side of the equation to a positive integer power that will eliminate the radical or rational exponent. Solve the resulting equation. (If a root is still present after Step 2, repeat Steps 1 and 2.) Check each proposed solution in the original equation.
REVIEW EXERCISES
For each rational function, do the following. (a) Explain how the graph of the function can be obtained from the graph of y = 1x or y = x12. (b) Sketch the graph by hand. (c) Use a graphing calculator to obtain an accurate depiction of the graph. 1 1. y =  + 6 x
4 2. y =  3 x
1 3. y = (x  2) 2
2 4. y = 2 + 1 x
10. ƒ(x) =
x2  1 x
11. ƒ(x) =
12. ƒ(x) =
x2  1 x + 1
13. ƒ(x) =
4x  3 6. ƒ(x) = 2x  1
6x 7. ƒ(x) = (x  1)(x + 2)
2x 8. ƒ(x) = 2 x  1
x2 + 4 9. ƒ(x) = x + 2
2 + 1
x2  x  2 x 2 + 3x + 2
Find an equation for each rational function graphed. (Answers may vary.) 14.
15.
y
y
(–2, 14 ) 1 (2, 14 )
9 5. Under what conditions will the graph of a rational function defined by an expression written in lowest terms have an oblique asymptote? Sketch a graph of each rational function. Your graph should include all asymptotes.
x2
–3
–1 0
2
3
–3
–3
16.
x
0 1
x 1
x=1
y = –3
–6
17.
y
y 2
4 2 –2 0
x 2
(
)
–2, – 15 1 –2
0
(2, – 15) x 2
(0, –1) –2
314
Rational, Power, and Root Functions
CHAPTER 4
Solve the rational equation in part (a) analytically. Then, use a graph to determine the solution sets of the associated inequalities in parts (b) and (c). 5 18. (a) = 2x + 5 x 5 (b) 6 2x + 5 x (c)
3x  2 19. (a) = 0 x + 1 3x  2 (b) 6 0 x + 1
3 + 2 3 + 2
5 3 7 2x + 5 x + 2
(c)
3x  2 7 0 x + 1
Points are awarded for the general attractiveness of the car. The function defined by 10x 49(101  x) expresses the cost, in thousands of dollars, of restoring a car so that it will win x points. (a) Use a graphing calculator to graph the function in the window 30, 1014 by 30, 54. (b) How much would an owner expect to pay to restore a car in order to earn 95 points? C(x) =
6 5 + 2 = 0 x x 6 5 (b) 1  + 2 … 0 x x 5 6 (c) 1  + 2 Ú 0 x x
20. (a) 1 
3 1 1 + = 2 x  2 x + 1 x  x  2 3 1 1 (b) + … 2 x  2 x + 1 x  x  2 3 1 1 (c) + Ú 2 x  2 x + 1 x  x  2
21. (a)
28. Concentration of a Drug The concentration of a drug in a patient’s bloodstream is given by ƒ(t) =
A comprehensive graph of a rational function ƒ is shown. Use the graph to find each solution set. 22. ƒ(x) = 0
3.1
23. ƒ(x) 6 0 24. ƒ(x) 7 0
–4.7
4.7
25.  ƒ(x)  7 0 –3.1 The graph has vertical asymptote x = –1.
26. Environmental Pollution In situations involving environmental pollution, a cost–benefit model expresses cost as a function of the percentage of pollutant removed from the environment. Suppose a cost–benefit model is expressed as C(x) =
6.7x , 100  x
where C(x) is the cost, in thousands of dollars, of removing x percent of a certain pollutant. (a) Use a graphing calculator to graph the function in the window 30, 1004 by 30, 1504. (b) Find the cost to remove 95% of the pollutant. 27. AntiqueCar Competition Antiquecar owners often enter their cars in a concours d’elegance in which a maximum of 100 points can be awarded to a particular car.
5 , + 1
where t Ú 0 is in hours and ƒ(t) is in milligrams per liter. (a) Does the concentration of the drug increase or decrease? (b) The patient should not take a second dose until the concentration is below 1.5 milligrams per liter. How long should the patient wait before taking a second dose? 29. Waiting in Line A parking lot attendant can wait on 40 cars per hour. If cars arrive randomly at a rate of x cars per hour, then the average line length to enter the ramp is given by ƒ(x) =
(Modeling) Solve each problem.
t2
x2 , where 0 … x 6 40. 1600  40x
(a) Solve the inequality ƒ(x) … 8.
9 (b) Interpret your answer from part (a). 30. Slippery Roads The coefficient of friction x measures the friction between the tires of a car and the road, where 0 6 x … 1. A smaller value of x indicates that the road is more slippery. If a car is traveling at 60 mph, then braking distance D in feet is given by D(x) =
120 . x
(a) What happens to the braking distance as the coefficient of friction becomes smaller? (b) Find the interval of values for the coefficient x that corresponds to a braking distance of 400 feet or more. 31. Let y be inversely proportional to x. When x = 6, y = 5. Find y when x = 15.
CHAPTER 4
32. Let z be inversely proportional to the third power of t. When t = 5, z = 0.08. Find z when t = 2. 33. Suppose m varies jointly as n and the square of p, and inversely as q. If m = 20 when n = 5, p = 6, and q = 18, find m when n = 7, p = 11, and q = 2. 34. The formula for the height of a right circular cone with volume 100 is h =
300 p
r2
Review Exercises
315
39. Sports Arena Construction The roof of a new sports arena rests on round concrete pillars. The maximum load a cylindrical column of circular cross section can hold varies directly as the fourth power of the diameter and inversely as the square of the height. The arena has 9metertall columns that are 1 meter in diameter and will support a load of 8 metric tons. How many metric tons will be supported by a column 12 meters high and 23 meter in diameter?
.
The height of this cone varies as the of the of its base. The constant of variation is given by . 9m
1m Volume = 100
h
Load = 8 metric tons r
35. Illumination The illumination produced by a light source varies inversely as the square of the distance from the source. The illumination of a light source at 5 meters is 70 candela. Approximate the illumination 12 meters from the source. 36. Current Flow In electric current flow, it is found that the resistance (measured in units called ohms) offered by a fixed length of wire of a given material varies inversely as the square of the diameter of the wire. If a wire 0.01 inch in diameter has a resistance of 0.4 ohm, what is the resistance of a wire of the same length and material with a diameter of 0.03 inch?
40. Sports Arena Construction A sports arena requires a beam 16 meters long, 24 centimeters wide, and 8 centimeters high. The maximum load of a horizontal beam that is supported at both ends varies directly as the width and square of the height of the beam and inversely as the length between supports. If a beam of the same material 8 meters long, 12 centimeters wide, and 15 centimeters high can support a maximum of 400 kilograms, what is the maximum load the beam in the arena will support? 41. Weight of an Object The weight w of an object varies inversely as the square of the distance d between the object and the center of Earth. If a man weighs 90 kilograms on the surface of Earth, how much would he weigh 800 kilometers above the surface? (The radius of Earth is about 6400 kilometers.)
37. Simple Interest Simple interest varies jointly as principal and time. If $1000 left at interest for 2 years earned $110, find the amount of interest earned by $5000 for 5 years.
42. Rate of Work Louise and Keith are working on a building site cleanup. Keith can clean up all the trash in the area in 7 hours, while Louise can do the same job in 5 hours. How long will it take them if they work together?
38. Car Skidding The force needed to keep a car from skidding on a curve varies inversely as the radius of the curve and jointly as the weight of the car and the square of the speed. It takes 3000 pounds of force to keep a 2000pound car from skidding on a curve of radius 500 feet at 30 mph. What force is needed to keep the same car from skidding on a curve of radius 800 feet at 60 mph?
43. Rate of Work Suppose that Terry and Carrie can clean their entire house in 7 hours, while their toddler, Joey, just by being around, can completely mess it up in only 2 hours. If Terry and Carrie clean the house while Joey is at his grandma’s, and then start cleaning up after Joey the minute he gets home, how long does it take from the time Joey gets home until the whole place resembles a disaster area? 44. Rate of Work Suppose it takes Jack 35 minutes to do the dishes, and together he and his wife Jill can do them in 15 minutes. How long will it take Jill to do them alone?
r
Suppose that a and b are positive numbers. Sketch the general shape of the graph of each function. 45. y = 2 3x + a
46. y =  a 2x
47. y = 2 3x + b
48. y = 2 3x  a
49. y =  a 2 3x  b
50. y = 2x + a + b
316
CHAPTER 4
Rational, Power, and Root Functions
Evaluate each expression without the use of a calculator. Then, check your results with a calculator. 51. 2144
52. 2 3  64
1 53. 3 B 27
54. 2 4 81
32 55. 5 B 243
56.  ( 32) 1/5
57. 363/2
58.  1000 2/3
59. ( 27) 4/3
60. 163/4
Use your calculator to find each root or power. Give as many digits as your display shows. 1 61. 2 62. 4 5 84.6 B 16 1 4/3 63. a b 64. 12 1/3 8 Consider the function defined by ƒ(x) =  22x  4 for Exercises 65–67. 65. Find the domain of ƒ analytically. 66. Use a graph to determine the range of ƒ. 67. Give the largest interval (if any) over which the function is (a) increasing. (b) decreasing. 68. Determine the expressions for y1 and y2 that are required to graph the circle x 2 + y 2 = 25 with a graphing calculator in function mode. Then graph the circle in a square window. Solve each equation involving radicals or rational exponents in part (a) analytically. Then, use a graph to determine the solution sets of the associated inequalities in parts (b) and (c). 69. (a) 25 + 2x = x + 1 (b) 25 + 2x 7 x + 1 (c) 25 + 2x 6 x + 1 70. (a) 22x + 1  2x = 1
78. 2x  2 = 14  x 79. 2 3 2x  3 + 1 = 4 80. x 1/3 + 3x 1/3 =  2 81. 2x 2  5x 1 = 3 82. x 3 + 2x 2 + x 1 = 0 83. x 2/3  4x 1/3  5 = 0 84. x 3/4  16x 1/4 = 0 85. 2x + 1 + 1 = 22x 86. 2x  2 = 5  2x + 3 (Modeling) Solve each problem. 87. Swing of a Pendulum A simple pendulum will swing back and forth in regular time intervals (periods). Grandfather clocks use pendulums to keep accurate time. The relationship between the length L of a pendulum and the time T for one complete oscillation can be expressed by the equation L = kT n, where k is a constant and n is a positive integer to be determined. The following data were taken for different lengths of pendulums. T (in seconds) 1.11 1.36 1.57 1.76 1.92 2.08 2.22
L (in feet) 1.0 1.5 2.0 2.5 3.0 3.5 4.0
(a) As L increases, what happens to T?
9 (b) Discuss how n and k can be found.
(b) 22x + 1  2x 7 1 (c) 22x + 1  2x 6 1
(c) Use the data to approximate k, and determine the best value for n. (d) Using the values of k and n from part (c), predict T for a pendulum having a length of 5 feet. (e) If the length L of a pendulum doubles, what happens to the period T ?
71. (a) 2 3 6x + 2 = 2 3 4x (b) 2 3 6x + 2 Ú 2 3 4x (c) 2 3 6x + 2 … 2 3 4x 72. (a) (x  2) 2/3  x 1/3 = 0 (b) (x  2) 2/3  x 1/3 Ú 0 (c) (x  2) 2/3  x 1/3 … 0 Solve each equation. Check your results. 73. x 5 = 1024
74. x 1/3 = 4
75. 2x  2 = x  4
76. x 3/2 = 27
77. 2x 1/4 + 3 = 6
88. Volume of a Cylindrical Package A company plans to package its product in a cylinder that is open at one end. The cylinder is to have a volume of 27p cubic inches. What radius should the circular bottom of the cylinder have to minimize the cost of the material? (Hint: The volume V of a circular cylinder is V = pr 2h, where r is the radius of the circular base and h is the height; the surface area S of a cylinder open at one end is S = 2prh + pr 2.)
CHAPTER 4
4
317
TEST
For each rational function, do the following. (a) Sketch its graph. (b) Explain how the graph is obtained from the graph of y = 1x or y = x12. (c) Use a graphing calculator to obtain an accurate depiction of the graph. 1. ƒ(x) = 
Test
1 x
2. ƒ(x) = 
1  3 x2
3. Consider the rational function defined by ƒ(x) =
(x + 3)(x  2) x2 + x  6 . = 2 (x  4)(x + 1) x  3x  4
Determine the answers to (a)–(e) analytically. Find (a) equations of the vertical asymptotes. (b) the equation of the horizontal asymptote. (c) the yintercept. (d) xintercepts, if any. (e) the coordinates of the point where the graph of ƒ intersects its horizontal asymptote. (f ) Now sketch a comprehensive graph of ƒ. 4. Find the equation of the oblique asymptote of the graph of the rational function defined by ƒ(x) =
2x 2 + x  3 . x  2
Then graph the function and its asymptotes.
rate of x vehicles per minute, then the average wait W in minutes for a car to pass through this section of highway is approximated by W(x) =
1 , where 0 … x 6 40. 40  x
9 (a)
Evaluate W(30), W(39), and W(39.9). Interpret the results. (b) Graph W, using the window 30, 404 by 3  0.5, 14. Identify the vertical asymptote. What happens to W as x approaches 40? (c) Find x when the wait is 5 minutes.
8. (Modeling) Measure of Malnutrition The pelidisi, a measure of malnutrition, varies directly as the cube root of a person’s weight in grams and inversely as the person’s sitting height in centimeters. A person with a pelidisi below 100 is considered to be undernourished, while a pelidisi greater than 100 indicates overfeeding. A person who weighs 48,820 grams and has a sitting height of 78.7 centimeters has a pelidisi of 100. Find the pelidisi (to the nearest whole number) of a person whose weight is 54,430 grams and whose sitting height is 88.9 centimeters. Is this individual undernourished or overfed?
88.9 cm 78.7 cm
5. Consider the rational function defined by ƒ(x) =
x 2  16 . x + 4
(a) For what value of x does the graph exhibit a “hole”? (b) Graph the function ƒ and show the “hole” in the graph.
Weight: 48,820 g
Weight: 54,430 g
6. (a) Solve the following rational equation analytically. 21 3 14 + 2 = x  2 x + 2 x  4 (b) Use the result of part (a) and a graph to find the solution set of 3 21 14 + 2 . Ú x  2 x + 2 x  4 7. (Modeling) Waiting in Line A small section of highway that is under construction can accommodate at most 40 cars per minute. If cars arrive randomly at an average
9. (Modeling) Volume of a Cylindrical Can A manufacturer wants to construct a cylindrical aluminum can with a volume of 4000 cubic centimeters. If x represents the radius of the circular top and bottom of the can, the surface area S as a function of x is given by S(x) =
8000 + 2px 3 . x
Use a graph to find the radius that will minimize the amount of aluminum needed, and determine what this amount will be. (Hint: Use the window 3 5, 424 by 3 1000, 10,0004.)
318
CHAPTER 4
Rational, Power, and Root Functions
10. Graph the function defined by ƒ(x) =  25  x in the standard viewing window. Then, do the following. (a) Determine the domain analytically. (b) Use the graph to find the range. (c) This function over its entire (increases/decreases) domain. (d) Solve the equation ƒ(x) = 0 graphically. (e) Solve the inequality ƒ(x) 6 0 graphically. 11. (a) Solve the equation 24  x = x + 2 analytically, and support the solution(s) with a graph. (b) Use the graph in part (a) to find the solution set of 24  x 7 x + 2. (c) Use the graph in part (a) to find the solution set of 24  x … x + 2.
12. (Modeling) Laying a Telephone Cable A telephone company wishes to minimize the cost of laying a cable from point P to point R. Points P and Q are directly opposite each other along the banks of a straight river 300 yards wide. Point R lies on the same side of the river as point Q, but 600 yards away. If the cost per yard for the cable is $125 per yard under the water and $100 per yard on land, how should the company lay the cable to minimize the cost? P
300 yd
Q
River
600 yd
R
5 Inverse, Exponential, and Logarithmic Functions IN DECEMBER 2004, a massive earthquake struck the ocean floor off the west coast of Sumatra. Registering magnitude 9.0 on the Richter scale, the earthquake generated a tsunami in the Indian Ocean that killed 283,000 people in several coastal countries. The Sumatra earthquake occurred along the interface of the India and Burma tectonic plates. Pressure gradually built up due to friction as the India plate subducted under the Burma plate at a rate of 2.5 inches per year. When the pressure reached a high enough level, it overcame friction and generated the earthquake as the India plate thrust forward. The Burma plate displaced an enormous volume of water as it pushed upward 30 feet during the earthquake, resulting in the devastating tsunami. The magnitude of an earthquake is an indication of its intensity and is based on the amplitude, or height, of seismic waves measured at seismograph sites. Magnitude is measured with the Richter scale, an application of logarithmic functions. (See Section 5.6, Exercises 9 and 10.) Logarithmic functions and their inverses, exponential functions, have many other applications in science and business, as we will see in this chapter. Source: U.S. Geological Survey, British Geological Survey, and Canada Geological Survey.
CHAPTER OUTLINE 5.1 Inverse Functions 5.2 Exponential Functions 5.3 Logarithms and Their Properties 5.4 Logarithmic Functions 5.5 Exponential and Logarithmic Equations and Inequalities 5.6 Further Applications and Modeling with Exponential and Logarithmic Functions Summary Exercises on Functions: Domains, Defining Equations, and Composition From Chapter 5 of A Graphical Approach to Precalculus with Limits: A Unit Circle Approach, Fifth Edition. John Hornsby, Margaret Lial, Gark K. Rockswold. Copyright © 2011 by Pearson Education, Inc. All rights reserved.
320
CHAPTER 5
5.1
Inverse, Exponential, and Logarithmic Functions
Inverse Functions
Inverse Operations • OnetoOne Functions • Inverse Functions and Their Graphs • Equations of Inverse Functions • An Application of Inverse Functions to Cryptography
Inverse Operations Addition and subtraction are inverse operations: Starting with a number x, adding 5, and subtracting 5 gives x back as the result. Similarly, some functions are inverses of each other. For example, the functions defined by ƒ(x) = 8x and
g(x) =
1 x 8
are inverses of each other with respect to function composition. This means that if a value of x such as x = 12 is chosen, then ƒ(12) = 8 # 12 = 96. Calculating g(96) gives g (96) =
1 8
#
96 = 12.
Thus, (g ⴰ ƒ)(12) = g(ƒ(12)) = 12. Also, (ƒ ⴰ g)(12) = ƒ( g(12)) = 12. (Verify this.) For any value of x, it can be shown that (ƒ ⴰ g)(x) = ƒ( g (x)) = x and
(g ⴰ ƒ)(x) = g (ƒ(x)) = x.
OnetoOne Functions For the function y = ƒ(x) = 5x  8, any two different values of x produce two different values of y. On the other hand, for the function y = g (x) = x 2, two different values of x can lead to the same value of y; for example, both x = 4 and x =  4 give y = 42 = ( 4)2 = 16. A function such as y = ƒ(x) = 5x  8, for which different elements from the domain always lead to different elements from the range, is called a onetoone function.
OnetoOne Function A function ƒ is a onetoone function if, for elements a and b from the domain of ƒ, a ⴝ b implies
ƒ(a) ⴝ ƒ(b).
Using the concept of the contrapositive from the study of logic, the last line in the preceding box is equivalent to ƒ(a) ⴝ ƒ(b) implies
a ⴝ b.
We use this statement to show that a function ƒ is onetoone in the next example.
5.1
Inverse Functions
321
EXAMPLE 1
Deciding whether Functions Are OnetoOne Decide whether each function is onetoone. (b) ƒ(x) = 225  x 2
(a) ƒ(x) =  4x + 12 Solution (a)
ƒ(a)  4a + 12  4a a
y
f(x) = √ 25 – x 2
5
= = = =
ƒ(b)  4b + 12  4b b
Show that f (a) ⴝ f (b) leads to the result a ⴝ b. ƒ( x ) =  4x + 12 Subtract 12. Divide by  4.
By the definition, ƒ(x) =  4x + 12 is onetoone. (b) If we choose a = 3 and b =  3, then 3 Z  3, but
–5 –3
0
x 3
ƒ(3) = 225  32 = 225  9 = 216 = 4
5
ƒ( 3) = 225  ( 3)2 = 225  9 = 4.
and The graph of the function in Example 1(b) is a semicircle. The horizontal line test shows that the function is not onetoone. FIGURE 1
Even though 3 Z  3, ƒ(3) = ƒ( 3) = 4. This is not a onetoone function. As shown in Example 1(b), a way to demonstrate that a function is not onetoone is to produce a pair of unequal numbers that lead to the same function value. The horizontal line test also tells us whether a function is onetoone. (See FIGURE 1.)
Horizontal Line Test If every horizontal line intersects the graph of a function at no more than one point, then the function is onetoone.
EXAMPLE 2
Using the Horizontal Line Test Determine whether each graph is the graph of a onetoone function. (a) (b) y y y1
(x1, y1 ) (x2 , y1 )
y1
FOR DISCUSSION Based on your knowledge of the basic functions studied so far in this text, answer each question. In each case, assume that the function has the largest possible domain. 1. Is a nonconstant linear function always onetoone? 2. Is an odddegree polynomial function always onetoone? Why or why not? 3. Is an evendegree polynomial function ever onetoone? Why or why not?
(x 3, y1 ) y2
0
x1
x2
x3
x
x1 x 2 0 y3
x3
x
Solution (a) Each point where the horizontal line y = y1 intersects the graph has the same value of y but a different value of x. Since more than one different value of x (here, x1, x2, and x3) leads to the same value of y, the function is not onetoone. (b) Every horizontal line will intersect the graph at exactly one point. This function is onetoone. NOTE A function that is either always increasing or always decreasing on its entire domain, such as ƒ( x ) = x 3 or g ( x ) = 2x, must be onetoone.
322
CHAPTER 5
Inverse, Exponential, and Logarithmic Functions
Inverse Functions and Their Graphs Certain pairs of onetoone functions “undo” one another. For example, if ƒ(x) = 8x + 5 and g (x) = then
ƒ(10) = 8 # 10 + 5 = 85
x  5 , 8
and g (85) =
85  5 = 10. 8
Starting with 10, we “applied” function ƒ and then “applied” function g to the result, which gave back the number 10. See FIGURE 2. f
g
10
85
x–5 8
8x + 5
10
FIGURE 2
Similarly, for these same functions, check that ƒ(3) = 29 and ƒ( 5) =  35 and g (2) = 
3 8
and
g (29) = 3, g ( 35) =  5, 3 ƒ a  b = 2. 8
In particular, for these functions, (ƒ ⴰ g)(2) = ƒ( g (2)) = 2 and
(g ⴰ ƒ)(2) = g ( ƒ(2)) = 2.
In fact for any value of x, (ƒ ⴰ g)(x) = x and
(g ⴰ ƒ)(x) = x.
Because of this property, g is called the inverse of ƒ.
Inverse Function Let ƒ be a onetoone function. Then g is the inverse function of ƒ and ƒ is the inverse function of g if (ƒ ⴰ g)(x) ⴝ x for every x in the domain of g, and
(g ⴰ ƒ)(x) ⴝ x for every x in the domain of ƒ.
EXAMPLE 3
Deciding whether Two Functions Are Inverses Let ƒ be defined by ƒ(x) = x 3  1 and g by g (x) = 2 3 x + 1. Is g the inverse function of ƒ? Solution As shown in FIGURE 3, the horizontal line test applied to the graph indicates that ƒ is onetoone, so it does have an inverse. Now find (ƒ ⴰ g)(x) and (g ⴰ ƒ)(x).
y
3 (x 3  1) + 1 3 x + 1 B  1 (g ⴰ ƒ)(x) = g (ƒ(x)) = 2 (ƒ ⴰ g)(x) = ƒ(g (x)) = A 2 3
x
ƒ(x) = x 3 – 1
FIGURE 3
= x + 1  1 = x
= 2 3 x3 = x
Since (ƒ ⴰ g)(x) = x and (g ⴰ ƒ)(x) = x, function g is the inverse of function ƒ. Also, note that ƒ is the inverse of g.
5.1
Inverse Functions
323
A special notation is often used for inverse functions: If g is the inverse function of ƒ, then g can be written as ƒⴚ1 (read “ƒinverse”). In Example 3, ƒ 1(x) = 2 3 x + 1. CAUTION Do not confuse the ⴚ1 in ƒⴚ1 with a negative exponent. The symbol ƒ 1( x ) does not represent
1 ƒ(x) ;
rather, it represents the inverse function of ƒ.
By the definition of inverse function, the domain of ƒ equals the range of ƒⴚ1, and the range of ƒ equals the domain of ƒⴚ1. See FIGURE 4. Y
X Domain of f
Range of f f x
y f
Range of
–1
f –1
Domain of f –1 FIGURE 4
EXAMPLE 4
Finding Inverses of OnetoOne Functions Determine whether each function is onetoone. If so, find its inverse. (a) ƒ = {( 2, 1), ( 1, 0), (0, 1), (1, 2), (2, 2)} (b) g = {(3, 1), (0, 2), (2, 3), (4, 0)} Solution (a) Each xvalue in ƒ corresponds to just one yvalue. However, the yvalue 2 corresponds to two xvalues: 1 and 2. Also, the yvalue 1 corresponds to both  2 and 0. Because some yvalues correspond to more than one xvalue, ƒ is not onetoone and does not have an inverse. (b) Every xvalue in g corresponds to only one yvalue, and every yvalue corresponds to only one xvalue, so g is a onetoone function. The inverse function is found by interchanging the x and yvalues in each ordered pair. g 1 = {(1, 3), (2, 0), (3, 2), (0, 4)} The domain and range of g become the range and domain, respectively, of g 1.
Equations of Inverse Functions Finding the Equation of the Inverse of y ⴝ ƒ( x ) For a onetoone function ƒ defined by an equation y = ƒ(x), find the defining equation of the inverse as follows. (You may need to replace ƒ(x) with y first. Any restrictions on x and y should be considered.) Step 1 Interchange x and y. Step 2 Solve for y. Step 3 Replace y with ƒ1(x).
324
Inverse, Exponential, and Logarithmic Functions
CHAPTER 5
EXAMPLE 5
Finding Equations of Inverses Decide whether each equation defines a onetoone function. If so, find the equation of the inverse. (a) ƒ(x) = 2x + 5 (b) y = x 2 + 2 (c) ƒ(x) = (x  2)3 Solution (a) The graph of ƒ(x) = 2x + 5 is a nonhorizontal line, so by the horizontal line test, ƒ is a onetoone function. Follow the steps in the preceding box. 2x + 5 2x + 5 Replace ƒ( x ) with y. 2y + 5 Interchange x and y. (Step 1) 2y Subtract 5. Solve for y. r (Step 2) x  5 y = Divide by 2; rewrite. 2 x  5 ƒ 1(x) = Replace y with ƒ 1( x ). (Step 3) 2 (b) The equation y = x 2 + 2 has a parabola opening up as its graph, so some horizontal lines will intersect the graph at two points. This graph fails the horizontal line test. Thus, the function defined by y = x 2 + 2 is not onetoone and does not have an inverse. ƒ(x) y x x  5
Using the method of Example 5, we see that the inverse of Y1 = (4X + 6)>5 is Y2 = (5X  6)>4. In the tables of selected points for these functions, notice that the X and Yvalues are interchanged. This is typical for inverse functions.
= = = =
(c) Translations of the graph of the cubing function are onetoone. y = (x  2)3 x = ( y  2)3 3 (y 2 3x = 2
Replace ƒ(x) with y. Interchange x and y.
2)3
Take the cube root of each side.
2 3x = y  2
Simplify on the right.
2 3x + 2 = y
Solve for y by adding 2.
ƒ 1(x) = 2 3x + 2
Rewrite; replace y with ƒ 1( x ).
Suppose ƒ and ƒ 1 are inverse functions and ƒ(a) = b for real numbers a and b. Then, by the definition of inverse function, ƒ 1(b) = a. This means that if a point (a, b) is on the graph of ƒ, then (b, a) will belong to the graph of ƒ 1. As shown in FIGURE 5, the points (a, b) and (b, a) are reflections of one another across the line y = x.
y
(a, b) b
y=x
Geometric Relationship between the Graphs of ƒ and ƒⴚ1 (b, a)
a 0
x b
a
If a function ƒ is onetoone, then the graph of its inverse ƒ 1 is a reflection of the graph of ƒ across the line y = x.
FIGURE 5 FIGURE 6
illustrates this idea for each function ƒ (in blue) and its inverse ƒ 1 (in red). y
y
f –1 f
y=x f 0
f –1
x
0
FIGURE 6
y=x
x
5.1
GCM
325
EXAMPLE 6
Finding the Inverse of a Function with a Restricted Domain Let ƒ(x) = 2x + 5. Find ƒ 1(x). Solution The domain of ƒ is restricted to the interval 3 5, q). Since ƒ is always increasing on its domain, as shown in FIGURE 7, ƒ is onetoone and thus has an inverse function.
y 5
f(x) = √x + 5 x
0
–5
Inverse Functions
5
y=x
y = 2x + 5,
x Ú 5
Let y = ƒ( x ).
x = 2y + 5, x2 = y + 5 y = x2  5
y Ú 5
Interchange x and y. Square both sides. Solve for y.
We cannot simply give ƒ 1(x) as x 2  5. The domain of ƒ is 3 5, q) and its range is 30, q). The range of ƒ is the domain of ƒ 1, so ƒ 1 must be defined as
–5
f –1 (x) = x 2 – 5, x ≥ 0 FIGURE 7
ƒ 1(x) = x 2  5,
x Ú 0.
3 5, q), is the domain of ƒ. Graphs of ƒ and ƒ 1 are shown in FIGURE 7. The range of The line y = x is included to show that the graphs of ƒ and ƒ 1 are reflections of one another—that is, mirror images with respect to this line. ƒ 1,
EXAMPLE 7
Finding the Inverse of a Rational Function 1 The rational function defined by ƒ(x) = x 2x  1 is onetoone. Find ƒ (x). Solution y = x =
Move terms with y to one side and factor to isolate y.
TECHNOLOGY NOTE When graphing inverse functions on a calculator, a square window sometimes work best.
x ( y  1) xy  x xy  2y y (x  2)
= = = =
y = Thus ƒ 1(x) =
2x x  1 2y y  1 2y 2y x x x x  2
Let y = ƒ( x ). Interchange x and y. Multiply by y  1. Distributive property Add x ; subtract 2y. Factor out y. Divide by x  2.
x x  2.
Important Facts about Inverses y = x2
x = y2 10
–10
10
1. If ƒ is onetoone, then ƒ 1 exists. 2. The domain of ƒ is equal to the range of ƒ 1, and the range of ƒ is equal to the domain of ƒ 1. 3. If the point (a, b) lies on the graph of f, then (b, a) lies on the graph of ƒ 1. The graphs of ƒ and ƒ 1 are reflections of each other across the line y = x.
–10 Despite the fact that y = x 2 is not onetoone, the calculator will draw its “inverse,” x = y 2. FIGURE 8
CAUTION Some graphing calculators have the capability of “drawing the inverse” of a function. This feature does not require that the function be onetoone, however, so the resulting figure may not be the graph of a function. See FIGURE 8. Again, it is necessary to understand the mathematics to interpret results correctly.
326
Inverse, Exponential, and Logarithmic Functions
CHAPTER 5
An Application of Inverse Functions to Cryptography A onetoone function and its inverse can be used to make information secure. The function is used to encode a message, and its inverse is used to decode the coded message. In practice, complicated functions are used. We illustrate the process with a simple function in Example 8. A B C D E F G H I J K L M
N O P Q R S T U V W X Y Z
1 2 3 4 5 6 7 8 9 10 11 12 13
14 15 16 17 18 19 20 21 22 23 24 25 26
Using Functions to Encode and Decode a Message Use the onetoone function defined by ƒ(x) = 3x + 1 and the numerical values assigned to each letter of the alphabet, as shown in the margin, to encode and decode the message BE VERY CAREFUL. Solution The message BE VERY CAREFUL would be encoded as 7
16
67
16
55
76
10
4
55
16
19
64
37
because B corresponds to 2
and
ƒ(2) = 3(2) + 1 = 7,
E corresponds to 5
and
ƒ(5) = 3(5) + 1 = 16,
and so on. Using the method of earlier examples, we find that ƒ 1(x) = ƒ 1 to decode yields ƒ 1(7) =
7  1 = 2, 3
ƒ 1(16) =
16  1 = 5, 3
x  1 3 .
which corresponds to B, which corresponds to E,
EXERCISES
Checking Analytic Skills Decide whether each function is onetoone. Do not use a calculator. 1. ƒ(x) =  3x + 5
2. ƒ(x) =  5x + 2
3. ƒ(x) = x 2
4. ƒ(x) =  x 2
5. ƒ(x) = 236  x 2
6. ƒ(x) =  2100  x 2
7. ƒ(x) = x 3
8. ƒ(x) = 2 3x
9. ƒ(x) =  2x + 1 
Decide whether each function graphed or defined is onetoone. 10.
11.
y
0
Using
and so on. The table feature of a graphing calculator can be useful in this procedure. FIGURE 9 shows how decoding can be accomplished for the words BE VERY.
FIGURE 9
5.1
EXAMPLE 8
x
12.
y
0
x
y
0
x
5.1
13.
14.
y
16. ƒ(x) = b
9 25.
y
0
3 if x Ú 0  x if x 6 0
1 x + 2
17. y = (x  2)2
18. y =  (x + 3)2  8
20. y =  2x 5  4
21. y =  2 3x + 5
23. y =
4 x  8
Explain why a polynomial function of even degree cannot be onetoone.
Concept Check
327
0
19. y = 2x 3 + 1 22. y =
15.
y
x
0
Inverse Functions
24. ƒ(x) =  7
9 26.
Explain why in some cases a polynomial function of odd degree is not onetoone.
Answer the following.
27. For a function to have an inverse, it must be 29. If ƒ and g are inverses, then (ƒ ⴰ g)(x) = = x.
28. For a function ƒ to be of the type mentioned in Exercise 27, if a Z b, then .
.
30. The domain of ƒ is equal to the range of ƒ is equal to the
and
of ƒ 1, and the of ƒ 1.
31. If the point (a, b) lies on the graph of ƒ and ƒ has an inverse, then the point lies on the graph of ƒ 1.
32. If ƒ(x) = x, then for any function g, (ƒ ⴰ g)(x) = (g ⴰ ƒ)(x) = .
33. If a function ƒ has an inverse, then the graph of ƒ 1 may be obtained by reflecting the graph of ƒ across the line with equation .
34. If a function ƒ has an inverse and ƒ( 3) = 6, then ƒ 1(6) = .
35. If ƒ( 4) = 16 and ƒ(4) = 16, then ƒ
36. If ƒ is a function that has an inverse and the graph of ƒ lies completely within the second quadrant, then the graph of ƒ 1 lies completely within the quadrant.
an inverse because
(does/does not)
have
.
Use the definition of inverse functions to show analytically that ƒ and g are inverses. x + 7 3
37. ƒ(x) = 3x  7,
g (x) =
40. ƒ(x) = x 3  7,
g (x) = 2 3x + 7
38. ƒ(x) = 4x + 3, 41. ƒ(x) =  x 5,
g (x) =
x  3 4
g (x) =  2 5x
39. ƒ(x) = x 3 + 4, 42. ƒ(x) =  x 7,
g (x) = 2 3x  4
g (x) =  2 7x
Determine whether each function is onetoone. If so, find its inverse. 43. ƒ = {(10, 4), (20, 5), (30, 6), (40, 7)}
44. g = {(5, 12), (10, 22), (15, 32), (20, 42)}
45. ƒ = {(1, 5), (2, 6), (3, 5), (4, 8)}
46. g = {(0, 10), (1, 20), (2, 10), (3, 40)}
47. ƒ = {(0, 02), (1, 12), (2, 22), (3, 32), (4, 42)}
48. g = {(0, 04), ( 1, ( 1)4), ( 2, ( 2)4), ( 3, ( 3)4)}
Concept Check In Exercises 49–54, an everyday activity is described. Keeping in mind that an inverse operation “undoes” what an operation does, describe the inverse activity. 49. tying your shoelaces
50. pressing a car’s accelerator
51. entering a room
52. climbing the stairs
53. wrapping a package
54. putting on a coat
328
CHAPTER 5
Inverse, Exponential, and Logarithmic Functions
For each function defined that is onetoone, write an equation for the inverse function in the form y = ƒ 1(x), and then graph ƒ and ƒ 1 on the same axes. Give the domain and range of ƒ and ƒ 1. If the function is not onetoone, say so. 55. y = 3x  4
56. y = 4x  5
57. y = x 3 + 1
58. y =  x 3  2
59. y = x 2
60. y =  x 2 + 2
61. y =
1 x
62. y =
64. y =
3 x  4
65. ƒ(x) = 26 + x, x Ú  6
4 x
63. y =
2 x + 3
66. ƒ(x) =  2x 2  16, x Ú 4
The given function ƒ is onetoone. Find ƒ 1(x). 3x 5  x
69. ƒ(x) =
1  2x 3x
70. ƒ(x) = 2x  8, x Ú 8
71. ƒ(x) = 2x 2  4, x Ú 2
72. ƒ(x) =
4  x 5x
73. ƒ(x) = 5x 3  7
74. ƒ(x) = 4  3x 3
75. ƒ(x) =
67. ƒ(x) =
76. ƒ(x) =
4x x + 1
68. ƒ(x) =
2x + 1 x  1
Concept Check
3  4x 3 x 78. ƒ(x) = 1  3x
3  x 2x + 1
77. ƒ(x) =
Let ƒ(x) = x 3. Evaluate each expression.
79. ƒ(2)
80. ƒ(0)
81. ƒ( 2)
82. ƒ 1(8)
83. ƒ 1(0)
84. ƒ 1( 8)
Concept Check
The graph of a function ƒ is shown in the figure. Use the graph to find each value.
85. ƒ 1(4)
86. ƒ 1(2)
87. ƒ 1(0)
88. ƒ 1( 2)
4
89. ƒ 1( 3)
90. ƒ 1 ( 4)
2
y
–4
–2
2
0
4
x
–4
Decide whether the pair of functions ƒ and g are inverses. In Exercises 91–93, assume axes have equal scales. 91.
92.
y
93.
y
y
y = f (x)
y = f (x) y=x y = f(x) y=x
94. ƒ(x) = 
y = g(x)
y = g(x) 0
y=x
x
0
0
x
x
y = g(x)
3 11 x, g (x) =  x 11 3
95. ƒ(x) = 2x + 4,
g (x) =
1 x  2 2
96. ƒ(x) = 5x  5, g (x) =
1 x + 5 5
5.1
329
Inverse Functions
Indicate whether the screens suggest that Y1 and Y2 are inverse functions.
Concept Check 97.
98.
Graph the inverse of each onetoone function. 99.
100.
y
0
102.
x
0
Concept Check
0
103.
y
x
101.
y
x
0
x
0
104.
y
y
x
y
x
0
Answer the following.
105. Suppose ƒ(x) is the number of cars that can be built for x dollars. What does ƒ 1(1000) represent?
106. Suppose ƒ(r) is the volume (in cubic inches) of a sphere of radius r inches. What does ƒ 1(5) represent?
107. If a line has nonzero slope a, what is the slope of its reflection across the line y = x?
108. Find ƒ 1( ƒ(2)), where ƒ(2) = 3.
Use a graph with the given viewing window to decide which functions are onetoone. If a function is onetoone, give the equation of its inverse function. Check your work by graphing the inverse function on the same coordinate axes. 109. ƒ(x) = 6x 3 + 11x 2  x  6; 3 3, 24 by 3 10, 104 111. ƒ(x) =
x  5 ; 3 9.4, 9.44 by 3 6.2, 6.24 x + 3
110. ƒ(x) = x 4  5x 2 + 6; 3 3, 34 by 3 1, 84 112. ƒ(x) =
x ; 3 4.7, 9.44 by 3 6.2, 6.24 x  4
While a function may not be onetoone when defined over its “natural” domain, it may be possible to restrict the domain in such a way that it is onetoone and the range of the function is unchanged. For example, if we restrict the domain of the function ƒ(x) = x 2 (which is not onetoone over ( q, q)) to 30, q), we obtain a onetoone function whose range is still 30, q). See the figure at the right. Notice that we could choose to restrict the domain of ƒ(x) = x 2 to ( q, 04 and also obtain the graph of a onetoone function, except that it would be the left half of the parabola.
y
f (x) = x 2, x ≥ 0 4
x 2
(continued)
330
CHAPTER 5
Inverse, Exponential, and Logarithmic Functions
For each function defined in Exercises 113–118, restrict the domain so that the function is onetoone and the range is not changed. You may wish to use a graph to help decide. Answers may vary. 113. ƒ(x) =  x 2 + 4
114. ƒ(x) = (x  1)2
115. ƒ(x) =  x  6 
116. ƒ(x) = x 4
117. ƒ(x) = x 4 + x 2  6
118. ƒ(x) =  2x 2  16
Using the restrictions on the functions defined in Exercises 119–122, find a formula for ƒ 1. 119. ƒ(x) =  x 2 + 4, x Ú 0
120. ƒ(x) = (x  1)2, x Ú 1
121. ƒ(x) =  x  6 , x Ú 6
122. ƒ(x) = x 4, x Ú 0
Use the alphabet coding assignment given in Example 8 for Exercises 123–126. 123. The function defined by ƒ(x) = 4x  1 was used to encode the following message: 79
71
19
3
75
83
71
19
31
83
55
79
35
75
59
55.
Find the inverse function and decode the message. 124. The function defined by ƒ(x) = 3x  3 was used to encode the following message: 54
12
57
54
0
24
33
57
42
9
0
72.
Find the inverse function and decode the message. 125. Encode the message NO PROBLEM, using the onetoone function defined by ƒ(x) = x 3 + 1. Give the inverse function the decoder would need when the message is received. 126. Encode the message BEGIN OPERATIONS, using the onetoone function defined by ƒ(x) = (x + 2)3. Give the inverse function the decoder would need when the message is received.
5.2
Exponential Functions
RealNumber Exponents • Graphs of Exponential Functions • Exponential Equations (Type 1) • Compound Interest • The Number e and Continuous Compounding • An Application of Exponential Functions
RealNumber Exponents Recall that if r = and n,
m n
is a rational number, then, for appropriate integer values of m a m/n ⴝ A 2a B . n
For example, 27 ⴚ 1/3 =
m
163/4 = A 216 B = 23 = 8, 4
1 1 1 = 3 = , and 271/3 3 227
3
64 ⴚ 1/2 =
1 1 1 = = . 641/2 8 264
In this section, we extend the definition of a r to include all real (not just rational) values of the exponent r. For example, the symbol 213 might be evaluated by approximating the exponent 23 by the numbers 1.7, 1.73, 1.732, and so on. Since these decimals approach 23 more and more closely, it seems reasonable that 213 should be approximated more and more closely by the numbers 21.7, 21.73, 21.732, and so on. 17 10 2 B .) In fact, this is exactly how 213 (Recall, for example, that 21.7 = 217/10 = A 2 is defined (in a more advanced course).
5.2
Exponential Functions
331
With this interpretation of real exponents, all rules and theorems for exponents are valid for real number exponents as well as rational ones. In addition to the usual rules for exponents (see Chapter R), we use several new properties in this chapter. For example, if y = 2x, then each value of x leads to exactly one value of y; therefore, y = 2x defines a function. Furthermore, if 2x = 24,
then
x = 4,
x = 4,
then
2x = 24.
42 6 43,
but
1 2 1 3 a b 7 a b . 4 4
if
and Also,
In general, when a>1, increasing the exponent on a leads to a larger number, but when 01
(a, 1) 0
(1, 0)
x
(a –1, –1) Y1 = log 2 X =
log X log 2
FIGURE 35 ■ ■ ■
ƒ(x) = log a x, a 7 1, is increasing and continuous on its entire domain, (0, q). The yaxis is the vertical asymptote as x : 0 from the right. The graph passes through the points (a 1,  1), (1, 0), and (a, 1).
354
CHAPTER 5
Inverse, Exponential, and Logarithmic Functions
FUNCTION CAPSULE ƒ(x) ⴝ log a x,
LOGARITHMIC FUNCTION Domain: (0, q)
00 True or False?
A: ( q,  2)
3
( 3)2  4 7 0 ? 5 7 0 True
B: ( 2, 2)
0
02  4 7 0 ?  4 7 0 False
C: (2, q)
3
32  4 7 0 ? 5 7 0 True
The domain of ƒ(x) = ln(x 2  4) is the union of the values in Intervals A and C, ( q,  2) ´ (2, q). EXAMPLE 3
Graphing Translated Logarithmic Functions Graph each function. Give the asymptote, xintercept, domain, and range, and tell whether the function is increasing or decreasing on its domain. (a) y = log 2(x  1) (b) y = (log 3 x)  1 Solution (a) Because the argument is x  1, the graph of y = log 2(x  1) is the graph of y = log 2 x shifted 1 unit to the right. The vertical asymptote also moves 1 unit to the right, so its equation is x = 1. The xintercept is 2. The domain is (1, q), as found in Example 2(a), and the range is ( q, q). The function is always increasing on its domain. See FIGURE 38. This graph can also be plotted by using the equivalent exponential form. This table numerically supports the result in Example 3(a): the domain of y = log2(x – 1), entered here as Y1 = log(X – 1)/log 2, consists of real numbers greater than 1.
y = log 2(x  1) x  1 = 2y Write in exponential form. y x = 2 + 1 Add 1. Using this equation, we choose values for y and calculate the corresponding xvalues. x
y
5 4 3 2
2
2 3 5
1 0 1 2
In the equation x = 2y + 1, choose values for y and calculate x.
y = log 2(x – 1)
y
6 4
y = log 2 (x – 1)
2
–2
10
x –2
2
4
6
8
–2
–6 x=1 This calculator graph was obtained by using log(x – 1) . y= log 2 FIGURE 38
(continued)
356
CHAPTER 5
Inverse, Exponential, and Logarithmic Functions
(b) Here, 1 is subtracted from log 3 x, so the graph of y = log 3 x is shifted 1 unit downward. The vertical asymptote is the yaxis, or x = 0, and is not affected. The xintercept of y = (log 3 x)  1 is 3. The domain is (0, q), from Example 2(b), and the range is ( q, q). The function is always increasing. See FIGURE 39. The alternative method described in part (a) can also be used. y = (log 3 x)  1 y + 1 = log 3 x
Add 1.
x = 3y + 1
Write in exponential form.
Again, choose yvalues and calculate the corresponding xvalues. x
y
y
1 3
2
1 3 9
1 0 1
y = (log 3 x) – 1 4
y = (log 3 x) – 1
1
x
0
3
6
–2
9
10
–2
In the equation x = 3y + 1, choose values for y and calculate x.
This calculator graph was obtained by using log x – 1. y= log 3 FIGURE 39
EXAMPLE 4
Determining Symmetry Show that the graph of each function is symmetric with respect to the yaxis. (a) ƒ(x) = log 3 x  (b) ƒ(x) = ln(x 2  4) Solution (a) Since   x  =  x  for all x, ƒ( x) = log 3  x  = log 3 x  = ƒ(x). Thus ƒ( x) = ƒ(x), and the graph is symmetric with respect to the yaxis. See FIGURE 40. y y
4
–2 1
0
–3
0
2 x
x 3
f (x) = log 3⏐x⏐ FIGURE 40
2 1
–2
y = ln(x 2 – 4) FIGURE 41
(b) Since x 2 = ( x)2 for all x, ƒ( x) = ln3( x)2  44 = ln(x 2  4) = ƒ(x). Thus, ƒ( x) = ƒ(x), and the graph is symmetric with respect to the yaxis. See FIGURE 41.
5.4
Logarithmic Functions
357
WHAT WENT WRONG? A student wanted to support the power rule for logarithms (Property 6 from Section 5.3): log a x r = r log a x. The student defined Y1 and Y2, as shown in the screen on the left, and expected the two graphs to be the same. However, the graph of Y1 was different from the graph of Y2, as shown. Y2 = 2 log (X)
Y1 = log (X)2
2
2
–2
4
–2
4
–2
–2
What Went Wrong? How can the student change the input for Y1 to obtain the same graph as that of Y2? y1 = log x
y2 = log 8x
5
Applying Earlier Work to Logarithmic Functions –2
10
EXAMPLE 5
Using a Property of Logarithms to Translate a Graph Describe how to graph ƒ(x) = log 8x by translation. Solution By the product rule for logarithms, log 8x = log 8 + log x, so we can obtain the graph of y2 = log 8x by shifting the graph of y1 = log x upward log 8 L 0.90309 unit. FIGURE 42 shows the graph.
–5 The vertical distance between y2 and y1 is log 8. FIGURE 42
EXAMPLE 6
Finding the Inverse of an Exponential Function Find analytically the inverse of the function defined by ƒ(x) =  2x + 3. Then graph both functions, and discuss the relationship between the two graphs. f –1(x)
Solution The function is onetoone because it is a reflection and translation of the onetoone function defined by y = 2x. Find the equation of the inverse function.
= log 2(–x + 3) 3.1
–4.7
ƒ(x) y x y 2 y ƒ 1(x)
4.7
–3.1
f (x) = –2 x + 3 FIGURE 43
= = = = = =
 2x + 3  2x + 3  2y + 3 x + 3 log 2( x + 3) log 2( x + 3)
Replace ƒ( x) with y. Interchange x and y. Add 2 y; subtract x. Write in logarithmic form. Replace y with ƒ 1( x ).
shows the graphs of both ƒ and ƒ 1. The table compares some of the features of these inverses. Notice how the roles of x and y are reversed in ƒ and ƒ 1.
FIGURE 43
Function
Domain
Range
xintercept
yintercept
Asymptote
ƒ(x) =  2x + 3 ƒ 1(x) = log 2( x + 3)
( q, q) ( q, 3)
( q, 3) ( q, q)
log 2 3 L 1.58 2
2 log 2 3 L 1.58
Horizontal: y = 3 Vertical: x = 3
Answer to What Went Wrong? In this case, log(X)2 is being calculated as log(X)*log(X), not 2 log(X). With some calculators, the first equation should be entered as Y1 = (log(X2))(X 7 0).
358
Inverse, Exponential, and Logarithmic Functions
CHAPTER 5
A Logarithmic Model EXAMPLE 7
Modeling Drug Concentration The concentration of a drug injected into the bloodstream decreases with time. The intervals of time in hours when the drug should be administered are given by T =
1 C2 ln , k C1
where k is a constant determined by the drug in use, C2 is the concentration at which the drug is harmful, and C1 is the concentration below which the drug is ineffective. (Source: Horelick, B. and S. Koont, “Applications of Calculus to Medicine: Prescribing Safe and Effective Dosage,” UMAP Module 202.) Thus, if T = 4, the drug should be administered every 4 hours. For a certain drug, k = 13, C2 = 5, and C1 = 2. How often should the drug be administered? Solution Substitute the given values in the equation of the function. T =
1 C2 1 5 ln = 1 ln L 2.75 k C1 2 3
The drug should be given about every 2 34 hours.
5.4
EXERCISES
Checking Analytic Skills The graph of an exponential function ƒ is given, with three points labeled. Sketch the graph of ƒ 1 by hand, labeling three points on the graph. For ƒ 1, also state the domain, the range, whether it increases or decreases on its domain, and the equation of its vertical asymptote. Do not use a calculator. 1.
2.
y
3.
y
(1, 4)
(1, 3) f(x) =
(–1, 13)
(–1, 3)
3x
f(x) =
(0, 1)
(–1, 14 ) x
0
4.
6.
f (x) = 2 x + 1
y
(1, 3) f(x) =
(0, 1)
1 x 4
()
(–1, 32 )
0
f(x) =
y=1 x
x
0
7. In Exercises 1–6, each function ƒ is an exponential function. Therefore, each function ƒ 1 is a(n) function.
( 12 )x – 3 x
0
(0, 2)
(1, 14 )
x
0
x
y
( 13 )x
(1, 13)
(0, 1)
0
5.
y
f(x) =
4x
(0, 1)
(–1, 4)
y
(–1, –1) (0, –2)
(1, – 52 ) y = –3
9 8.
Compare the characteristics of the graph of ƒ(x) = log a x with those of the graph of ƒ(x) = a x in Section 5.2. Make a list of characteristics that reinforce the idea that these are inverse functions.
5.4
Logarithmic Functions
359
Find the domain of each logarithmic function analytically. You may wish to support your answer graphically. 1 9. y = log 2x 10. y = log x 11. y = log( x) 2 1 12. y = log a xb 13. y = ln(x 2 + 7) 14. y = ln(x 4 + 8) 2 15. y = ln( x 2 + 4)
16. y = ln( x 2 + 16)
17. y = log 4(x 2  4x  21)
18. y = log 6(2x 2  7x  4)
19. y = log(x 3  x)
20. y = log(x 3  81x)
21. y = log a
x + 3 b x  4
22. y = loga
23. y = log  3x  7 
x + 1 b x  5
24. y = log  6x + 6 
Sketch the graph of ƒ(x) = log 2 x. Then refer to it and use the techniques of Chapter 2 to graph each function. 25. ƒ(x) = (log 2 x) + 3
26. ƒ(x) = log 2(x + 3)
27. ƒ(x) =  log 2(x + 3) 
Sketch the graph of ƒ(x) = log 1/2 x. Then refer to it and use the techniques of Chapter 2 to graph each function. 28. ƒ(x) = (log 1/2 x)  2
29. ƒ(x) = log 1/2(x  2)
30. ƒ(x) =  log 1/2(x  2) 
In Exercises 31–38, match the correct graph in choices A–H to each equation.
Concept Check
32. y = e x  3
33. y = e x + 3
34. y = e x  3
35. y = ln x + 3
36. y = ln x  3
37. y = ln(x  3)
38. y = ln(x + 3)
A.
B.
C.
D.
31. y =
ex
+ 3
y
y
y
y=3
(1, 3)
(0, e 3) x
0
y
x
0
x
0
x
0
x=3
E.
F.
y
G.
y
H.
y
y
(3, 1) x
0
0
x
0
(1, –3)
x
y = –3
0
x
x = –3
Graph each function. 39. ƒ(x) = log 5 x
40. ƒ(x) = log 10 x
41. ƒ(x) = log 1>2(1  x)
42. ƒ(x) = log 1>3(3  x)
43. ƒ(x) = log 3(x  1)
44. ƒ(x) = log 2(x 2)
In Exercises 45–50, (a) explain how the graph of the given function can be obtained from the graph of y = log 2 x, and (b) graph the function. 45. y = log 2(x + 4)
46. y = log 2(x  6)
47. y = 3 log 2 x + 1
48. y =  4 log 2 x  8
49. y = log 2( x) + 1
50. y =  log 2( x)
9 51.
Graph y = log x 2 and y = 2 log x on separate viewing screens. It would seem, at first glance, that by applying the power rule for logarithms, these graphs should be the same. Are they? If not, why not? (Hint: Consider the domain in each case.)
9 52.
Graph ƒ(x) = log 3 x  in the window 3 4, 44 by 3  4, 44, and compare ƒ(x) with the traditional graph in FIGURE 40. How might one easily misinterpret the domain of the function simply by observing the calculator graph? What is the domain of this function?
360
CHAPTER 5
Inverse, Exponential, and Logarithmic Functions
Evaluate each logarithm in three ways: (a) Use the definition of logarithm in Section 5.3 to find the exact value analytically. (b) Support the result of part (a) by using the changeofbase rule and common logarithms on your calculator. (c) Support the result of part (a) by locating the appropriate point on the graph of the function y = log a x. 1 1 53. log 9 27 54. log 4 a b 55. log 16 a b 56. log 2 28 8 8 For each exponential function ƒ, find ƒ 1 analytically and graph both ƒ and ƒ 1 in the same viewing window. 1 x 58. ƒ(x) = a b  5 2
57. ƒ(x) = 4x  3
(a) ƒ A
60. ƒ(x) =  e x + 6
Suppose ƒ(x) = log a x and ƒ(3) = 2. Determine each function value.
61. Concept Check 1 9
59. ƒ(x) =  10x + 4
B
(c) ƒ 1( 2)
(b) ƒ(27)
(d) ƒ 1(0)
Use a graphing calculator to solve each equation. Give solutions to the nearest hundredth. 63. 2x = log 10 x
62. log 10 x = x  2
64. e x = x 2
(Modeling) Solve each problem. 65. Height of the Eiffel Tower The right side of Paris’s Eiffel Tower has a shape that can be approximated by the graph of the function defined by ƒ(x) =  301 ln
x . 207
(Source: Banks, Robert B., Towing Icebergs, Falling Dominoes, and Other Adventures in Applied Mathematics, Princeton University Press.) y
f (x) = –301 ln
x 207
(b) Estimate the length of a female blue whale that is 4 years old. 9 (c) The equation that defines t has domain 24 6 L 6 87. Explain why. 67. Barometric Pressure
The function defined by
ƒ(x) = 27 + 1.105 log(x + 1)
100 x 100
9 (a)
Explain why the shape of the left side of the Eiffel Tower has the formula given by ƒ( x). (b) The short horizontal line at the top of the figure has length 15.7488 feet. Approximately how tall is the Eiffel Tower? (c) Approximately how far from the center of the tower is the point on the right side that is 500 feet above the ground?
66. Age of a Whale The age in years of a female blue whale is approximated by t =  2.57 ln a
87  L b, 63
where L is its length in feet. (a) How old is a female blue whale that measures 80 feet?
approximates the barometric pressure in inches of mercury at a distance of x miles from the eye of a typical hurricane. (Source: Miller, A. and R. Anthes, Meteorology, Fifth Edition, Charles E. Merrill.) (a) Approximate the pressure 9 miles from the eye of the hurricane. 9 (b) The ordered pair (99, 29.21) belongs to this function. What information does it convey? 68. Race Speed At the IAAF Grand Prix final races held in Paris on September 14, 2002, American sprinter Tim Montgomery ran the 100meter race in 9.78 seconds. His speed in meters per second after t seconds is closely modeled by the function defined by ƒ(t) = 11.65(1  e  t/1.27). (Source: Banks, Robert B., Towing Icebergs, Falling Dominoes, and Other Adventures in Applied Mathematics, Princeton University Press; USA Today.) (a) How fast was he running of the start of the race? (b) How fast was he running as he crossed the finish line?
5.5
5.5
Exponential and Logarithmic Equations and Inequalities
361
Exponential and Logarithmic Equations and Inequalities
Exponential Equations and Inequalities (Type 2) • Logarithmic Equations and Inequalities • Equations Involving Exponentials and Logarithms • Formulas Involving Exponentials and Logarithms
General methods for solving exponential and logarithmic equations depend on the following properties, which are based on the fact that exponential and logarithmic functions are, in general, onetoone. Property 1 below was used in Section 5.2 to solve Type 1 exponential equations.
Properties of Logarithmic and Exponential Functions For b 7 0 and b Z 1, 1. b x ⴝ b y if and only if x ⴝ y. 2. If x 7 0 and y 7 0, then log b x ⴝ log b y if and only if
x ⴝ y.
Exponential Equations and Inequalities (Type 2) Unlike a Type 1 exponential equation (or inequality)—see Section 5.2 for examples— a Type 2 exponential equation (or inequality) is one in which the exponential expressions cannot easily be written as powers of the same base, such as 23x + 1 = 34  x.
7x = 12 and EXAMPLE 1
Solve
7x
Solving a Type 2 Exponential Equation
= 12.
Analytic Solution Property 1 cannot be used to solve this equation, so we apply Property 2. While any appropriate base b can be used to apply Property 2, the most practical is base 10 or base e. 7x = 12 ln 7x = ln 12 x ln 7 = ln 12 ln 12 x = ln 7
Take base e logarithms.
Graphing Calculator Solution Using the intersectionofgraphs method, we graph y1 = 7x and y2 = 12. The xcoordinate of the point of intersection is approximately 1.277, as seen in FIGURE 44(a). FIGURE 44(b) illustrates the xintercept method of solution. The xintercept of y = 7x  12 is also approximately 1.277. y1 = 7 x
y 2 = 12
Power rule
y = 7 x – 12
15
5
Divide by ln 7.
–2
x The expression lnln12 7 is the exact solution of 7 = 12. Had we used common logarithms, the solution would have the log 12 form log 7 . In either case, we use a calculator to find a decimal approximation.
log 12 ln 12 = L 1.277 ln 7 log 7
The exact solution set is E F , or E is an approximate solution set. ln 12 ln 7
–5
2
5 –2
–15
(a)
(b) FIGURE 44
Nearest thousandth log 12 log 7
F , while 51.2776
To check the analytic solution, we can also raise 7 to log 12 either of the powers lnln12 7 or log 7 to obtain a result of 12. Verify this with a calculator.
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CHAPTER 5
Inverse, Exponential, and Logarithmic Functions
y1 = 7 x
y 2 = 12
EXAMPLE 2
15
Solving Type 2 Exponential Inequalities
(a) Use FIGURE 44(a) to solve 7x 6 12.
–5
5 –2
Solution (a) Because the graph of y1 = 7x is below the graph of y2 = 12 for all xvalues less q ln 12 q than lnln12 7 L 1.277, the solution set is A  , ln 7 B , or ( , 1.277).
(b) This inequality is equivalent to 7x  12 7 0. Because the graph of y = 7x  12 is above the xaxis for all values of x greater than lnln12 7 L 1.277, the solution set is ln 12 q q A ln 7 , B , or (1.277, ).
(a)
y = 7 x – 12 5 –2
(b) Use FIGURE 44(b) to solve 7x 7 12.
2
EXAMPLE 3
Solve –15
23x + 1
=
Solving a Type 2 Exponential Equation
34  x.
Solution
(b)
23x + 1 = 34  x
FIGURE 44 (repeated)
log 23x + 1 = log 34  x
Take common logarithms.
(3x + 1) log 2 = (4  x) log 3 3x log 2 + log 2 = 4 log 3  x log 3 3x log 2 + x log 3 = 4 log 3  log 2 x(3 log 2 + log 3) = 4 log 3  log 2 4 log 3  log 2 x = Exact solution 3 log 2 + log 3 Approximate solution
x L 1.165
Power rule Distributive property Move all xterms to one side. Factor out x on the left. Divide by 3 log 2 + log 3. Nearest thousandth
The exact solution can be written in a more compact form. y1 = 2 3x + 1
y2 = 3 4 – x 30
–5
5
x =
4 log 3  log 2 3 log 2 + log 3
Exact solution
x =
log 34  log 2 log 23 + log 3
Power rule
x =
log 81  log 2 log 8 + log 3
34 = 81; 23 = 8
–10 FIGURE 45
FOR DISCUSSION 1. Use FIGURE 45 to determine the solution sets of 23x + 1 6 34x
x =
log 81 2 log 24
Quotient and product rules
The exact solution set is E log 24 F . An approximate solution set is 51.1656. As seen in FIGURE 45, the xcoordinate is about 1.165, supporting the solution. log (81>2)
and 23x + 1 7 34x. 2. Discuss the difference between the exact solution and an approximate solution of an exponential equation. In general, can you find an exact solution from a graph?
Logarithmic Equations and Inequalities In Example 4 on the next page, we use the quotient property of logarithms from Section 5.3 to combine terms on the left side and then apply Property 2 from the preceding page. We could also solve this equation by transforming so that all logarithmic terms appear on one side. Then, by applying the properties of logarithms and rewriting in exponential form, the same solution set would result. This would also allow for solving graphically by the xintercept method.
5.5
363
Exponential and Logarithmic Equations and Inequalities
EXAMPLE 4
Solving a Logarithmic Equation Solve log 3(x + 6)  log 3(x + 2) = log 3 x. Analytic Solution The domain must satisfy x + 6 7 0, x + 2 7 0, and x 7 0. The intersection of their solution sets gives the domain (0, q). log 3(x + 6)  log 3(x + 2) x + 6 log 3 x + 2 x + 6 x + 2 x + 6 x + 6 2 x + x  6 (x + 3)(x  2) x =  3 or x
Graphing Calculator Solution FIGURE 46 shows that the xcoordinate of the point of intersection of the graphs of y1 = log 3(x + 6)  log 3(x + 2)
= log 3 x = log 3 x
Quotient rule
= x
Property 2
= = = = =
Multiply by x + 2.
and y2 = log 3 x
x(x + 2) x 2 + 2x 0 0 2
Distributive property Standard form
is 2, which agrees with our analytic solution. Notice that the graphs do not intersect when x =  3, supporting the conclusion in the analytic solution that  3 is an extraneous value.
Factor.
y1 = log 3(x + 6) – log 3(x + 2) 5
Zeroproduct property
The value x =  3 is not in the domain of log 3 x or log 3(x + 2), so the only valid solution is 2, giving the solution set 526. Check:
log 3(2 + 6)  log 3(2 + 2) = log 3 2 ? Let x = 2. 8 log 3 = log 3 2 ? 4 log 3 2 = log 3 2 True
–5
10 –2
y2 = log 3 x FIGURE 46
EXAMPLE 5
Solving a Logarithmic Equation Solve log(3x + 2) + log(x  1) = 1. Analytic Solution Find the domain of each logarithm. Solve 3x + 2 7 0 to find x 7  23, and solve x  1 7 0 to find x 7 1. Thus the domain is (1, q). Recall from Section 5.3 that log x means log 10 x. log(3x + 2) + log(x  1) log3(3x + 2)(x  1)4 (3x + 2)(x  1) 3x 2  x  2 3x 2  x  12
= = = = =
1 1 101 10 0
Product rule Write in exponential form.
Graphing Calculator Solution We arbitrarily choose to use the xintercept method of solution. As seen in FIGURE 47, the xintercept is approximately 2.174, and since 1 + 2145 L 2.174, this supports 6
the analytic result.
Multiply. Standard form
y = log(3x + 2) + log(x – 1) – 1 5
The equation 3x 2  x  12 = 0 cannot be solved by factoring. Use the quadratic formula. x =
 ( 1) ⫾ 2( 1)2  4(3)( 12)  b ⫾ 2b 2  4ac 1 ⫾ 2145 = = 2a 2(3) 6
a = 3, b =  1, c =  12 1  2145 Since is less than 1, it is not in the domain and must be rejected, 6 1 + 2145 f. giving the solution set e 6
–2
10
–5 FIGURE 47
364
CHAPTER 5
Inverse, Exponential, and Logarithmic Functions
EXAMPLE 6
Solving an Equation and Associated Inequalities Let ƒ(x) = 3 log 2(2x)  15. Solve the equation in (a) analytically and solve the inequalities in (b) and (c) graphically. (a) ƒ(x) = 0 (b) ƒ(x) 6 0 (c) ƒ(x) Ú 0 Solution (a)
Y1 = 3log(2X)/log(2) – 15 10
0
32
–10 FIGURE 48
FOR DISCUSSION Use FIGURE 47 from Example 5 on the preceding page to answer each item. 1. Give the exact solution set of log(3x + 2) + log(x  1)  1 Ú 0. 2. Give the exact solution set of log(3x + 2) + log(x  1)  1 … 0. (Hint: Pay attention to the domain.)
3 log 2(2x)  15 3 log 2(2x) log 2(2x) 2x x The solution set is {16}.
= = = = =
0 15 5 25 16
Add 15 to each side. Divide each side by 3. Write in exponential form. Divide by 2 and simplify.
(b) To graph y = ƒ(x), we use the changeofbase formula and graph the equation Y1 = 3 log(2X)>log(2)  15, as shown in FIGURE 48. We must have x 7 0, due to the domain of the logarithmic function. We see that ƒ(x) 6 0 for 0 6 x 6 16, or (0, 16). (c) From FIGURE 48, ƒ(x) Ú 0 for x Ú 16. Thus, the solution set is [16, q).
Equations Involving Exponentials and Logarithms EXAMPLE 7
Solve e 2 ln x =
Solving a More Complicated Exponential Equation
1 . 16
Solution 1 16 1 = 16 1 = 16 = 42
e  2 ln x = e ln x
2
x 2 x 2
x = 4
The domain is (0, q). Power rule a loga x = x 1 16
=
1 2 4
= 42;  4 is not a solution, since  4 6 0 and x 7 0.
Properties of exponents
An analytic check shows that 4 is indeed a solution. The solution set is 546. EXAMPLE 8
Solve ln
e ln x
Solution
Solving a More Complicated Logarithmic Equation  ln(x  3) = ln 2. ln e ln x  ln(x  3) ln x  ln(x  3) x ln x  3 x x  3 x x x
= ln 2 = ln 2
The domain is (3, q).
= ln 2
Quotient rule
= 2
Property 2
= 2x  6 = 6 = 6
Multiply by x  3. Subtract 2x. Multiply by  1.
e ln x = x
5.5
Exponential and Logarithmic Equations and Inequalities
ln e ln x  ln(x  3) ln e ln 6  ln(6  3) ln 6  ln 3 x 6 ln e = x ln 3 ln 2
Check:
The solution set is 566.
= ln 2 = ln 2 ? = ln 2 ?
Let x = 6.
= ln 2
Quotient rule
?
= ln 2
365
Original equation Simplify.
True
EXAMPLE 9
Solve
e 2x

4e x
Solving an Equation Quadratic in Form + 3 = 0 for x.
Solution Because e 2x = (e x)2, we can write and solve the given equation as follows. e 2x  4e x + 3 (e x )2  4(e x) + 3 u2  4u + 3 (u  1)(u  3) u = 1 or u x e = 1 or e x x = ln 1 or x x = 0 or x
= = = = = = = =
0 0 0 0 3 3 ln 3 ln 3
Original equation e2x = (e x )2 Substitute u for e x. Factor the quadratic expression. Zeroproduct property Substitute e x for u. Write in logarithmic form. loga 1 = 0
Both values check, and the solution set is 50, ln 36.
Solving Exponential and Logarithmic Equations An exponential or logarithmic equation can be solved by changing the equation into one of the following forms, where a and b are real numbers, a 7 0, and a Z 1: 1. aƒ(x) ⴝ b Solve by taking a logarithm of each side. (The natural logarithm is the best choice if a = e.) 2. log a ƒ(x) ⴝ log a g(x) From the given equation, ƒ(x) = g(x), which is solved analytically. 3. log a ƒ(x) ⴝ b Solve by changing to exponential form, ƒ(x) = a b.
Formulas Involving Exponentials and Logarithms EXAMPLE 10
Solving an Exponential Formula from Psychology The strength of a habit is a function of the number of times the habit is repeated. If N is the number of repetitions and H is the strength of the habit, then, according to psychologist C. L. Hull, H = 1000(1  e kN ), where k is a constant. Solve this formula for k. (continued)
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CHAPTER 5
Inverse, Exponential, and Logarithmic Functions
Solution H = 1000(1  e kN ) H = 1  e kN 1000
Divide by 1000.
H  1 =  e kN 1000 e kN = 1 
Subtract 1.
H 1000
Multiply by  1; rewrite.
ln e kN = ln a1 
H b 1000
Take a logarithm of each side.
 kN = ln a1 
H b 1000
ln e x = x
k = 
1 H lna1 b N 1000
Multiply by  N1 .
EXAMPLE 11
Modeling the Life Span of a Robin A study monitored the life spans of 129 robins over a 4year period. The equation y =
2  log(100  x) 0.42
was developed to calculate the number of years y it takes for x percent of the robin population to die. (Source: Lack, D., The Life of a Robin, Collins.) How many robins had died after 6 months? Solution Let y =
6 12
= 0.5 year, and solve the equation for x.
y = 0.5 = 0.21 log(100  x) 101.79 x x
= = = = L
2  log(100  x) 0.42 2  log(100  x) 0.42 2  log(100  x) 1.79 100  x 100  101.79 38.3
Let y = 0.5. Multiply by 0.42. Add log(100  x ); subtract 0.21. Write in exponential form. Add x; subtract 101.79. Use a calculator.
About 38% of the robins had died after 6 months.
5.5
EXERCISES
Checking Analytic Skills Solve each equation. Express all solutions in exact form. Do not use a calculator. 1. 2(10x) = 14 5.
1 3 log 2 x = 2 4
2.
1 x e = 13 2
6. 2 log 3 x =
4 5
3. 3e 2x + 1 = 5
4. 5(103x )  4 = 6
7. 4 ln 3x = 8
8. 7 log 2x = 10
5.5
Exponential and Logarithmic Equations and Inequalities
367
Solve each exponential equation. Express the solution set so that (a) solutions are in exact form and, if irrational, (b) solutions are approximated to the nearest thousandth. Support your solutions by using a calculator. 1 x 9. 3x = 7 10. 5x = 13 11. a b = 5 2 1 x x 12. a b = 6 13. 0.8 = 4 14. 0.6x = 3 3 15. 4x  1 = 32x
16. 2x + 3 = 5x
17. 6x + 1 = 42x  1
18. 3x  4 = 72x + 5
19. 2x =  5
20. 3x =  4
21. e x  3 = 23x
22. e 0.5x = 31  2x
1 x 23. a b =  3 3
1 x 24. a b =  9 9
25. 0.05(1.15)x = 5
26. 1.2(0.9)x = 0.6
27. 3(2)x  2 + 1 = 100
28. 5(1.2)3x  2 + 1 = 7
29. 2(1.05)x + 3 = 10
30. 3(1.4)x  4 = 60
31. 5(1.015)x  1980 = 8
32. 30  3(0.75)x  1 = 29
Solve each logarithmic equation. Express all solutions in exact form. Support your solutions by using a calculator. 33. 5 ln x = 10
34. 3 log x = 2
35. ln(4x) = 1.5
36. ln(2x) = 5
37. log(2  x) = 0.5
38. ln(1  x) =
39. log 6(2x + 4) = 2
40. log 5(8  3x) = 3
1 2 41. log 4(x 3 + 37) = 3
43. ln x + ln x 2 = 3
44. log x + log x 2 = 3
45. log x + log(x  21) = 2
46. log x + log(3x  13) = 1
47. ln(4x  2)  ln 4 =  ln(x  2)
48. ln(5 + 4x)  ln(3 + x)  ln 3 = 0
49. log 5(x + 2) + log 5(x  2) = 1
50. log 2(x  7) + log 2 x = 3
51. log 7(4x)  log 7(x + 3) = log 7 x
52. log 2(2x) + log 2(x + 2) = log 2 16
53. ln e x  2 ln e = ln e 4
54. log 2(log 2 x) = 1
55. log x = 2log x
56. ln(ln x) = 0
42. log 7
(x 3
+ 65) = 0
57. In Example 7, we found that the solution of the equation 1 e 2 ln x = 16 is 4. Complete the following: (a) Solve the equation graphically. 1 , using the graph (b) Solve the inequality e 2 ln x 6 16 from part (a). 1 , using the graph (c) Solve the inequality e 2 ln x 7 16 from part (a).
9 59.
A student told a friend, “You must reject any negative solution of an equation involving logarithms.” Is this correct? Write an explanation of your answer.
58. In Example 8, we found that the solution of the equation ln e ln x  ln(x  3) = ln 2 is 6. Complete the following: (a) Solve the equation graphically. (b) Solve the inequality ln e ln x  ln(x  3) 7 ln 2, using the graph from part (a). (c) Solve the inequality ln e ln x  ln(x  3) 6 ln 2, using the graph from part (a).
9 60.
Use a graph to explain why the logarithmic equation ln x  ln(x + 1) = ln 5 has no solution.
Use a graphing calculator to find the solution of each equation. Round your result to the nearest thousandth. 61. 1.5log x = e 0.5
62. 1.5ln x = 100.5
Solve each formula for the indicated variable. 63. r = p  k ln t, for t
64. p = a +
65. T = T0 + (T1  T0)10kt, for t
66. A =
k , for x ln x
Pi , for n 1  (1 + i)n
368
CHAPTER 5
Inverse, Exponential, and Logarithmic Functions
K , for b 1 + ae bx
67. A = T0 + Ce kt, for k
68. y =
69. y = A + B(1  e Cx), for x
70. m = 6  2.5 loga
71. log A = log B  C log x, for A
I 72. d = 10 loga b , for I I0
73. A = Pa 1 +
r nt b , for t n
M b, for M M0
74. D = 160 + 10 log x, for x
The given equations are quadratic in form. Solve each and give exact solutions. Refer to Example 9. 75. e 2x  6e x + 8 = 0 81. 32x + 35 = 12(3x)
76. e 2x  8e x + 15 = 0 1 79. e 2x + e x = 1 2 82. 52x + 3(5x) = 28
77. 2e 2x + e x = 6 1 80. e 2x + 2e x = 3 4 83. (log 2 x)2 + log 2 x = 2
84. (log x)2  6 log x = 7
85. (ln x)2 + 16 = 10 ln x
86. 2 (ln x)2 + 9 ln x = 5
78. 3e 2x + 2e x = 1
For the given ƒ(x), solve the equation ƒ(x) = 0 analytically and then use a graph of y = ƒ(x) to solve the inequalities ƒ(x) 6 0 and ƒ(x) Ú 0. 87. ƒ(x) =  2e x + 5
88. ƒ(x) =  3e x + 7
89. ƒ(x) = 2(3x)  18
90. ƒ(x) = 4x  2  2
91. ƒ(x) = 32x  9x + 1
92. ƒ(x) = 23x  8x  3
93. ƒ(x) = 8  4 log 5 x
94. ƒ(x) = 9 log 3(3x)  18
95. ƒ(x) = ln(x + 2)
96. ƒ(x) = ln(x  1)  ln(x + 1)
97. ƒ(x) = 7  5 log x
98. ƒ(x) = 3  2 log 4(x  5)
In general, it is not possible to find exact solutions analytically for equations that involve exponential or logarithmic functions together with polynomial, radical, and rational functions. Solve each equation using a graphical method, and express solutions to the nearest thousandth if an approximation is appropriate. 100. x 2  4 = e x  4 + 4 1 103. e x = x + 2
99. x 2 = 2x 102. ln x =  2 3x + 3
Use any method (analytic or graphical) to solve each equation. 2 2 105. log 22x 2  1 = 0.5 106. log x = (log x) 2
(Modeling) Life Span of Robins y =
Use the equation
2  log(100  x) 0.42
107. ln(ln e  x ) = ln 3
101. log x = x 2  8x + 14 104. 3x = 2x + 5
108. e x + ln 3 = 4e x
models salinity to depths of 1000 meters at a latitude of 57.5°N. The variable x is the depth in meters, and ƒ(x) is in grams of salt per kilogram of seawater. (Source: Hartman, D., Global Physical Climatology, Academic Press.)
from Example 11 for Exercises 109 and 110. 109. Estimate analytically the percentage of robins that died after 2 years. 110. Estimate the number of years elapsed for 75% of the robins to die. (Modeling) Salinity The salinity of the oceans changes with latitude and depth. In the tropics, the salinity increases on the surface of the ocean due to rapid evaporation. In the higher latitudes, there is less evaporation and rainfall causes the salinity to be less on the surface than at lower depths. The function given by ƒ(x) = 31.5 + 1.1 log(x + 1)
111. Approximate analytically the depth where the salinity equals 33. 112. Estimate the salinity at a depth of 500 meters.
5.6
SECTIONS 5.4 AND 5.5
Further Applications and Modeling with Exponential and Logarithmic Functions
REVIEWING BASIC CONCEPTS
1. Fill in the blanks: If ƒ(x) = 3x and g(x) = log 3 x, then functions ƒ and g are functions, and their graphs are with respect to the line with equation . The domain of ƒ is the of g and vice versa. Let ƒ(x) = 2  log 2 (x  1) in Exercises 2–5.
Solve each equation. 6. 32x  1 = 4x 105 log x
8.
7. ln 5x  ln(x + 2) = ln 3 9. H = 1000(1  e kN ), for N
= 32
10. (Modeling) Caloric Intake
2. Graph ƒ(x).
The function defined by
ƒ(x) = 280 ln(x + 1) + 1925
3. Give the equation of the asymptote of the graph of ƒ(x) and any intercepts.
models the number of calories consumed daily by a person owning x acres of land in a developing country. (Source: Grigg, D., The World Food Problem, Blackwell Publishers.) Estimate the number of acres owned for the average intake to be 2300 calories per day.
4. Compare the graph of ƒ(x) with the graph of g(x) = log 2 x. 5. Find the inverse of the function defined by ƒ(x).
5.6
369
Further Applications and Modeling with Exponential and Logarithmic Functions
Physical Science Applications • Financial Applications • Population Growth and Medical Applications • Modeling Data with Exponential and Logarithmic Functions
Looking Ahead to Calculus
The exponential growth and decay function formulas are studied in calculus in conjunction with the topic known as differential equations.
Physical Science Applications A function of the form A(t) ⴝ A0e kt,
Exponential growth function
where A0 represents the initial quantity present, t represents time elapsed, k 7 0 represents the growth constant associated with the quantity, and A(t) represents the amount present at time t, is called an exponential growth function. (The formula A = Pe rt for continuous compounding of interest is an example of an exponential growth function.) This is an increasing function, because e 7 1 and k 7 0. On the other hand, for k 7 0, a function of the form A(t) ⴝ A0e ⴚkt
Exponential decay function
is an exponential decay function. It is a decreasing function because 1 k e k = (e 1)k = a b e
and 0 6
1 6 1. e
In both cases, we usually restrict t to be nonnegative, giving domain 30, q). (Why?) FIGURES 49 and 50 respectively show graphs of typical growth and decay functions. y
y
A(t) = A0 e kt, k > 0
A(t) = A0 e –kt, k > 0 A0
A0
t
0
Exponential growth function FIGURE 49
t
0
Exponential decay function FIGURE 50
370
CHAPTER 5
Inverse, Exponential, and Logarithmic Functions
If a quantity decays exponentially, the amount of time that it takes to reach onehalf its original amount is called the halflife. EXAMPLE 1
Finding the Age of a Fossil Carbon 14 is a radioactive form of carbon that is found in all living plants and animals. After a plant or animal dies, the radiocarbon disintegrates. Using a technique called carbon dating, scientists determine the age of the remains by comparing the amount of carbon 14 present with the amount found in living plants and animals. The amount of carbon 14 present after t years is given by the exponential function defined by ln 2 A(t) = A0e kt, with k = 5700 . The halflife of carbon 14 is 5700 years. If a fossil contains onefifth of the carbon 14 contained in a contemporary living sample, estimate its age. A(t) = A0e kt
Solution
1 A 5 0 1 5 1 ln 5 1 ln 5 5700 1 ln ln 2 5 t
ln 2
= A0e  5700 t ln 2
= e  5700 t ln 2
= ln e  5700 t = 
ln 2 t 5700
Exponential decay function Let A( t ) =
1 5
A0 and k =
ln 2 5700 .
Divide by A 0. Take a logarithm of each side. ln e x = x
= t
Multiply by  5700 ln 2 .
L 13,235
Approximate.
The age of the fossil is about 13,000 years.
EXAMPLE 2
Finding an Exponential Decay Function Given Halflife Radium 226, which decays according to the function defined by A(t) = A0e kt, where t is time in years, has a halflife of about 1612 years. Find the exact value of k and then approximate k. Solution The halflife tells us that when t = 1612, A(t) = A(t) = A0 1 A 2 0 1 2 1 ln 2 1 ln 2
A0.
e kt
= A0 e k (1612)
Substitute.
= e 1612k
Divide by A 0.
= ln e 1612k
Take logarithms of each side.
=  1612k
ln e x = x
k =
ln 12  1612
L 0.00043
Divide by  1612; approximate.
Thus, radium 226 decays according to the equation A(t) = A0e 0.00043t.
1 2
5.6
Further Applications and Modeling with Exponential and Logarithmic Functions
371
EXAMPLE 3
Measuring Sound Intensity The loudness of sounds is measured in a unit called a decibel. To measure with this unit, we first assign an intensity of I0 to a very faint sound, called the threshold sound. If a particular sound has intensity I, then the decibel rating of this louder sound is d ⴝ 10 log
I . I0
In the movie The Lord of the Rings: The Two Towers, action sequences reached levels of 109.5I0. (Source: Canadian Broadcast Company News.) Find the decibel rating of these sequences, and compare it with the average rating of 80 decibels for a ringing telephone. Solution
d = 10 log
109.5I0
I0 d = 10 log 109.5 d = 10(9.5) d = 95
Substitute I = 109.5I0 in the equation. Divide out I0. log 109.5 = 9.5 Multiply.
The action sequences reached 95 decibels. Since 95 decibels is equivalent to 109.5I0 and 80 decibels is equivalent to 108.0I0, their ratio is 109.5I0 108.0I0
= 101.5 L 31.6.
These action sequences are 31.6 times the intensity of a ringing telephone.
Financial Applications If a quantity grows exponentially, the amount of time that it takes to become twice its original amount is called the doubling time. This is analogous to halflife for quantities that decay exponentially. The initial amount present does not affect either the doubling time or halflife. EXAMPLE 4
Solving Compound Interest Formulas for t
(a) How long will it take $1000 invested at 6% interest compounded quarterly to grow to $2700? (b) How long will it take for the money in an account that is compounded continuously at 3% interest to double? Solution (a)
A = Pa 1 +
r nt b n
2700 = 1000a1 +
This table numerically supports the result of Example 4(a). Notice that Y1 = 2700 for some Xvalue between 16 and 17. We found analytically that this value is approximately 16.678.
2700 2.7 log 2.7 log 2.7 log 2.7 4 log 1.015 t It will take about
= = = =
0.06 4t b 4
1000(1.015)4t 1.0154t log 1.0154t 4t log 1.015
Compound interest formula (Section 5.2) A = 2700, P = 1000, r = 0.06, n = 4 Simplify. Divide by 1000. Take a logarithm of each side. Power rule
= t
Divide by 4 log1.015.
L 16.678
Use a calculator.
16 34
years for the initial amount to grow to $2700. (continued)
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Inverse, Exponential, and Logarithmic Functions
(b)
A 2P 2 ln 2 ln 2 ln 2 0.03 t
= = = = =
Pe rt Pe 0.03t e 0.03t ln e 0.03t 0.03t
Continuous compounding formula (Section 5.2) A = 2P, r = 0.03 Divide by P. Take a logarithm of each side. ln e x = x
= t
Divide by 0.03.
L 23.1
Use a calculator.
The initial amount will double in about 23.1 years. GCM
A loan is amortized if both the principal and interest are paid by a sequence of equal periodic payments.
TECHNOLOGY NOTE
Amortization Payments A loan of P dollars at interest rate i per period (as a decimal) may be amortized in n equal periodic payments of R dollars made at the end of each period, where Graphing calculators such as the TI83/84 Plus are capable of financial calculations. Consult the graphing calculator manual that accompanies this text or your owner's manual.
P . 1 ⴚ (1 ⴙ i)ⴚn c d i
Rⴝ
The total interest I that will be paid during the term of the loan is I ⴝ nR ⴚ P.
GCM
EXAMPLE 5
Using Amortization to Finance a Camper Trailer You purchase a camper trailer for $24,000. After a down payment of $4000, the balance will be paid off in 36 equal monthly payments at 8.5% interest per year. Find the amount of each payment. How much interest will you pay over the life of the loan? Solution Use the amortization formula. R =
20,000 c
0.007083)36
1  (1 + 0.007083
d
L 631.35
P = 24,000  4000 = 20,000, i =
0.085 12
L 0.007083, n = 36
The monthly payment will be about $631.35. The total interest paid will be about 36($631.35)  $20,000 = $2728.60. In Example 5, the approximate unpaid balance y after x payments is given by y ⴝ Rc
1 ⴚ (1 ⴙ i) ⴚ (nⴚ x) d. i
For example, the unpaid balance after 12 payments is y = $631.35c
1  (1 + 0.007083)24 d L $13,889.41. 0.007083
You may be surprised that the remaining balance on a $20,000 loan after 12 payments is as large as $13,889.41. This is because most of each early payment on a loan goes toward interest.
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Further Applications and Modeling with Exponential and Logarithmic Functions
373
Population Growth and Medical Applications EXAMPLE 6
Modeling World Population World population in billions during year x can be modeled by P(x) = 3(1.017)x  1960. Solve the equation 3(1.017)x  1960 = 7 analytically to estimate the year when world population reached 7 billion. Solution 3(1.017)x  1960 = 7 7 (1.017)x  1960 = 3 The natural logarithm could also be used.
Divide by 3.
7 3 7 (x  1960) log 1.017 = log 3 log(1.017)x  1960 = log
x  1960 =
Take the common logarithm. Power rule: log m r = r log m
log 73
Divide by log 1.017.
log 1 .017
x = 1960 +
log 73 log 1.017
x L 2010.3
Add 1960. Approximate.
This model predicts that world population reached 7 billion during 2010. EXAMPLE 7
Determining the Amount of Medication When physicians prescribe medication, they must consider how the drug’s effectiveness decreases over time. If, each hour, a drug is only 90% as effective as the previous hour, at some point the patient will not be receiving enough medication and must receive another dose. For example, if the initial dose was 200 milligrams and the drug was administered 3 hours ago, the exponential expression 200(0.90)2 represents the amount of effective medication still available. Thus, 200(0.90)2 = 162 milligrams are still in the system. (The exponent here is the number of hours since the drug was administered, less 1 because it takes 1 hour for the drug to be absorbed into the bloodstream.) How long will it take for this initial dose to reach the dangerously low level of 50 milligrams? Analytic Solution We must solve the equation 200(0.90)x = 50. 200(0.90)x (0.90)x log(0.90)x x log 0.90
= = = =
50 0.25 log 0.25 log 0.25 log 0.25 L 13.16 x = log 0.90
Graphing Calculator Solution The graph in FIGURE 51 confirms the result that y = 50 when x L 13.16. y = 200(0.90) x
Divide by 200. Take logarithms.
100
Power rule Divide by log 0.90.
Since x represents one less than the number of hours from when the drug was administered, the drug will reach a level of 50 milligrams in about 14 hours.
0 –5
20 FIGURE 51
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Inverse, Exponential, and Logarithmic Functions
Modeling Data with Exponential and Logarithmic Functions Year
CO2 (ppm)
2000 2050 2100 2150 2200
364 467 600 769 987
GCM
Source: Turner, R., Environmental Economics, an Elementary Approach, Johns Hopkins University Press.
EXAMPLE 8
Modeling Atmospheric CO2 Concentrations Future concentrations of atmospheric carbon dioxide (CO2 ) in parts per million (ppm) are shown in the table. (These concentrations assume that recent trends continue.) CO2 levels in the year 2000 were greater than they had been anytime in the previous 160,000 years. The increase in concentration of CO2 has been accelerated by the burning of fossil fuels and by deforestation. (a) Let x = 0 correspond to 2000 and x = 200 to 2200. Find values for C and a so that ƒ(x) = Ca x models these data. (b) Use a graphing calculator with regression capability to find an exponential function g that models all the data, and graph it with the data points. How does function g compare with function ƒ from part (a)? (c) Estimate CO2 concentrations for the year 2025. Solution (a) The concentration is 364 when x = 0, so C = 364. This gives ƒ(x) = Ca x = 364a x. One possibility for determining a is to require that the graph of ƒ pass through the point (2200, 987), where x = 200. Thus, ƒ(200) = 987. 364a x = ƒ(x) 364 # a 200 = 987 987 a 200 = 364 987 1/200 (a 200)1/200 = a b 364 987 1/200 a = a b 364 a L 1.005
(a)
g(x) = 364.0(1.005) x 1000
–25
250
ƒ(200) = 987 Divide by 364. Raise to the
1 200 th
power.
Property of exponents Use a calculator.
Hence, ƒ(x) = 364(1.005)x. (Answers may vary slightly.)
300 (b)
(b)
FIGURE 52
shows the coefficients given by the calculator for exponential regression. With these values, g(x) L 364.0(1.005)x, which is essentially the same as ƒ(x) from part (a). The data points and the function are graphed in FIGURE 52(b). FIGURE 52(a)
(c) Since 2025 corresponds to x = 25, evaluate ƒ(25). ƒ(25) = 364(1.005)25 L 412. The concentration of carbon dioxide could reach 412 ppm by 2025.
GCM
EXAMPLE 9
Modeling Interest Rates The table lists the interest rates for certificates of deposit during January 2005. Rates are for deposits of $10,000 or more. Time
1 year
2 years
3 years
4 years
5 years
Yield (%)
2.326
2.685
3.148
3.354
3.665
Source: Union Bank of California.
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Further Applications and Modeling with Exponential and Logarithmic Functions
375
(a) Make a scatter diagram of the data. What type of function might model these data? (b) Use leastsquares regression to obtain a formula ƒ(x) = a + b ln x that models the data. (c) Graph ƒ and the data in the same viewing window. 6
Solution (a) Enter the data points (1, 2.326), (2, 2.685), (3, 3.148), (4, 3.354), and (5, 3.665) into a calculator. A scatter diagram of the data is shown in FIGURE 53. A logarithmic function model may be appropriate.
–1
6 –1
(b) In FIGURES 54(a) and 54(b), leastsquares regression gives a logarithmic function defined (approximately) by ƒ(x) = 2.244 + 0.8263 ln x.
FIGURE 53
(c) A graph of ƒ and the data are shown in FIGURE 54(c). f(x) = 2.244 + 0.8263 ln x 6
–1
6 –1
(a)
(b)
(c)
FIGURE 54
5.6
EXERCISES
(Modeling) The exercises in this set are grouped according to discipline; they involve exponential or logarithmic models. Physical Sciences (Exercises 1–16) The information in Example 1 allows us to use the function defined by A(t) = A0e 0.0001216t to approximate the amount of carbon 14 remaining in a sample, where t is in years. Use this ln 2 function in Exercises 1–4. A Note:  0.0001216 L  5700 .B 1. Carbon 14 Dating Suppose an Egyptian mummy is discovered in which the amount of carbon 14 present is only about onethird the amount found in the atmosphere. About how long ago did the Egyptian die?
3. Carbon 14 Dating Paint from the Lascaux caves of France contains 15% of the normal amount of carbon 14. Estimate the age of the paint. 4. Carbon 14 Dating Estimate the age of a specimen that contains 20% of the carbon 14 of a comparable living specimen. 5. Radioactive Lead 210 The halflife of radioactive lead 210 is 21.7 years. (a) Find an exponential decay model for lead 210. (b) Estimate how long will it take a sample of 500 grams to decay to 400 grams. (c) Estimate how much of the sample of 500 grams will remain after 10 years. 6. Radioactive Cesium 137 Radioactive cesium 137 was emitted in large amounts in the Chernobyl nuclear power station accident in Russia on April 26, 1986. The amount of cesium 137 remaining after x years in an initial sample of 100 milligrams can be described by A(x) = 100e 0.02295x.
2. Carbon 14 Dating A sample from a refuse deposit near the Strait of Magellan had 60% of the carbon 14 of a contemporary living sample. Estimate the age of the sample.
(Source: Mason, C., Biology of Freshwater Pollution, John Wiley and Sons.) (a) Estimate how much is remaining after 50 years. Is the halflife of cesium 137 greater or less than 50 years? (b) Estimate the halflife of cesium 137.
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Inverse, Exponential, and Logarithmic Functions
7. Radioactive Polonium 210 The table shows the amount y of polonium 210 remaining after t days from an initial sample of 2 milligrams. t (days)
0
100
y (milligrams)
2
1.22
200
300
0.743 0.453
(a) Use the table to determine whether the halflife of polonium 210 is greater or less than 200 days. (b) Find a formula that models the amount A of polonium 210 in the table after t days. (c) Estimate the halflife of polonium 210. 8. Sound Intensity
Use the formula d = 10 log
I I0
ƒ(t) = T0 + Ce kt,
to estimate the average decibel level for each sound with the given intensity I. For comparison, conversational speech has a sound level of about 60 decibels. (a) Jackhammer: 31,620,000,000I0 (b) Movie: Harry Potter and the Chamber of Secrets: 1.995 * 109I0 (c) Rock singer screaming into microphone: 1014 I0 The intensity I of an earthquake, measured on the Richter scale, is given by log II , where I0 is the intensity of an earthquake of 0 a certain small size. Use this information in Exercises 9 and 10. 9. Earthquake Intensity In 2008 the earthquake in Rhodes, Greece, measured 6.7 on the Richter scale. (a) Express this intensity I in terms of I0. (b) In 2008 the earthquake in Sichuan, China, measured 8.0 on the Richter scale. Express this intensity I in terms of I0. (c) How many times more intense was the earthquake in China than the earthquake in Greece? 10. Earthquake Intensity The earthquake off the coast of Northern Sumatra on Dec. 26, 2004, had a Richter scale rating of 9.0. (a) Express the intensity of this earthquake in terms of I0. (b) On Dec. 14, 2004, the Cayman Islands experienced an earthquake with a Richter scale rating of 6.8. Express the intensity of this earthquake in terms of I0. (c) Compare the intensities of the Sumatran and Cayman Islands earthquakes. 11. Magnitude of a Star by the equation
where C and k are constants. Use this result in Exercises 12–14. 12. Newton’s Law of Cooling Boiling water at 100°C is placed in a freezer at 0°C. The temperature of the water is 50°C after 24 minutes. Approximate the temperature of the water after 96 minutes. 13. Newton’s Law of Cooling A pot of coffee with a temperature of 100°C is set down in a room with a temperature of 20°C. The coffee cools to 60°C after 1 hour. (a) Write an equation to model the data. (b) Estimate the temperature after a half hour. (c) About how long will it take for the coffee to cool to 50°C? Support your answer graphically. 14. Newton’s Law of Cooling A piece of metal is heated to 300°C and then placed in a cooling liquid at 50°C. After 4 minutes, the metal has cooled to 175°C. Estimate its temperature after 12 minutes. 15. Greenhouse Gases Chlorofluorocarbons (CFCs) are gases that increase the greenhouse effect and damage the ozone layer. CFC 12 is one type of chlorofluorocarbon used in refrigeration, air conditioning, and foam insulation. The table lists future concentrations of CFC 12 in parts per billion (ppb) if current trends continue.
I , I0
where I0 is the measure of a justvisible star and I is the actual intensity of the star being measured. The dimmest stars are of magnitude 6, and the brightest are of magnitude 1. Determine the ratio of light intensities between a star of magnitude 1 and a star of magnitude 3.
Year
2000
2005
2010
2015
2020
CFC 12 (ppb)
0.72
0.88
1.07
1.31
1.60
Source: Turner, R., Environmental Economics, an Elementary Approach, Johns Hopkins University Press.
The magnitude of a star is defined
M = 6  2.5 log
Newton’s law of cooling says that the rate at which an object cools is proportional to the difference C in temperature between the object and the environment around it. The temperature ƒ(t) of the object at time t in appropriate units after being introduced into an environment with a constant temperature T0 is
(a) Let x = 0 correspond to 2000 and x = 20 to 2020. Find values for C and a so that ƒ(x) = Ca x models these data. (b) Estimate the CFC 12 concentration in 2013.
9 16.
Reducing Carbon Emissions Governments could reduce carbon emissions by placing a tax on fossil fuels. The cost–benefit equation ln(1  P) =  0.0034  0.0053x (continued)
5.6
Further Applications and Modeling with Exponential and Logarithmic Functions
estimates the relationship between a tax of x dollars per ton of carbon and the percent P reduction in emissions of carbon, where P is in decimal form. Determine P when x = 60. Interpret the result. (Source: Clime, W., The Economics of Global Warming, Institute for International Economics.) Finance
(Exercises 17–34)
377
Use the future value formula in Exercises 25–28. 25. Present Value Find the present value of $10,000 5 years from now if interest is compounded semiannually at 3%. 26. Present Value Find the present value of $25,000 2.75 years from now if interest is compounded quarterly at 6%.
17. Interest on an Account Estimate how long it will take for $1000 to grow to $5000 at an interest rate of 3.5% if interest is compounded (a) quarterly; (b) continuously.
27. Future Value Estimate the interest rate necessary for a present value of $25,000 to grow to a future value of $30,416 if interest is compounded annually for 5 years.
18. Interest on an Account Estimate how long it will take for $5000 to grow to $8400 at an interest rate of 6% if interest is compounded (a) semiannually; (b) continuously.
28. Future Value Estimate the interest rate necessary for a present value of $1200 to grow to a future value of $1408 if interest is compounded quarterly for 8 years.
19. Interest on an Account Tom Tupper wants to buy a $30,000 car. He has saved $27,000. Find the number of years (to the nearest tenth) it will take for his $27,000 to grow to $30,000 at 6% interest compounded quarterly. 20. Doubling Time Estimate the doubling time of an investment earning 2.5% interest if interest is compounded (a) quarterly; (b) continuously. 21. Comparison of Investment A construction worker wants to invest $60,000 in a pension plan. One investment offers 7% compounded quarterly. Another offers 6.75% compounded continuously. Which investment will earn more interest in 5 years? How much more will the better plan earn? 22. Growth of an Account See Exercise 21. If the worker chooses the plan with continuous compounding, estimate how long it will take for her $60,000 to grow to $80,000. The interest rate stated by a financial institution is sometimes called the nominal rate. If interest is compounded, the actual rate is, in general, higher than the nominal rate, and is called the effective rate. If r is the nominal rate and n is the number of times interest is compounded annually, then R = a1 +
r n b  1 n
is the effective rate. Here, R represents the annual rate that the investment would earn if simple interest were paid. Use this formula in Exercises 23 and 24. 23. Effective Rate Find the effective rate to the nearest hundredth of a percent if the nominal rate is 3% and interest is compounded quarterly. 24. Effective Rate Estimate the effective rate if the nominal rate is 4.5% and interest is compounded daily (n = 365). In the formula A = P A 1 + nr B , we can interpret P as the present value of A dollars t years from now, earning annual interest r compounded n times per year. In this context, A is called the future value. If we solve the formula for P, we obtain nt
P = A a1 +
r nt b . n
In Exercises 29–31, use the amortization formulas given in this section to find (a) the monthly payment on a loan with the given conditions and (b) the total interest that will be paid during the term of the loan. 29. Amortization $8500 is amortized over 4 years with an interest rate of 7.5%. 30. Amortization $9600 is amortized over 5 years with an interest rate of 5.2%. 31. Amortization $125,000 is amortized over 30 years with an interest rate of 7.25%. 32. Amortization Payments y = Rc
Use the formula
1  (1 + i)(n  x) d, i
where y is the unpaid balance after x payments have been made on a loan with n payments of R dollars. Find the balance after 120 payments have been made on the loan in Exercise 31. 33. Growth of an Account Use the table feature of your graphing calculator to work parts (a) and (b). (a) Find how long it will take $1500 invested at 5.75%, compounded daily, to triple in value. Locate the solution by systematically decreasing ¢Tbl. Find the answer to the nearest day. (Find your answer to 1 the nearest day by eventually letting ¢Tbl = 365 . The decimal part of the solution can be multiplied by 365 to determine the number of days greater than the nearest year. For example, if the solution is determined to be 16.2027 years, then multiply 0.2027 by 365 to get 73.9855. The solution is then, to the nearest day, 16 years and 74 days.) Confirm your answer analytically. (b) Find how long it will take $2000 invested at 8%, compounded daily, to be worth $5000. 34. Growth of an Account If interest is compounded continuously and the interest rate is tripled, what effect will this have on the time required for an investment to double?
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Population Growth and Medicine
(Exercises 35–46)
35. Growth of Bacteria Escherichia coli is a strain of bacteria that occurs naturally in many organisms. Under certain conditions, the number of bacteria present in a colony is approximated by A(t) = A0e 0.023t, where t is in minutes. If A0 = 2,400,000, find the number of bacteria at each time. Round to the nearest hundred thousand. (a) 5 minutes (b) 10 minutes (c) 60 minutes 36. Growth of Bacteria The growth of bacteria in food products makes it necessary to date some products (such as milk) so that they will be sold and consumed before the bacteria count becomes too high. Suppose that, under certain storage conditions, the number of bacteria present in a product is given by ƒ(t) = 500e 0.1t, where t is time in days after packing of the product and the value of ƒ(t) is in millions.
39. Population of Pakistan In 2007 the population of Pakistan was 164 million, and it is expected to be 250 million in 2025. (Source: United Nations.) (a) Approximate C and a so that P(x) = Ca x  2007 models these data, where P is in millions and x is the year. (b) Estimate the population of Pakistan in 2015, and compare your estimate to the predicted value of 204 million. (c) Estimate when this population could reach 212 million. 40. Population Growth In 2000 India’s population reached 1.003 billion. It is projected to be 1.362 billion in 2025. (Source: U.S. Census Bureau.) (a) Find values for C and a so that ƒ(x) = Ca x  2000 models the population of India in year x. (b) Estimate India’s population in 2020. (c) Use ƒ to determine the year when India’s population might reach 1.5 billion. 41. Bacteria Growth Suppose that the concentration of a bacteria sample is 100,000 bacteria per milliliter. If the concentration doubles every 2 hours, how long will it take for the concentration to reach 350,000 bacteria per milliliter? 42. Bacteria Growth Suppose that the concentration of a bacteria sample is 50,000 bacteria per milliliter. If the concentration triples in 4 days, how long will it take for the concentration to reach 85,000 bacteria per milliliter? For Exercises 43 and 44, refer to Example 7. 43. Drug Level If 250 milligrams of a drug are administered and the drug is only 75% as effective each subsequent hour, how much effective medicine will remain in the person’s system after 6 hours?
(a) If the product cannot be safely eaten after the bacteria count reaches 3,000,000,000, how long will this take? (b) If t = 0 corresponds to January 1, what date should be placed on the product? 37. Population Growth lions is given by
The population of Tennessee in mil
P(x) = 5.33e 0.0133x where x = 0 corresponds to 1995, x = 5 to 2000, and so on. (Source: U.S. Census Bureau.) (a) Determine analytically the year when the population of Tennessee was 5.9 million. (b) Solve part (a) graphically. (c) According to this model, when did the population reach 6.25 million? 38. U.S. Liver Transplants The number of individuals waiting for liver transplants can be modeled by the function ƒ(x) = 2340(1.124)x  1988, where x is the year. (Source: United Network for Organ Sharing.) (a) Estimate this number in 2005. (b) Determine when this number reached 30,000.
44. Drug Level A new drug has been introduced that is 80% as effective each hour as the previous hour. A minimum of 20 milligrams must remain in the patient’s bloodstream during the course of treatment. If 100 milligrams are administered, how many hours may elapse before another dose is necessary? 45. Epidemics In 1666 the village of Eyam, located in England, experienced an outbreak of the Great Plague. Out of 261 people in the community, only 83 survived. The table shows a function ƒ that computes the number of people who had not (yet) been infected after x days. x ƒ(x) x ƒ(x)
0
15
30
45
254
240
204
150
60
75
90
125
125
103
97
83
Source: Raggett, G., “Modeling the Eyam Plague,” The Institute of Mathematics and Its Applications 18.
(continued)
5.6
Further Applications and Modeling with Exponential and Logarithmic Functions
45. (a) Use a table to represent a function g that computes the number of people in Eyam who were infected after x days. (b) Write an equation that shows the relationship between ƒ(x) and g (x). (c) Use graphing to decide which equation represents g(x) better, y1 =
171 1 + 18.6e 0.0747x
(Modeling) In real life, populations of bacteria, insects, and animals do not continue to grow indefinitely. Initially, population growth may be slow. Then, as their numbers increase, so does the rate of growth. After a region has become heavily populated or saturated, the population usually levels off because of limited resources. This type of growth may be modeled by a logistic function represented by
or y2 = 18.3(1.024)x.
(d) Use your results from parts (b) and (c) to find a formula for ƒ(x). 46. Heart Disease Death Rates The table lists heart disease death rates per 100,000 people for selected ages. Age
30
40
50
60
70
Death Rate
8.0
29.6
92.9
246.9
635.1
ƒ(x) =
Refer to this information for Exercises 49 and 50. 49. Heart Disease As age increases, so does the likelihood of coronary heart disease (CHD). The fraction of people x years old with some CHD is approximated by ƒ(x) =
47. Midair Near Collisions The table shows the number of aircraft midair near collisions y in year x. x
2000
2001
2002
2003
y
239
205
180
162
Source: Federal Aviation Administration.
(a) Use exponential regression to approximate constants C and a so that ƒ(x) = Ca x  2000 models the data. (b) Support your answer by graphing ƒ and the data.
0.9 . 1 + 271e 0.122x
(Source: Hosmer, D. and S. Lemeshow, Applied Logistic Regression, John Wiley and Sons.) 9 (a) Evaluate ƒ(25) and ƒ(65). Interpret the results. (b) At what age does this likelihood equal 50%? 50. Tree Growth modeled by
Social Sciences (Exercises 47 and 48)
c , 1 + ae bx
where a, b, and c are positive constants.
Source: Center for Disease Control.
(a) Make a scatter diagram of the data in the window 325, 754 by 3 100, 7004. (b) Find an exponential function ƒ that models the data. (c) Estimate the heart disease death rate for people who are 80 years old.
379
The height of a tree in feet after x years is
ƒ(x) =
50 . 1 + 47.5e 0.22x
(a) Make a table for ƒ starting at x = 10 and incrementing by 10. What appears to be the maximum height of the tree? (b) Graph ƒ and identify the horizontal asymptote. 9 Explain its significance. (c) After how long was the tree 30 feet tall?
48. Telecommuting Some workers use technology such as smart phones, email, and computers to work at home rather than in the office. The table lists telecommuters in millions from 2001 to 2006. Year
2001
2002
2003
2004
Telecommuters
19.7
32.7
41.3
44.4
Year
2005
2006
Telecommuters
49.5
52.8
Source: International Telework Association and Council (ITAC).
Find values for a and b so that ƒ(x) = a + b ln x models the data, where x = 1 corresponds to 2001, x = 2 to 2002, and so on.
380
CHAPTER 5
Inverse, Exponential, and Logarithmic Functions
Summary Exercises on Functions: Domains, Defining Equations, and Composition Finding the Domain of a Function: A Summary • Determining Whether an Equation Defines y as a Function of x • Composite Functions and Their Domains
The authors thank Professor Mohammad Maslam of Georgia Perimeter College for his suggestion to include this summary.
Finding the Domain of a Function: A Summary To find the domain of a function, given the equation that defines the function, remember that the value of x input into the equation must yield a real number for y when the function is evaluated. For the functions studied so far in this book, there are three cases to consider when determining domains. 1. No input value can lead to 0 in a denominator, because division by 0 is undefined. 2. No input value can lead to an even root of a negative number, because this situation does not yield a real number. 3. No input value can lead to the logarithm of a negative number or 0, because this situation does not yield a real number. Unless domains are otherwise specified, we can determine domains as follows: The domain of a polynomial function will be all real numbers. ■ The domain of an absolute value function will be all real numbers for which the expression inside the absolute value bars (the argument) is defined. ■ If a function is defined by a rational expression, the domain will be all real numbers for which the denominator is not 0. ■ The domain of a function defined by a radical with even root index is all numbers that make the radicand greater than or equal to 0; if the root index is odd, the domain is the set of all real numbers for which the radicand is itself a real number. ■ For an exponential function with constant base, the domain is the set of all real numbers for which the exponent is a real number. ■ For a logarithmic function, the domain is the set of all real numbers that make the argument of the logarithm greater than 0. ■
Determining Whether an Equation Defines y as a Function of x For y to be a function of x, it is necessary that every input value of x in the domain leads to one and only one value of y. To determine whether an equation such as x  y3 = 0
or x  y 2 = 0
represents a function, solve the equation for y. In the first equation above, doing so leads to y = 2 3 x. Notice that every value of x in the domain (that is, all real numbers) leads to one and only one value of y. So in the first equation, we can write y as a function of x. However, in the second equation above, solving for y leads to y = ⫾2x. If we let x = 4, for example, we get two values of y:  2 and 2. Thus, in the second equation, we cannot write y as a function of x.
Summary Exercises on Functions: Domains, Defining Equations, and Composition
381
Composite Functions and Their Domains EXAMPLE 1
Determining Composite Functions and Their Domains Given that ƒ(x) = 2x and g(x) = 4x + 2, find each of the following. (a) (ƒ ⴰ g)(x) and its domain (b) ( g ⴰ ƒ)(x) and its domain Solution (a) (ƒ ⴰ g)(x) = ƒ(g (x)) = ƒ(4x + 2) = 24x + 2 The domain and range of g are both the set of all real numbers, ( q, q). However, the domain of ƒ is the set of all nonnegative real numbers, 30, q). Therefore g (x), which is defined as 4x + 2, must be greater than or equal to zero: 4x + 2 Ú 0 4x Ú  2 1 x Ú  . 2
The radicand must be nonnegative.
Solve the inequality. Subtract 2. Divide by 4.
Therefore, the domain of ƒ ⴰ g is C  12, q B .
(b) (g ⴰ ƒ)(x) = g(ƒ(x)) = g A 2x B = 4 2x + 2
The domain and range of ƒ are both the set of all nonnegative real numbers, 30, q). The domain of g is the set of all real numbers, ( q, q). Therefore, the domain of g ⴰ ƒ is [0, q).
EXAMPLE 2
Given that ƒ(x) =
Determining Composite Functions and Their Domains 6 x  3
and g (x) = 1x , find each of the following.
(a) (ƒ ⴰ g)(x) and its domain
(b) (g ⴰ ƒ)(x) and its domain
Solution 1 (a) (ƒ ⴰ g)(x) = ƒ(g (x)) = ƒ a b x = =
6 1 x
Note that this is meaningless if x = 0.
 3
6x 1  3x
Multiply the numerator and denominator by x.
The domain of g is all real numbers except 0, which makes g (x) undefined. The domain of ƒ is all real numbers except 3. The expression for g (x), therefore, cannot equal 3; we determine the value that makes g (x) = 3 and exclude it from the domain of ƒ ⴰ g. 1 = 3 x 1 = 3x 1 x = 3
The solution must be excluded. Multiply by x. Divide by 3.
Therefore, the domain of ƒ ⴰ g is the set of all real numbers except 0 and 13, written in interval notation as ( q, 0) ´ A 0, 13 B ´ A 13, q B . (continued)
382
Inverse, Exponential, and Logarithmic Functions
CHAPTER 5
(b) (g ⴰ ƒ)(x) = g(ƒ(x)) = g a
1
= =
6 b x  3 Note that this is meaningless if x = 3.
6 x  3
x  3 6
Simplify the complex fraction.
The domain of ƒ is all real numbers except 3, and the domain of g is all real numbers except 0. The expression for ƒ(x), which is x 6 3, is never zero, since the numerator is the nonzero number 6. Therefore, the domain of g ⴰ ƒ is the set of all real numbers except 3, written ( q, 3) ´ (3, q).
EXERCISES Find the domain of each function. Write answers using interval notation. 1. ƒ(x) = 3x  6 x + 2 x  6 x2 + 7 7. ƒ(x) = 2 x  9 4. ƒ(x) =
2. ƒ(x) = 22x  7 5. ƒ(x) =
x2
3. ƒ(x) =  x + 4 
2 + 7
6. ƒ(x) = 2x 2  9
8. ƒ(x) = 2 3 x 3 + 7x  4
10. ƒ(x) = log
x + 7 x  3
11. ƒ(x) = 2x 2  7x  8
13. ƒ(x) =
1  x + 7
14. ƒ(x) =
2x 2
16. ƒ(x) = ln  x 2  5 
19. ƒ(x) =
A
x3
1  1
9. ƒ(x) = log 5(16  x 2) 12. ƒ(x) = 21>x
x 2  25 x + 5
15. ƒ(x) = 2x 3  1
2+x+4
17. ƒ(x) = e x
20. ƒ(x) = 3
A
x3
18. ƒ(x) =
x3  1 x2  1
1  8
21. ƒ(x) = ln(x 2 + 1)
x + 2 2 b x  3
24. ƒ(x) = 2 12 (4  x)2(x + 3)
22. ƒ(x) = 2(x  3)(x + 2)(x  4)
23. ƒ(x) = log a
25. ƒ(x) = e 1>x
26. ƒ(x) =
28. ƒ(x) = 2  x 2  9
29. ƒ(x) = 2 4 16  x 4
30. ƒ(x) = 2 3 16  x 4
32. ƒ(x) = 2 55  x
33. ƒ(x) =  25  x 
35. ƒ(x) = log `
36. ƒ(x) = 6x
31. ƒ(x) =
34. ƒ(x) =
x 2  2x  63 A x 2 + x  12 1
Ax  3 2  25
37. ƒ(x) = 62x 39. ƒ(x) = ln a
3 b (x + 2)(x  6)
1  x2  7 
27. ƒ(x) = x 100  x 50 + x 2 + 5
1 ` 4  x
29
38. ƒ(x) = 623 x
2  25
40. ƒ(x) =
2 log x
Summary Exercises on Functions: Domains, Defining Equations, and Composition
383
Determine which one of the choices A, B, C, or D is an equation in which y can be written as a function of x. 41. A. 3x + 2y = 6
B. x = 2 y 
C. x =  y + 3 
D. x 2 + y 2 = 9
42. A. 3x 2 + 2y 2 = 36
B. x 2 + y  2 = 0
C. x   y  = 0
D. x = y 2  4
43. A. x = 2y 2
B. x = log y 2
C. x 3 + y 3 = 5
D. x =
B. x = 5y 2  3
C.
B. x = ln( y + 1)2
C. 2x =  y + 1 
D. 2 4 x = y2
46. A. e y = x
B. e y + 2 = x
C. e  y = x
D. 10 y + 2  = x
1 y2
B. x + 2 =
44. A.
y2 x2 + = 1 4 4 2  y y + 3
45. A. x = 2
47. A. x 2 =
48. A.  x  =  y  49. A.
1 y2
B. x =  y 2 
y2 x2 = 1 4 9
B.
50. A. y 2  2(x + 2)2 = 0
y2 x2 = 1 4 9
B. y  2(x + 2)2 = 0
y2 x2 = 1 4 9
C. 3x = C. x = C.
y2
1 + 3
D. x = 10y
1 y4
D. 2x =
1 y
1 y3
D. x 4 + y 4 = 81
y x  = 0 4 9
D.
C. y 6  2(x + 1)2 = 0
y2 x2 = 0 4 9
D. y 4  2x 2 = 0
Given functions ƒ and g, find (a) (ƒ ⴰ g)(x) and its domain, and (b) (g ⴰ ƒ)(x) and its domain. 51. ƒ(x) =  6x + 9, g (x) = 5x + 7
52. ƒ(x) = 8x + 12, g (x) = 3x  1
53. ƒ(x) = 2x, g (x) = x + 3
54. ƒ(x) = 2x, g (x) = x  1
55. ƒ(x) = x 3, g (x) = x 2 + 3x  1
56. ƒ(x) = x + 2, g (x) = x 4 + x 2  3x  4
57. ƒ(x) = 2x  1, g (x) = 3x
58. ƒ(x) = 2x  2, g (x) = 2x
59. ƒ(x) =
2 , g (x) = x + 1 x
61. ƒ(x) = 2x + 2, g (x) = 63. ƒ(x) = 2x, g (x) = 65. ƒ(x) =
1 x + 5
1 1 , g (x) = x  2 x
60. ƒ(x) = 1 x
4 , g (x) = x + 4 x
62. ƒ(x) = 2x + 4, g (x) = 64. ƒ(x) = 2x, g (x) = 66. ƒ(x) =
2 x
3 x + 6
1 1 , g (x) = x + 4 x
67. ƒ(x) = log x, g (x) = 2x
68. ƒ(x) = ln x, g (x) = 2 4x
69. ƒ(x) = e x, g (x) = 2x
70. ƒ(x) = e x, g (x) = 2 3x
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CHAPTER 5
5
Inverse, Exponential, and Logarithmic Functions
SUMMARY
KEY TERMS & SYMBOLS
5.1 Inverse Functions onetoone function contrapositive inverse function ƒ 1 ( ƒinverse)
KEY CONCEPTS ONETOONE FUNCTION A function ƒ is a onetoone function if, for elements a and b from the domain of ƒ, aⴝb
implies ƒ(a) ⴝ ƒ(b).
HORIZONTAL LINE TEST If every horizontal line intersects the graph of a function at no more than one point, then the function is onetoone. INVERSE FUNCTION Let ƒ be a onetoone function. Then g is the inverse function of ƒ and ƒ is the inverse function of g if ( ƒ ⴰ g)(x) ⴝ x for every x in the domain of g, ( g ⴰ ƒ )(x) ⴝ x for every x in the domain of ƒ.
and
FINDING THE EQUATION OF THE INVERSE OF y ⴝ ƒ(x) For a onetoone function ƒ defined by an equation y = ƒ(x), find the defining equation of the inverse as follows. (You may need to replace ƒ(x) with y first. Any restrictions on x and y should be considered.) Step 1 Step 2 Step 3
Interchange x and y. Solve for y. Replace y with ƒ 1 (x).
IMPORTANT FACTS ABOUT INVERSES 1. If ƒ is onetoone, then ƒ 1 exists. 2. The domain of ƒ is equal to the range of ƒ 1, and the range of ƒ is equal to the domain of ƒ 1. 3. If the point (a, b) lies on the graph of ƒ, then (b, a) lies on the graph of ƒ 1. The graphs of ƒ and ƒ 1 are reflections of each other across the line y = x.
5.2 Exponential Functions
EXPONENTIAL FUNCTION If a 7 0 and a Z 1, then ƒ(x) ⴝ a x is the exponential function with base a.
exponential function the number e
CHARACTERISTICS OF EXPONENTIAL FUNCTIONS ƒ(x) ⴝ a x, ■ ■ ■
This function is increasing and continuous on its entire domain, ( q, q). The xaxis is the horizontal asymptote as x :  q. The graph passes through the points ( 1, a 1), (0, 1), and (1, a).
ƒ(x) ⴝ a x, ■ ■ ■
a>1
01 ■ This function is increasing and continuous on its entire domain, (0, q). ■ The yaxis is the vertical asymptote as x : 0 from the right. ■ The graph passes through the points (a 1,  1), (1, 0), and (a, 1). ƒ(x) ⴝ log a x, ■ ■ ■
5.5 Exponential and Logarithmic Equations and Inequalities
00
SOUND INTENSITY
d ⴝ 10 log
AMORTIZATION PAYMENTS
I I0
Rⴝ
P 1 ⴚ (1 ⴙ i )ⴚn d c i
CHAPTER 5
5
387
Review Exercises
REVIEW EXERCISES
Determine whether each function is onetoone. Assume that the graphs shown are comprehensive. 1.
2.
y
3.
6.2
–9.4
9.4
4
–2
2
x
0
–6.2
4. ƒ(x) = 23x + 2
–4
5. ƒ(x) = 3x 2 + 1
6. ƒ(x) =  2x + 7 
For Exercises 7–12, consider the function defined by ƒ(x) = 2 3 2x  7. 7. What is the domain of ƒ?
9 9 11.
8. What is the range of ƒ?
9. Explain why ƒ 1 exists.
10. Find a formula for ƒ 1(x).
Graph both ƒ and ƒ  1 in a square viewing window, along with the line y = x. Describe how the graphs of ƒ and ƒ  1 are related.
12. Verify analytically that ( ƒ ⴰ ƒ 1)(x) = x
and
( ƒ 1 ⴰ ƒ )(x) = x.
Concept Check Match each equation with the graph that most closely resembles its graph. Assume that a 7 1. 14. y = a x + 2 B. y
13. y = a x + 2 y A.
a2
3
3 x
0
x
0
16. y = a x + 2 D. y
15. y =  a x + 2 C. y
1
x
0
Concept Check Consider the exponential function defined by ƒ(x) = a x, graphed at the right. Answer the following based on the graph. 17. What is true about the value of a in comparison to 0 and 1?
18. What is the domain of ƒ?
19. What is the range of ƒ?
20. What is the value of ƒ(0)?
21. Graph y = ƒ 1 (x) by hand.
22. What expression defines ƒ 1 (x)?
x
0
y
f(x) = a x
0
x
Graph each exponential function. Give the domain and range. 1 x1 23. ƒ(x) = a b 2 26. Concept Check
24. ƒ(x) = e x + 2
25. ƒ(x) =  4x
If ƒ(x) = a x and 0 6 a 6 1, is ƒ increasing or decreasing on its domain?
Solve the equation in part (a) analytically. Then use a graph to solve the inequality in part (b). 1 2x 27. (a) a b = 2x + 3 8 1 2x Ú 2x + 3 (b) a b 8
28. (a) 3x = a (b) 3x
1 12x b 27
1 12x 6 a b 27
29. (a) 0.5x = 0.25x + 1
30. (a) 0.4x = 2.51  x
(b) 0.5x 7 0.25x + 1
(b) 0.4x 6 2.51  x
388
CHAPTER 5
Inverse, Exponential, and Logarithmic Functions
31. The graphs of y = x 2 and y = 2x have the points (2, 4) and (4, 16) in common. There is a third point in common whose coordinates can be approximated by using a graphing calculator. Find the coordinates, giving as many decimal places as your calculator will display. 32. Solve 3x = p by the intersectionofgraphs method. Express the solution to the nearest thousandth. Use a calculator to find an approximation for each logarithm to four decimal places. 33. log 58.3
34. log 0.00233
35. ln 58.3
36. log 2 0.00233
Evaluate each expression, giving an exact value. 37. log 13 1
38. ln e 16
39. log 5 512
40. 7log7 13
41. 3log3 5
42. log 4 64
Concept Check In Exercises 43–48, identify the corresponding graph in choices A–F for each function. 43. ƒ(x) = log 2 x
44. ƒ(x) = log 2 (2x)
1 45. ƒ(x) = log 2 a b x
x 46. ƒ(x) = log 2 a b 2
47. ƒ(x) = log 2 (x  1)
48. ƒ(x) = log 2 ( x)
A.
B.
C.
y 2
y
1
E.
y
1
x
1
1
D.
1
x
x
y
x 1
4
F.
y
y
1
1
x
x 1
1
Graph each logarithmic function. Give the domain and range. (Hint: See Exercises 23–25, as those exponential functions are closely related to these.) 49. g (x) = 1 + log 1/2 x
51. g (x) = log 4 ( x)
50. g (x) = ln(x  2)
52. Concept Check How do the functions in Exercises 49–51 relate to those in Exercises 23–25? 53. Concept Check What is the base of the logarithmic function whose graph contains the point (81, 4)?
54. Concept Check What is the base of the exponential func1 tion whose graph contains the point A  4, 16 B?
Write each logarithm as a sum, difference, or product of logarithms if possible. Assume variables represent positive numbers. 55. log 3
mn 5r
57. log 5 A x 2 y 4 2 5 m 3p B
56. log 2
27 15
58. log 7 A 7k + 5r 2 B
Solve the equation in part (a) analytically. Then use a graph to solve the inequality in part (b). 59. (a) log(x + 3) + log x = 1 (b) log(x + 3) + log x 7 1
60. (a) ln e ln x  ln(x  4) = ln 3 (b) ln e ln x  ln(x  4) … ln 3
61. (a) ln e ln 2  ln(x  1) = ln 5 (b) ln e ln 2  ln(x  1) Ú ln 5
Solve each equation. Express the solution set so that (a) solutions are in exact form and, if irrational, (b) solutions are approximated to the nearest thousandth. Support your solutions by using a calculator. 8 64. 102x  3 = 17 62. 8x = 32 63. = x 3 27
CHAPTER 5
1 3
Review Exercises
389
67. ln(6x)  ln(x + 1) = ln 4
65. e x + 1 = 10
66. log 64 x =
68. log 12(2x) + log 12(x  1) = 1
69. log 16 2x + 1 =
71. ln(ln (e x)) = ln 3
72. ln e x  ln e 3 = ln e 5
2 70. ln x + 3 ln 2 = ln a b x
1 4
73. 2x =  3
Solve each formula for the indicated variable. c 74. N = a + b ln a b, for c d
75. y = y0 e kt, for t
Use the xintercept method to estimate the solution(s) of each equation. Round to the nearest thousandth. 77. 2x = log 10 x
76. log 10 x = x  2 Solve each application.
84. Caloric Intake
79. Interest Rate To the nearest tenth, what interest rate, compounded annually, will produce $4613 if $3500 is left at this interest for 10 years? 80. Growth of an Account Find the number of years (to the nearest tenth) needed for $48,000 to become $58,344 at 5% interest compounded semiannually. 81. Growth of an Account Lateisha Shaw deposits $12,000 for 8 years in an account paying 5% compounded annually. She then leaves the money alone, with no further deposits, at 6% compounded annually for an additional 6 years. Approximate the total amount on deposit after the entire 14year period. 82. Growth of an Account Suppose that $2000 is invested in an account that pays 3% annually and is left for 5 years. (a) How much will be in the account if interest is compounded quarterly (4 times per year)? (b) How much will be in the account if interest is compounded continuously? (c) To the nearest tenth of a year, how long will it take the $2000 to triple if interest is compounded continuously? (Modeling) Solve each problem.
9 83.
78. x 2  3 = log x
Runway Length There is a mathematical relationship between an airplane’s weight x and the runway length required at takeoff. For some airplanes, the minimum runway length in thousands of feet may be modeled by L(x) = 3 log x, where x is measured in thousands of pounds. (Source: Haefner, L., Introduction to Transportation Systems, Holt, Rinehart and Winston.) (a) Graph L in the window 30, 504 by 30, 64. Interpret the graph. (b) If the weight of an airplane increases tenfold from 10,000 to 100,000 pounds, does the length of the required runway also increase by a factor of 10? Explain.
The function defined by
ƒ(x) = 280 ln(x + 1) + 1925 models the number of calories consumed daily by a person owning x acres of land in a developing country. Find the number of acres owned by someone whose average calorie intake is 2200 calories per day. (Source: Grigg, D., The World Food Problem, Blackwell Publishers.) 85. Pollutant Concentration The concentration of a pollutant, in grams per liter, in the east fork of the Big Weasel River is approximated by P(x) = 0.04e 4x, where x is the number of miles downstream from a paper mill where the measurement is taken. Find the following: (a) P(0.5) (b) P(1) (c) The concentration of the pollutant 2 miles downstream (d) The number of miles downstream where the concentration of the pollutant is 0.002 gram per liter. 86. Repetitive Skills A person learning certain skills involving repetition tends to learn quickly at first. Then, learning tapers off and approaches some upper limit. Suppose the number of symbols per minute that a textbook typesetter can produce is given by p(x) = 250  120(2.8)0.5x, where x is the number of months the typesetter has been in training. Find the following to the nearest whole number: (a) p(2) (b) p(10) (c) Graph y = p(x) in the window 30, 104 by 30, 3004, and support the answer to part (a). 87. Free Fall of a Skydiver a speed of
A skydiver in free fall travels at
v(t) = 176(1  e 0.18t ) feet per second after t seconds. How long will it take for the skydiver to attain a speed of 147 feet per second (100 mph)?
390
Inverse, Exponential, and Logarithmic Functions
CHAPTER 5
5
TEST
1. Match each equation in parts (a)–(d) with its graph from A–D. (a) y = log 1/2 x (b) y = e x 1 x (d) y = a b 2 (e) Which pairs of functions in parts (a)–(d) are inverses? (c) y = ln x
A.
B.
y
x
0
C.
D.
y
x
0
x
y
0
7. Use the properties of logarithms to write log
x
2. Consider the function defined by ƒ(x) =  2x  1 + 8. (a) Graph ƒ in the standard viewing window of your calculator. (b) Give the domain and range of ƒ. (c) Does the graph have an asymptote? If so, is it vertical or horizontal, and what is its equation? (d) Find the x and yintercepts analytically, and use the graph from part (a) to support your answers graphically. (e) Find ƒ 1(x).
A 18 B
2x  3
= 16x + 1 analytically.
4. Growth of an Account Suppose that $10,000 is invested at 3.5% for 4 years. Find the total amount present at the end of that period if interest is compounded (a) quarterly and (b) continuously.
9 5.
2y
as an
equivalent expression.
9. Consider the equation log 2 x + log 2(x + 2) = 3. (a) Solve the equation analytically. If there is an extraneous value, what is it? (b) To support the solution in part (a), we can graph y1 = log 2 x + log 2(x + 2)  3 and find the xintercept. Write an expression for y1, using the changeofbase rule with base 10, and graph the function. (c) Use the graph to solve the logarithmic inequality log 2 x + log 2(x + 2) 7 3. Solve each equation. Give the solution set (a) with an exact value and (b) with an approximation to the nearest thousandth. 10. 2e 5x + 2 = 8
3. Solve the equation
m 3n
8. Solve A = Pe rt for t.
y
0
6. Use a calculator to find an approximation of each logarithm to the nearest thousandth. (a) log 45.6 (b) ln 470 (c) log 3 769
One of your friends tells you, “I have no idea what an expression like log 5 27 really means.” Write an explanation of what it means, and tell how you can find an approximation for it with a calculator.
11. 62  x = 23x + 1
12. log(ln x) = 1 13. (Modeling) Population Growth ing according to the equation
A population is increas
y = 2e 0.02t, where y is in millions and t is in years. Match each question in parts (a)–(d) with one of the solution methods A–D. (a) How long will it take for the population to triple? (b) When will the population reach 3 million? (c) How large will the population be in 3 years? (d) How large will the population be in 4 months? A. Evaluate 2e 0.02(1/3). B. Solve 2e 0.02t = 3 # 2 for t. 0 . 02(3) C. Evaluate 2e . D. Solve 2e 0.02t = 3 for t. 14. (Modeling) Drug Level in the Bloodstream After a medical drug is injected into the bloodstream, it is gradually eliminated from the body. Graph each function in parts (a)–(d) on the interval 30, 104. Use 30, 5004 as the range of A(t). Use a graphing calculator to determine the function that best models the amount A(t) (in milligrams) of a drug remaining in the body after t hours if 350 milligrams were initially injected. (a) A(t) = t 2  t + 350 (b) A(t) = 350 log(t + 1) (c) A(t) = 350(0.75)t (d) A(t) = 100(0.95)t 15. (Modeling) Decay of Radium Radium 226 has a halflife of about 1600 years. An initial sample weighs 2 grams. (a) Find a formula for the decay function. (b) Find the amount left after 9600 years. (c) Find the time for the initial amount to decay to 0.5 gram.
6 Analytic Geometry ASTEROIDS CAN BE likened to interplanetary garbage. These large rocks hurtle through space, following an orbit around the sun. When an asteroid strikes a planet or moon, the destruction can be devastating. In June 2004, astronomers working at Kitt Peak National Observatory near Tucson, Arizona, discovered an asteroid onequarter mile in diameter that appeared to be heading toward Earth. In December, the asteroid, named 2004 MN4, rose from level 2 to level 4 on the Torino probabilityofimpact scale, making it the first nearEarth asteroid to reach the latter level. The projected path showed that 2004 MN4 could hit Earth on April 13, 2029, releasing energy equivalent to a 1000to1500megaton bomb. After exhaustive research and refined calculations of the asteroid’s orbit, astronomers determined in January 2005 that 2004 MN4 would not hit Earth, but would likely pass within 41,000 miles of it. Paths of planets, asteroids, comets, and satellites launched into space can be modeled by equations and graphs of conic sections. (See Section 6.2, Exercise 97.) We study conic sections in this chapter. Source: http://impact.arc.nasa.gov
CHAPTER OUTLINE 6.1 Circles and Parabolas 6.2 Ellipses and Hyperbolas 6.3 Summary of the Conic Sections 6.4 Parametric Equations
From Chapter 6 of A Graphical Approach to Precalculus with Limits: A Unit Circle Approach, Fifth Edition. John Hornsby, Margaret Lial, Gark K. Rockswold. Copyright © 2011 by Pearson Education, Inc. All rights reserved.
392
CHAPTER 6
6.1
Analytic Geometry
Circles and Parabolas
Conic Sections • Equations and Graphs of Circles • Equations and Graphs of Parabolas • Translations of Parabolas • An Application of Parabolas
Conic Sections Parabolas, circles, ellipses, and hyperbolas form a group of curves known as the conic sections, because they are the result of intersecting a cone with a plane. FIGURE 1 illustrates these curves, which all can be defined mathematically by using the distance formula from Section 1.1, d(P, R) = 2(x2  x1) 2 + ( y2  y1) 2. Axis
Axis
Distance formula
Axis
Hyperbola Ellipse
Circle
Parabola FIGURE 1
If the plane intersects the cone at the vertex, the intersections illustrated in FIGURE 1 are reduced respectively to a point, a line, and two intersecting lines. These are called degenerate conic sections.
Equations and Graphs of Circles Circle
y
A circle is a set of points in a plane that are equidistant from a fixed point. The distance is called the radius of the circle, and the fixed point is called the center. √(x – h) 2 + ( y – k) 2 = r (x – h) 2 + ( y – k) 2 = r 2 (x, y)
Suppose a circle has center (h, k) and radius r 7 0. See FIGURE 2. Then the distance between the center (h, k) and any point (x, y) on the circle must equal r. Thus, an equation of the circle is
r (h, k) x
0
FIGURE 2
2(x  h) 2 + ( y  k) 2 = r (x 
h) 2
+ (y 
k) 2
=
r 2.
Distance formula Square both sides.
6.1
Circles and Parabolas
393
Center–Radius Form of the Equation of a Circle The center–radius form of the equation of a circle with center (h, k) and radius r is (x ⴚ h)2 ⴙ ( y ⴚ k)2 ⴝ r 2. Notice that a circle is the graph of a relation that is not a function, since it does not pass the vertical line test. EXAMPLE 1
Finding the Equation of a Circle Find the center–radius form of the equation of a circle with radius 6 and center ( 3, 4). Graph the circle by hand. Give the domain and range of the relation. Solution Using the center–radius form with h =  3, k = 4, and r = 6, we find that the equation of the circle is 3x  ( 3)42 + ( y  4) 2 = 62, or
(x + 3) 2 + ( y  4) 2 = 36.
The graph is shown in FIGURE 3. Because the center is ( 3, 4) and the radius is 6, the circle must pass through the four points ( 3 ⫾ 6, 4) and ( 3, 4 ⫾ 6), as illustrated in the table. The domain is 3  9, 34 and the range is 3 2, 104. y
x y –9 4
(3, 10) 12
3 4 3 2 3 10
8 (9, 4)
(3, 4)
4
(3, 4)
y
12 (0, r) x2 + y2 = r 2 r 0
(–r, 0)
(r, 0)
(x
8 (3, 2) 4
+ 3) 2 + (y – 4) 2
4
x
36
x
FIGURE 3
(0, –r)
If a circle has center at the origin (0, 0), then its equation is found by using h = 0 and k = 0 in the center–radius form. See FIGURE 4.
FIGURE 4
Equation of a Circle with Center at the Origin A circle with center (0, 0) and radius r has equation x 2 ⴙ y 2 ⴝ r 2.
y
x2 + y2 = 9
Range
–3
EXAMPLE 2
3
0
x 3
–3 (x, y)
Domain FIGURE 5
3
Finding the Equation of a Circle with Center at the Origin Find the equation of a circle with center at the origin and radius 3. Graph the relation, and state the domain and range. Solution Using the form x 2 + y 2 = r 2 with r = 3, we find that the equation of the circle is x 2 + y 2 = 9. See FIGURE 5. The domain and range are both 3 3, 34.
394
CHAPTER 6
Analytic Geometry
As mentioned in Section 4.4, a graphing calculator in function mode cannot directly graph a circle. We must solve the equation of the circle for y, obtaining two functions y1 and y2. The union of these two graphs is the graph of the circle. TECHNOLOGY NOTE
GCM
To obtain an undistorted graph of a circle in function mode on a graphing calculator screen, a square viewing window must be used. For example, the graph below of x 2 + y 2 = 9 in a rectangular window looks like an ellipse.
EXAMPLE 3
Graphing Circles Use a graphing calculator to graph each circle in a square viewing window. (a) x 2 + y 2 = 9 (b) (x + 3) 2 + ( y  4) 2 = 36 Solution In both cases, first solve for y. Recall that, by the square root property, if k 7 0, then y 2 = k has two real solutions: 2k and  2k.
–5
x2 + y2 = 9
(a)
5
Remember both the positive and negative square roots. 5
y2 = 9  x2
Subtract x 2.
y = ⫾29  x 2
Take square roots.
Graph y1 = 29  x 2 and y2 =  29  x 2. See FIGURE 5 on the preceding page.
–5
y1 = √9 – x2
FIGURE 6,
y1 = 4 + √36 – (x + 3)2 10.4
3.1
–4.7
TECHNOLOGY NOTE If we graph the circle in FIGURE 6 without using a decimal window, we may get a figure similar to the one shown below, where the two graphs do not meet at the points ( 3, 0) and (3, 0). In some cases, a decimal window does not correct this problem. Although the technology is deceiving, mathematically this is a complete circle.
and compare with
4.7 –14.4
4.4
–3.1
–2
y2 = – √9 – x2
y 2 = 4 – √36 – (x + 3)2
FIGURE 6
FIGURE 7
(b) (x + 3) 2 + ( y  4) 2 = 36 ( y  4) 2 = 36  (x + 3) 2
Subtract ( x + 3)2.
y  4 = ⫾236  (x + 3) 2 y = 4 ⫾ 236  (x + 3) 2
4
Take square roots. Add 4.
The two functions graphed in FIGURE 7 are –6
6
–4
y1 = 4 + 236  (x + 3) 2 and
y2 = 4  236  (x + 3) 2.
Suppose that we start with the center–radius form of the equation of a circle and rewrite it.
Remember the middle term when squaring.
(x  h) 2 + ( y  k) 2 = r 2 x 2  2xh + h 2 + y 2  2yk + k 2  r 2 = 0
x 2 + y 2 + ( 2h)x + ( 2k)y + (h 2 + k 2  r 2) = 0 c
d
Center–radius form Square each binomial; subtract r 2. Properties of real numbers
e
If r 7 0, then the graph of this equation is a circle with center (h, k) and radius r, as seen earlier. This is the general form of the equation of a circle.
6.1
Circles and Parabolas
395
General Form of the Equation of a Circle For real numbers c, d, and e, the equation x 2 ⴙ y 2 ⴙ cx ⴙ dy ⴙ e ⴝ 0 can have a graph that is a circle, a point, or is empty.
Starting with an equation in this general form, we can work in reverse by completing the square to get an equation of the form (x  h) 2 + ( y  k) 2 = m, Looking Ahead to Calculus
for some number m.
There are three possibilities for the graph, based on the value of m:
The circle x 2 + y 2 = 1 is called the unit circle. It is important in interpreting the trigonometric or circular functions in calculus.
1. If m 7 0, then r 2 = m, and the equation represents a circle with radius 2m. 2. If m = 0, then the equation represents the single point (h, k). 3. If m 6 0, then no points satisfy the equation and the graph is empty.
EXAMPLE 4
Finding the Center and Radius by Completing the Square Decide whether each equation has a circle as its graph. Sketch the graphs. (a) x 2  6x + y 2 + 10y + 25 = 0 (b) x 2 + 10x + y 2  4y + 33 = 0 (c) 2x 2 + 2y 2  6x + 10y = 1 Solution (a)
x 2  6x + y 2 + 10y + 25 = 0
(x 2  6x y
2
2 1 c ( 6) d = ( 3) 2 = 9 2 2
4
6
8
–2
(3, –5)
) =  25 and
Add 9 and 25 to each side.
Add  25.
2 1 c (10) d = 52 = 25 2
(x 2  6x + 9) + ( y 2 + 10y + 25) =  25 + 9 + 25
x
0
–4
) + ( y 2 + 10y
(x  3) 2 + ( y + 5) 2 = 9
Complete the square. Factor; add.
–6 –8
Since 9 7 0, the equation represents a circle with center at (3,  5) and radius 3. See FIGURE 8.
(x – 3) 2 + ( y + 5) 2 = 9 FIGURE 8
(b)
x 2 + 10x + y 2  4y + 33 = 0 (x 2 + 10x
) + ( y 2  4y
) =  33
2 1 c (10) d = 25 and 2
Add  33.
2 1 c ( 4) d = 4 2
(x 2 + 10x + 25) + ( y 2  4y + 4) =  33 + 25 + 4 (x + 5) 2 + ( y  2) 2 =  4
Complete the square. Factor; add.
Because  4 6 0, there are no ordered pairs (x, y), with x and y both real numbers, satisfying the equation. The graph is empty; it contains no points. (continued)
396
Analytic Geometry
CHAPTER 6
(c) 2x 2 + 2y 2  6x + 10y = 1 Coefficients of the x 2and y 2terms must be 1.
To complete the square, make the coefficients of the x 2 and y 2terms 1. 2(x 2  3x) + 2( y 2 + 5y) = 1 2 a x 2  3x +
y
(x – 32 ) 2 + (y + 52 ) 2 = 9 –4 –2
2
2
4
2 ax 
x
3 2 5 2 b + 2 a y + b = 18 2 2
ax 
( 32 , – 52 )
–6
9 25 9 25 b + 2 a y 2 + 5y + b = 1 + 2a b + 2a b 4 4 4 4
–8
Be careful here.
Complete the square.
6
0
Group the terms: factor out 2.
3 2 5 2 b + ay + b = 9 2 2
Factor; simplify on the right.
Divide each side by 2.
The equation is a circle with center at A 32,  52 B and radius 3. See FIGURE 9.
FIGURE 9
Equations and Graphs of Parabolas Like a circle, the definition of a parabola is based on distance.
y
Axis of symmetry
Focus
Parabola x
Vertex
A parabola is a set of points in a plane equidistant from a fixed point and a fixed line. The fixed point is called the focus, and the fixed line the directrix, of the parabola.
Directrix
FIGURE 10
y
Equal distance F(0, c) Focal chord y = – c V(0, 0)
P(x, y) x
D(x, – c)
As shown in FIGURE 10, the axis of symmetry of a parabola passes through the focus and is perpendicular to the directrix. The vertex is the midpoint of the line segment joining the focus and directrix on the axis. We can find an equation of a parabola from the preceding definition. Let the directrix be the line y =  c and the focus be the point F with coordinates (0, c). See FIGURE 11. To find the equation of the set of points that are the same distance from the line y =  c and the point (0, c), choose one such point P and give it coordinates (x, y). By the definition of a parabola, d(P, F) and d(P, D) must be equal, so using the distance formula gives the following result.
d(P, F) = d(P, D) for all P on the parabola. FIGURE 11
d(P, F) = d(P, D)
2(x  0) 2 + ( y  c) 2 = 2(x  x) 2 + 3y  ( c)42 2x 2 + ( y  c) 2 = 2( y + c) 2 x2
+ (y 
c) 2
= (y +
c) 2
x 2 + y 2  2yc + c 2 = y 2 + 2yc + c 2 Remember the middle x 2 = 4cy term when squaring a binomial.
See FIGURE 11. Distance formula Simplify. Square each side. Square both binomials. Subtract y 2 and c 2; add 2yc.
This discussion is summarized in the box on the next page.
6.1
Circles and Parabolas
397
Parabola with a Vertical Axis and Vertex (0, 0) The parabola with focus (0, c) and directrix y =  c has equation x 2 ⴝ 4cy. The parabola has vertex (0, 0), vertical axis x = 0, and opens upward if c 7 0 or downward if c 6 0. y
y
x2 = 4cy, c>0
F(0, c)
y = –c x
0
x
0
y = –c x2 = 4cy, c0
F(c, 0) y2 = 4cx, c0 0
V(h, k)
(y – k)2 = 4c(x – h), c>0 F(h + c, k)
F(h, k + c) x
x
0
V(h, k) y=k–c
x=h–c
Vertical axis
Horizontal axis
EXAMPLE 7
Graphing a Parabola with Vertex ( h, k) Graph the parabola given by the equation x =  18 ( y + 3) 2 + 2. Label the vertex, focus, and directrix. Solution Rewrite the equation in the form ( y  k) 2 = 4c(x  h). 1 x =  ( y + 3) 2 + 2 8 1 x  2 =  ( y + 3) 2 8  8(x  2) = ( y + 3) 2 ( y + 3) 2 =  8(x  2)
Subtract 2. Multiply by  8. Rewrite.
The vertex is (2,  3). Since 4c =  8, we have c =  2, and the parabola opens to the left. The focus is 2 units to the left of the vertex, and the directrix is 2 units to the right of the vertex. Thus, the focus is (0,  3) and the directrix is x = 4. See FIGURE 15.
x y –9
–2.5 0 2 0 –2.5
Choose values for y and calculate x in the original equation.
x
–7 –3 1 3
 81 (y + 3) 2 + 2 y
4 8
4 2 2
F(0, 3)
6 10
x
6
V(2, 3)
x 4
FIGURE 15
400
Analytic Geometry
CHAPTER 6 y
EXAMPLE 8
Writing an Equation of a Parabola Write an equation for the parabola with focus (2, 2) and vertex ( 1, 2), and graph it. Give the domain and range.
6
( y – 2) 2 = 12(x + 1) V (–1, 2) –4
Solution Since the focus is to the right of the vertex, the axis is horizontal and the parabola opens to the right. See FIGURE 16. The distance between the focus and the vertex is 2  ( 1) = 3, so c = 3. The equation of the parabola is
F (2, 2) 0
–2
x 2
4
( y  k) 2 = 4c(x  h) ( y  2) 2 = 4(3)(x + 1) ( y  2) 2 = 12(x + 1).
FIGURE 16
Parabola with horizontal axis Substitute for c, h, and k. Multiply.
From FIGURE 16, the domain is 3 1, q) and the range is ( q, q). EXAMPLE 9
Completing the Square to Graph a Horizontal Parabola 2 Graph x =  2y + 4y  3. Give the domain and range. Analytic Solution x =  2y 2 + 4y  3 x + 3 =  2y 2 + 4y x + 3 =  2( y 2  2y) x + 3  2 =  2( y 2  2y + 1) x + 1 =  2( y  1) 2
Add 3. Factor out  2. Complete the square. Add; factor the trinomial.
1 ( y  1) 2 =  (x + 1) (*) 2
Multiply by  12; rewrite.
The parabola opens to the left, is narrower than the graph of x = y 2, and has vertex ( 1, 1). The domain is ( q,  14 and the range is ( q, q). See FIGURE 17.
Graphing Calculator Solution To obtain a calculator graph, we begin by solving the equation ( y  1) 2 =  12 (x + 1) (*) from the analytic solution for y. y  1 = ⫾
x + 1 B 2
y = 1 ⫾
x + 1 B 2
y
y
3 3 1
0 2 1
Add 1.
Graphing these two equations as y1 and y2 results in FIGURE 18. y1 = 1 +
x
Take square roots.
x+1 –2
5
(–1, 1) –8.6
y=1 –3
x=
–2y 2
–1 0
+ 4y – 3
( y – 1) 2 = –
1 2
(x + 1)
3.6
x
y2 = 1 –
FIGURE 17
x+1 –2
–3
FIGURE 18
An Application of Parabolas Focus
Parabolic reflector FIGURE 19
Axis
The geometric properties of parabolas lead to many practical applications. For example, if a light source is placed at the focus of a parabolic reflector, as in FIGURE 19, light rays reflect parallel to the axis, making a spotlight or flashlight. The process also works in reverse. Light rays from a distant source come in parallel to the axis and are reflected to a point at the focus. This use of parabolic reflection is seen in the satellite dishes used to pick up signals from communications satellites.
6.1
Focus
Modeling the Reflective Property of Parabolas The Parkes radio telescope has a parabolic dish shape with diameter 210 feet and depth 32 feet. Because of this parabolic shape, distant rays hitting the dish are reflected directly toward the focus. A cross section of the dish is shown in FIGURE 20. (Source: Mar, J. and H. Liebowitz, Structure Technology for Large Radio and Radar Telescope Systems, MIT Press.) (a) Determine an equation describing this cross section by placing the vertex at the origin with the parabola opening upward.
32 ft
FIGURE 20
(b) The receiver must be placed at the focus of the parabola. How far from the vertex of the parabolic dish should the receiver be located? Solution (a) Locate the vertex at the origin, as shown in FIGURE 21. The form of the parabola is x 2 = 4cy. The parabola must pass through the point A 210 2 , 32 B = (105, 32). Thus,
y
50
(105, 32)
x 2 = 4cy (105) 2 = 4c(32) 11,025 = 128c 11,025 c = . 128
25 –100
401
EXAMPLE 10
210 ft
(–105, 32)
Circles and Parabolas
x
0
–50
50
100
FIGURE 21
Vertical parabola Let x = 105 and y = 32. Multiply. Solve for c.
The cross section can be modeled by the equation x 2 = 4cy 11,025 x2 = 4a by 128 11,025 x2 = y. 32
Substitute for c. Simplify.
(b) The distance between the vertex and the focus is c. In part (a), we found that c = 11,025 128 L 86.1, so the receiver should be located at (0, 86.1), or 86.1 ft above the vertex.
6.1
EXERCISES
Checking Analytic Skills Match each equation in Column I with the appropriate description in Column II. Do not use a calculator. I 1. x = 2y 2
II A. Circle; center (3,  4); radius 5
2. y = 2x 2
B. Parabola; opens left
3. x 2 =  3y
C. Parabola; opens upward
4. y 2 =  3x
D. Circle; center ( 3, 4); radius 5
5. x 2 + y 2 = 5
E. Parabola; opens right
6. (x  3) 2 + ( y + 4) 2 = 25
F. Circle; center (0, 0); radius 25
7. (x + 3) 2 + ( y  4) 2 = 25
G. No points on its graph
8. x 2 + y 2 =  4
H. Parabola; opens downward
402
Analytic Geometry
CHAPTER 6
Find the center–radius form for each circle satisfying the given conditions. 9. Center (1, 4); radius 3
10. Center ( 2, 5); radius 4
11. Center (0, 0); radius 1
12. Center (0, 0); radius 5
2 4 3 13. Center a ,  b; radius 3 5 7
12 1 1 14. Center a  ,  b; radius 2 4 5
15. Center ( 1, 2); passing through (2, 6)
16. Center (2,  7); passing through ( 2,  4)
17. Center ( 3,  2); tangent to the xaxis (Hint: Tangent to means touching at one point.)
18. Center (5,  1); tangent to the yaxis
9 19.
Describe the graph of the equation (x 
3) 2
+ (y 
3) 2
= 0.
9 20.
Describe the graph of the equation (x  3) 2 + ( y  3) 2 =  1.
We can find an equation of a circle if we know the coordinates of the endpoints of a diameter of the circle. First, find the midpoint of the diameter, which is the center of the circle. Then find the radius, which is the distance from the center to either endpoint of the diameter. Finally use the center–radius form to find the equation. Find the center–radius form for each circle having the given endpoints of a diameter. 21. ( 1, 3) and (5,  9)
22. ( 4, 5) and (6,  9)
23. ( 5,  7) and (1, 1)
24. ( 3,  2) and (1,  4)
25. ( 5, 0) and (5, 0)
26. (0, 9) and (0,  9)
9 27.
Explain why, in Exercises 21–26, either endpoint can be used (along with the coordinates of the center) to find the radius.
28. Concept Check diameter.
Refer to any of Exercises 21–26, and show that the radius is half the length of the
Graph each circle by hand if possible. Give the domain and range. 29. x 2 + y 2 = 4
30. x 2 + y 2 = 36
31. x 2 + y 2 = 0
32. x 2 + y 2 =  9
33. (x  2) 2 + y 2 = 36
34. (x + 2) 2 + ( y  5) 2 = 16
35. (x  5) 2 + ( y + 4) 2 = 49
36. (x  4) 2 + ( y  3) 2 = 25
37. (x + 3) 2 + ( y + 2) 2 = 36
38. x 2 + ( y  2) 2 = 9
39. (x  1) 2 + ( y + 2) 2 = 16
40. (x + 1) 2 + y 2 + 2 = 0
Graph each circle using a graphing calculator. Use a square viewing window. Give the domain and range. 41. x 2 + y 2 = 81
42. x 2 + ( y + 3) 2 = 49
43. (x  3) 2 + ( y  2) 2 = 25
44. (x + 2) 2 + ( y + 3) 2 = 36
Decide whether each equation has a circle as its graph. If it does, give the center and radius. 45. x 2 + 6x + y 2 + 8y + 9 = 0
46. x 2 + 8x + y 2  6y + 16 = 0
47. x 2  4x + y 2 + 12y =  4
48. x 2  12x + y 2 + 10y =  25
49. 4x 2 + 4x + 4y 2  16y  19 = 0
50. 9x 2 + 12x + 9y 2  18y  23 = 0
51. x 2 + 2x + y 2  6y + 14 = 0
52. x 2 + 4x + y 2  8y + 32 = 0
53. x 2  2x + y 2 + 4y + 9 = 0
54. 4x 2 + 4x + 4y 2  4y  3 = 0
55. 9x 2 + 36x + 9y 2 =  32
56. 9x 2 + 9y 2 + 54y =  72
6.1
Circles and Parabolas
403
Concept Check Each equation in Exercises 57–64 defines a parabola. Without actually graphing, match the equation in Column I with its description in Column II. II
I 57. (x 
4) 2
= y + 2
58. (x 
2) 2
= y + 4
59. y + 2 =  (x  4) 2
60. y =  (x  2) 2  4
61. ( y  4) 2 = x + 2
62. ( y  2) 2 = x + 4
63. x + 2 =  ( y  4) 2
64. x =  ( y  2) 2  4
A. Vertex (2,  4); opens downward B. Vertex (2,  4); opens upward C. Vertex (4,  2); opens downward D. Vertex (4,  2); opens upward E. Vertex ( 2, 4); opens left F. Vertex ( 2, 4); opens right G. Vertex ( 4, 2); opens left H. Vertex ( 4, 2); opens right
65. Concept Check For the graph of (x  h) 2 = 4c( y  k), in what quadrant is the vertex if: (a) h 6 0, k 6 0; (b) h 6 0, k 7 0; (c) h 7 0, k 6 0; (d) h 7 0, k 7 0? 66. Concept Check Repeat parts (a)–(d) of Exercise 65 for the graph of ( y  k) 2 = 4c(x  h). Give the focus, directrix, and axis of each parabola. 67. x 2 = 16y
68. x 2 = 4y
1 x 16
72. y 2 = 
71. y 2 =
1 x 32
1 69. x 2 =  y 2
70. x 2 =
1 y 9
73. y 2 =  16x
74. y 2 =  4x
1 77. Focus a  , 0b 2
1 78. Focus a0, b 4
Write an equation for each parabola with vertex at the origin. 75. Focus (0,  2)
76. Focus (5, 0)
79. Through A 2,  222 B ; opening to the right 81. Through A 210,  5 B ; opening downward
80. Through A 23, 3 B ; opening upward
82. Through ( 3, 3); opening to the left
83. Through (2,  4); symmetric with respect to the yaxis
84. Through (3, 2); symmetric with respect to the xaxis
Find an equation of a parabola that satisfies the given conditions. 85. Focus (0, 2); vertex (0, 1)
86. Focus ( 1, 2); vertex (3, 2)
87. Focus (0, 0); directrix x =  2
88. Focus (2, 1); directrix x =  1
89. Focus ( 1, 3); directrix y = 7
90. Focus (1, 2); directrix y = 4
91. Horizontal axis; vertex ( 2, 3); passing through ( 4, 0)
92. Horizontal axis; vertex ( 1, 2); passing through (2, 3)
Graph each parabola by hand, and check using a graphing calculator. Give the vertex, axis, domain, and range. 93. y = (x + 3) 2  4
94. y = (x  5) 2  4
95. y =  2(x + 3) 2 + 2
97. y = x 2  2x + 3
98. y = x 2 + 6x + 5
99. y = 2x 2  4x + 5
3 ( y + 1) 2 100. y =  3x 2 + 24x  46 96. (x  2) 2 =
104. ( y + 2) 2 = x + 1
110. x = 2y 2  4y + 6
103. x = ( y  3) 2 2 107. x = y 2  4y + 8 3 111. x =  2y 2 + 2y  3
114. x =  3y 2 + 6y  1
115. y 2  4y + 4 = 4x + 4
116. y 2 + 2y + 1 =  2x + 4
101. x = y 2 + 2
102. x = ( y + 1) 2
105. x = ( y  4) 2 + 2
106. x =  2( y + 3) 2
109. x =  4y 2  4y  3 113. 2x = y 2  2y + 9
108. x = y 2 + 2y  8 112. 2x = y 2  4y + 6
117. Prove that the parabola with focus (c, 0) and directrix x =  c has equation y 2 = 4cx.
404
CHAPTER 6
Analytic Geometry
(Modeling) Solve each problem. 118. Path of an Object on a Planet When an object moves under the influence of a constant force (without air resistance), its path is parabolic. This is the path of a ball thrown near the surface of a planet or other celestial object. Suppose two balls are simultaneously thrown upward at a 45° angle on two different planets. If their initial velocities are both 30 mph, then their xycoordinates in feet can be expressed by the equation y = x 
g 2 x , 1922
where g is the acceleration due to gravity. The value of g will vary with the mass and size of the planet. (Source: Zeilik, M., S. Gregory, and E. Smith, Introductory Astronomy and Astrophysics, Saunders College Publishers.)
121. Path of an Alpha Particle When an alpha particle (a subatomic particle) is moving in a horizontal path along the positive xaxis and passes between charged plates, it is deflected in a parabolic path. If the plate is charged with 2000 volts and is 0.4 meter long, then an alpha particle’s path can be described by the equation y = 
k 2 x , 2v0
where k = 5 * 10 9 is constant and v0 is the initial velocity of the particle. If v0 = 10 7 meters per second, what is the deflection of the alpha particle’s path in the ydirection when x = 0.4 meter? (Source: Semat, H. and J. Albright, Introduction to Atomic and Nuclear Physics, Holt, Rinehart and Winston.) 122. Height of a Bridge’s Cable Supports The cable in the center portion of a bridge is supported as shown in the figure to form a parabola. The center support is 10 feet high, the tallest supports are 210 feet high, and the distance between the two tallest supports is 400 feet. Find the height of the remaining supports if the supports are evenly spaced. (Ignore the width of the supports.)
210 10
(a) On Earth, g = 32.2; on Mars, g = 12.6. Find the two equations, and use the same screen of a graphing calculator to graph the paths of the two balls thrown on Earth and Mars. Use the window 30, 1804 by 30, 1204. (Hint: If possible, set the mode on your graphing calculator to simultaneous.) (b) Determine the difference in the horizontal distances traveled by the two balls.
400
123. Parabolic Arch An arch in the shape of a parabola has the dimensions shown in the figure. How wide is the arch 9 feet up?
119. Path of an Object on a Planet Refer to Exercise 118. Suppose the two balls are now thrown upward at a 60° angle on Mars and the moon. If their initial velocities are 60 mph, then their xycoordinates in feet can be expressed by the equation g 2 19 y = x x . 11 3872 (a) Graph the paths of the balls if g = 5.2 for the moon. Use the window 30, 15004 by 30, 10004. (b) Determine the maximum height of each ball to the nearest foot.
12 ft
12 ft
124. Headlight A headlight is being constructed in the shape of a paraboloid with depth 4 inches and diameter 5 inches, as illustrated in the figure. Determine the distance d that the bulb should be from the vertex in order to have the beam of light shine straight ahead. 4 in.
120. Design of a Radio Telescope The U.S. Naval Research Laboratory designed a giant radio telescope weighing 3450 tons. Its parabolic dish had a diameter of 300 feet, with a focal length (the distance from the focus to the parabolic surface) of 128.5 feet. Determine the maximum depth of the 300foot dish. (Source: Mar, J. and H. Liebowitz, Structure Technology for Large Radio and Radar Telescope Systems, MIT Press.)
d
5 in.
6.2
6.2
Ellipses and Hyperbolas
405
Ellipses and Hyperbolas
Equations and Graphs of Ellipses • Translations of Ellipses • An Application of Ellipses • Equations and Graphs of Hyperbolas • Translations of Hyperbolas
Equations and Graphs of Ellipses The ellipse is another relation whose equation is based on the distance formula. y
Minor axis Vertex
Focus
Ellipse An ellipse is the set of all points in a plane, the sum of whose distances from two fixed points is constant. Each fixed point is called a focus (plural, foci) of the ellipse.
Focus Vertex x Center
Major axis FIGURE 22
y
(0, b) B (–a, 0) V' F'(–c, 0)
P(x, y) V (a, 0)x F(c, 0)
(0, –b) B'
As shown in FIGURE 22, an ellipse has two axes of symmetry: the major axis (the longer one) and the minor axis (the shorter one). The foci are always located on the major axis. The midpoint of the major axis is the center of the ellipse, and the endpoints of the major axis are the vertices of the ellipse. A chord through a focus and perpendicular to the major axis is called a latus rectum. (There are two.) The graph of an ellipse is not the graph of a function. (Why?) The ellipse in FIGURE 23 has its center at the origin, foci F(c, 0) and F¿( c, 0), and vertices V(a, 0) and V¿( a, 0). From the figure, the distance from V to F is a  c and the distance from V to F¿ is a + c. The sum of these distances is 2a. Since V is on the ellipse, this sum is the constant referred to in the definition of an ellipse. Thus, for any point P(x, y) on the ellipse, d(P, F ) + d(P, F¿) = 2a.
FIGURE 23
By the distance formula, d(P, F) = 2(x  c) 2 + y 2, and d(P, F¿) = 23x  ( c)42 + y 2 = 2(x + c) 2 + y 2. Thus, 2(x  c) 2 + y 2 + 2(x + c) 2 + y 2 = 2a 2(x  c) 2 + y 2 = 2a  2(x + c) 2 + y 2
Isolate 2(x  c)2 + y 2.
(x  c) 2 + y 2 = 4a 2  4a 2(x + c) 2 + y 2 + (x + c) 2 + y 2 Square each side.
x2 Divide each term by 4.
 2cx +
c2
+
y2
=
4a 2
 4a 2(x +
c) 2
4a 2(x + c) 2 + y 2 = 4a 2 + 4cx
a 2(x + c) 2 + y 2 a 2(x 2 + 2cx + c 2 + y 2) a 2x 2 + 2ca 2x + a 2c 2 + a 2y 2 a 2x 2 + a 2c 2 + a 2y 2 a 2x 2  c 2x 2 + a 2y 2 (a 2  c 2)x 2 + a 2y 2 y2 x2 + a2 a2  c2
= = = = = =
a 2 + cx a 4 + 2ca 2x + c 2x 2 a 4 + 2ca 2x + c 2x 2 a 4 + c 2x 2 a 4  a 2c 2 a 2(a 2  c 2)
= 1. (*)
+
y2
+ x 2 + 2cx + c 2 + y 2
Square x  c; square x + c. Isolate 4a 2(x + c)2 + y 2. Divide by 4.
Square each side and ( x + c ). Distributive property Subtract 2ca 2x. Rearrange terms. Factor. Divide by a 2(a 2  c 2).
406
CHAPTER 6
Analytic Geometry
Since B(0, b) is on the ellipse in FIGURE 23 on the preceding page, d(B, F) + d(B, F¿) = 2a 2( c) 2 + b 2 + 2c 2 + b 2 = 2a 2 2c 2 + b 2 = 2a 2c 2 c2
+ = a 2 + b = a2 b 2 = a 2  c 2. b2
Combine like terms. Divide by 2. Square each side. Subtract c 2.
Replacing a 2  c 2 with b 2 in equation (*) gives y2 x2 ⴙ ⴝ 1, a2 b2 the standard form of the equation of an ellipse centered at the origin and with foci on the xaxis. If the vertices and foci were on the yaxis, an almost identical derivation could be used to get the standard form y2 x2 ⴙ 2 ⴝ 1. 2 b a
Standard Forms of Equations for Ellipses y
The ellipse with center at the origin and equation y2 x2 ⴙ ⴝ1 a2 b2
b
(a 7 b 7 0) –a –c
has vertices (⫾a, 0), endpoints of the minor axis (0, ⫾b), and foci (⫾c, 0), where c 2 = a 2  b 2.
c
0
x
a
–b Major axis on xaxis
The ellipse with center at the origin and equation y2 x2 ⴙ ⴝ1 b2 a2
y
a
(a 7 b 7 0)
has vertices (0, ⫾a), endpoints of the minor axis (⫾b, 0), and foci (0, ⫾c), where c 2 = a 2  b 2.
c –b
0
b
x
–c –a Major axis on yaxis
NOTE The relationship among a, b, and c in the definition of an ellipse alternatively can be remembered as a 2 = b 2 + c 2, because a, and thus a 2, is the greatest of the three values.
Do not be confused by the two standard forms—in one case a 2 is associated with in the other case a 2 is associated with y 2. In practice, we need only find the intercepts of the graph—if the positive xintercept is greater than the positive yintercept, then the major axis is horizontal; otherwise, it is vertical. When using a 2  c 2 = b 2, or a 2  b 2 = c 2, choose a 2 and b 2 so that a2>b 2. x 2;
GCM
EXAMPLE 1
Graphing an Ellipse Centered at the Origin Graph each ellipse, and find the foci. Give the domain and range. (a) 4x 2 + 9y 2 = 36 (b) 4x 2 + y 2 = 64
6.2
4x 2 + 9y 2 36 2 y2 y x
+
9
3
( 5, 0)
1
4
4x 2 + 9y 2 = 36 y2 x2 + = 1 9 4
Divide each term by 36.
( 5, 0)
1 4
Solution (a)
2 1
2
(By definition, c 7 0. See FIGURE 23 on page 405.) The major axis is along the xaxis, so the foci have coordinates A  25, 0 B and A 25, 0 B . The domain of this relation is 3 3, 34, and the range is 3 2, 24.
FIGURE 24
(0, 4 3) 8
4x 2 x2 16
+ + 64
y2 y2
2
4x 2 + y 2 = 64 y2 x2 + = 1 16 64
(b) 64 1
Divide by 64 to write in standard form.
The xintercepts are ⫾4 and the yintercepts are ⫾8. See FIGURE 25. Here 64 7 16, so a 2 = 64 and b 2 = 16. Thus,
4
4
Divide by 36 to write in standard form.
c 2 = 9  4 = 5, so c = 25.
3
y
407
Thus, the xintercepts are ⫾3, and the yintercepts are ⫾2. See FIGURE 24. Since 9 7 4, we find the foci by letting a 2 = 9 and b 2 = 4 in c 2 = a 2  b 2.
x
4
Ellipses and Hyperbolas
2
4
c 2 = 64  16 = 48, so c = 248 = 216 # 3 = 4 23.
x
The major axis is on the yaxis, so the coordinates of the foci are (0,  423) and (0, 4 23). The domain of the relation is 3 4, 44 and the range is 3 8, 84.
4
The graph of an ellipse is not the graph of a function. To graph the ellipse in Example 1(a) with a graphing calculator, solve for y to get equations of two functions.
(0, 4 3)
8
4x 2 + 9y 2 = 36
FIGURE 25
x2 9
y1 = 2 3.1
+
2 1– x 9
Remember both the positive and negative square roots.
–4.7
4.7
Original equation
y2
= 1 4 y2 x2 = 1 4 9 y2 = 4a1 y = ⫾2
A
Divide by 36. Subtract
x2 b 9
1 
x2 9
x2 9.
Multiply by 4.
Take square roots.
The ellipse is obtained by graphing both equations, as shown in FIGURE 26. –3.1
y2 = –2
2 1– x 9
Since y2 = – y1, we can enter – y1 to get the portion of the graph below the xaxis. FIGURE 26
EXAMPLE 2
Writing the Equation of an Ellipse and Graphing It Write the equation of the ellipse having center at the origin, foci at (0, 3) and (0,  3), and major axis 8 units long. Graph the ellipse, and give the domain and range. Solution Since the major axis is 8 units long, 2a = 8 and thus a = 4. To find b 2, use the relationship a 2  b 2 = c 2, with a = 4 and c = 3. a2  b2 = 42  b 2 = 16  b 2 = b2 =
c2 32 9 7
Substitute for a and c. Apply the exponents. Solve for b 2.
(continued)
408
Analytic Geometry
CHAPTER 6
x 2 y2 + =1 7 16
y
x2
0
–√7
y2
Since the foci are on the yaxis, our equation will be of the form 2 + 2 = 1. We use a b the larger intercept, a = 4, to find the denominator for y 2, giving the equation in standard form as y2 x2 + = 1. b 2 = 7; a 2 = 16 7 16
4
x √7
A graph of this ellipse is shown in the margin in FIGURE 27. The domain of this relation is C  27, 27 D and the range is 3 4, 44.
–4
FIGURE 27
Translations of Ellipses As with a circle, an ellipse also may have its center translated from the origin.
Standard Forms for Ellipses Centered at (h, k) An ellipse with center at (h, k) and either a horizontal or vertical major axis satisfies one of the following equations, where a 7 b 7 0 and c 2 = a 2  b 2 with c 7 0: ( y ⴚ k)2 (x ⴚ h)2 ⴙ ⴝ1 a2 b2
Major axis: horizontal; foci: ( h ⫾ c, k ); vertices: ( h ⫾ a, k )
(x ⴚ h)2 ( y ⴚ k)2 ⴙ ⴝ 1. 2 b a2
Major axis: vertical; foci: ( h, k ⫾ c ); vertices: ( h, k ⫾ a)
EXAMPLE 3
GCM
Graph
(x 9
2) 2
Graphing an Ellipse Translated from the Origin ( y + 1) 2 + = 1. Give the domain and range. 16
Analytic Solution The graph of this equation is an ellipse centered at (2,  1). For this ellipse, a = 4 and b = 3. Since a = 4 is associated with y 2, the vertices of the ellipse are on the vertical line through (2,  1). Find the vertices by locating two points on the vertical line through (2,  1), one 4 units up from (2,  1) and one 4 units down. The vertices are (2, 3) and (2,  5). Locate two other points on the ellipse by locating points on the horizontal line through (2,  1), one 3 units to the right and one 3 units to the left. The graph is shown in FIGURE 28. The domain is 3 1, 54 and the range is 3 5, 34.
Graphing Calculator Solution Solve for y in the equation of the ellipse to obtain y = 1 ⫾ 4
A
1 
(x  2) 2 . 9
The + sign yields the top half of the ellipse, while the  sign yields the bottom half. See FIGURE 29. y1 = –1 + 4
1–
6.2
(x – 2)2 9
y
(x – 2)2 ( y + 1)2 =1 + 9 16
–9.4
9.4
3 –1
5 x
(2, –1)
–6.2
y2 = –1 – 4
1–
–5
FIGURE 28
FIGURE 29
(x – 2)2 9
6.2
Ellipses and Hyperbolas
409
NOTE An ellipse is symmetric with respect to its major axis, its minor axis, and its center. If a ⴝ b in the equation of an ellipse, the graph is that of a circle. EXAMPLE 4
Finding the Standard Form for an Ellipse Write  16x + 9y 2 + 54y + 61 = 0 in the standard form for an ellipse centered at (h, k). Identify the center and vertices. 4x 2
Solution 4x 2  16x + 9y 2 + 54y + 61 = 0 4(x 2  4x + ) + 9( y 2 + 6y + ) =  61
Focusing reflector
Emitter
4(x 2  4x + 4) + 9( y 2 + 6y + 9) =  61 + 4(4) + 9(9) 4(x  2) 2 + 9( y + 3) 2 = 36 ( y + 3) 2 (x  2) 2 + = 1 9 4
Distributive property; add  61. Complete the square. Factor; add. Divide each side by 36.
The center is (2,  3). Because a = 3 and the major axis is horizontal, the vertices of the ellipse are (2 ⫾ 3,  3), or (5,  3) and ( 1,  3).
An Application of Ellipses
Shock waves Kidney stone The top of an ellipse is illustrated in this depiction of how a lithotripter crushes a kidney stone. FIGURE 30
Ellipses have many useful applications. As Earth makes its yearlong journey around the sun, it traces an ellipse. Spacecraft travel around Earth in elliptical orbits, and planets make elliptical orbits around the sun. An application from medicine is a lithotripter, a machine used to crush kidney stones via shock waves. The patient is placed in an elliptical tub with the kidney stone at one focus of the ellipse. A beam is projected from the other focus to the tub so that it reflects to hit the kidney stone. See FIGURE 30 and Exercise 91. EXAMPLE 5
Modeling the Reflective Property of Ellipses If a lithotripter is based on the ellipse
Kidney stone √27
–6
y
Wave path
Wave source
x
(–3, 0)
– √27
(3, 0)
6
y2 x2 + =1 36 27
FIGURE 31
y2 x2 + = 1, 36 27 determine how many units the kidney stone and the wave source must be placed from the center of the ellipse. Solution The kidney stone and the source of the beam must be placed at the foci, (c, 0) and ( c, 0). Here a 2 = 36 and b 2 = 27, so c = 2a 2  b 2 = 236  27 = 29 = 3. Thus, the foci are ( 3, 0) and (3, 0), so the kidney stone and the source both must be placed on a line 3 units from the center. See FIGURE 31.
Equations and Graphs of Hyperbolas An ellipse was defined as the set of all points in a plane, the sum of whose distances from two fixed points is constant. A hyperbola is defined similarly.
410
Analytic Geometry
CHAPTER 6
Hyperbola A hyperbola is the set of all points in a plane such that the absolute value of the difference of the distances from two fixed points is constant. The two fixed points are called the foci of the hyperbola. y
P(x, y) Vertices V(a, 0) F′(–c, 0)
V′(–a, 0) Transverse axis
FIGURE 32
x
F(c, 0)
Suppose a hyperbola has center at the origin and foci at F¿( c, 0) and F(c, 0). See The midpoint of the segment F¿F is the center of the hyperbola, and the points V¿( a, 0) and V(a, 0) are the vertices of the hyperbola. The line segment V¿V is the transverse axis of the hyperbola. A chord through a focus and perpendicular to an extension of the transverse axis is, again, a latus rectum. (There are two.) Using the distance formula in conjunction with the definition of a hyperbola, we can verify the following standard forms of the equations for hyperbolas centered at the origin. (See Exercise 100.) FIGURE 32.
Standard Forms of Equations for Hyperbolas The hyperbola with center at the origin and equation
y
y2 x2 ⴚ ⴝ1 a2 b2
b
has vertices (⫾a, 0), asymptotes y = ⫾ ba x, and foci (⫾c, 0), where c 2 = a 2 + b 2.
–a
a
–c
c
x
–b Transverse axis on xaxis y
The hyperbola with center at the origin and equation c
y2 x2 ⴚ 2ⴝ1 2 a b
a
has vertices (0, ⫾a), asymptotes y = ⫾ ab x, and foci (0, ⫾c), where c 2 = a 2 + b 2.
–b
–a
b
x
–c Transverse axis on yaxis
NOTE The relationship among a, b, and c in the definition of a hyperbola can be remembered correctly as c 2 = a 2 + b 2, because c is the greatest of the three values.
To explain the concept of asymptotes, we can start with the first equation for a hyperbola, where the foci are on the xaxis, and solve for y to obtain y2 x2  2 = 1 2 a b 2 y2 x 1 = a2 b2 y2 x2  a2 = Remember both the a2 b2 positive and negative b square roots. y = ⫾ 2x 2  a 2. (*) a
Subtract 1; add
y2 b
2
.
Write the left side as a single fraction. Take square roots; multiply by b.
6.2
Ellipses and Hyperbolas
411
If x 2 is very large in comparison to a 2, the difference x 2  a 2 would be relatively close to x 2. If this happens, then the points satisfying equation (*) would approach one of the lines b y ⴝ ⴞ x. a Thus, as  x  gets larger and larger, the points of the hyperbola ⫾ ba x.
x2 a2

y2 b2
= 1 become
closer to the lines y = These lines, called the asymptotes of the hyperbola, are helpful in sketching the graph. The lines are the extended diagonals of the rectangle whose vertices are (a, b), ( a, b), (a,  b), and ( a,  b). This rectangle is called the fundamental rectangle of the hyperbola. For hyperbolas, it is possible that a 7 b, a 6 b, or a = b. If the foci are on the y2
yaxis, the equation of the hyperbola has the form a 2 
x2 b2
= 1, with asymptotes
a y ⴝ ⴞ x. b GCM
EXAMPLE 6
Using Asymptotes to Graph a Hyperbola y2 x2 = 1. Give the foci, domain, Sketch the asymptotes and graph the hyperbola 25 49 and range.
Analytic Solution For this hyperbola, a = 5 and b = 7. With these values, y = ⫾ ba x becomes y = ⫾ 75 x. Choosing x = 5 gives y = ⫾7. Choosing x =  5 also gives y = ⫾7. These four points, (⫾5, ⫾7), are the corners of the fundamental rectangle shown in FIGURE 33.
(5, 7)
y2
x2
y
25 – 49
1
(5, 7)
2 22 2
( 74, 0)
6 (5, 7) (5, 7)
(5, 0) 8
x
y = ⫾7
( 74, 0)
The extended diagonals of this rectangle are the asymptotes of the hyperbola. See FIGURE 33. Note that the hyperbola has vertices ( 5, 0) and (5, 0). We find the foci by letting =
a2
+
b2
= 25 + 49 = 74,
Subtract
x2 25 .
Multiply by  1. Take square roots.
x2  1 A 25
Multiply by 7.
The graphs of the hyperbola and its asymptotes are shown in FIGURE 34. The graph of y1 creates the upper half of the hyperbola, while the graph of y2 creates the lower half.
FIGURE 33
c2
y2 x2 = 1 25 49 y2 x2 = 1 49 25 2 2 y x =  1 49 25 y x2 = ⫾  1 7 A 25
6 (5, 0)
Graphing Calculator Solution Solve the given equation for y.
so
y4 = – 75 x
10
y3 = 75 x
–9.4
9.4
c = 274.
Therefore, the foci are ( 274, 0) and (274, 0). The domain of this relation is ( q,  54 ´ 35, q), and the range is ( q, q).
–10
y1 = 7
x 2 – 1 y = –7 2 25
x2 – 1 25
FIGURE 34
412
Analytic Geometry
CHAPTER 6
EXAMPLE 7
Graph
25y 2

4x 2
Graphing a Hyperbola Centered at the Origin = 9. Give the domain and range. 25y 2  4x 2 = 9
Solution
y
(– 32 ,
3 5
)
25y 2
2 1
–
4x 2
Be sure to divide each term by 9.
9
( ) 3, 3 2 5
2
2
25y 2 4x 2 = 1 9 9
Divide by 9.
To determine the values of a and b, write the equation y2 9 25
x

x2 9 4
= 1.
a b
=
25y 2 9

4x 2 9
= 1 as
1 b a
This hyperbola is centered at the origin, has foci on the yaxis, and has yintercepts  35 and 35. Use the four points A ⫾ 32, ⫾ 35 B to get the fundamental rectangle shown in FIGURE 35. Use the diagonals of this rectangle to determine the asymptotes for the graph. The domain is ( q, q) and the range is A  q,  35 D ´ C 35, q B .
1
(– 32 , – 35)2 ( 32 , – 35) FIGURE 35
Translations of Hyperbolas Standard Forms for Hyperbolas Centered at (h, k) A hyperbola with center (h, k) and either a horizontal or vertical transverse axis satisfies one of the following equations, where c 2 = a 2 + b 2. (x ⴚ h)2 ( y ⴚ k)2 ⴚ ⴝ1 2 a b2
Transverse axis: horizontal; vertices: ( h ⫾ a, k); foci: ( h ⫾ c, k); asymptotes: y = ⫾ ba ( x  h) + k
(x ⴚ h)2 ( y ⴚ k)2 ⴚ ⴝ1 a2 b2
Transverse axis: vertical; vertices: ( h, k ⫾ a); foci: ( h, k ⫾ c); asymptotes: y = ⫾ ab ( x  h) + k
NOTE The asymptotes for a hyperbola always pass through the center ( h, k ). By the point–slope form of a line, the equation of any asymptote is y = m( x  h) + k. If the transverse axis is horizontal, then m = ⫾ ba ; if it is vertical, then m = ⫾ ba .
GCM
EXAMPLE 8
Graph y
(–3, 1)
2 0
(–3, –2)
x 2
(–3, –5) (x + 3)2 ( y + 2)2 – =1 9 4 FIGURE 36
(y + 9
2) 2
Graphing a Hyperbola Translated from the Origin (x + 3) 2 = 1, and identify the center. Give the domain and range. 4
Solution This equation represents a hyperbola centered at ( 3,  2). For this vertical hyperbola, a = 3 and b = 2. The xvalues of the vertices are  3. Locate the yvalues of the vertices by taking the yvalue of the center,  2, and adding and subtracting 3. Thus, the vertices are ( 3, 1) and ( 3,  5). The asymptotes have slopes ⫾ 32 and pass through the center (  3,  2). The equations of the asymptotes, 3 y = ⫾ (x + 3)  2, 2 can be found by using the point–slope form of the equation of a line. The graph is shown in FIGURE 36. The domain of the relation is ( q, q) and the range is ( q,  5] ´ [1, q).
6.2
FOR DISCUSSION
Ellipses and Hyperbolas
413
EXAMPLE 9
Writing the Equation of a Hyperbola Write an equation of the hyperbola having center (1, 2), focus (6, 2), and vertex (4, 2).
How would you graph the hyperbola in Example 8,
Solution Because the center, focus, and vertex lie on the line y = 2, the hyperbola has a horizontal transverse axis. The center is (1, 2), so the standard form of the equation is
( y + 2) 2 (x + 3) 2 = 1, 9 4 on your graphing calculator? Duplicate the calculator graph of that hyperbola shown in FIGURE 37. The two branches of a hyperbola are reflections about two different axes and also about a point. What are the axes and the point for the hyperbola shown in the figure?
( y  2) 2 (x  1) 2 = 1, 2 a b2 with a and b to be determined. The distance from the focus to the center is 6  1 = 5, and the distance from the vertex to the center is 4  1 = 3, so c = 5 and a = 3. Because a 2 + b 2 = c 2, it follows that b = 4. Thus, the equation of the hyperbola is (x  1) 2 ( y  2) 2 = 1. 9 16
4 –12
6
EXAMPLE 10
Finding the Standard Form for a Hyperbola Write  18x  4y 2  16y = 43 in the standard form for a hyperbola centered at (h, k). Identify the center and vertices. 9x 2
–8
Solution
FIGURE 37
9x 2  18x  4y 2  16y 9(x 2  2x + )  4( y 2 + 4y + ) 9(x 2  2x + 1)  4( y 2 + 4y + 4) 9(x  1) 2  4( y + 2) 2 (x  1) 2 ( y + 2) 2 4 9
= = = =
43 43 43 + 9(1)  4(4) 36
= 1
Distributive property Complete the square. Factor; add. Divide each side by 36.
The center is (1,  2). Because a = 2 and the transverse axis is horizontal, the vertices of the hyperbola are (1 ⫾ 2,  2), or (3,  2) and ( 1,  2).
6.2
EXERCISES
Checking Analytic Skills Match each equation in Column I with the appropriate description in Column II. Do not use a calculator. I
II
1.
y2 x2 + = 1 4 16
2.
y2 x2 + = 1 16 4
3.
y2 x2 = 1 4 16
4.
y2 x2 = 1 4 16
5.
(x + 2) 2 ( y  4) 2 + = 1 9 25
6.
( y + 4) 2 (x  2) 2 + = 1 9 25
7.
(x + 2) 2 ( y  4) 2 = 1 9 25
8.
(x  2) 2 ( y  4) 2 = 1 9 25
A. B. C. D. E. F. G. H.
Hyperbola; center (2, 4) Ellipse; foci (⫾223, 0) Hyperbola; foci (0, ⫾225) Hyperbola; center ( 2, 4) Ellipse; center ( 2, 4) Center (0, 0); horizontal transverse axis Ellipse; foci (0, ⫾2 23) Vertical major axis; center (2,  4)
414
9 9.
CHAPTER 6
Analytic Geometry
Explain how a circle can be interpreted as a special case of an ellipse.
10. Concept Check If an ellipse has endpoints of the minor axis and vertices at ( 3, 0), (3, 0), (0, 5), and (0,  5), what is its domain? What is its range?
Checking Analytic Skills Graph each ellipse by hand. Give the domain and range. Give the foci in Exercises 11–14 and identify the center in Exercises 17–22. Do not use a calculator. y2 x2 + = 1 9 4
12.
y2 x2 + = 1 16 36
13. 9x 2 + 6y 2 = 54
14. 12x 2 + 8y 2 = 96
15.
25y 2 64x 2 + = 1 36 9
16.
16y 2 121x 2 + = 1 9 25
11.
17.
(x  1) 2 ( y + 3) 2 + = 1 9 25
18.
(x + 3) 2 ( y  2) 2 + = 1 16 36
19.
(x  2) 2 ( y  1) 2 + = 1 16 9
20.
(x + 3) 2 ( y + 2) 2 + = 1 25 36
21.
(x + 1) 2 ( y  2) 2 + = 1 64 49
22.
(x  4) 2 ( y + 2) 2 + = 1 9 4
Find an equation for each ellipse. 24. yintercepts ⫾3; foci A 0, 23 B and A 0,  23 B
23. xintercepts ⫾4; foci ( 2, 0) and (2, 0) 25. yintercepts ⫾2 22; foci (0, 2) and (0,  2)
26. xintercepts ⫾3 22; foci A 223, 0 B and A  223, 0 B
27. xintercepts ⫾4; yintercepts ⫾2
28. xintercepts ⫾3; yintercepts ⫾6
29. Endpoints of major axis at (6, 0) and ( 6, 0); c = 4
30. Vertices (0, 5) and (0,  5); b = 2
31. Center (3,  2); a = 5; c = 3; major axis vertical
32. Center (2, 0); minor axis of length 6; major axis horizontal and of length 9
33. Major axis of length 6; foci (0, 2) and (0,  2)
34. Minor axis of length 4; foci ( 5, 0) and (5, 0)
35. Center (5, 2); minor axis vertical, with length 8; c = 3
36. Center ( 3, 6); major axis vertical, with length 10; c = 2
37. Vertices (4, 9) and (4, 1); minor axis of length 6
38. Foci at ( 3,  3) and (7,  3); the point (2, 1) on ellipse
Write the equation in standard form for an ellipse centered at (h, k). Identify the center and vertices. 39. 9x 2 + 18x + 4y 2  8y  23 = 0
40. 9x 2  36x + 16y 2  64y  44 = 0
41. 4x 2 + 8x + y 2 + 2y + 1 = 0
42. x 2  6x + 9y 2 = 0
43. 4x 2 + 16x + 5y 2  10y + 1 = 0
44. 2x 2 + 4x + 3y 2  18y + 23 = 0
45. 16x 2  16x + 4y 2 + 12y = 51
46. 16x 2 + 48x + 4y 2  20y + 57 = 0
Checking Analytic Skills Graph each hyberbola by hand. Give the domain and range. Give the center in Exercises 55–61. Do not use a calculator. 48.
y2 x2 = 1 9 9
49. 49y 2  36x 2 = 1764
50. 144x 2  49y 2 = 7056
51.
25y 2 4x 2 = 1 9 16
52. x 2  y 2 = 1
53. 9x 2  4y 2 = 1
54. 25y 2  9x 2 = 1
47.
y2 x2 = 1 16 9
55.
(x  1) 2 ( y + 3) 2 = 1 9 25
6.2
56.
(x + 3) 2 ( y  2) 2 = 1 16 36
57.
59. 16(x + 5) 2  ( y  3) 2 = 1
( y  5) 2 (x + 1) 2 = 1 4 9
60. 4(x + 9) 2  25( y + 6) 2 = 100
Ellipses and Hyperbolas
58.
415
(x  3) 2 ( y + 1) 2 = 1 25 36
61. 9(x  2) 2  4( y + 1) 2 = 36
62. The graph of the rational function y = 1x is a hyperbola that is rotated. Experiment with a graphing calculator to determine the vertices of its graph. Find an equation for each hyperbola. 63. xintercepts ⫾3; foci ( 4, 0) and (4, 0)
64. yintercepts ⫾5; foci A 0, 323 B and A 0,  323 B
65. Asymptotes y = ⫾ 35 x; yintercepts 3 and  3
66. yintercept  2; center at origin; passing through (2, 3)
67. Vertices (0, 6) and (0,  6); asymptotes y = ⫾12 x
68. Vertices ( 10, 0) and (10, 0); asymptotes y = ⫾5x
69. Vertices ( 3, 0) and (3, 0); passing through (6, 1)
70. Vertices (0, 5) and (0,  5); passing through (3, 10)
71. Foci (0, 213) and (0,  213); asymptotes y = ⫾5x
72. Foci ( 325, 0) and (325, 0); asymptotes y = ⫾2x
73. Vertices (4, 5) and (4, 1); asymptotes y = ⫾7(x  4) + 3
74. Vertices (5,  2) and (1,  2); asymptotes y = ⫾32 (x  3)  2
75. Center (1,  2); focus (4,  2); vertex (3,  2)
76. Center (9,  7); focus (9, 3); vertex (9,  1)
Write the equation in standard form for a hyperbola centered at (h, k). Identify the center and vertices. 77. x 2  2x  y 2 + 2y = 4
78. y 2 + 4y  x 2 + 2x = 6
79. 3y 2 + 24y  2x 2 + 12x + 24 = 0
80. 4x 2 + 16x  9y 2 + 18y = 29
81. x 2  6x  2y 2 + 7 = 0
82. y 2 + 8y  3x 2 + 13 = 0
83. 4y 2 + 32y  5x 2  10x + 39 = 0
84. 5x 2 + 10x  7y 2 + 28y = 58
RELATING CONCEPTS For individual or group investigation (Exercises 85–90)
Consider the ellipse and hyperbola defined by y2 x2 + = 1 16 12
and
y2 x2 = 1, 4 12
respectively. Work Exercises 85–90 in order. 85. Find the foci of the ellipse. Call them F1 and F2. 86. Graph the ellipse with your calculator, and trace to find the coordinates of several points on the ellipse. 87. For each of the points P, verify that
3distance of P from F14 + 3distance of P from F24 = 8.
88. Repeat Exercises 85 and 86 for the hyperbola. 89. For each of the points P from Exercise 88, verify that
9
3distance of P from F14  3distance of P from F24  = 4.
90. How do Exercises 87 and 89 relate to the definitions of the ellipse and the hyperbola given in this section?
(Modeling) Solve each problem. 91. Shape of a Lithotripter A patient’s kidney stone is placed 12 units away from the source of the shock waves of a lithotripter. The lithotripter is based on an ellipse with a minor axis that measures 16 units. Find an equation of an ellipse that would satisfy this situation. 92. Orbit of Venus The orbit of Venus is an ellipse, with the sun at one focus. An approximate equation for the orbit is y2 x2 + = 1, 5013 4970 where x and y are measured in millions of miles. (a) Approximate the length of the major axis. (b) Approximate the length of the minor axis. 93. The Roman Colosseum The Roman Colosseum is an ellipse with major axis 620 feet and minor axis 513 feet. Approximate the distance between the foci of this ellipse.
416
CHAPTER 6
Analytic Geometry
94. The Roman Colosseum A formula for the approximate perimeter of an ellipse is P L 2p
a2 + b2 , A 2
98. Sound Detection Microphones are placed at points ( c, 0) and (c, 0). An explosion occurs at point P(x, y) having positive xcoordinate. y
where a and b are the lengths shown in the figure. Use this formula to find the approximate perimeter of the Roman Colosseum. (See Exercise 93.) a
b
P(x, y)
95. Height of an Overpass A oneway road passes under an overpass in the form of half of an ellipse 15 feet high at the center and 20 feet wide. Assuming that a truck is 12 feet wide, what is the height of the tallest truck that can pass under the overpass?
15 ft 20 ft
96. Design of a Sports Complex Two buildings in a sports complex are shaped and positioned like a portion of the branches of the hyperbola 400x 2  625y 2 = 250,000. In this equation, x and y are in meters. 50 m d
x
0
(–c, 0)
(c, 0)
The sound is detected at the closer microphone t seconds before being detected at the farther microphone. Assume that sound travels at a speed of 330 meters per second, and show that P must be on the hyperbola y2 x2 1 = . 2 2 2 2 2 4 330 t 4c  330 t 99. Structure of an Atom In 1911, Ernest Rutherford discovered the basic structure of the atom by “shooting” positively charged alpha particles with a speed of 10 7 meters per second at a piece of gold foil 6 * 10  7 meter thick. Only a small percentage of the alpha particles struck a gold nucleus headon and were deflected directly back toward their source. The rest of the particles often followed a hyperbolic trajectory because they were repelled by positively charged gold nuclei. Thus, Rutherford proposed that the atom was composed of mostly empty space and a small, dense nucleus. The figure shows an alpha particle A initially approaching a gold nucleus N and being deflected at an angle u = 90°. N is located at a focus of the hyperbola, and the trajectory of A passes through a vertex of the hyperbola. (Source: Semat, H. and J. Albright, Introduction to Atomic and Nuclear Physics, Holt, Rinehart and Winston.) y
NOT TO SCALE A
(a) How far apart are the buildings at their closest point? (b) Find the distance d in the figure to the nearest tenth.
N
x
d
97. Orbit of a Satellite The coordinates in miles for the orbit of the artificial satellite Explorer VII can be described by the equation y2 x2 + = 1, a2 b2 where a = 4465 and b = 4462. Earth’s center is located at one focus of its elliptical orbit. (Source: Loh, W., Dynamics and Thermodynamics of Planetary Entry, PrenticeHall; Thomson, W., Introduction to Space Dynamics, John Wiley and Sons.) (a) Graph both the orbit of Explorer VII and that of Earth on the same coordinate axes if the average radius of Earth is 3960 miles. Use the window 3 6750, 67504 by 3 4500, 45004. (b) Determine the maximum and minimum heights of the satellite above Earth’s surface.
y=x
y = –x
(a) Determine the equation of the trajectory of the alpha particle if d = 5 * 10  14 meter. (b) Approximate the minimum distance between the centers of the alpha particle and the gold nucleus. 100. Suppose a hyperbola has center at the origin, foci at F¿( c, 0) and F(c, 0), and equation
 d(P, F¿)  d(P, F )  = 2a. Let b 2 = c 2  a 2, and show that the points on the hyperbola satisfy the equation y2 x2 = 1. a2 b2
6.3
417
102. Use the definition of a hyperbola to find an equation of a hyperbola with center at the origin, foci ( 2, 0) and (2, 0), and the absolute value of the difference of the distances from any point of the hyperbola to the two foci equal to 2.
101. Use the definition of an ellipse to find an equation of an ellipse with foci (3, 0) and ( 3, 0) where the sum of the distances from any point of the ellipse to the two foci is 10.
SECTIONS 6.1 AND 6.2
Summary of the Conic Sections
REVIEWING BASIC CONCEPTS
1. Match each definition in A–D with the appropriate conic section. (a) Circle (b) Parabola (c) Ellipse (d) Hyperbola A. A set of points in a plane such that the sum of their distances from two fixed points is constant B. A set of points in a plane that are equidistant from a fixed point C. A set of points in a plane such that the absolute value of the difference of their distances from two fixed points is constant D. A set of points in a plane equidistant from a fixed point and a fixed line Graph each conic section by hand and check with your calculator. 2. 12x 2  4y 2 = 48
3. y = 2x 2 + 3x  1
4. x 2 + y 2  2x + 2y  2 = 0
5. 4x 2 + 9y 2 = 36
9 6.
Given the two vertices and two foci of a conic section, how can you tell whether it is an ellipse or a hyperbola?
Write an equation for each conic section. 7. Center (2,  1); radius 3
8. Foci (⫾4, 0); major axis length 12
10. Single focus A 0, 12 B ; vertex at the origin
9. Vertices (0, ⫾2); foci (0, ⫾4)
6.3
Summary of the Conic Sections
Characteristics • Identifying Conic Sections • Eccentricity
Characteristics The conic sections in this chapter have equations that can be written in the form Ax 2 ⴙ Dx ⴙ Cy 2 ⴙ Ey ⴙ F ⴝ 0, where either A or C must be nonzero. The special characteristics of each conic section Ax 2 + Dx + Cy 2 + Ey + F = 0 are summarized in the following table. Conic Section
Characteristic
Examples
Parabola
Either A = 0 or C = 0, but not both
y = x2 x = 3y 2 + 2y  4
Circle
A = C Z 0
x 2 + y 2 = 16
Ellipse
A Z C, AC 7 0
y2 x2 + = 1 16 25
Hyperbola
AC 6 0
x2  y2 = 1
418
CHAPTER 6
Analytic Geometry
The chart summarizes our work with conic sections. Equation
Graph
(x ⴚ h)2 ⴝ 4c( y ⴚ k)
Description
y
Identification
Opens up if c 7 0, down if c 6 0. Vertex is (h, k).
x 2term is present. y is not squared.
Opens to the right if c 7 0, to the left if c 6 0. Vertex is (h, k).
y 2term is present. x is not squared.
Center is (h, k), radius is r.
x 2 and y 2terms have the same positive coefficient.
xintercepts are a and  a. yintercepts are b and  b.
x 2 and y 2terms have different positive coefficients.
xintercepts are b and  b. yintercepts are a and  a.
x 2 and y 2terms have different positive coefficients.
xintercepts are a and  a. Asymptotes are diagonals of the rectangle with corners at (⫾a, ⫾b).
x 2term has a positive coefficient. y 2term has a negative coefficient.
yintercepts are a and  a. Asymptotes are diagonals of the rectangle with corners at (⫾b, ⫾a).
y 2term has a positive coefficient. x 2term has a negative coefficient.
Parabola (h, k) x
0 y
( y ⴚ k)2 ⴝ 4c(x ⴚ h)
(h, k) Parabola
x
0 y
(x ⴚ h)2 ⴙ ( y ⴚ k)2 ⴝ r 2
(h, k) r
Circle
x
0
y2 x2 ⴙ 2ⴝ1 2 a b
y
(a>b>0)
(0, b) (–a, 0)
(a, 0) x
0
(0, –b)
y2 x2 ⴙ 2ⴝ1 2 b a
Ellipse
y
(a>b>0)
(0, a) (b, 0)
(–b, 0)
x
0
Ellipse (0, –a) y
y2 x2 ⴚ ⴝ1 a2 b2 (– a, 0)
(a, 0) 0
x
Hyperbola
y2 x2 ⴚ 2ⴝ1 2 a b
y
(0, a)
0
(0, –a) Hyperbola
x
6.3
Summary of the Conic Sections
419
To find the value of c for an ellipse, use the equation c 2 = a 2  b 2; for a hyperbola, use the equation c 2 = a 2 + b 2. If the center is (0, 0), then the foci of these conic sections are located either at (⫾c, 0) or at (0, ⫾c ).
NOTE
Identifying Conic Sections To recognize the type of graph that a given conic section has, we sometimes need to transform the equation into a more familiar form. EXAMPLE 1
Determining Types of Conic Sections from Equations Decide on the type of conic section represented by each equation, and give each graph. (a) x 2 = 25 + 5y 2 (b) x 2  8x + y 2 + 10y =  41 (c) 4x 2  16x + 9y 2 + 54y =  61
(d) x 2  6x + 8y  7 = 0
(e) 4y 2  16y  9x 2 + 18x =  43 y
Solution x 2 = 25 + 5y 2
(a)
x2 = 25 + 5y2 2 x –5
5 –2
Be sure to divide each term by 25.
x 2  5y 2 = 25 y2 x2 = 1 25 5
Subtract 5y 2. Divide by 25.
The equation represents a hyperbola centered at the origin and with asymptotes y = ⫾
FIGURE 38
25 x. 5
The vertices are ( 5, 0) and (5, 0). See FIGURE 38.
(b) x 2  8x + y 2 + 10y =  41 (x 2  8x + 16) + ( y 2 + 10y + 25) =  41 + 16 + 25
y
(x  4) 2 + ( y + 5) 2 = 0 4
(4, –5)
–5
x 2 – 8x + y2 + 10y = –41
(c) Since the coefficients of the x 2 and y 2terms of 4x 2  16x + 9y 2 + 54y =  61 are unequal and both positive, this equation may represent an ellipse, but not a circle. (It may also represent a single point or no points at all.)
FIGURE 39
y 2 0
x 5
–1
4x 2  16x + 9y 2 + 54y =  61 4(x 2  4x ) + 9( y 2 + 6y ) =  61 4(x 2
 4x + 4) +
9( y 2
+ 6y + 9) =  61 + 4(4) + 9(9)
(2, –3)
4(x  2) 2 + 9( y + 3) 2 = 36 ( y + 3) 2 (x  2) 2 + = 1 9 4
–6
4x 2 – 16x + 9y 2 + 54y = –61 FIGURE 40
Factor; add.
This result shows that the equation is that of a circle with radius 0—that is, the point (4,  5). See FIGURE 39. Had a negative number (instead of 0) been obtained on the right, the equation would have represented no points at all, and there would be no points on its graph.
x
0
Complete the square on both x and y.
Factor out 4 and 9. Complete the square. Factor; simplify. Divide by 36.
This equation represents an ellipse having center (2,  3). See FIGURE 40. (continued)
420
CHAPTER 6
Analytic Geometry
(d) Since only one variable of x 2  6x + 8y  7 = 0 is squared (x, and not y), the equation represents a parabola. Rearrange the terms so that the term 8y (with the variable that is not squared) is alone on one side. x 2  6x + 8y  7 8y 8y 8y 8y 2 Be careful (x  3) with signs. (x  3) 2
y
(3, 2) 2
x –1
3
–2
6
x2 – 6x + 8y – 7 = 0
= = = = = =
0  x 2 + 6x + 7  (x 2  6x ) + 7 2  (x  6x + 9) + 7 + 9  (x  3) 2 + 16  8y + 16
=  8( y  2)
Isolate the term with y. Regroup; factor out  1. Complete the square. Factor; add. Add ( x  3)2; add  8y. Factor.
The parabola has vertex (3, 2) and opens downward, as shown in FIGURE 41.
FIGURE 41
(e) 4( y 2
4y 2  16y  9x 2 + 18x =  43  4y )  9(x 2  2x ) =  43
4( y 2  4y + 4  4)  9(x 2  2x + 1  1) =  43 4( y 2  4y + 4)  16  9(x 2  2x + 1) + 9 =  43 4( y  2) 2  9(x  1) 2 =  36 y
(–1, 2) 0
Complete the square. Distributive property Factor; add 16 and subtract 9.
Because of the  36, we might think that this equation does not have a graph. However, the minus sign in the middle on the left shows that the graph is that of a hyperbola. Be careful here! ( y  2) 2 (x  1) 2 = 1 Divide by  36; rearrange terms. 4 9
(3, 2)
–2
Factor out 4; factor out  9.
x 3
4y2 – 16y – 9x 2 + 18x = –43
This hyperbola has center (1, 2). The graph is shown in FIGURE 42.
FIGURE 42
Eccentricity In Sections 6.1 and 6.2, we introduced definitions of the conic sections. The conic sections (or conics) can all be characterized by one general definition.
Conic A conic is the set of all points P(x, y) in a plane such that the ratio of the distance from P to a fixed point and the distance from P to a fixed line is constant.
y
x = –c D(–c, y)
P(x, y) d(P, F ) e= d(P, D) 0
x
F(c, 0)
FIGURE 43
For a parabola, the fixed line is the directrix and the fixed point is the focus. In the focus is F(c, 0) and the directrix is the line x =  c. The constant ratio is called the eccentricity of the conic, written e. This is not the same e as the base of natural logarithms. If the conic is a parabola, then by definition, the distances d(P, F) and d(P, D) in FIGURE 43 are equal. Thus, every parabola has eccentricity 1. FIGURE 43,
6.3 y
x
e=1 4
0 6 c 6 a c 0 6 6 1 a 0 6 e 6 1.
x
y
x
e=3 4 FIGURE 44
421
For an ellipse, eccentricity is a measure of its “roundness.” The constant ratio in the definition is e = ac , where (as before) c is the distance from the center of the figure to a focus and a is the distance from the center to a vertex. By the definition of an ellipse, a 2 7 b 2 and c = 2a 2  b 2. Thus, for the ellipse,
y
e=1 2
Summary of the Conic Sections
Divide by a. e =
c a
If a is constant, letting c approach 0 would cause the ratio ac to approach 0, which also causes b to approach a (so that 2a 2  b 2 = c would approach 0). Since b determines the endpoints of the minor axis, the lengths of the major and minor axes are almost the same, producing an ellipse very close in shape to a circle when e is very close to 0. In a similar manner, if e approaches 1, then b will approach 0. The path of Earth around the sun is an ellipse that is very nearly circular. In fact, for this ellipse, e L 0.017. By contrast, the path of Halley’s comet is a very flat ellipse, with e L 0.97. FIGURE 44 compares ellipses with different eccentricities. The equation of a circle can be written (x  h) 2 + ( y  k) 2 = r 2 ( y  k) 2 (x  h) 2 + = 1. r2 r2
Divide by r 2.
In a circle, the foci coincide with the center, so a = b and c = 2a 2  b 2 = 0. Thus, it follows that e = 0. EXAMPLE 2
Finding Eccentricity from Equations of Ellipses Find the eccentricity of each ellipse. y2 x2 + = 1 (a) (b) 5x 2 + 10y 2 = 50 9 16 Solution (a) Since 16 7 9, let a 2 = 16, which gives a = 4. Also,
y
c = 2a 2  b 2 = 216  9 = 27.
x
e = 1.1
Finally, e =
c a
=
27 4
L 0.66.
(b) Divide 5x 2 + 10y 2 = 50 by 50 to obtain
y
x2 10
+
y2 5
= 1. Here, a 2 = 10, with
a = 210. Now find c. x
c = 210  5 = 25 and
e=2
e =
25 210
=
1 22
L 0.71.
The hyperbola in standard form
y
x
e=5
FIGURE 45
y2 x2 = 1 or a2 b2
y2 x2 = 1, a2 b2
where c = 2a 2 + b 2, also has eccentricity e = ac . By definition, c = 2a 2 + b 2 7 a, so ac 7 1, and for a hyperbola, e 7 1. Narrow hyperbolas have e near 1, and wide hyperbolas have large e. See FIGURE 45.
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CHAPTER 6
Analytic Geometry
EXAMPLE 3
Finding Eccentricity from the Equation of a Hyperbola
Find the eccentricity of the hyperbola
x2 9

y2 4
= 1.
Solution Here a 2 = 9; thus, a = 3, c = 29 + 4 = 213, and e =
213 c = L 1.2. a 3
The table summarizes this discussion of eccentricity. Conic
Eccentricity
Circle Parabola Ellipse
e = 0 e = 1 e = ac and 0 6 e 6 1
Hyperbola
e =
c a
and e 7 1
EXAMPLE 4
Finding Equations of Conics by Using Eccentricity Find an equation for each conic with center at the origin and eccentricity e. (a) Focus (3, 0); e = 2 (b) Vertex (0,  8); e = 12 Solution (a) Since e = 2, which is greater than 1, the conic is a hyperbola with c = 3. c a 3 2 = a 3 a = 2 e =
Let e = 2 and c = 3. Solve for a.
Now we find b using the equation a 2 + b 2 = c 2, or b 2 = c 2  a 2. 9 27 2 = c 2 = 32 = 9; a 2 = A 32 B = 94 4 4 The given focus is on the xaxis, so the x 2term is positive, and the equation is b2 = c2  a2 = 9 
x2 9 4

y2 27 4
= 1, or
4y 2 4x 2 = 1. 9 27
Simplify the complex fractions.
(b) The graph of the conic is an ellipse because e = 12 6 1. From the given vertex (0,  8), we know that the vertices are on the yaxis and a = 8. c e = a 1 c = Let e = 12 and a = 8. 2 8 Solve for c. c = 4 Since b 2 = a 2  c 2 = 64  16 = 48, the equation is y2 x2 + = 1. 48 64
6.3
Summary of the Conic Sections
423
EXAMPLE 5
Applying an Ellipse to the Orbit of a Planet The orbit of the planet Mars is an ellipse with the sun at one focus. The eccentricity of the ellipse is 0.0935, and the closest Mars comes to the sun is 128.5 million miles. (Source: The World Almanac and Book of Facts.) Find the maximum distance of Mars from the sun. y
Mars Sun a–c a+c
c
a
x
Solution FIGURE 46 shows the orbit of Mars with the origin at the center of the ellipse and the sun at one focus. Mars is closest to the sun when Mars is at the right endpoint of the major axis and farthest from the sun when Mars is at the left endpoint. Therefore, the smallest distance is a  c, and the greatest distance is a + c. Since a  c = 128.5, c = a  128.5. Using e = 0.0935, we can find a. e = 0.0935 = 0.0935a =  0.9065a = a L
Not to scale FIGURE 46
c a a  128.5 a a  128.5  128.5 141.8
Substitute for e and c. Multiply by a. Subtract a. Divide by  0.9065.
c L 141.8  128.5 = 13.3
Then
a + c L 141.8 + 13.3 = 155.1.
and
Thus, the maximum distance of Mars from the sun is about 155.1 million miles.
6.3
EXERCISES
Checking Analytic Skills The equation of a conic section is given in a familiar form. Identify the type of graph (if any) that each equation has, without actually graphing. See the summary chart in this section. Do not use a calculator. 1. x 2 + y 2 = 144
2. (x  2) 2 + ( y + 3) 2 = 25
3. y = 2x 2 + 3x  4
4. x = 3y 2 + 5y  6
5. x  1 =  3( y  4) 2
6.
y2 x2 + = 1 25 36
8. x 2  y 2 = 1
9.
y2 x2 = 1 4 16
7.
y2 x4 + = 1 49 100
10.
(x + 2) 2 ( y  4) 2 + = 1 9 16
11.
y2 x2 = 1 25 25
12. y + 7 = 4(x + 3) 2
13.
y2 x2 = 1 4 9
14.
y2 x2 = 1 + 4 9
15.
16. x 2 = 25  y 2
17. x 2  6x + y = 0
( y  2) 2 (x + 3) 2 + = 1 16 16
18. 11  3x = 2y 2  8y
19. 4(x  3) 2 + 3( y + 4) 2 = 0
20. 2x 2  8x + 2y 2 + 20y = 12
21. x  4y 2  8y = 0
22. x 2 + 2x =  4y
23. 6x 2  12x + 6y 2  18y + 25 = 0
24. 4x 2  24x + 5y 2 + 10y + 41 = 0
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Analytic Geometry
CHAPTER 6
Determine the type of conic section represented by each equation, and graph it, provided a graph exists. 25. x 2 = 4y  8
26.
y2 x2 + = 1 4 4
27. x 2 = 25 + y 2
28. 9x 2 + 36y 2 = 36
29.
y2 x2 + = 1 4 4
30.
31. y 2  4y = x + 4
32. (x + 7) 2 + ( y  5) 2 + 4 = 0
33. 3x 2 + 6x + 3y 2  12y = 12
34.  4x 2 + 8x + y 2 + 6y =  6
35. 4x 2  8x + 9y 2  36y =  4
36. 3x 2 + 12x + 3y 2 = 0
( y + 1) 2 (x  4) 2 + = 0 8 2
37. Concept Check Identify the type of conic section consisting of the set of all points in the plane for which the sum of the distances from the points (5, 0) and ( 5, 0) is 14.
38. Concept Check Identify the type of conic section consisting of the set of all points in the plane for which the absolute value of the difference of the distances from the points (3, 0) and ( 3, 0) is 2.
39. Concept Check Identify the type of conic section consisting of the set of all points in the plane for which the distance from the point (3, 0) is one and onehalf times the distance from the line x = 43.
40. Concept Check Identify the type of conic section consisting of the set of all points in the plane for which the distance from the point (2, 0) is onethird the distance from the line x = 10.
Find the eccentricity of each ellipse or hyperbola. 41. 12x 2 + 9y 2 = 36
42. 8x 2  y 2 = 16
43. x 2  y 2 = 4
44. x 2 + 2y 2 = 8
45. 4x 2 + 7y 2 = 28
46. 9x 2  y 2 = 1
47. x 2  9y 2 = 18
48. x 2 + 10y 2 = 10
Write an equation for each conic. Each parabola has vertex at the origin, and each ellipse or hyperbola is centered at the origin. 49. Focus (0, 8); e = 1 52. Focus (0,  2); e =
50. Focus ( 2, 0); e = 1 2 3
53. Vertex ( 6, 0); e = 2
55. Focus (0,  1) ; e = 1
54. Vertex (0, 4); e
56. Focus (2, 0); e =
57. Vertical major axis of length 6; e =
4 5
1 2 = 53
51. Focus (3, 0); e = 6 5
58. yintercepts  4 and 4; e =
7 3
Find the eccentricity of each conic section. The point shown on the xaxis is a focus and the line shown is a directrix. y
59.
y
60.
y
61. x = –√ 2
x = –9 x = 27 (–3, 8)
(4, 103 )
x
(√ 2 , 0)
x
x
(3, 0) (–4, 0)
62.
63.
(–27, 48.75) y x = 4
y
64. x = –20
x=4
y
(5, 20) (9, 0) x
x
x
(27, 0)
(9, –7.5)
(20, 0)
6.4
(Modeling) Solve each application. 65. Orbit of Mars The orbit of Mars around the sun is an ellipse with equation y2 x2 + = 1, 5013 4970 where x and y are measured in millions of miles. Approximate the eccentricity of this ellipse. 66. Orbits of Neptune and Pluto Neptune and Pluto both have elliptical orbits with the sun at one focus. Neptune’s orbit has a = 30.1 astronomical units (AU) and eccentricity e = 0.009, whereas Pluto’s orbit has a = 39.4 and e = 0.249. (1 AU is equal to the average distance from Earth to the sun and is approximately 149,600,000 kilometers.) (Source: Zeilik, M., S. Gregory, and E. Smith, Introductory Astronomy and Astrophysics, Saunders College Publishers.) (a) Position the sun at the origin, and determine an equation for each orbit. (b) Graph both equations on the same coordinate axes. Use the window 3 60, 604 by 3 40, 404. 67. Velocity of a Planet in Orbit The maximum and minimum velocities in kilometers per second of a planet moving in an elliptical orbit can be calculated with the equations vmax =
1 + e 2pa P B1  e
and
vmin =
1  e 2pa , P B1 + e
where a is in kilometers, P is its orbital period in seconds, and e is the eccentricity of the orbit. (Source: Zeilik, M., S. Gregory, and E. Smith, Introductory Astronomy and Astrophysics, Saunders College Publishers.)
6.4
Parametric Equations
425
(a) Calculate vmax and vmin for Earth if a = 1.496 * 10 8 kilometers and e = 0.0167. 9 (b) If an object has a circular orbit, what can be said about its orbital velocity? 9 (c) Kepler showed that the sun is located at a focus of a planet’s elliptical orbit. He also showed that a planet has minimum velocity when its distance from the sun is maximum, and a planet has maximum velocity when its distance from the sun is minimum. Where do the maximum and minimum velocities occur in an elliptical orbit? 68. Distance between Halley’s Comet and the Sun The famous Halley’s comet last passed Earth in February 1986 and will next return in 2062. Halley’s comet has an elliptical orbit of eccentricity 0.9673 with the sun at one of the foci. The greatest distance of the comet from the sun is 3281 million miles. (Source: The World Almanac and Book of Facts.) Find the least distance between Halley’s comet and the sun. 69. Orbit of Earth The orbit of Earth is an ellipse with the sun at one focus. The distance between Earth and the sun ranges from 91.4 to 94.6 million miles. (Source: The World Almanac and Book of Facts.) Estimate the eccentricity of Earth’s orbit. 70. Distance Traveled by Mercury Use the formula of Exercise 94 in Section 6.2 to estimate the distance that is traveled by the planet Mercury in one orbit of the sun, if a = 36.0, b = 35.2, and the units are in millions of miles.
Parametric Equations
Graphs of Parametric Equations and Their Rectangular Equivalents • Alternative Forms of Parametric Equations • An Application of Parametric Equations
GCM
TECHNOLOGY NOTE In addition to graphing rectangular equations, graphing calculators are capable of graphing plane curves defined by parametric equations. The calculator must be set in parametric mode, and the window requires intervals for the parameter t, as well as for x and y. Consult the graphing calculator manual that accompanies this text or your owner’s manual.
We have graphed sets of ordered pairs that correspond to functions of the form y = ƒ(x) or to relations of the form Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0. Another way to determine a set of ordered pairs involves two functions ƒ and g defined by x = ƒ(t) and y = g (t), where t is a real number in some interval I. Each value of t leads to a corresponding xvalue and a corresponding yvalue, and thus to an ordered pair (x, y).
Parametric Equations of a Plane Curve A plane curve is a set of points (x, y) such that x = ƒ(t), y = g (t), and ƒ and g are both continuous on an interval I. The equations x = ƒ(t) and y = g (t) are parametric equations with parameter t.
426
CHAPTER 6
Analytic Geometry
Graphs of Parametric Equations and Their Rectangular Equivalents GCM
EXAMPLE 1
Graphing a Plane Curve Defined Parametrically Graph the plane curve defined by the parametric equations x = t 2,
y = 2t + 3, for t in 3 3, 34,
and then find an equivalent rectangular equation.
Analytic Solution Make a table of corresponding values of t, x, and y over the domain of t. Then plot the points. The graph is a portion of a parabola with horizontal axis y = 3. See FIGURE 47. The arrowheads indicate the direction the curve traces as t increases. y
t x y –3 –2 1 0 1 2 3
Graphing Calculator Solution FIGURE 48 shows a TI83/84 Plus table of values. The calculator is in parametric mode.
9 –3 4 –1 1 1 0 3 1 5 4 7 9 9
(9, 9)
9
Parametric mode
6
x t2 y 2t
3
for t in + 3 [–3, 3]
FIGURE 48
shows a graph. The formation of the curve as T increases can be seen by using the pause feature and carefully observing the path of the curve. See the settings in the margin on page 427.
FIGURE 49
6 3
9
x
(9, 3) FIGURE 47
X1T = T 2, Y1T = 2T + 3 T in [–3, 3]
To find an equivalent rectangular equation, we eliminate the parameter t. For this curve, we begin by solving for t in the second equation, y = 2t + 3, because it leads to a unique solution. y = 2t + 3 2t = y  3 y  3 t = 2
Solve for t.
Divide by 2.
y  3 2 ( y  3) 2 1 b = = ( y  3) 2. 2 4 4
1 ( y  3) 2, 4
10
–10 Parametric mode
This is indeed an equation of a horizontal parabola that opens to the right. Because t is in 3 3, 34, x is in 30, 94 and y is in 3 3, 94. The rectangular equation must be given with its restricted domain as x =
–10
Subtract 3; rewrite.
Now we substitute the result in the first equation to get x = t2 = a
10
for x in 30, 94.
FIGURE 49
The displays at the bottom of FIGURES 48 and 49 indicate particular values of T, X, and Y. How do these relate to the parametric and rectangular forms of the equation of this parabola? How would you graph this curve if the calculator were in function mode?
6.4
Parametric Equations
427
EXAMPLE 2
Graphing a Plane Curve Defined Parametrically Graph the plane curve defined by x = 2t + 5, y = 24  t 2,
for t in 3 2, 24,
and then find an equivalent rectangular equation. For the parametric equations in Example 1, the screens show the entries that produce the standard window shown in FIGURE 49 on the previous page.
x = 2t + 5, y = √4 – t 2, t in [–2, 2]
Solution A graph is shown in FIGURE 50. To get an equivalent rectangular equation, solve the first equation, which does not involve a radical, for t, and then substitute into the second equation. Since t is in 3 2, 24, x is in the interval 31, 94. 2t + 5 = x,
t =
x  5 . 2
The rectangular equation is y = 24  t 2
6.7
y = –10.1
so
10.1
A
4  a
y2 = 4  a
x  5 2 b 2
x  5 2 b 2
(x  5) 2 4 2 4y = 16  (x  5) 2
–6.7
y2 = 4 
FIGURE 50
4y 2
+ (x 
5) 2
= 16
Substitute.
Square each side. 22 = 4 Multiply by 4. Add (x  5)2.
y2 (x  5) 2 + = 1, 4 16
Divide by 16.
which represents the complete ellipse. (By definition, the parametric graph has y Ú 0, so it is only the upper half of the ellipse.) EXAMPLE 3
Graphing a Line Defined Parametrically Graph the plane curve defined by x = t 2, y = t 2
x = t 2, y = t 2,
10
and then find an equivalent rectangular equation. –10
10
–10
Solution FIGURE 51 shows the graph of the curve in a standard window with both x and y in 3 10, 104. The graph is a ray, because both x and y must be greater than or equal to 0. Since both x and y equal t 2, y = x. To be equivalent, however, the rectangular equation must be given as y = x,
FIGURE 51
x Ú 0.
Alternative Forms of Parametric Equations Parametric representations of a curve are not unique. In fact, there are infinitely many parametric representations of a given curve. If the curve can be described by a rectangular equation y = ƒ(x) with domain X, then one simple parametric representation is x = t,
y = ƒ(t),
for t in X.
428
CHAPTER 6
Analytic Geometry
FOR DISCUSSION
EXAMPLE 4
Finding Alternative Parametric Equation Forms Give two parametric representations for the parabola y = (x  2) 2 + 1.
Consider the rectangular equation y  1 = (x  2) 2.
Solution The simplest choice is to let x = t, y = (t  2) 2 + 1,
Why are the parametric equations
for t in ( q, q).
Another choice that leads to a simpler equation for y is
x = 2t + 2, y = t + 1,
x = t + 2, y = t 2 + 1,
for t in 30, q) not equivalent to the rectangular equation?
for t in ( q, q).
An Application of Parametric Equations One application of parametric equations is to determine the path of a moving object whose position is given by the function defined by x = ƒ(t),
y = g (t), where t represents time.
The parametric equations give the position of the object at any time t. EXAMPLE 5
Using Parametric Equations to Define the Position of an Object in Motion The motion of a projectile moving in a direction at an angle u = 45° with the horizontal (neglecting air resistance) is given by x = v0
y = v0
22 t  16t 2, for t in 30, k4, 2
where t is time in seconds, v0 is the initial speed of the projectile in feet per second, x and y are in feet, and k is a positive real number. See FIGURE 52. Find the rectangular form of the equation.
y
v0 √2 t – 16t 2 2
v0
0
22 t, 2
Solution Solving the first equation for t and substituting the result into the second equation gives (after simplification) y = x 
x
v0 √2 t 2
the equation of a vertical parabola opening downward, as shown in FIGURE 52.
FIGURE 52
6.4
32 2 x , v0 2
EXERCISES
Checking Analytic Skills Graph each pair of parametric equations by hand, using values of t in 3 2, 24. Make a table of t, x, and yvalues, using t =  2,  1, 0, 1, and 2. Then plot the points and join them with a line or smooth curve for all values of t in 3 2, 24. Do not use a calculator. 1. x = 2t + 1, y = t  2
2. x =  t + 1, y = 3t + 2
3. x = t + 1, y = t 2  1
4. x = t  1, y = t 2 + 2
5. x = t 2 + 2, y =  t + 1
6. x =  t 2 + 2, y = t + 1
For each plane curve, use a graphing calculator to generate the curve over the interval for the parameter t, in the window specified. Then, find a rectangular equation for the curve. 7. x = 2t, y = t + 1, for t in 3 2, 34; window: 3 8, 84 by 3 8, 84
8. x = t + 2, y = t 2, for t in 3 1, 14; window: 30, 44 by 3 2, 24
6.4
9. x = 2t, y = 3t  4, for t in 30, 44; window: 3 6, 64 by 3 6, 104
window: 3 2, 304 by 3 2, 104
429
10. x = t 2, y = 2t, for t in 30, 44; window: 3 2, 204 by 30, 44
11. x = t 3 + 1, y = t 3  1, for t in 3 3, 34; window: 3 30, 304 by 3 30, 304 13. x = 2 t, y = 23t  1, for t in
Parametric Equations
C 13, 4 D ;
12. x = 2t  1, y = t 2 + 2, for t in 3 10, 104; window: 3 20, 204 by 30, 1204 14. x = ln(t  1), y = 2t  1, for t in (1, 104; window: 3 5, 54 by 3 2, 204
15. x = t + 2, y =  12 29  t 2, for t in 3 3, 34; window: 3 6, 64 by 3 4, 44
16. x = t, y = 24  t 2, for t in 3 2, 24; window: 3 6, 64 by 3 4, 44
17. x = t, y = 1t , for t in ( q, 0) ´ (0, q); window: 3 6, 64 by 3 4, 44
18. x = 2t  1, y = 1t , for t in ( q, 0) ´ (0, q); window: 3 6, 64 by 3 4, 44
For each plane curve, find a rectangular equation. State the appropriate interval for x or y. 19. x = 3t, y = t  1, for t in ( q, q)
20. x = t + 3, y = 2t, for t in ( q, q)
21. x = 3t 2, y = t + 1, for t in ( q, q)
22. x = t  2, y = 12 t 2 + 1, for t in ( q, q)
23. x = 3t 2, y = 4t 3, for t in ( q, q)
24. x = 2t 3, y =  t 2, for t in ( q, q)
25. x = t, y = 2t 2 + 2, for t in ( q, q)
26. x = 2t, y = t 2  1, for t in 30, q)
27. x = e t, y = e  t, for t in ( q, q)
28. x = e 2t, y = e t, for t in ( q, q)
29. x =
1 2t + 2
,y =
31. x = t + 2, y =
t , for t in ( 2, q) t + 2
1 , for t Z  2 t + 2
30. x =
t 1 ,y = , for t in (1, q) t  1 2t  1
32. x = t  3, y =
2 , for t Z 3 t  3
34. x = ln t, y = 3 ln t, for t in (0, q)
33. x = t 2, y = 2 ln t, for t in (0, q)
Give two parametric representations for each plane curve. Use your calculator to verify your results. 3 x  4 2
35. y = 2x + 3
36. y =
39. x = y 3 + 1
40. x = 2( y  3) 2  4
(Modeling) Solve each application. 43. Firing a Projectile A projectile is fired with an initial velocity of 400 feet per second at an angle of 45° with the horizontal. (See Example 5.) (a) Find the time to the nearest tenth when it strikes the ground. (b) Find the range (horizontal distance covered). (c) What is the maximum altitude? 44. Firing a Projectile If a projectile is fired at an angle of 30° with the horizontal, the parametric equations that describe its motion are v0 23 x = v0 t, y = t  16t 2, for t in 30, q). 2 2
37. y = 23x + 2, x in C  23, q B
38. y = (x + 1) 2 + 1
41. x = 2y + 1
42. x =
1 y + 1
Repeat Exercise 43 if the projectile is fired at 800 feet per second. 45. Path of a Projectile A projectile moves so that its position at any time t is given by the equations x = 60t and
y = 80t  16t 2.
Graph the path of the projectile, and find the equivalent rectangular equation. Use the window 30, 3004 by 30, 2004. 46. Path of a Projectile Repeat Exercise 45, using the window 30, 3004 by 30, 2004, if the path is given by the equations x = t 2 and
y =  16t + 64 2t.
430
Analytic Geometry
CHAPTER 6
47. Show that the rectangular equation for the curve defined by the equations x = v0
22 22 t, y = v0 t  16t 2, 2 2
is y = x 
for t in 30, k4,
32 2 x . (See Example 5.) v02
SECTIONS 6.3 AND 6.4
48. Find the vertex of the parabola given by the rectangular equation of Exercise 47. 49. Give two parametric representations of the line through the point (x1, y1) with slope m. 50. Give two parametric representations of the parabola y = a(x  h) 2 + k.
REVIEWING BASIC CONCEPTS
Write each equation in standard form, and name the type of conic section defined. 1. 3x 2 + y 2  6x + 6y = 0
2. y 2  2x 2 + 8y  8x  4 = 0
3. 3y 2 + 12y + 5x = 3
5. 8y 2  4x 2 = 8
6. 3x 2 + 4y 2 = 108
Find the eccentricity of each conic section. 4. x 2 + 25y 2 = 25
Find an equation of each conic. (Hint: A sketch may be helpful.) 7. Focus ( 2, 0); vertex (0, 0); eccentricity 1
8. Foci (⫾3, 0); major axis length 10 10. Vertices (⫾3, 0); asymptotes y = ⫾23 x
9. Foci (0, ⫾5); vertices (0, ⫾4) Solve each problem. 11. The figure represents an elliptical stone arch with the dimensions (in feet) indicated. Find the height y of the arch 6 feet from the center of the base.
6
x = 2t, y = 2t 2 + 1, 9
y 6
for t in ( q, q).
(a) Graph the curve by hand and support your graph with your calculator. (b) Find an equivalent rectangular equation for the curve.
30
SUMMARY
KEY TERMS & SYMBOLS
KEY CONCEPTS
6.1 Circles and Parabolas
CIRCLE A circle is a set of points in a plane that are equidistant from a fixed point. The distance is called the radius of the circle, and the fixed point is called the center.
conic sections degenerate conic sections circle radius center parabola focus directrix latus rectum
12. A plane curve is defined by
CENTER–RADIUS FORM OF THE EQUATION OF A CIRCLE The circle with center (h, k) and radius r has equation (x ⴚ h)2 ⴙ ( y ⴚ k)2 ⴝ r 2. A circle with center (0, 0) and radius r has equation x 2 ⴙ y 2 ⴝ r 2. GENERAL FORM OF THE EQUATION OF A CIRCLE For real numbers c, d, and e, the general form of the equation of a circle is x 2 ⴙ y 2 ⴙ cx ⴙ dy ⴙ e ⴝ 0.
CHAPTER 6
KEY TERMS & SYMBOLS
Summary
431
KEY CONCEPTS PARABOLA A parabola is a set of points in a plane equidistant from a fixed point and a fixed line. The fixed point is called the focus, and the fixed line the directrix, of the parabola. PARABOLA WITH A VERTICAL AXIS The parabola with focus (0, c) and directrix y =  c has equation x 2 ⴝ 4cy. The parabola has vertical axis x = 0 and opens upward if c 7 0 or downward if c 6 0. PARABOLA WITH A HORIZONTAL AXIS The parabola with focus (c, 0) and directrix x =  c has equation y 2 ⴝ 4cx. The parabola has horizontal axis y = 0 and opens to the right if c 7 0 or to the left if c 6 0. TRANSLATION OF A PARABOLA The parabola with vertex (h, k) and vertical line x = h as axis has an equation of the form (x ⴚ h)2 ⴝ 4c( y ⴚ k). The parabola opens upward if c 7 0 or downward if c 6 0. The parabola with vertex (h, k) and horizontal line y = k as axis has an equation of the form ( y ⴚ k)2 ⴝ 4c(x ⴚ h). The parabola opens to the right if c 7 0 or to the left if c 6 0.
6.2 Ellipses and Hyperbolas ellipse focus (foci) major axis minor axis center vertices hyperbola transverse axis asymptotes fundamental rectangle
ELLIPSE An ellipse is the set of all points in a plane, the sum of whose distances from two fixed points is constant. Each fixed point is called a focus (plural foci) of the ellipse. STANDARD FORMS OF EQUATIONS FOR ELLIPSES y2 x2 The ellipse with center at the origin and equation 2 ⴙ 2 ⴝ 1 has vertices (⫾a, 0), enda b points of the minor axis (0, ⫾b), and foci (⫾c, 0), where c 2 = a 2  b 2 and a 7 b 7 0. y2 x2 The ellipse with center at the origin and equation 2 ⴙ 2 ⴝ 1 has vertices (0, ⫾a), endb a points of the minor axis (⫾b, 0), and foci (0, ⫾c), where c 2 = a 2  b 2 and a 7 b 7 0. TRANSLATED ELLIPSES The preceding equations can be extended to ellipses having center (h, k) by replacing x and y with x  h and y  k, respectively. HYPERBOLA A hyperbola is the set of all points in a plane such that the absolute value of the difference of the distances from two fixed points is constant. The two fixed points are called the foci of the hyperbola. STANDARD FORMS OF EQUATIONS FOR HYPERBOLAS y2 x2 The hyperbola with center at the origin and equation 2 ⴚ 2 ⴝ 1 has vertices (⫾a, 0) and a b foci (⫾c, 0), where c 2 = a 2 + b 2. The hyperbola with center at the origin and equation y2 x2 ⴚ ⴝ 1 has vertices (0, ⫾a) and foci (0, ⫾c), where c 2 = a 2 + b 2. a2 b2 TRANSLATED HYPERBOLAS The preceding equations can be extended to hyperbolas having center (h, k) by replacing x and y with x  h and y  k, respectively. (continued)
432
Analytic Geometry
CHAPTER 6
KEY TERMS & SYMBOLS
KEY CONCEPTS
6.3 Summary of the Conic Sections
The conic sections in this chapter have equations that can be written in the form Ax 2 ⴙ Dx ⴙ Cy 2 ⴙ Ey ⴙ F ⴝ 0.
conic eccentricity, e
Conic Section
Characteristic
Examples
Parabola
Either A = 0 or C = 0, but not both
y = x2 x = 3y 2 + 2y  4
Circle
A = C Z 0
x 2 + y 2 = 16
Ellipse
A Z C, AC 7 0
y2 x2 + = 1 16 25
Hyperbola
AC 6 0
x2  y2 = 1
CONIC A conic is the set of all points P (x, y) in a plane such that the ratio of the distance from P to a fixed point and the distance from P to a fixed line is constant. This constant ratio is called the eccentricity of the conic. Conic
6.4 Parametric Equations plane curve parametric equations parameter
6
Eccentricity
Circle Parabola Ellipse
e = 0 e = 1 e = ac and 0 6 e 6 1
Hyperbola
e =
c a
and e 7 1
PARAMETRIC EQUATIONS OF A PLANE CURVE A plane curve is a set of points (x, y) such that x = ƒ(t), y = g (t), and ƒ and g are both continuous on an interval I. The equations x = ƒ(t) and y = g (t) are parametric equations with parameter t.
REVIEW EXERCISES
Write an equation for the circle satisfying the given conditions. Graph it by hand, and give the domain and range. 1. Center ( 2, 3); radius 5
2. Center A 25,  27 B ; radius 23
3. Center ( 8, 1); passing through (0, 16)
4. Center (3,  6); tangent to the xaxis
Find the center and radius of each circle. 5. x 2  4x + y 2 + 6y + 12 = 0
6. x 2  6x + y 2  10y + 30 = 0
7. 2x 2 + 14x + 2y 2 + 6y =  2
8. 3x 2 + 3y 2 + 33x  15y = 0
9 9.
Describe the graph of (x  4) 2 + ( y  5) 2 = 0.
CHAPTER 6
Review Exercises
433
Give the focus, directrix, and axis for each parabola, graph it by hand, and give the domain and range. 2 10. y 2 =  x 3
11. y 2 = 2x
12. 3x 2  y = 0
13. x 2 + 2y = 0
Write an equation for each parabola with vertex at the origin. 14. Focus (4, 0)
15. Through (2, 5); opening to the right
16. Through (3,  4); opening downward
17. Focus (0,  3)
9 18.
Explain how to determine just by looking at its equation whether a parabola has a vertical or a horizontal axis, and whether it opens up, down, to the left, or to the right.
Write an equation for each parabola. 19. Vertex ( 5, 6); focus (2, 6)
20. Vertex (4, 3); focus (4, 5)
Graph each ellipse or hyperbola by hand, and give the domain, range, and coordinates of the vertices. 21.
y2 x2 + = 1 5 9
22.
y2 x2 + = 1 16 4
23.
y2 x2 = 1 64 36
24.
y2 x2 = 1 25 9
25.
(x  3) 2 + ( y + 1) 2 = 1 4
26.
( y + 3) 2 (x  2) 2 + = 1 9 4
27.
( y + 2) 2 (x + 3) 2 = 1 4 9
28.
(x + 1) 2 ( y  2) 2 = 1 16 4
Write an equation for each conic section with center at the origin. 29. Ellipse: vertex (0, 4); focus (0, 2)
30. Ellipse: xintercept 6; focus ( 2, 0)
31. Hyperbola: focus (0,  5); yintercepts  4 and 4
32. Hyperbola: yintercept  2; passing through (2, 3)
33. Focus (0,  3); e =
2 3
34. Focus (5, 0); e =
5 2
35. Consider the circle with equation x 2 + y 2 + 2x + 6y  15 = 0. (a) What are the coordinates of the center? (b) What is the radius? (c) What two functions must be graphed to graph this circle with your calculator in function mode?
Match each equation in Column I with the appropriate description in Column II.
Concept Check
II
I 36.
4x 2
+
y2
37. x =
2y 2
38. (x 
1) 2
39.
x2 36
+
y2 9
A. Circle; center (1,  2); radius 6
= 36 + 3 + (y +
B. Hyperbola; center (2, 1) 2) 2
= 36
= 1
40. ( y  1) 2  (x  2) 2 = 36 41.
y2
= 36 +
4x 2
C. Ellipse; major axis on xaxis D. Ellipse; major axis on yaxis E. Parabola; opens right F. Hyperbola; transverse axis on yaxis
434
CHAPTER 6
Analytic Geometry
Write the equation in standard form for an ellipse or a hyperbola centered at (h, k). Identify the center and the vertices. 42. 4x 2 + 8x + 25y 2  250y =  529
43. 5x 2 + 20x + 2y 2  8y =  18
44. x 2 + 4x  4y 2 + 24y = 36
45. 4y 2 + 8y  3x 2 + 6x = 11
Find the eccentricity of each ellipse or hyperbola. 46. 9x 2 + 25y 2 = 225
47. 4x 2 + 9y 2 = 36
48. 9x 2  y 2 = 9
Write an equation for each conic section. 49. Parabola with vertex ( 3, 2) and yintercepts 5 and  1
50. Hyperbola with foci (0, 12) and (0,  12); asymptotes y = ⫾x
51. Ellipse consisting of all points in the plane, the sum of whose distances from (0, 0) and (4, 0) is 8
52. Hyperbola consisting of all points in the plane for which the absolute value of the difference of the distances from (0, 0) and (0, 4) is 2
Use a graphing calculator to graph each plane curve in the specified window. 53. x = 4t  3, y = t 2, for t in 3 3, 44; window: 3 20, 204 by 3 20, 204
54. x = t 2, y = t 3, for t in 3 2, 24;
window: 3 15, 154 by 3 10, 104
Find a rectangular equation for each plane curve. State the interval for x. 55. x = 2t  1, y = 2t, for t in 31, q)
56. x = 3t + 2, y = t  1, for t in 3 5, 54
(Modeling) Solve each application.
Trajectory of a Satellite When a satellite is near Earth, its orbital trajectory may trace out a hyperbola, a parabola, or an ellipse. The type of trajectory depends on the satellite’s velocity V in meters per second. It will be hyperbolic if V 7 k> 2D,
57. Orbit of Venus The orbit of Venus is an ellipse with the sun at one focus. The eccentricity of the orbit is e = 0.006775, and the major axis has length 134.5 million miles. (Source: The World Almanac and Book of Facts.) Estimate the least and greatest distances of Venus from the sun.
parabolic if V = k> 2D, and elliptic if V 6 k> 2D, where k = 2.82 * 10 7 is a constant and D is the distance in meters from the satellite to the center of Earth. (Source: Loh, W., Dynamics and Thermodynamics of Planetary Entry, PrenticeHall; Thomson, W., Introduction to Space Dynamics, John Wiley and Sons.) Use this information in Exercises 59–61. 59. When the artificial satellite Explorer IV was at a maximum distance D of 42.5 * 10 6 meters from Earth’s center, it had a velocity V of 2090 meters per second. Determine the shape of its trajectory. 60. If a satellite is scheduled to leave Earth’s gravitational influence, its velocity must be increased so that its trajectory changes from elliptic to hyperbolic. Determine the minimum increase in velocity necessary for Explorer IV to escape Earth’s gravitational influence given that D = 42.5 * 10 6 meters.
9 61. 58. Orbit of the Comet Swift–Tuttle Comet Swift–Tuttle has an elliptical orbit of eccentricity e = 0.964 with the sun at one focus. Find the equation of the comet, given that the closest it comes to the sun is 89 million miles.
Explain why it is easier to change a satellite’s trajectory from an ellipse to a hyperbola when D is maximum rather than minimum.
62. Center of an Ellipse If Ax 2 + Cy 2 + Dx + Ey + F = 0 is the general equation of an ellipse, find its center point by completing the square.
CHAPTER 6
6
Test
435
TEST
1. Match each equation in Column I with the appropriate description in Column II. II
I (x + 3) 2 ( y + 2) 2 = 1 4 16 2 2 (b) (x  3) + ( y  2) = 16
A. Circle; center (3, 2); radius 4
(a)
(c) (x +
3) 2
+ (y 
(d) (x +
3) 2
=  ( y  4)
2) 2
B. Hyperbola; center ( 3,  2) C. Ellipse; center ( 3,  2)
= 16
D. Circle; center ( 3, 2); radius 4
(e) x  4 = ( y  2) 2 (x + 3) 2 ( y + 2) 2 (f) + = 1 4 16
E. Parabola; opens downward F. Parabola; opens right
2. Give the coordinates of the focus, and the equation of the directrix, for the parabola with equation y 2 = 18 x. 3. Graph y = 
A
1 
x2 . Is it the graph of a function? Find the domain and range. 36
4. What two functions are used to graph
y2 x2 = 1 on a calculator in function mode? 25 49
Graph each relation by hand. Identify the graph, and give the radius, center, vertices, and foci, as applicable. 5.
y2 x2 = 1 4 9
8. x 2 + ( y  4) 2 = 9
6. x 2 + 4y 2 + 2x  16y + 17 = 0 9.
(x  3) 2 ( y + 1) 2 + = 1 49 16
7. y 2  8y  2x + 22 = 0 10. x = 4t 2  4, y = t  1, for t in 3  1.5, 1.54
11. Write an equation for each conic. (a) Center at the origin; focus (0,  2); e = 1 (b) Center at the origin; vertical major axis of length 6; e =
5 6
12. (Modeling) Height of a Bridge’s Arch An arch of a bridge has the shape of the top half of an ellipse. The arch is 40 feet wide and 12 feet high at the center. Find the equation of the complete ellipse. Find the height of the arch 10 feet from the center at the bottom.
12 ft 40 ft
13. Use a graphing calculator to graph x = t + ln t, y = t + e t, for t in (0, 2]. Use the window [ 5, 5] by [0, 10]. 14. Find a rectangular equation for x =
1 , y = t + 3, for t Z  3. t + 3
7 Systems of Equations and Inequalities; Matrices IMAGINE THAT WE are developing coffee blends for Starbucks Corporation, the leading roaster and retailer of specialty coffees in the world. Three varieties of coffee—Arabian Mocha Sanani, Organic ShadeGrown Mexico, and Guatemala Antigua—are combined and roasted, yielding 50pound batches of coffee beans. To develop the perfect balance of flavor, boldness, and acidity, we combine twice as many pounds of Guatemala Antigua, which retails for $10.19 per pound, as Arabian Mocha Sanani, which sells for $15.99 per pound. Organic ShadeGrown Mexico rounds out the blend at $12.99 per pound. How many pounds of each variety must we use to create a blend that retails for $12.37 per pound? We can answer this question by setting up a system of equations in three variables, a + b + c = 50 c = 2a 15.99a + 12.99b + 10.19c = 12.37(50), where a represents the number of pounds of Arabian Mocha Sanani, b the number of pounds of Organic ShadeGrown Mexico, and c the number of pounds of Guatemala Antigua. (See Exercise 77 in Section 7.3.) In this chapter, we learn to write and solve systems of equations and inequalities in two, three, or four variables for many business and science applications. Source: Starbucks Corp.
CHAPTER OUTLINE 7.1 Systems of Equations 7.2 Solution of Linear Systems in Three Variables 7.3 Solution of Linear Systems by Row Transformations 7.4 Matrix Properties and Operations 7.5 Determinants and Cramer’s Rule 7.6 Solution of Linear Systems by Matrix Inverses 7.7 Systems of Inequalities and Linear Programming 7.8 Partial Fractions
From Chapter 7 of A Graphical Approach to Precalculus with Limits: A Unit Circle Approach, Fifth Edition. John Hornsby, Margaret Lial, Gark K. Rockswold. Copyright © 2011 by Pearson Education, Inc. All rights reserved.
7.1
7.1
437
Systems of Equations
Systems of Equations
Linear Systems • Substitution Method • Elimination Method • Special Systems • Nonlinear Systems • Applications of Systems
Linear Systems The definition of a linear equation given in Chapter 1 can be extended to more variables; any equation of the form a 1 x1 ⴙ a 2 x2 ⴙ Á ⴙ a nxn ⴝ b, for real numbers a1, a2, Á , an (not all of which are 0) and b, is a linear equation, or a firstdegree equation in n unknowns. A set of equations is called a system of equations. The solutions of a system of equations must satisfy every equation in the system. If all the equations in a system are linear, the system is a system of linear equations, or a linear system. The solution set of a linear equation in two variables is an infinite set of ordered pairs. Since the graph of such an equation is a straight line, there are three possible outcomes for the graph of a system of two linear equations in two variables, as shown in FIGURE 1. They are as follows: 1. The graphs intersect at exactly one point, which gives the (single) orderedpair solution of the system. The system is consistent and the equations are independent. See FIGURE 1(a). 2. The graphs are parallel lines, so there is no solution and the solution set is 0. The system is inconsistent and the equations are independent. See FIGURE 1(b). 3. The graphs are the same line, and there are infinitely many solutions. The system is consistent and the equations are dependent. See FIGURE 1(c). y
y
2x – y = 2 x + y = –2
y
3x + 2y = 3 3x + 2y = –4 x
0
x
0
x – y = –3 2x – 2y = –6
(0, –2)
One solution (a)
x
0
No solutions (b)
Infinitely many solutions (c)
FIGURE 1
Substitution Method Although the number of solutions of a linear system can usually be seen from the graph of the equations of the system, determining an exact solution from a graph is often difficult. In such cases, we use algebraic methods of solution. In a system of two equations with two variables, the substitution method involves using one equation to find an expression for one variable in terms of the other, and then substituting into the other equation of the system.
438
CHAPTER 7
Systems of Equations and Inequalities; Matrices
EXAMPLE 1
GCM
Solving a System by Substitution
Solve the system. 3x + 2y = 11 x + y = 3 Analytic Solution One way to solve this system is to solve equation (2) for y, getting y = x + 3. Then substitute x + 3 for y in equation (1) and solve for x. 3x + 2y 3x + 2(x + 3) 3x + 2x + 6 5x + 6 5x x
= = = = = =
11 11 11 11 5 1
(1) Let y = x + 3. Distributive property
(1) (2)
Graphing Calculator Solution To find the solution graphically, first solve equations (1) and (2) for y to get Y1 =  1.5X + 5.5 and Y2 = X + 3. Graph Y1 and Y2 in the standard viewing window to determine that the point of intersection is (1, 4), as seen in FIGURE 2. The table in FIGURE 3 shows numerically that when X = 1, both Y1 and Y2 are 4. Y2 = X + 3
Combine like terms. 10
Subtract 6. Divide by 5.
Replace x with 1 in y = x + 3 to determine that y = 1 + 3 = 4, so the solution is (1, 4). Check by substituting 1 for x and 4 for y in both equations of the original system. The solution set is 5(1, 4)6.
–10
10
–10
Y1 = –1.5X + 5.5 FIGURE 3
FIGURE 2
Looking Ahead to Calculus
In calculus, the definite integral (see Looking Ahead to Calculus, p. 168), b
La
3 ƒ( x )  g( x )4 dx,
often gives the area between the graphs of ƒ and g from x = a to x = b. A system of equations is used to find the xvalues a and b where the two graphs intersect.
Elimination Method Another way to solve a system of two equations, the elimination method, uses multiplication and addition to eliminate a variable from one equation. In this process, the given system is replaced by new systems that have the same solution set as the original system. Systems that have the same solution set are called equivalent systems.
EXAMPLE 2
Solving a System by Elimination
Solve the system. 3x  4y = 1 2x + 3y = 12
(1) (2)
Solution To eliminate x, multiply each side of equation (1) by  2 and each side of equation (2) by 3 to get  6x + 8y =  2 6x + 9y = 36.
Multiply (1) by  2. (3) Multiply (2) by 3.
(4)
This new system will have the same solution set as the given system. We add equations (3) and (4) to eliminate x and solve the result for y.  6x + 8y 6x + 9y 17y y
= = = =
2 36 34 2
(3) (4) Add. Divide by 17.
7.1
Systems of Equations
439
Substitute 2 for y in equation (1) or (2). Write the xvalue 3x  4(2) = 1 first in the 3x = 9 ordered pair.
(1) Multiply; add 8.
x = 3
Divide by 3.
The solution is (3, 2), which can be checked by substituting 3 for x and 2 for y in both equations of the original system. Check:
3x  4y = 1 3(3)  4(2) = 1 1 = 1
2x + 3y = 12 2(3) + 3(2) = 12 12 = 12
(1) ? True
(2) ? True
Both check; the solution set is 5(3, 2)6. The graph in FIGURE 4 supports this result and that the system is consistent.
y
3x – 4y = 1 4
(3, 2) 0 –1
x 3
6
2x + 3y = 12
Consistent system FIGURE 4
WHAT WENT WRONG? To find the solution in Example 2 graphically, a student solved each equation for y, entered the functions shown on the left, and got the incorrect graph shown on the right. 15
–10
10
–10
What Went Wrong?
What can be done to correct the graph?
Special Systems The examples presented so far in this section have all been consistent systems with a single solution. This is not always the case.
Answer to What Went Wrong? The student neglected to insert parentheses as necessary. Y1 should be entered as (3X  1)>4 and Y2 should be entered as (12  2X)>3.
440
CHAPTER 7
Systems of Equations and Inequalities; Matrices
EXAMPLE 3
Solving an Inconsistent System
Solve the system. 3x  2y = 4  6x + 4y = 7 Analytic Solution Eliminate the variable x by multiplying each side of equation (1) by 2 and then adding the result to equation (2). 6x  4y = 8  6x + 4y = 7 0 = 15
Multiply (1) by 2. (2)
(1) (2)
Graphing Calculator Solution In slope–intercept form, equation (1) is y = 1.5x  2 and equation (2) is y = 1.5x + 1.75. Since the lines have the same slope, but different yintercepts, they are parallel and have no point of intersection, which supports the analytic solution. See FIGURE 5.
False
y = 1.5x + 1.75
y = 1.5x – 2
10
Both variables were eliminated, leaving the false statement 0 = 15. This contradiction indicates that these two equations have no solutions in common. The system is inconsistent, and the solution set is 0.
–10
10
–10 Parallel lines FIGURE 5
EXAMPLE 4
Solving a System with Dependent Equations
Solve the system. 8x  2y =  4  4x + y = 2 Analytic Solution Divide each side of equation (1) by 2, and add the result to equation (2). 4x  y =  2  4x + y = 2 0 = 0
Divide (1) by 2. (2) True
The result, 0 = 0, is a true statement, which indicates that the equations of the original system are equivalent. Any ordered pair (x, y) that satisfies either equation will satisfy the system. From equation (2),  4x + y = 2,
or y = 2 + 4x.
The solutions of the system can be written as a set of ordered pairs (x, 2 + 4x), for any real number x. Some ordered pairs in the solution set are (0, 2 + 4 # 0) = (0, 2), (1, 2 + 4 # 1) = (1, 6), (3, 14), A  12, 0 B , and ( 2,  6). The solution set is 5(x, 2 + 4x)6, where x is any real number.
(1) (2)
Graphing Calculator Solution Solving the equations for y gives Y1 = 8X 2+ 4 and Y2 = 2 + 4X. The graphs of the two equations coincide, as seen in FIGURE 6. The table indicates that Y1 = Y2 for arbitrarily selected values of X, providing another way to show that the two equations lead to the same graph. Y1 = 8X + 4 and Y2 = 2 + 4X 2 10
–10
10
–10 Identical lines FIGURE 6
Refer to the analytic solution to see how the solution set can be written in terms of an arbitrary variable.
7.1
NOTE arbitrary:
Systems of Equations
441
We could solve equation (2) in Example 4 for x and write the solution set with y
EA
y  2 4 ,
y B F . Choose values for y and verify that the same ordered pairs result.
Nonlinear Systems A nonlinear system of equations is a system in which at least one of the equations is not a linear equation. The number of possible solutions to a nonlinear system can sometimes be determined by thinking graphically. For example, the equations x + y = 3 (line) and x 2 + y 2 = 5 (circle) can have zero, one, or two solutions, because a line can intersect a circle at 0, 1, or 2 points. See Exercises 51–68.
EXAMPLE 5
Solving a Nonlinear System by Substitution
Solve the system. 3x 2  2y = 5 x + 3y =  4 Analytic Solution The graph of equation (1) is a parabola, and that of equation (2) is a line. A line can intersect a parabola at 0, 1, or 2 points, so there could be 0, 1, or 2 solutions to this nonlinear system. Although either equation could be solved for either variable, we choose here to solve the linear equation (2) for y. x + 3y =  4 3y =  4  x 4  x y = 3
(1) (2)
Graphing Calculator Solution Graph equation (1) as y1 = 1.5x 2  2.5 and equation (2) as y2 =  13 x  43. FIGURE 7 indicates the points of intersection, ( 1,  1) and (0.7, 1.592), supporting the analytic solution. x + 3y = –4
(2) Subtract x. Divide by 3. (3)
–10
10
Substitute this expression for y into equation (1). 4  x b 3 9x 2  2( 4  x) 9x 2 + 8 + 2x Be careful with signs. 9x 2 + 2x  7 (9x  7)(x + 1) 7 or x x = 9 3x 2  2 a
y =
3
7 9
= 
(1)
= 15 = 15
Multiply each term by 3.
= 0 = 0
Standard form
= 1
Solve for x.
43 27
–5 (a)
= 5
3x 2 – 2y = 5
x + 3y = –4 10
Distributive property
Factor.
–10
10 –5 (b)
Substitute both xvalues into equation (3) to find the yvalues. 4 
2 10 3x – 2y = 5
y =
 4  ( 1) = 1 3
The solution set is E A 79,  43 27 B , ( 1,  1) F .
FIGURE 7
The decimal values for x and y displayed at the bottom of FIGURE 7(b) are approximations for the repeating decimal forms for 79 and  43 27 , respectively.
442
CHAPTER 7
Systems of Equations and Inequalities; Matrices
EXAMPLE 6
GCM
Solving a Nonlinear System by Elimination
Solve the system. x2 + y2 = 4 
2x 2 2x 2 – y 2 = 8 3.1
= 8
x2 + y2 = 4
4.7
2x 2 Remember both the positive and negative square roots.
–3.1
2x 2 – y 2 = 8
x2 + y2 = 4
(2)

y2
3x 2 x2 x = 2
(1)
= 8
(2)
= 12
Add.
= 4
Divide by 3.
or x =  2
Square root property
Substituting into equation (1) gives the corresponding values of y.
3.1
If x = 2, then –4.7
(1)
Solution The graph of equation (1) is a circle and that of equation (2) is a hyperbola. Visualizing them suggests that there may be 0, 1, 2, 3, or 4 points of intersection. Add the two equations to eliminate y 2.
x2 + y2 = 4
–4.7
y2
22
4.7
+
If x =  2, then
= 4 2 y = 0 y = 0.
( 2) 2 + y 2 = 4 y2 = 0 y = 0.
y2
Check that the solution set of the system is 5(2, 0), (  2, 0)6. See FIGURE 8.
–3.1 The two points of intersection are (2, 0) and ( 2, 0).
Many systems are difficult or even impossible to solve analytically. Graphing calculators allow us to solve some of these systems graphically.
FIGURE 8
EXAMPLE 7 Solve the system.
Solving a Nonlinear System Graphically y = 2x
(1)
x + 2  y = 0
(2)
Solution Enter equation (1) as y1 = 2 x and equation (2) as y2 =  x + 2 . As seen in FIGURE 9, the various windows indicate that there are three points of intersection of the graphs and thus three solutions. Using the capabilities of the calculator, we find that (2, 4) is an exact solution and both ( 2.22, 0.22) and ( 1.69, 0.31) are approximate solutions. Therefore, the solution set is 5(2, 4), ( 2.22, 0.22), ( 1.69, 0.31)6.
TECHNOLOGY NOTE Example 7 illustrates why proficiency in setting various windows is important when studying algebra with a graphing calculator.
y = 2x
⏐x + 2⏐ – y = 0 10
⏐x + 2⏐ – y = 0
1
–4
0
y = 2x 1
–4
0
10
–5
–1
–4 FIGURE 9
⏐x + 2⏐ – y = 0
y = 2x
–1
7.1
Systems of Equations
443
Applications of Systems We solve an applied problem with a system of equations as follows.
Solving Application Problems Step 1 Read the problem carefully and assign a variable to represent each quantity. Step 2 Write a system of equations involving these variables. Step 3 Solve the system of equations and determine the solution. Step 4 Look back and check your solution. Does it seem reasonable?
EXAMPLE 8
Using a Linear System to Solve an Application Landmark Title IX legislation prohibits sex discrimination in sports programs. In a recent year, the national average spent on two varsity athletes, one female and one male, was $6050 for Division IA schools. However, average expenditures for a male athlete exceeded those for a female athlete by $3900. Determine how much was spent per varsity athlete for each gender. (Source: USA Today.) Solution Step 1 Let x represent average expenditures per male athlete and y represent average expenditures per female athlete. Step 2 Since the average amount spent on one female and one male athlete was $6050, one equation is x + y = 6050. 2
(1)
The expenditures for a male athlete exceeded those for a female athlete by $3900. Thus, another equation is x  y = 3900. Step 3
(2)
Multiply equation (1) by 2 and then add the resulting equations. x + y = 12,100 x  y = 3900 2x = 16,000 x = 8000
Multiply (1) by 2. (2) Add. Divide by 2.
The average expenditure per male athlete was $8000. Now determine y. x  y = 3900 8000  y = 3900 y = 4100
(2) Let x = 8000. Solve for y.
The average amount spent on each female athlete was $4100. Step 4
The average of $8000 and $4100 is given by $8000 $8000  $4100 = $3900. The solution checks.
+ $4100 2
= $6050, and
444
Systems of Equations and Inequalities; Matrices
CHAPTER 7
EXAMPLE 9
Using a Nonlinear System to Find the Dimensions of a Box A box with an open top has a square base and four sides of equal height. The volume of the box is 75 cubic inches, and the surface area is 85 square inches. What are the dimensions of the box? Solution FIGURE 10 shows a diagram of such a box. If each side of the square base measures x inches and the height measures y inches, then the volume is
y x
x 2y = 75
x
and the surface area is
FIGURE 10
5
x 2 + 4xy = 85.
y = x 3 – 85x + 300
We solve equation (1) for y to get y =
10
0
–5
y = x 3 – 85x + 300
0
10
–5 FIGURE 11
7.1
Sum of areas of base and four sides (2) 75 x2
75 b x2 300 x2 + x x 3 + 300 3 x  85x + 300 x 2 + 4xa
5
Volume formula (1)
and substitute this into equation (2).
= 85
(2)
= 85
Multiply.
= 85x = 0
Multiply by x. Subtract 85x.
The solutions are xintercepts of the graph of y = x 3  85x + 300. A comprehensive graph indicates that there are three real solutions. However, one of them is negative and must be rejected. As FIGURE 11 indicates, one positive solution is 5 and the other is approximately 5.64. By substituting back into equation (1), we find that when x = 5, y = 3, and when x L 5.64, y L 2.36. Therefore, this problem has two solutions: The box may have base 5 inches by 5 inches and height 3 inches, or it may have base 5.64 inches by 5.64 inches and height 2.36 inches (approximately).
EXERCISES
Changes in Ozone Varying weather conditions affect ozone levels in Earth’s lower atmosphere, which in turn affect air quality. The graph shows seasonal average 8hour ozone concentrations in 53 urban areas during the summer months of May–September for the years 1997–2005.
Seasonal Average 8Hour Ozone Air Quality (53 Urban Areas)
2. At the point where the trend “Unadjusted for Weather” intersects the trend “Adjusted for Weather,” in the year found in Exercise 1, what was the ozone level? 3. Express as ordered pairs the five solutions of the system containing the graphs of the two trends. 4. Use the terms increasing and decreasing to describe the trends for the “Unadjusted for Weather” graph.
Ozone, ppm
0.062
1. In what year after 2002 does the trend “Unadjusted for Weather” show a higher ozone level than the trend “Adjusted for Weather”?
0.058 0.054 0.05 0 1997 1998 1999 2000 2001 2002 2003 2004 2005
Year Adjusted for Weather
Source: http://www.epa.gov/airtrends/weather_html
5. If equations of the form y = ƒ(t) were determined that modeled either of the two graphs, then the variable t would represent and the variable y would represent .
9 6.
Explain why each graph is that of a function.
Unadjusted for Weather
7.1
445
Systems of Equations
Checking Analytic Skills Use each graph to estimate the solution of the system of equations. Then solve the system analytically. Do not use a calculator. 7.
8.
y
9.
y
10.
y
y
6x + 4y = –5 x−y = 0
−x + y = 3 2
3
x+y=4 1
3
–2 –1
x
3
2
2
5x – 2y = –4
1
1 1
3
1
–3 –2 –1
x
1
–x + 2y = 0
x
2
–3 –2
1
2
3
x
–1 –2
x + y = −1
2x – 3y = 7
–2
–3
–3
Solve each system by substitution. 11. 6x  y = 5 y = x
12. 5x + y = 2 y =  3x
13. x + 2y =  1 2x + y = 4
14. 2x + y =  11 x + 3y =  8
15. y = 2x + 3 3x + 4y = 78
16. y = 4x  6 2x + 5y =  8
17. 3x  2y = 12 5x = 4  2y
18. 8x + 3y = 2 5x = 17 + 6y
19. 4x  5y =  11 2x + y = 5
20. 7x  y =  10 3y  x = 10
21. 4x + 5y = 7 9y = 31 + 2x
22.  2x = 6y + 18  29 = 5y  3x
23. 3x  7y = 15 3x + 7y = 15
24. 3y = 5x + 6 x + y = 2
25.
26.
0.6x  0.2y = 2  1.2x + 0.4y = 3
27.
x  2y = 4  2x + 4y =  8
2x 
7y = 8 21 y = 5  3x + 2
28.  3x + 2y =  10 9x  6y = 30
Solve each system by elimination. 29. 3x  y =  4 x + 3y = 12
30. 2x  3y =  7 5x + 4y = 17
31. 4x + 3y =  1 2x + 5y = 3
33. 12x  5y = 9 3x  8y =  18
34. 6x + 7y =  2 7x  6y = 26
35.
38. 3x + 2y = 5 6x + 4y = 8
39. 3x + y = 6 6x + 2y = 1
37.
41.
9x  5y = 1  18x + 10y = 1 y x + = 8 2 3 3y 2x + = 17 3 2
42.
x + 3y = 31 5 y 2x  = 8 5
43.
4x  y = 9  8x + 2y =  18
y + 2 2x  1 + = 4 3 4 x  y x + 3 = 3 2 3
32. 5x + 7y = 6 10x  3y = 46 36. x + y = 4 3x + 3y = 12 40. 3x + 5y =  2 9x + 15y =  6 44.
2y  x x + 6 + = 1 5 10 3y + 2 x + 2 + = 3 4 5
Use a graphing calculator to solve each system. Express solutions with approximations to the nearest thousandth. 11 3 6y = 9 47. 25 x + 2 48. px + ey = 3 x + y = 0.5 46. 45. 23 x  y = 5 3 ex + py = 4 100x + y = 9 22 x + 2 5 9 y = 12 0.6x  y = 3
446
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Explain how one can determine whether a system is inconsistent or has dependent equations when using the substitution or elimination method.
50. Concept Check For what value(s) of k will the following system of linear equations have no solution? infinitely many solutions? x  2y = 3  2x + 4y = k
Draw a sketch of the two graphs described with the indicated number of points of intersection. (There may be more than one way to do this.) 51. A line and a circle; no points
52. A line and a circle; one point
53. A line and a circle; two points
54. A line and a hyperbola; no points
55. A line and a hyperbola; one point
56. A line and a hyperbola; two points
57. A circle and an ellipse; four points
58. A parabola and an ellipse; one point
59. A parabola and a hyperbola; four points
60. A circle and a hyperbola; two points
Checking Analytic Skills Solve each system graphically. Check your solutions. Do not use a calculator. 61.
x + y = 3 2x  y = 0
65. x 2  y = 0 x2 + y2 = 2
62. 3x  y = 4 x + y = 0
63. x 2 + y 2 = 5 x + y = 3
64. x  y = 3 x2 + y2 = 9
66. x  y 2 = 1 x2 + y2 = 5
67. x 2  y 2 = 4 x + y = 2
68. x 2  y = 0 x + y2 = 0
Solve each nonlinear system of equations analytically. 69. y =  x 2 + 2 x  y = 0
70. y = (x  1) 2 x  3y =  1
71. 3x 2 + 2y 2 = 5 x  y = 2
72.
x2 + y2 = 5  3x + 4y = 2
73.
x 2 + y 2 = 10 2x 2  y 2 = 17
74. x 2 + y 2 = 4 2x 2  3y 2 =  12
75.
x 2 + 2y 2 = 9 3x 2  4y 2 = 27
76. 2x 2 + 3y 2 = 5 3x 2  4y 2 =  1
77. 2x 2 + 2y 2 = 20 3x 2 + 3y 2 = 30
78.
x2 + y2 = 4 5x 2 + 5y 2 = 28
79. y =  x  1  y = x2  4
80. 2x 2  y 2 = 4  x = y
Solve each system graphically. Give x and ycoordinates correct to the nearest hundredth. 81. y = log(x + 5) y = x2
82.
y = 5x xy = 1
83. y = e x + 1 2x + y = 3
84. y = 2 3x  4 x2 + y2 = 6
(Modeling) Solve each problem. 85. Robberies The total number of robberies during 2000 and 2001 was 831,000. From 2000 to 2001, the number of robberies declined by 15,000. (Source: FBI.) (a) Write a system of equations whose solution represents the number of robberies committed in each of these years. (b) Solve the system. 9 (c) Interpret the solution.
86. Smallpox Vaccinations During the first four months of 2004, a total of 6767 people were vaccinated for smallpox in Florida and Texas. There were 273 more people vaccinated in Florida than in Texas. (Source: FBI.) (a) Write a system of linear equations whose solution represents the number of vaccinations given in each state. (b) Solve the system. 9 (c) Interpret the solution.
7.1
87. Time on the Internet From 2001 to 2005 the average number of hours that a user spent on the Internet each week increased by 110%. This percent increase amounted to 11 hours. Find the average number of hours that a user spent on the Internet each week in 2001 and 2005. (Source: UCLA Center for Communications Policy.)
Systems of Equations
447
(a) Write a system of equations whose solution gives the consumption of natural gas and petroleum (in quadrillion Btu). (b) Solve the system of equations. (c) Is your system consistent or inconsistent? State whether the equations are dependent or independent. 94. Populations of Minorities in the United States The current and estimated resident populations (in percent) of blacks and Hispanics in the United States for the years 1990–2050 are modeled by the linear functions defined by y = 0.0515x + 12.3 y = 0.255x + 9.01.
88. Dimensions of a Box A box has an open top, rectangular sides, and a square base. The volume of the box is 576 cubic inches, and the surface area of the outside of the box is 336 square inches. Find the dimensions of the box. 89. Dimensions of a Box A box has rectangular sides and a rectangular top and base that are twice as long as they are wide. The volume of the box is 588 cubic inches, and the surface area of the outside of the box is 448 square inches. Find the dimensions of the box. 90. Investments A student invests $5000 at 5% and 7% annually. After 1 year, the student receives a total of $325 in interest. How much did the student invest at each interest rate?
9 91.
Heart Rate In one study, a group of conditioned athletes was exercised to exhaustion. Let x represent an athlete’s heart rate 5 seconds after stopping exercise and y the rate after 10 seconds. It was found that the maximum heart rate H for these athletes satisfied the two equations H = 0.491x + 0.468y + 11.2 H =  0.981x + 1.872y + 26.4. If an athlete had maximum heart rate H = 180, determine x and y graphically. Interpret your answer. (Source: Thomas, V., Science and Sport, Faber and Faber.)
9 92.
Heart Rate Repeat Exercise 91 for an athlete with a maximum heart rate of 195.
93. U.S. Energy Consumption In 2006, the commercial sector used 3.74 quadrillion (10 15) Btu of energy from natural gas and petroleum. It used 3.02 quadrillion Btu more natural gas than petroleum. (Source: Department of Energy.)
Blacks Hispanics
In each case, x represents the number of years since 1990. (Source: U.S. Census Bureau.) (a) Solve the system to find the year when these population percents were equal. (b) What percent of the U.S. resident population will be black or Hispanic in the year found in part (a)? (c) Graphically support the analytic solution in part (a). (d) Which population is increasing more rapidly? 95. Geometry Approximate graphically the radius and height of a cylindrical container with volume 50 cubic inches and lateral (side) surface area 65 square inches.
2πr r
h
h
96. Geometry Determine graphically whether it is possible to construct a cylindrical container, including the top and bottom, with volume 38 cubic inches and surface area 38 square inches. 97. Roof Truss The forces or weights W1 and W2 exerted on each rafter for the roof truss shown in the figure are determined by the following system of linear equations. Solve the system to the nearest tenth. W1 + 22W2 = 300 23W1  22W2 = 0 150 lb
W1
W2 30°
45°
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98. Height and Weight The relationship between a professional basketball player’s height h in inches and weight w in pounds was modeled by using two samples of players. The resulting equations were
150
w = 7.46h  374 w = 7.93h  405.
500 cars/hr
Assume that 65 … h … 85. (a) Use each equation to predict the weight to the nearest pound of a professional basketball player who is 6 feet 11 inches. (b) Determine graphically the height at which the two models give the same weight. (c) For each model, what change in weight is associated with a 1inch increase in height? 99. Traffic Control The figure shows two intersections labeled A and B that involve oneway streets. The numbers and variables represent the average traffic flow rates measured in vehicles per hour. For example, an average of 500 vehicles per hour enter intersection A from the west, whereas 150 vehicles per hour enter this intersection from the north. A stoplight will control the unknown traffic flow denoted by the variables x and y. Use the fact that the number of vehicles entering an intersection must equal the number leaving to determine x and y.
RELATING CONCEPTS
N
cars/hr
A
50 cars/hr
y
x
400
B
cars/hr
100 cars/hr
(Modeling) BreakEven Point The breakeven point for a company is the point where costs equal revenues. If both cost and revenue are expressed as linear equations, the breakeven point is the solution of a linear system. In each exercise, C represents cost in dollars to produce x items, and R represents revenue in dollars from the sale of x items. Use the substitution method to find the breakeven point in each case—that is, the point where C = R. Then find the value of C and R at that point. 100. C = 20x + 10,000 R = 30x  11,000
101. C = 4x + 125 R = 9x  200
For individual or group investigation (Exercises 102–107)
Because variables appear in denominators, the system 15 5 + = 16 x y 5 4 + = 5 x y is not a linear system. However, we can solve it in a manner similar to the method for solving a linear system by using a substitutionofvariable technique. Let t = 1x and let u = 1y . Work Exercises 102–107 in order. 102. Write a system of equations in t and u by making the appropriate substitutions.
103. Solve the system in Exercise 102 for t and u.
104. Solve the given system for x and y by using the equations relating t to x and u to y.
105. Refer to the first equation in the given system, and solve for y in terms of x to obtain a rational function.
106. Repeat Exercise 105 for the second equation in the given system.
107. Using a viewing window of 30, 104 by 30, 24, show that the point of intersection of the graphs of the functions in Exercises 105 and 106 has the same x and yvalues as found in Exercise 104.
Use the substitutionofvariable technique from the preceding Relating Concepts exercises to solve each system analytically. 108.
1 3 2 + = x y 2 3 1  = 1 x y
109.
1 2 + = 11 x y 3 5  = 10 x y
110.
3 2 + = 18 x y 4 5  = 8 x y
111.
3 16 1 + = x y 5 5 4 + = 5 x y
7.2
7.2
Solution of Linear Systems in Three Variables
449
Solution of Linear Systems in Three Variables
Geometric Considerations • Analytic Solution of Systems in Three Variables • Applications of Systems • Curve Fitting Using a System
Geometric Considerations We can extend the ideas of systems of equations in two variables to linear equations of the form Ax ⴙ By ⴙ Cz ⴝ D,
which has an ordered triple (x, y, z)
as its solution. For example, (1, 2,  4) is a solution of 2x + 5y  3z = 24. The solution set of such an equation is an infinite set of ordered triples. In geometry, the graph of a linear equation in three variables is a plane in threedimensional space. Considering the possible intersections of the planes representing three equations in three unknowns shows that the solution set of such a system may be either a single ordered triple (x, y, z), an infinite set of ordered triples (dependent equations), or the empty set (an inconsistent system). See FIGURE 12.
I I
P
II
III
II III
I, II, III Points of a line in common
A single solution
II
III
All points in common
III
I I
I
II
II
III
III No points in common
No points in common
I, II
No points in common
No points in common
FIGURE 12
Analytic Solution of Systems in Three Variables To solve a linear system with three unknowns, first eliminate a variable from any two of the equations. Second, eliminate the same variable from a different pair of equations. Third, eliminate a second variable using the resulting two equations in two variables to get an equation with just one variable whose value you can now determine. Finally, find the values of the remaining variables by substitution.
450
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GCM
EXAMPLE 1
Solving a System of Three Equations in Three Variables
Solve the system. 3x + 9y + 6z = 3 2x + y  z = 2 x + y + z = 2
(1) (2) (3)
Solution First, eliminate z by adding equations (2) and (3) to get 3x + 2y = 4.
(4)
Second, eliminate z from another pair of equations by multiplying each side of equation (2) by 6 and adding the result to equation (1). 12x + 6y  6z = 12 3x + 9y + 6z = 3 15x + 15y = 15
Make sure equation (5) has the same two variables as equation (4).
Multiply (2) by 6. (1) Add to obtain (5).
Third, eliminate x from equations (4) and (5) by multiplying each side of equation (4) by  5 and adding the result to equation (5). Solve the resulting equation for y.  15x  10y 15x + 15y 5y y
=  20 = 15 = 5 = 1
Multiply (4) by  5. (5) Add. Divide by 5.
Finally, using y =  1, find x from equation (4) by substitution. 3x + 2( 1) = 4 3x = 6 x = 2
Let y =  1 in (4). Multiply; add 2 to each side. Divide by 3.
Substitute 2 for x and  1 for y in equation (3) to find z. Write the values in the correct order.
2 + ( 1) + z = 2 z = 1
(3) Add; subtract 1.
Verify that the ordered triple (2,  1, 1) satisfies all three equations in the original system. The solution set is 5(2,  1, 1)6. CAUTION Be careful not to end up with two equations that still have three variables. Eliminate the same variable from each pair of equations.
EXAMPLE 2
Solving a System of Two Equations in Three Variables
Solve the system. x + 2y + z = 4 3x  y  4z =  9
(1) (2)
Solution Geometrically, the solution is the intersection of the two planes given by equations (1) and (2). The intersection of two nonparallel planes is a line. Thus, there will be infinitely many ordered triples in the solution set.
7.2
Solution of Linear Systems in Three Variables
451
To eliminate x, multiply both sides of equation (1) by  3 and add the result to equation (2). (Either y or z could have been eliminated instead.)  3x  6y 3x  y  7y Solve this equation for z.
 3z  4z  7z  7z z
= = = = =
 12 9  21 7y  21 y + 3
Multiply (1) by  3. (2) Add to obtain (3). Add 7y. Divide by  7.
This gives z in terms of y. Express x in terms of y by solving equation (1) for x and substituting  y + 3 for z in the result. x + 2y + z x x Be careful with signs. x
= 4 =  2y  z + 4 =  2y  ( y + 3) + 4
(1)
= y + 1
Simplify.
Solve for x. Substitute  y + 3 for z.
The system has infinitely many solutions. For any value of y, the value of x is  y + 1 and the value of z is  y + 3. For example, if y = 1, then x =  1 + 1 = 0 and z =  1 + 3 = 2, giving the solution (0, 1, 2). Another solution is ( 1, 2, 1). With y arbitrary, the solution set is of the form 5( y + 1, y,  y + 3)6. Had we solved equation (3) in Example 2 for y instead of z, the solution would have had a different form, but would have led to the same set of solutions. In that case we would have z arbitrary, and the solution set would be of the form {( z  2,  z + 3, z )}, or {( 2 + z, 3  z, z)}. By choosing z = 2, one solution would be (0, 1, 2), which was found in Example 2.
NOTE
A system of three linear equations in three variables can be inconsistent. For example, x + y + z = 1 x + y + z = 2 x + y + z = 3 clearly has no solutions because the sum of x, y, and z cannot equal three different numbers: 1, 2, and 3. This situation corresponds to the three parallel planes in FIGURE 12 (on the first page of this section), where there are no points in common.
Applications of Systems EXAMPLE 3
Solving a System to Satisfy Feed Requirements An animal feed is made from three ingredients: corn, soybeans, and cottonseed. One unit of each ingredient provides units of protein, fat, and fiber, as shown in the table. How many units of each ingredient should be used to make a feed that contains 22 units of protein, 28 units of fat, and 18 units of fiber?
Protein Fat Fiber
Corn
Soybeans
Cottonseed
Total
0.25 0.4 0.3
0.4 0.2 0.2
0.2 0.3 0.1
22 28 18
(continued)
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CHAPTER 7
Systems of Equations and Inequalities; Matrices
Solution Let x represent the number of units of corn, y the number of units of soybeans, and z the number of units of cottonseed that are required. Since the total amount of protein must be 22 units, the first row of the table yields 0.25x + 0.4y + 0.2z = 22.
(1)
Also, for the 28 units of fat, 0.4x + 0.2y + 0.3z = 28,
(2)
and for the 18 units of fiber, 0.3x + 0.2y + 0.1z = 18.
(3)
To clear decimals, multiply each side of the first equation by 100 and each side of the second and third equations by 10, to get the system 25x + 40y + 20z = 2200 4x + 2y + 3z = 280 3x + 2y + z = 180.
(4) (5) (6)
Using the method described in Example 1 shows that x = 40,
y = 15, and z = 30.
The feed should contain 40 units of corn, 15 units of soybeans, and 30 units of cottonseed to fulfill the given requirements.
EXAMPLE 4
Solving a Feed Requirements Application with Fewer Equations than Variables In Example 3, suppose that only the fat and fiber content of the feed are of interest. How would the solution be changed? Solution We would need to solve the system 4x + 2y + 3z = 280 3x + 2y + z = 180,
(5) (6)
which does not have a unique solution, since there are two equations with three variables. Following the procedure outlined in Example 2, we find that the solution set of this system is 5(100  2z, 2.5z  60, z)6. In this applied problem, however, all three variables must be nonnegative, so z must satisfy the conditions 100  2z Ú 0, 2.5z  60 Ú 0, z Ú 0. From the first inequality, z … 50; from the second inequality, z Ú 24; thus, 24 … z … 50. Only solutions with z in this range are usable.
Curve Fitting Using a System Recall from Chapter 3 that the graph of a quadratic function is a parabola with a vertical axis. Given three noncollinear points, we can find the equation of a parabola that passes through them.
7.2
7.1
–4.7
GCM
4.7 –1.1 FIGURE 13
Solution of Linear Systems in Three Variables
453
EXAMPLE 5
Using a System to Fit a Parabola to Three Data Points FIGURE 13 shows three data points: (2, 4), ( 1, 1), and ( 2, 5). Find the equation of the parabola (with vertical axis) that passes through these points. Solution The three points lie on the graph of the equation y = ax 2 + bx + c and must satisfy the equation. Substitute each ordered pair into the equation. 4 = a(2) 2 + b(2) + c, 1 = a( 1) 2 + b( 1) + c, 5 = a( 2) 2 + b( 2) + c,
or 4 = 4a + 2b + c or 1 = a  b + c or 5 = 4a  2b + c
(1) (2) (3)
To solve this system, first eliminate c, using equations (1) and (2). 4 = 4a + 2b + c 1 = a + b  c 3 = 3a + 3b
(1) Multiply (2) by  1. Add to obtain (4).
Now, use equations (2) and (3) to also eliminate c. Equation (5) must have the same two variables as equation (4).
1 = a  b + c  5 =  4a + 2b  c  4 =  3a + b
(2) Multiply (3) by  1. Add to obtain (5).
Solve the system of equations (4) and (5) in two variables by eliminating a. 3 4 1 1 4
= 3a + 3b =  3a + b = 4b
(4)
= b
Solve for b.
(5) Add.
Find a by substituting  14 for b in transformed equation (4). 1 = a + b 1 1 = a 4 5 = a 4 Finally, find c by substituting a =
4.7 –1.1
The figure shows that the parabola contains the three data points. A calculator with regression capability can also be used to find the quadratic function that exactly fits these data points.
Let b =  14. Solve for a.
and b =  14 in equation (2).
1 = a  b + c 5 1 1 =  a b + c 4 4 6 1 = + c 4 1  = c 2
7.1
–4.7
5 4
Equation (4) divided by 3
(2) Let a = 54, b =  14. Simplify. Subtract 64.
An equation of the parabola given by y = ax 2 + bx + c is y =
5 2 1 1 x  x  , 4 4 2
or equivalently,
y = 1.25x 2  0.25x  0.5.
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Systems of Equations and Inequalities; Matrices
CHAPTER 7
7.2
EXERCISES
Checking Analytic Skills Verify that the given ordered triple is a solution of the system. Do not use a calculator. 1. ( 3, 6, 1) 2x + y  z =  1 x  y + 3z =  6  4x + y + z = 19 4. ( 1,  2,  3)
1 3 1 2. a ,  , b 2 4 6 2x + 8y  6z =  6 1 x + y + z = 12 x + 3z = 1 5. ( 2,  1, 3)
x  y + z = 2 x  2y + z = 0 y  z = 1
x  y + z = 2 3x  2y + z =  1 x + y = 3
3. ( 0.2, 0.4, 0.5) 5x  y + 2z =  0.4 x + 4z = 1.8  3y + z =  0.7 1 1 6. a , ,  2 b 2 2 3x + y + z = 0 4x + 2y + z = 1 2x  2y  z = 2
Solve each system analytically. If the equations are dependent, write the solution set in terms of the variable z. ( Hint: In Exercises 33–36, let t = 1x , u = 1y , and v = 1z . Solve for t, u, and v, and then find x, y, and z.) 7.
10.
x + y + z = 2 2x + y  z = 5 x  y + z = 2 4x  y + 3z =  2 3x + 5y  z = 15  2x + y + 4z = 14
8. 2x + y + z = 9 x  y + z = 1 3x  y + z = 9 11.
x + 2y + 3z = 8 3x  y + 2z = 5  2x  4y  6z = 5
9.
x + 3y + 4z = 14 2x  3y + 2z = 10 3x  y + z = 9
12. 3x  2y  8z = 1 9x  6y  24z =  2 x  y + z = 1
13. x + 4y  z = 6 2x  y + z = 3 3x + 2y + 3z = 16
14. 4x  3y + z = 9 3x + 2y  2z = 4 x  y + 3z = 5
16. 2x  5y + 4z =  35 5x + 3y  z = 1 x + y + z = 1
17.
19. x  2y + 3z = 6 2x  y + 2z = 5
20. 3x + 4y  z = 13 x + y + 2z = 15
21. 5x  4y + z = 9 x + y = 15
22. 2x + 6y  z = 6 4x  3y + 5z =  5 6x + 9y  2z = 11
23. 8x  3y + 6z =  2 4x + 9y + 4z = 18 12x  3y + 8z =  2
24. x + z = 4 x + y = 4 y + z = 4
25. x  z = 2 x + y = 3 y  z = 1
26. 2x + y  z =  4 y + 2z = 12 2x  z =  4
27. 3x + 2y  z =  1 3y + z = 12 x  3z =  3
28. 3x  5y  4z =  7 y  z =  13
29.
31. 2x + 3y + 4z = 3 6x + 3y + 8z = 6 6y  4z = 1
32. 10x + 2y  3z = 0 5x + 4y + 6z =  1 6y + 3z = 2
x  3y  2z =  3 3x + 2y  z = 12  x  y + 4z = 3
x  y + z = 6 4x + y + z = 7
15.
5x + y  3z =  6 2x + 3y + z = 5  3x  2y + 4z = 3
18. x + y + z = 3 3x  3y  4z =  1 x + y + 3z = 11
30. 3x  2y + z = 15 x + 4y  z = 11
33.
1 1 1 +  = x y z 1 3 2  + = x y z 1 2 4   + = x y z
1 4 9 4 1
7.2
34.
2 1 11 3 +  = x y z 6 1 3 11 1  + = x y z 12 2 1 1 7 + + = x y z 12
35.
Solution of Linear Systems in Three Variables
2 1 2  + = 1 x y z 1 2 4 +  = 9 x y z 1 1 3 +  = 9 x y z
36.
37. x  4y + 2z =  2 x + 2y  2z =  3 x  y = 4
38.
40. x + 3y + z = 6 3x + y  z = 6 x  y  z = 0
41. 2x  y + 2z = 6 x + y + z = 0  x  3z =  6
2x + y + 3z = 4  3x  y  4z = 5 x + y + 2z = 0
43. Concept Check Consider the linear equation in three variables x + y + z = 4. Find a pair of linear equations that, when considered together with the given equation, will form a system having (a) exactly one solution, (b) no solution, (c) infinitely many solutions. (Modeling) Solve each problem. 45. Mixing Waters A sparklingwater distributor wants to make up 300 gallons of sparkling water to sell for $6.00 per gallon. She wishes to mix three grades of water selling for $9.00, $3.00, and $4.50 per gallon, respectively. She must use twice as much of the $4.50 water as the $3.00 water. How many gallons of each should she use? 46. Coin Collecting A coin collection made up of pennies, nickels, and quarters contains a total of 29 coins. The number of quarters is 8 less than the number of pennies. The total face value of the coins is $1.77. How many of each denomination are there? 47. Pricing Concert Tickets Frank Capek and his Generation Gap group sell three kinds of concert tickets: “up close,” “middle,” and “farther back.” “Up close” tickets cost $6 more than “middle” tickets, while “middle” tickets cost $3 more than “farther back” tickets. Twice the cost of an “up close” ticket is $3 more than 3 times the cost of a “farther back” seat. Find the price of each kind of ticket.
9 44.
455
1 2 5   = 6 x y z 3 3 1  +  =  12 x y z 2 1 1   = 6 x y z
39. x + y + z = 0 x  y  z = 3 x + 3y + 3z = 5 42.
x + 2y + z = 0 3x + 2y  z = 4  x + 2y + 3z =  4
Using your immediate surroundings, give an example of (a) three planes that intersect in a single point. (b) three planes that intersect in a line.
48. Mixing Glue A glue company needs to make some glue that it can sell for $120 per barrel. It wants to use 150 barrels of glue worth $100 per barrel, along with some glue worth $150 per barrel and glue worth $190 per barrel. It must use the same number of barrels of $150 and $190 glue. How much of the $150 and $190 glue will be needed? How many barrels of $120 glue will be produced? 49. Triangle Dimensions The sum of the measures of the angles of any triangle is 180°. In a certain triangle, the largest angle measures 55° less than twice the medium angle, and the smallest measures 25° less than the medium angle. Find the measures of the three angles. 50. Triangle Dimensions The perimeter of a triangle is 59 inches. The longest side is 11 inches longer than the medium side, and the medium side is 3 inches more than the shortest side. Find the length of each side.
x+?
x x+?
51. Investment Decisions A student invested $10,000 received in an inheritance in three parts. With one part, she bought mutual funds that offered a return of 4% per year. The second part, which amounted to twice the first, was used to buy government bonds paying 4.5% per year. She put the rest into a savings account that paid 2.5% annual interest. During the first year, the total interest was $415. How much did she invest at each rate?
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52. Investment Decisions Tom Accardo won $100,000 in the Louisiana state lottery. He invested part of the money in real estate with an annual return of 5% and another part in a money market account at 4.5% interest. He invested the rest, which amounted to $20,000 less than the sum of the other two parts, in certificates of deposit that pay 3.75%. If the total annual interest on the money was $4450, how much was invested at each rate? 53. Scheduling Deliveries Doctor Rug sells rugcleaning machines. The EZ model weighs 10 pounds and comes in a 10cubicfoot box. The compact model weighs 20 pounds and comes in an 8cubicfoot box. The commercial model weighs 60 pounds and comes in a 28cubicfoot box. Each of the company’s delivery vans has 248 cubic feet of space and can hold a maximum of 440 pounds. In order for a van to be fully loaded, how many of each model should it carry? 54. Scheduling Production Ciolino’s makes dining room furniture. A buffet requires 30 hours for construction and 10 hours for finishing, a chair 10 hours for construction and 10 hours for finishing, and a table 10 hours for construction and 30 hours for finishing. The construction department has 350 hours of labor and the finishing department has 150 hours of labor available each week. How many pieces of each type of furniture should be produced each week if the factory is to run at full capacity? 55. Predicting Home Prices The table shows the selling prices for three representative homes. Price P is given in thousands of dollars, age A in years, and home size S in thousands of square feet. These data may be modeled by the equation P = a + bA + cS. Price (P)
Age (A)
Size (S)
190 320 50
20 5 40
2 3 1
Curve Fitting Find the equation of the parabola (with vertical axis) that passes through the data points specified. Check your answer. 56. (2, 9), ( 2, 1), ( 3, 4) 57. (1.5, 6.25), (0,  2), ( 1.5, 3.25) 58. (2, 14), (0, 0), ( 1,  1) 59. ( 1, 4), (1, 2), (3, 8) 60. ( 2, 2), (0, 2), (2,  6) 61. (0, 1), (1, 0), (2,  5) 62. Curve Fitting The values in the table are from a quadratic function given by ƒ(x) = ax 2 + bx + c. Find a, b, and c. x
2
1
0
1
2
ƒ(x)
2.9
1.26
0.56
0.8
1.98
Curve Fitting Given three noncollinear points, there is one and only one circle that passes through them. Knowing that the equation of a circle may be written in the form x 2 + y 2 + ax + by + c = 0, find the equation of the circle passing through the given points. 63. ( 1, 3), (6, 2), and ( 2,  4) 64. ( 1, 5), (6, 6), and (7,  1) 65. (2, 1), ( 1, 0), and (3, 3) 66. ( 5,  2), (0, 3), and (4, 2) (Modeling) Position of a Particle Suppose that the position of a particle moving along a straight line is given by s(t) = at 2 + bt + c, where t is time in seconds and a, b, and c are real numbers.
(a) Write a system of linear equations whose solution gives a, b, and c. (b) Solve this system of linear equations. (c) Predict the price of a home that is 10 years old and has 2500 square feet.
67. If s(0) = 5, s(1) = 23, and s(2) = 37, find the equation that defines s(t). Then find s(8). 68. If s(0) =  10, s(1) = 6, and s(2) = 30, find the equation that defines s(t). Then find s(10).
7.3
7.3
Solution of Linear Systems by Row Transformations
457
Solution of Linear Systems by Row Transformations
Matrix Row Transformations • Row Echelon Method • Reduced Row Echelon Method • Special Cases • An Application of Matrices
Matrix Row Transformations Solving linear systems of equations can be streamlined by using matrices (singular: matrix). Consider a system of three equations and three unknowns. a1 x + b1 y + c1 z = d1 a2 x + b2 y + c2 z = d2 a3 x + b3 y + c3 z = d3 GCM
TECHNOLOGY NOTE The manner in which matrices are entered and displayed in a calculator can vary greatly among different manufacturers and even among the various models manufactured by the same company.
can be written as
a1 b1 C a2 b2 a3 b3
c1 c2 c3
d1 d2 S . d3
Matrix
Such a rectangular array of numbers enclosed by brackets is called a matrix. Each number in the array is an element or entry. The constants in the last column of the matrix can be set apart from the coefficients of the variables with a vertical line, as shown in the following augmented matrix.
Rows
a1 C a2 a3
b1 b2 b3
c1 d1 c2 3 d2 S c3 d3
Augmented matrix
Columns
As an example, the system on the left has the augmented matrix shown on the right. x + 3y + 2z = 1 2x + y  z = 2 x + y + z = 2
1 C2 1
3 1 1
2 1 1 3 2 S 1 2
Augmented matrix
FIGURE 14 shows how this matrix can be entered into a TI83/84 Plus calculator. The matrix has 3 rows (horizontal) and 4 columns (vertical), so it is a 3 * 4 (read “three by four”) matrix. The number of rows is always given first. To refer to a number in the matrix, use its row and column numbers. For example, the number 3 is in the first row, second column.
FIGURE 14
The rows of an augmented matrix can be treated just like the equations of the corresponding system of linear equations. Since the augmented matrix is nothing more than a short form of the system, any transformation of the matrix that results in an equivalent system of equations can be performed.
458
GCM
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Systems of Equations and Inequalities; Matrices
TECHNOLOGY NOTE
Matrix Row Transformations
Consult the graphing calculator manual that accompanies this text or your owner’s manual to see how these transformations are accomplished with your model. Also see Example 1.
For any augmented matrix of a system of linear equations, the following row transformations will result in the matrix of an equivalent system: 1. Any two rows may be interchanged. 2. The elements of any row may be multiplied by a nonzero real number. 3. Any row may be changed by adding to its elements a multiple of the corresponding elements of another row.
GCM
EXAMPLE 1
Using Row Transformations Use matrix row transformations to transform 1 C0 1
3 1 1
5 2 2 3 3S 4 2
to each matrix. 0 (a) C 1 1
1 3 1
2 2 3 5 3 2 S (b) C 0 1 2 4
Analytic Solution (a) Interchange the first two rows in the given matrix. 0 C1 1
1 3 1
2 3 5 3 2S 2 4
R2 : R1
6 1 1
1  10 4 2 3 3 S (c) C 0 0 2 4
6 1 1
 10 4 2 3 3S 2 4
R1 : R2 R3 is unchanged.
 2R1 R2 and R3 are unchanged.
FIGURE 15
(c) Multiply each element of the first row by  1, and add it to the corresponding element in the third row. For example, to obtain  4 in row 3, column 2, multiply 3 in row 1 by  1 and add to  1 in row 3: 3( 1) + ( 1) =  4. The final result is 1 C0 0
3 1 4
5 2 2 3 3 S. 7 2
R1 and R2 are unchanged.  1R1 + R3 FIGURE 16
5 2 2 3 3S 7 2
Graphing Calculator Solution FIGURE 15 shows the given matrix, designated 3A4. The TI83/84 Plus uses the commands indicated in FIGURE 16 to perform the transformations.
(b) Multiply the elements of the first row by  2. 2 C 0 1
3 1 4
7.3
Solution of Linear Systems by Row Transformations
459
Row Echelon Method 1 C0 0
5 1 0
Matrix row transformations are used to transform the augmented matrix of a system into one that is in echelon (or triangular) form. The echelon (triangular) form of an augmented matrix has 1s down the diagonal from upper left to lower right and 0s below each 1. The 1s lie on the main diagonal. See the matrix in the margin. Once a system of linear equations is in echelon form, backsubstitution can be used to find the solution set. The row echelon method uses matrices to solve a system of linear equations.
3 7 3 2 9S 1 4
Echelon (triangular) form
EXAMPLE 2
GCM
Solving by the Row Echelon Method
Solve the system. 5x + 2y = 1 2x  y = 4 Analytic Solution
B
5 2
Graphing Calculator Solution FIGURE 17 shows matrix 3A4.
2 1 ` R 1 4
Augmented matrix
Transform the first row so that the first entry is 1. To do this, multiply row 1 by 15. 1 B 2
` R
2 5
1 5
1 5 R1
1 4
FIGURE 17
Transform so that the entry below the main diagonal (that is, 2) is 0. Multiply row 1 by  2 and add to row 2. C
1
2 5
0  95
3
1 5 18 5
S
Using the ref( command on the TI83/84 Plus, we can find the row echelon form. See FIGURE 18.
R1 is unchanged.  2R1 + R2
Multiply row 2 by  59 to get 1 on the main diagonal.
B
1 1 25 ` 5R 0 1 2
FIGURE 18
 59 R2
This matrix represents the system of equations x +
2 1 y = 5 5 y =  2.
The decimal entries can be converted to fractions, as shown in FIGURE 19.
Since y =  2, use backsubstitution to find x. x + Write the x value first in the ordered pair.
2 1 ( 2) = 5 5 4 1 x  = 5 5 x = 1
The solution set is 5(1,  2)6.
Let y =  2. FIGURE 19
Multiply. Add 45.
The augmented matrix in FIGURE 19 corresponds to the last matrix in the analytic solution. The remainder of the solution process is the same.
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The row echelon method can be extended to larger systems. The final matrix will always have 0s below the main diagonal of 1s to the left of the vertical bar. To transform the matrix, work column by column from upper left to lower right. EXAMPLE 3 Solve the system.
Solving by the Row Echelon Method (Analytic ) x  y + 5z =  6 3x + 3y  z = 10 x + 3y + 2z = 5
Solution 1 C3 1
1 3 3
5 6  1 3 10 S 2 5
Augmented matrix
There is already a 1 in row 1, column 1. We must obtain 0s in the rest of column 1. 1 C0 1
1 6 3
5 6  16 3 28 S 2 5
1 C0 0
1 6 4
5 6  16 3 28 S  3 11
1 C0 0
1 1 4
5 6  83 3 14 3 S  3 11
1
1
5
D0
1
0
0
 83 23 3
1 C0 0
1 1 0
 1R1 + R3
1 6 R2
6 4
14 3 T 23 3
5 6  8 3 14 S 3
 3R1 + R2
3
1 1
 4R2 + R3
Row echelon form 3 23 R3
All integer elements can be obtained by multiplying row 2 by 3. However, the matrix is no longer in row echelon form. 1 C0 0
1 3 0
5 6  8 3 14 S 1 1
3R2
The final matrix corresponds to the system of equations x  y + 5z =  6 3y  8z = 14 z =  1.
(1) (2) (3)
Since z =  1 from equation (3), use backsubstitution into equation (2) to find that y = 2. Backsubstitute y = 2 and z =  1 into equation (1) to find that x = 1. The solution set is 5(1, 2,  1)6.
7.3
Solution of Linear Systems by Row Transformations
461
EXAMPLE 4
Solving by the Row Echelon Method (Graphical) Solve the system of Example 3 using a graphing calculator. x  y + 5z =  6 3x + 3y  z = 10 x + 3y + 2z = 5 Solution Enter the augmented matrix of the system as matrix 3A4. See the top screen in FIGURE 20. Then use the ref( command to obtain the row echelon form. The matrix in the bottom screen indicates that the row echelon form is 1 10 z = 3 3 8 14 y  z = 3 3 z =  1.
x + y 
The entries in row 1, columns 2, 3, and 4, are different from the corresponding entries in the row echelon form shown in Example 3, because the steps were performed in an alternative way. However, the solution set is the same.
Using backsubstitution, the solution set is 5(1, 2,  1)6, as in Example 3.
Reduced Row Echelon Method
FIGURE 20
Another matrix method for solving systems is the reduced row echelon method. Earlier, we saw that the row echelon form of a matrix has 1s along the main diagonal and 0s below. The reduced row echelon form has 1s along the main diagonal and 0s both below and above. For example, the augmented matrix of the system x + y + z = 6 2x  y + z = 5 3x + y  2z = 9
is
1 C2 3
1 1 1
1 6 1 3 5S . 2 9
By using row transformations, this augmented matrix can be transformed to 1 C0 0
0 1 0
0 3 0 3 2 S, 1 1
which represents the system
The solution set is 5(3, 2, 1)6.
GCM
x = 3 y = 2 z = 1.
EXAMPLE 5 Solving by the Reduced Row Echelon Method Use a graphing calculator with reduced row echelon capability to solve the system.
x + 2y + z = 30 3x + 2y + 2z = 56 2x + 3y + z = 44 Solution The top screen in FIGURE 21 shows the augmented matrix of the system, and the bottom screen shows its reduced row echelon form, which represents the system
FIGURE 21
and has solution set 5(8, 6, 10)6.
x = 8 y = 6 z = 10,
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Special Cases Whenever a row of the augmented matrix is of the form 0 0 0 Á  a, where a Z 0, the system is inconsistent and there will be no solution, since this row corresponds to the equation 0 = a. A row of the matrix of a linear system in the form 0 0 0 Á 0 indicates that the equations of the system are dependent.
EXAMPLE 6
Solving a Special System (No Solution)
Solve the system. x + 2y + z = 4 x  2y + 3z = 1 2x + 4y + 2z = 9 Analytic Solution 1 C1 2
2 2 4
1 4 3 3 1S 2 9
1 C1 0
2 1 4 2 3 3 1S 0 0 1
Graphing Calculator Solution Augmented matrix 3A4 and the row echelon form are shown in FIGURE 22. The final row indicates an inconsistent system with solution set 0. A similar conclusion can be made if the rref( command is used.
Augmented matrix
 2R1 + R3
The final row indicates that 0x + 0y + 0z = 1, which has no solutions. The system is inconsistent and has solution set 0. FIGURE 22
EXAMPLE 7
Solving a Special System (Dependent Equations)
Solve the system. 2x  5y + 3z = 1 x + 4y  2z = 8 4x  10y + 6z = 2 Solution
2 C1 4
5 4  10
2 C1 0
5 4 0
3 2 6
1 3 8S 2
3 1 2 3 8S 0 0
Augmented matrix
 2R1 + R3
7.3
Solution of Linear Systems by Row Transformations
463
The final row of 0s indicates that 0x + 0y + 0z = 0, which is true for all x, y, and z. The system has dependent equations. The first two rows represent the system 2x  5y + 3z = 1 x + 4y  2z = 8. Recall from Example 2 in Section 7.2 that a system with two equations and three variables may have infinitely many solutions.
B
2 1
5 4
3 1 ` R 2 8
B
1 2
4 5
2 8 ` R 3 1
B
1 0
4  13
B
1 0
4 1
Augmented matrix R2 : R1 R1 : R2
2 8 R ` 7  15
2 8 7 ` 15 R  13 13
 2R1 + R2
1  13 R2
This is as far as we go with the row echelon method. The equations that correspond to the final matrix are x + 4y  2z = 8 and Solve the second equation for y: y = first equation and solve for x.
15 13
+
7 13 z.
y 
7 15 z = . 13 13
Now substitute this result for y in the
x + 4y  2z = 8 x + 4a
15 7 + zb  2z = 8 13 13
x +
60 28 + z  2z = 8 13 13 60 2 + z = 8 x + 13 13 44 2 x = z 13 13
Let y =
15 13
+
7 13 z.
Distributive property 28 13 z
 2z =
2 13 z
Solve for x.
The solution set written with z arbitrary is
b¢
44  2z 15 + 7z , , z≤ r. 13 13
An Application of Matrices EXAMPLE 8 Determining a Model Using Given Data Three food shelters are operated by a charitable organization. Three different quantities are computed: monthly food costs F in dollars, number N of people served per month, and monthly charitable receipts R in dollars. The data are shown in the table on the next page.
(continued)
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FOR DISCUSSION The three screens shown here depict matrix 3A4 from Example 8, the row echelon form of 3A4, and the reduced row echelon form of 3A4.
Food Costs (F)
Number Served (N)
Charitable Receipts (R)
3000 4000 8000
2400 2600 5900
8000 10,000 14,000
Source: Sanders, D., Statistics: A First Course, McGrawHill.
(a) Model the data using the equation F = aN + bR + c, where a, b, and c are constants. (b) Predict the food costs for a shelter that serves 4000 people and receives charitable receipts of $12,000. Round your answer to the nearest hundred dollars. Solution (a) Since F = aN + bR + c, the constants a, b, and c satisfy the following three equations: 1. Use your calculator to recreate the screens shown here. (Note that only a portion of the matrix is visible.) 2. Which one of the two bottom screens provides an easier interpretation of the solution of the system? Why?
3000 = a(2400) + b(8000) + c 4000 = a(2600) + b(10,000) + c 8000 = a(5900) + b(14,000) + c. We can rewrite this system in equation form or as an augmented matrix: 2400a + 8000b + c = 3000 2600a + 10,000b + c = 4000 5900a + 14,000b + c = 8000
or
2400 C 2600 5900
8000 10,000 14,000
1 1 1
3000 3 4000 S . 8000
Using the reduced row echelon method (see the “For Discussion” box), we find that a L 0.6897, b L 0.4310, and c L  2103; thus, F L 0.6897N + 0.4310R  2103. (b) To predict food costs for a shelter that serves 4000 people and receives charitable receipts of $12,000, let N = 4000 and R = 12,000 and evaluate F. F L 0.6897(4000) + 0.4310(12,000)  2103 = 5827.8 This model predicts monthly food costs of approximately $5800.
7.3
EXERCISES
Checking Analytic Skills Use the given row transformation to transform each matrix. Do not use a calculator. 2 4
4 R 7
1 2 R1
2 4. C 4 3
5 1 7
1 2S 1
1. B
2. B
 2R1 + R2
1 7
3 5. C 2 10
4 R 0 1 1 5
7R1
4 3S 2
 5R2 + R3
1 3. C  2 4
5 3 7
4 6. C 7 1
10 4 1
6 1S 0 8 3S 0
2R1 + R2 4R3 + R1
7.3
Solution of Linear Systems by Row Transformations
465
Write the augmented matrix for each system. Do not solve the system. 7. 2x + 3y = 11 x + 2y = 8 11. 2x + y + z = 3 3x  4y + 2z =  7 x + y + z = 2
8. 3x + 5y =  13 2x + 3y =  9
9. x + 5y = 6 x = 3
12. 4x  2y + 3z = 4 3x + 5y + z = 7 5x  y + 4z = 7
13. x + y = 2 2y + z =  4 z = 2
10. 2x + 7y = 1 5x =  15 x = 6 y + 2z = 2 x  3z = 6
14.
Write the system of equations associated with each augmented matrix. 15. B
2 3
1 18. C 0 0
1 1 R ` 2 9
16. B
0 1 0
19.
1 0 1
4 3 2S 3
1 6
 5  18 R ` 2 20
1 17. C 0 0
0 1 0
0 2 3 0 3S 1 2
20.
Each augmented matrix is in row echelon form and represents a linear system. Use backsubstitution to solve the system if possible. 21. B
1 0
2 3 R ` 1 1
1 25. C 0 0
1 1 0
1 28. C 0 0
1 1 0
1 1 1
22. B
4 3 2S 1
2 8 3 4 2S 1 1
1 0
5 6 ` R 0 0
23. B
1 26. C 0 0
2 1 0
1 29. C 0 0
2 1 0
1 3 1
1 0
1 2 ` R 1 0
24. B
1 0
4 2 R ` 1 3
0 3 1S 2
1 27. C 0 0
2 1 0
1 2 0
1 3  3 3 12 S 0 4
1 30. C 0 0
0 1 0
3 4 4 2 3 1S 0 3
5 3 1S 0
Use row operations on an augmented matrix to solve each system of equations. Round to nearest thousandth when appropriate. 31. x + y = 5 x  y = 1
32. x + 2y = 5 2x + y =  2
33. x + y =  3 2x  5y =  6
34. 3x  2y = 4 3x + y =  2
35. 2x  3y = 10 2x + 2y = 5
36. 4x + y = 5 2x + y = 3
37. 2x  3y = 2 4x  6y = 1
38. x + 2y = 1 2x + 4y = 3
41. x + y =  1 y + z = 4 x + z = 1
42. x  z =  3 y + z = 9 x + z = 7
45.  x + y =  1 y  z = 6 x + z = 1
46. x + y = 1 2x  z = 0 y + 2z =  2
39.
6x  3y = 1  12x + 6y =  2
43. x + y  z = 6 2x  y + z =  9 x  2y + 3z = 1
40.
x  y = 1 x + y = 1
44. x + 3y  6z = 7 2x  y + 2z = 0 x + y + 2z =  1
47. 2x  y + 3z = 0 x + 2y  z = 5 2y + z = 1
48. 4x + 2y  3z = 6 x  4y + z =  4  x + 2z = 2
49. 2.1x + 0.5y + 1.7z = 4.9  2x + 1.5y  1.7z = 3.1 5.8x  4.6y + 0.8z = 9.3
50. 0.1x + 0.3y + 1.7z = 0.6 0.6x + 0.1y  3.1z = 6.2 2.4y + 0.9z = 3.5
51. 53x + 95y + 12z = 108 81x  57y  24z =  92  9x + 11y  78z = 21
52.
103x  886y + 431z = 1200  55x + 981y = 1108  327x + 421y + 337z = 99
466
9 53.
Systems of Equations and Inequalities; Matrices
CHAPTER 7
Compare the use of an augmented matrix as a shorthand way of writing a system of linear equations with the use of synthetic division as a shorthand way to divide polynomials.
9 54.
Compare the use of the third type of row transformation on a matrix with the elimination method of solving a system of linear equations.
Solve each system. Let z be the arbitrary variable if necessary. 55.
x  3y + 2z = 10 2x  y  z = 8
56. 3x + y  z = 12 x + 2y + z = 10
58.
3x + 5y  z = 0 4x  y + 2z = 1  6x  10y + 2z = 0
59.
x + 2y  z = 0 3x  y + z = 6  2x  4y + 2z = 0
57.
x  2y + z = 5  2x + 4y  2z = 2 2x + y  z = 2
60. 3x + 6y  3z = 12  x  2y + z = 16 x + y  2z = 20
Solve each system of four equations in four variables. Express the solutions in the form (x, y, z, w). 61.
x 4x  3x x
+ + 
3y y y y
+ + 
2z z z 3z
+ 
w 2w w 2w
= = = =
62.
9 2 5 2
(Modeling) Solve each application. 63. Food Shelter Costs Three food shelters have monthly food costs F in dollars, number N of people served per month, and monthly charitable receipts R in dollars, as shown in the table. Food Costs (F)
Number Served (N)
Charitable Receipts (R)
1300 5300 6500
1800 3200 4500
5000 12,000 13,000
3x 2x x 2x  y
+ 2y  w + z + 2w + 2y  z + z + w
= = = =
0 5 2 2
64. Paid Vacation for Employees The average number y of paid days off for fulltime workers at mediumtolarge companies after x years is listed in the table. x (years)
1
15
30
y (days)
9.4
18.8
21.9
Source: Bureau of Labor Statistics.
(a) Determine the coefficients for ƒ(x) = ax 2 + bx + c so that ƒ models these data. (b) Graph ƒ with the data in the window 3 4, 324 by 38, 234. (c) Estimate the number of paid days off after 3 years of experience. Compare it with the actual value of 11.2 days. 65. Scheduling Production A company produces two models of bicycles: model A and model B. Model A requires 2 hours of assembly time, and model B requires 3 hours of assembly time. The parts for model A cost $25 per bike; those for model B cost $30 per bike. If the company has a total of 34 hours of assembly time and $365 available per day for these two models, what is the maximum number of each model that can be made in a day and use all of the available resources?
(a) Model these data by using F = aN + bR + c, where a, b, and c are constants. (b) Predict food costs for a shelter that serves 3500 people and receives charitable receipts of $12,500. Round your answer to the nearest hundred dollars.
66. Scheduling Production Caltek Computer Company makes two products: computer monitors and printers. Both require time on two machines: monitors, 1 hour on machine A and 2 hours on machine B; printers, 3 hours on machine A and 1 hour on machine B. Both machines operate 15 hours per day. What is the maximum number of each product that can be produced per day under these conditions?
7.3
67. Financing Expansion To get funds necessary for a planned expansion, a small company took out three loans totaling $25,000. The company was able to borrow some of the money at 4%. It borrowed $2000 more than 12 the amount of the 4% loan at 6% and the rest at 5%. The total annual interest was $1220. How much did the company borrow at each rate? 68. Investment Decisions Rick Pal deposits some money in a bank account paying 3% per year. He uses some additional money, amounting to 13 the amount placed in the bank, to buy bonds paying 4% per year. With the balance of his funds, he buys a 4.5% certificate of deposit. The first year, his investments bring a return of $400. If the total of the investments is $10,000, how much did he invest at each rate? (Modeling) Each set of data in Exercises 69–72 can be modeled by ƒ(x) = ax 2 + bx + c. (a) Find a linear system whose solution represents values of a, b, and c. (b) Find ƒ(x) by using a method from this section. (c) Graph ƒ and the data in the same viewing window. (d) Make your own prediction using ƒ. 69. iPod Sales The table lists total iPod sales y in millions x years after 2004.
72. Carbon Dioxide Levels Carbon dioxide (CO2 ) is a greenhouse gas. Its concentration in parts per million ( ppm) has been measured at Mauna Loa, Hawaii, during past years. The table lists measurements for three selected years x. Year
1958
1973
2003
CO2 (ppm)
315
325
376
Source: Mauna Loa Observatory.
(Modeling) Traffic Flow Each figure in Exercises 73 and 74 shows three oneway streets with intersections A, B, and C. Numbers indicate the average traffic flow in vehicles per minute. The variables x, y, and z denote unknown traffic flows that need to be determined for timing of stoplights. (a) If the number of vehicles per minute entering an intersection must equal the number exiting an intersection, verify that the system of linear equations describes the traffic flow. (b) Rewrite the system and solve. 9 (c) Interpret your solution. 73. A: x + 5 = y + 7 B: z + 6 = x + 3 C: y + 3 = z + 4
5
x
0
2
4
y
3
55
150
3
cars/min
7
cars/min
A
cars/min
y
70. Head Start Enrollment The table lists annual enrollment in thousands for the Head Start program x years after 1980. 0
10
26
y
376
541
909
Source: Dept. of Health and Human Services.
71. Chronic Health Care A large percentage of the U.S. population will require chronic health care in the coming decades. The average caregiving age is 50–64, while the typical person needing chronic care is 85 or older. The ratio of potential caregivers to those needing chronic health care will shrink in the coming years x, as shown in the table.
6
B
x
Source: Apple Corporation.
x
467
Solution of Linear Systems by Row Transformations
cars/min
z C
3
4
cars/min
cars/min
74. A: x + 7 = y + 4 B: 4 + 5 = x + z C: y + 8 = 9 + 4
7
z
cars/min
4 cars/min
A
B
x
5 cars/min
4
y
cars/min
Year
1990
2010
2030
Ratio
11
10
6
Source: Robert Wood Johnson Foundation, Chronic Care in America: A 21st Century Challenge.
C 8
9
cars/min
cars/min
468
Systems of Equations and Inequalities; Matrices
CHAPTER 7
(Modeling) Solve each problem.
76. Weight of a Black Bear The table shows weight W, neck size N, overall length L, and chest size C for four bears.
75. Fawn Population To model spring fawn count F from adult antelope population A, precipitation P, and severity of winter W, environmentalists have used the equation F = a + bA + cP + dW, where the coefficients a, b, c, and d are constants that must be determined before using the equation. The table lists the results of four different (representative) years.
W (pounds)
N (inches)
L (inches)
C (inches)
125 316 436 514
19 26 30 30.5
57.5 65 72 75
32 42 48 54
Source: M. Triola, Elementary Statistics; Minitab, Inc.
Fawns
Adults
Precip. (in inches)
Winter Severity
239 234 192 343
871 847 685 969
11.5 12.2 10.6 14.2
3 2 5 1
(a) We can model these data with the equation W = a + bN + cL + dC, where a, b, c, and d are constants. To do so, represent a system of linear equations by a 4 * 5 augmented matrix whose solution gives values for a, b, c, and d. (b) Solve the system. Round each value to the nearest thousandth. (c) 9 Predict the weight of a bear with N = 24, L = 63, and C = 39. Interpret the result.
Source: Brase, C. and C. Brase, Understandable Statistics, D.C. Heath and Company; Bureau of Land Management.
(a) Substitute the values for F, A, P, and W from the table for each of the four years into the given equation F = a + bA + cP + dW to obtain four linear equations involving a, b, c, and d. (b) Write an augmented matrix representing the system, and solve for a, b, c, and d. (c) Write the equation for F, using the values from part (b) for the coefficients. (d) If a winter has severity 3, adult antelope population 960, and precipitation 12.6 inches, predict the spring fawn count. (Compare this with the actual count of 320.)
SECTIONS 7.1–7.3
77. Blending Coffee Beans Solve the system of linear equations discussed in the introduction to this chapter. a + b + c = 50 c = 2a 15.99a + 12.99b + 10.19c = 12.37(50)
9 78.
Blending Coffee Beans Rework Exercise 77 if Guatemala Antigua retails for $12.49 per pound instead of $10.19 per pound. Does your answer seem reasonable? Explain.
REVIEWING BASIC CONCEPTS
Solve each system, using the method indicated. 1. (Elimination) 2x  3y = 18 5x + 2y = 7
2. (Graphical) 2x + y =  4  x + 2y = 2
3. (Substitution) 5x + 10y = 10 x + 2y = 2
4. (Elimination) x  y = 6 x  y = 4
5. (Analytic, with graphical support) 6x + 2y = 10 2x 2  3y = 11
6. (Row echelon method) x + y + z = 1 x + y + z = 5 y + 2z = 5
7. (Reduced row echelon method) 2x + 4y + 4z = 4 x + 3y + z = 4  x + 3y + 2z =  1
8. Solve the system represented by the augmented matrix. 2 C1 1
1 0 2
2 10 2 3 5S 2 1
7.4
Matrix Properties and Operations
469
(Modeling) Solve each problem. 9. Sales of CRT and LCD Monitors In 2006 about 133 million monitors were sold. There were 43 million more LCD (liquid crystal display) monitors sold than CRT (cathode ray tube) monitors. (Source: International Data Corporation.) Determine about how many of each type of monitor were sold in 2006.
7.4
10. Investments A sum of $5000 is invested in three mutual funds that pay 3%, 4%, and 6% interest rates. The amount of money invested in the fund paying 6% equals the total amount of money invested in the other two funds, and the total annual interest from all three funds is $240. Find the amount invested at each rate.
Matrix Properties and Operations
Terminology of Matrices • Operations on Matrices • Applying Matrix Algebra
Terminology of Matrices Suppose that you are the manager of an electronics store and one day you receive the following products from two distributors: Wholesale Enterprises delivers 2 cell phones, 7 DVDs, and 5 video games; Discount Distributors delivers 4 cell phones, 6 DVDs, and 9 video games. We can organize the information in a table.
Product Distributor
Cell Phones
DVDs
Video Games
Wholesale Enterprises Discount Distributors
2 4
7 6
5 9
As long as we remember what each row and column represents, we can remove all the labels and write the numbers in the table as a matrix.
B TECHNOLOGY NOTE You should familiarize yourself with the matrix capabilities of your particular calculator. Refer to the graphing calculator manual that accompanies this text or your owner’s manual.
2 4
7 6
5 R 9
2 * 3 matrix
A matrix is classified by its dimension—that is, by the number of rows and columns it contains. For example, the preceding matrix has 2 rows and 3 columns, with dimension 2 * 3; a matrix with m rows and n columns has dimension m * n. The number of rows is always given first. Certain matrices have special names: an n * n matrix is a square matrix of order n. Also, a matrix with just one row is a row matrix, and a matrix with just one column is a column matrix. Two matrices are equal if they have the same dimension and if corresponding elements, position by position, are equal. Using this definition, the matrices
B
2 3
1 R 5
and
B
1 5
2 R 3
are not equal (even though they contain the same elements and are the same dimension), because the corresponding elements differ.
470
CHAPTER 7
Systems of Equations and Inequalities; Matrices
EXAMPLE 1
Classifying Matrices by Dimension Find the dimension of each matrix, and determine any special characteristics. 3 6 5 5 4S T (a) C 3 (b) D (c) 31 6 5  2 54 0 5 1 2 Solution (a) This is a 3 * 2 matrix, because it has 3 rows and 2 columns. (b) This matrix is a 4 * 1 column matrix. (c) 31
GCM
6
2
5
EXAMPLE 2
(a) If A = B
54 is a 1 * 5 row matrix. Determining Equality of Matrices
1 x y R and B = B R , determine x, y, p, and q such that A = B. q 1 0
2 p
(b) Are matrices 3A4 and 3B4 in FIGURE 23 equal?
Solution (a) From the definition of equality, the only way that the statement
B
FIGURE 23
1 x y R = B R q 1 0
2 p
can be true is if x = 2, y = 1, p =  1, and q = 0. (b) The matrices are not equal because corresponding entries in the third row are not all equal.
Operations on Matrices At the beginning of this section, we gave an example using the matrix
B
2 4
7 6
5 R, 9
where the columns represent the numbers of certain products (cell phones, DVDs, and video games, respectively) and the rows represent the two different distributors (Wholesale Enterprises and Discount Distributors, respectively). For example, the element 7 represents 7 DVDs received from Wholesale Enterprises. On another day, the shipments are described with the matrix
B
3 15
12 11
10 R. 8
Here, for example, 8 video games were received from Discount Distributors. The total number of products received from the distributors on the two days can be found by adding the corresponding elements of the two matrices.
B
2 4
7 6
5 3 R + B 9 15
12 11
10 2 + 3 R = B 8 4 + 15 = B
5 19
19 17
7 + 12 6 + 11 15 R 17
5 + 10 R 9 + 8
7.4
TECHNOLOGY NOTE Graphing calculators can perform operations on matrices, provided that the dimensions of the matrices are compatible for the operation. A dimension error message will occur if the operation cannot be performed.
Matrix Addition The sum of two m * n matrices A and B is the m * n matrix A + B in which each element is the sum of the corresponding elements of A and B.
Only matrices with the same dimension can be added.
EXAMPLE 3
Adding Matrices Find each sum if possible. 5 (a) B 8
6 4 R + B 9 8
(c) A + B if A = B Analytic Solution (a) B
5 8
471
The 5 in the sum indicates that, on the two days, 5 cell phones were received from Wholesale Enterprises. Generalizing from this example leads to the definition of matrix addition.
CAUTION
GCM
Matrix Properties and Operations
5 6
6 R 3
2 6 (b) C 5 S + C 3 S 8 12
8 3 R and B = B 2 4
9 2
1 R 5
Graphing Calculator Solution
6 4 R + B 9 8
(a) Using the TI83/84 Plus calculator, FIGURE 24(b) shows the sum of matrices A and B defined in FIGURE 24(a).
6 R 3
= B
5 + ( 4)  6 + 6 R 8 + 8 9 + ( 3)
= B
1 16
0 R 6
(a)
2 6 4 (b) C 5 S + C 3 S = C 8 S 8 12 20
(b) The screen in FIGURE 25 shows how the sum of two column matrices entered directly on the home screen is displayed. (c) A graphing calculator will return an ERROR message if it is directed to perform an operation on matrices that is not possible due to a dimension mismatch. See FIGURE 26.
(c) The matrices A = B
5 6
8 R 2
and B = B
3 4
9 2
2 * 2 matrix
1 R 5
(b) FIGURE 24
2 * 3 matrix
have different dimensions and so cannot be added; the sum A + B does not exist, or we say A + B is undefined.
FIGURE 25
FIGURE 26
A matrix containing only zeros as elements is called a zero matrix. 30
0
04
1 * 3 zero matrix
B
0 0
0 0
0 R 0
2 * 3 zero matrix
472
CHAPTER 7
Systems of Equations and Inequalities; Matrices
The additive inverse of a real number a is the unique real number  a such that a + ( a) = 0 and  a + a = 0. Given matrix A, the elements of matrix  A are the additive inverses of the corresponding elements of A. For example, if A = B
5 3
2 4
1 R, 6
A = B
then
5 3
2 4
1 R. 6
To check, show that A + ( A) equals the zero matrix O. A + ( A) = B
5 3
2 4
1 5 R + B 6 3
2 4
1 0 R = B 6 0
0 0
0 R = O 0
Then, show that  A + A is also O. Matrix  A is the additive inverse, or negative, of matrix A. Every matrix has a unique additive inverse.
WHAT WENT WRONG? A student entered 3A4 into his calculator as shown in FIGURE 27(a). He attempted to find the additive inverse of 3A4 and got the error message shown in FIGURE 27(b).
(a)
(b) FIGURE 27
What Went Wrong?
How can he correct his error?
Just as subtraction of real numbers is defined in terms of the additive inverse, subtraction of matrices is defined in the same way.
Matrix Subtraction If A and B are matrices with the same dimension, then A ⴚ B ⴝ A ⴙ (ⴚB).
GCM
EXAMPLE 4
Subtracting Matrices
Find each difference. (a) B
5 2
6 3 R  B 4 5
(c) A  B if A = B
2 0
2 R 8
(b) 38
6
 44  33
5
 84
5 3 R and B = B R 1 5
Answer to What Went Wrong? The student entered the subtraction symbol, rather than the negative symbol, preceding 3A4. He can obtain  3A4 by using the correct symbol.
7.4
Analytic Solution (a) B
5 2
473
Graphing Calculator Solution
6 3 R  B 4 5
(a) In FIGURE 28, the two matrices are defined as 3C4 and 3D4; the difference 3C4  3D4 is displayed in FIGURE 29.
2 R 8
= B
5 2
6 3 R + B 4 5
= B
2 3
4 R 12
(b) 38 6
Matrix Properties and Operations
 44  33 5
2 R 8
FIGURE 28
 84
= 35 1 44
FIGURE 29
FIGURE 30
(b) See FIGURE 30.
(c) Matrices A and B have different dimensions, so their difference does not exist.
(c) As in Example 3(c), the calculator returns a dimension error message.
If a matrix A is added to itself, each element in the sum is “twice as large” as the corresponding element of A. For example, 2 C1 4
5 2 3S + C1 6 4
5 4 3S = C2 6 8
10 2 6 S = 2C 1 12 4
5 3 S. 6
The number 2 in front of the last matrix is called a scalar to distinguish it from a matrix. A scalar (in this text) is just a special name for a real number.
Multiplication of a Matrix by a Scalar The product of a scalar k and a matrix A is the matrix kA, each of whose elements is k times the corresponding element of A.
GCM
EXAMPLE 5
Multiplying Matrices by Scalars Perform each multiplication. (a) 5 B
3 R 4
2 0
Analytic Solution (a) 5 B
2 0
3 5(2) 5( 3) R = B R 4 5(0) 5(4) = B
(b)
3 20 B 4 12
3
B 43
= B
3 20 B 4 12
36 R  16
Graphing Calculator Solution See FIGURES 31 and 32. Use parentheses carefully.
 15 R 20
10 0
36 R =  16
Multiply each element by 5.
(b)
3 4
(20)
4 (12)
15 9
3 4
(36)
( 16)
R FIGURE 31
FIGURE 32
27 R  12
474
CHAPTER 7
Systems of Equations and Inequalities; Matrices
Returning to the electronics store example, recall the matrix for the number of each type of product received from the two distributors.
B
2 4
7 6
5 R 9
Wholesale Enterprises Discount Distributors
Now suppose each cell phone costs the store $120, each DVD costs $18, and each video game costs $9. To find the total cost of the products from Wholesale Enterprises, we multiply as follows. Type of Product Cell phone DVD Video game
Number of Items
Cost per Item
Total Cost
2 7 5
$120 $ 18 $ 9
$240 $126 $ 45 $411
Total from Wholesale Enterprises
The products from Wholesale Enterprises cost a total of $411. This result is the sum of three products: 2($120) + 7($18) + 5($9) = $411. In the same way, using the second row of the matrix and the three product costs gives the total cost from Discount Distributors. 4($120) + 6($18) + 9($9) = $669 The product costs can be written as a column matrix. 120 C 18 S 9 The product of the matrices
2 B 4
7 6
2 B 4
7 6
Cell phone DVD Video game
5 R 9
and
120 C 18 S 9
can be written
120 5 2 # 120 + 7 # 18 + 5 # 9 411 R C 18 S = B # R = B R. 9 4 120 + 6 # 18 + 9 # 9 669 9
Total cost from each distributor
Each element of the product was found by multiplying the elements of the rows of the matrix on the left and the corresponding elements of the column of the matrix on the right and then finding the sum of these products. Notice that the product of a 2 * 3 matrix and a 3 * 1 matrix is a 2 * 1 matrix.
Matrix Multiplication The product AB of an m * n matrix A and an n * k matrix B is an m * k matrix and is found as follows: To find the ith row, jth column element of AB, multiply each element in the ith row of A by the corresponding element in the jth column of B. The sum of these products gives the element of row i, column j of AB. The product AB can be found only if the number of columns of A is the same as the number of rows of B. The final product will have as many rows as A and as many columns as B.
7.4
Matrix Properties and Operations
475
EXAMPLE 6
Deciding whether Two Matrices Can Be Multiplied Suppose matrix A has dimension 2 * 2, while matrix B has dimension 2 * 4. Can the product AB be calculated? If so, what is the dimension of the product? Can the product BA be calculated? Solution The following diagram helps answer these questions. Matrix A dimension
Matrix B dimension
2 * 2
2 * 4 must match dimension of AB 2 * 4
The product AB can be calculated because A has two columns and B has two rows. The dimension of the product is 2 * 4. The product BA cannot be found. To see why, consider the following diagram. The number of columns in B is different from the number of rows in A. Matrix B dimension
Matrix A dimension
2 * 4
2 * 2 different
EXAMPLE 7
Multiplying Matrices
Find the product AB of the two matrices A = B
3 5
4 0
6 2 R and B = C 2 4 3
4 3 S. 2
Solution Since A has dimension 2 * 3 and B has dimension 3 * 2, they can be multiplied. The product AB has dimension 2 * 2. A
B
Step 1
3 B 5
4 0
6 2 R C 2 4 3
Step 2
3 B 5
4 0
6 2 R C 2 4 3
4 3S 2
 3(4) + 4(3) + 2( 2) =  4
Step 3
3 B 5
4 0
6 2 R C 2 4 3
4 3S 2
5( 6) + 0(2) + 4(3) =  18
6 2 R C 2 4 3 Write the product.
4 3S 2
5(4) + 0(3) + 4( 2) = 12
Step 4 Step 5
B
3 5
4 0
3 B 5
4 0
4 3S 2
 3( 6) + 4(2) + 2(3) = 32
6 2 R C 2 4 3
4 32 3S = B  18 2
4 R 12
476
CHAPTER 7
Systems of Equations and Inequalities; Matrices
GCM
EXAMPLE 8
Multiplying Matrices Use a graphing calculator to find the product BA of the two matrices given in Example 7. Solution Define 3A4 and 3B4 as shown in FIGURE 33. Since 3B4 has dimension 3 * 2 and 3A4 has dimension 2 * 3, the dimension of the product BA is 3 * 3. As shown in Examples 7 and 8, AB ⴝ BA. In general, matrix multiplication is not commutative. In fact, in some cases, one of the products may be defined while the other is not. See Example 6.
FIGURE 33
GCM
EXAMPLE 9
Multiplying Square Matrices
Find AB and BA, given A = B
1 2
3 2 R and B = B 4 10
5 R. 3
Solution
FIGURE 34
AB = B
1 2
3 2 R B 4 10
5 2 + 30 R = B 3  4 + 40
BA = B
2 10
5 1 R B 3 2
3 2  10 R = B 4 10 + 6
FIGURE 34
5  9 32 R = B  10  12 36 6 + 20 8 R = B 30  12 16
4 R  22
26 R 18
shows matrices 3A4 and 3B4 and both products.
Applying Matrix Algebra EXAMPLE 10
Using Matrix Multiplication to Model Plans for a Subdivision A contractor builds three kinds of houses, models X, Y, and Z, with a choice of two styles, colonial or ranch. Matrix A shows the number of each kind of house the contractor is planning to build for a new 100home subdivision. The amounts for each of the main materials used depend on the style of the house. These amounts are shown in matrix B, while matrix C gives the cost in dollars for each kind of material. Concrete is measured here in cubic yards, lumber in 1000 board feet, brick in 1000s, and shingles in 100 square feet. Colonial Model X Model Y Model Z
0 C 10 20
Ranch
30 20 S = A 20 Cost per Unit
Concrete Colonial Ranch
10 B 50
Lumber Brick
2 1
0 20
Shingles
2 R = B 2
Concrete Lumber Brick Shingles
20 180 D T = C 60 25
7.4
477
Matrix Properties and Operations
(a) What is the total cost of materials for all houses of each model? (b) How much of each of the four kinds of material must be ordered? (c) Use a graphing calculator to find the total cost of the materials. Solution (a) To find the cost of materials for each model, first find matrix AB, which will give the total amount of each material needed for all houses of each model. Concrete
0 AB = C 10 20
30 10 20 S B 50 20
2 1
0 20
1500 2 R = C 1100 2 1200
Lumber
Brick
30 40 60
600 400 400
Shingles
60 Model X 60 S Model Y 80 Model Z
Multiplying AB and the cost matrix C gives the total cost of materials for each model. Cost
1500 (AB)C = C 1100 1200
30 40 60
600 400 400
20 60 72,900 Model X 180 60 S D T = C 54,700 S Model Y 60 80 60,800 Model Z 25
(b) The totals of the columns of matrix AB will give a matrix whose elements represent the amounts of each material needed for the subdivision. Call this matrix D, and write it as a row matrix. D = 33800
FIGURE 35
130
1400
2004
(c) The total cost of all the materials is given by the product of matrix C, the cost matrix, and matrix D, the total amounts matrix. To multiply these and get a 1 * 1 matrix representing the total cost requires multiplying a 1 * 4 matrix and a 4 * 1 matrix. This is why in part (b) a row matrix was written rather than a column matrix. The total materials cost is given by DC. FIGURE 35 shows how a graphing calculator computes this product. The total cost of the materials is $188,400. To help keep track of the quantities a matrix represents, let matrix A from Example 10 represent models/styles, matrix B represent styles/materials, and matrix C represent materials/cost. In each case, the meaning of the rows is written first and that of the columns second. When the product AB was found in Example 10, the rows of the matrix represented models and the columns represented materials. Therefore, the matrix product AB represents models/materials.
7.4
EXERCISES
Checking Analytic Skills Find the dimension of each matrix. Identify any square, column, or row matrices. Do not use a calculator. 1. B
3 7
2 5. B R 4
6 R 4
2 2. C 1 5
8 0 2
6. 34 94
6 5S 3
6 3. C 4 3 7. 3 94
8 1 5
0 9 7
0 2S 1
4. B
3 0
8. B
0 0
4 8
0 0
2 6
0 0
1 R 3
0 0
0 R 0
478
Systems of Equations and Inequalities; Matrices
CHAPTER 7
Checking Analytic Skills Find the value of each variable. Do not use a calculator. 9. B
w x 3 R = B y z 1
0 11. C  1 4
2 R 4
10. B
13. B
 7 + z 4r 8s 9 R + B 6p 2 5 2
14. B
a + 2 8k
9 15.
w + 3 3 1
x 0 y + 2 S = C 1 z 4
5 3 1
3z + 1 0
5 6 z R = B m n 1
3 + x 4 12. C 5 8  w 4 3
6 0S 8
8r 3 2 R = B 5 4 20
5m 3a R + B 3 2k
2 1
36 7
27 R 12a
2z 5m 10 R = B 5 6 10
 14 5
y w R 8 2 t 9 4 y + 1 S = Cz + 3 w 2r p q
6 9S r
80 R 9
9 16.
Your friend missed the lecture on adding matrices. In your own words, explain to her how to add two matrices.
Explain to a friend in your own words how to multiply a matrix by a scalar.
Perform each operation if possible. 17. B
6 4
9 1
4 6 3S  C 2 4 7
1 20. C 2 8 23. B
8 2
9 25. C 5 2
2 8 R + B 3 6
4 5
0 6 R + B 0 8
4 6 5
2 27. 2C 5 0
2 3 9 5S  12
1 5 1S + C7 3 1
18. B
9 8
21. B
6 5
4 3 R + B 2 4 2 1 R + B 4 7
3 R 9 5 1 3
1 3 2 0 2S + C4 2 1 2
2 29. 2 C  1 1
1 2 1
1 1  1 S + 3C 2 2 2
6 31. 3C 2 4
1 8 5
4 2 3 S + 5C 4 6 2
2 1 3
4 5 3
0 1 0
3 3S 1 8 1 1
6 3S 5
34.  33B4
36.  23A4 + 3B4
37. 53A4 + 0.53B4 39. Concept Check B = B
4 6
6 3
38.  43A4 + 1.53B4 Find matrix A if 5 R 2
and
6 0
8 0 R  B 0 4
0 R 2
7 R 4
22. B
12 10
5 6 R  B 3 2
9 R 0
A + B = B
2 5 0
28.  3 B
3 1
1 30. 3C 0 1
0 1 0
1 32. 4C 2 8
Matrices 3A4 and 3B4 are shown on the screen. Find each matrix. 35. 23A4  3B4
19. B
3 6 6 S 26. C  2 8 1
4 4S 6
0 9 3S  C4 1 1
33. 23A4
2 R 7
2 8 24. B R  B 3 9
2 2 4S  C 0 2 8
1 3 1
5 R 4
6  10
12 4
0 R. 11
1 R 4 4 3 8S + C1 2 6
8 4 R + 5B 9 1 3 2 3
0 2 9
8 4 4S  C 0 2 4
8 R 6
1 1  1 S  4C 0 1 2
4 6  3 S  3C  2 4 7
2 3 2
9 5S  12
0 1 0
0 3 1
4 2S 1
1 2S 0
7.4
40. Concept Check A = B
3 2
6 1
479
Matrix Properties and Operations
Find matrix B if 5 R 4
and A  B = B
9 4
0 6
5 R. 3
The dimensions of matrices A and B are given. Find the dimensions of the product AB and of the product BA if the products are defined. If they are not defined, say so. 41. A is 4 * 2; B is 2 * 4.
42. A is 3 * 1; B is 1 * 3.
43. A is 3 * 5; B is 5 * 2.
44. A is 7 * 3; B is 2 * 7.
45. A is 4 * 3; B is 2 * 5.
46. A is 1 * 6; B is 2 * 4.
48. Concept Check In finding the product AB of matrices A and B, the first row, second column, entry is found by multiplying the elements in A and the elements in B and then these products.
47. Concept Check The product MN of two matrices can be found only if the number of of M equals the number of of N.
If possible, find AB and BA. 49. A = B
1 2
1 2 R, B = B 0 1
51. A = B
5 0
7 1
3 53. A = C 1 2
50. A = B
3 2
5 1 R, B = B 7 0 1 1 1 S, B = C 2 1 3
8 1
7 R 2
2 52. A = C 0 3
1 2 0 S, B = B 9 4
5 7
3 R 0
1 54. A = B 4
0 2
2 2 R, B = C 5 1 0
2 56. A = C 4 2
1 1 0
5 1 6 S, B = C 1 9 2
0 1 5 S, B = B 7 4
3 7
1 R 1
1 2 2
2 R 7
2 9 R, B = B 5 1
1 1 1
1 55. A = C 2 6
3 R 2
0 1S 1 2 1S 1 2 1S 0
Find each matrix product if possible. 4 0
57. B
3 5
60. B
4 1
2 63. C 2 4 66. 30
3
1 1 R C 4S 2 2
0 2 R B 3 0 3 1 2
4 R 1
4 0 0S C1 3 3
2 6  44 C 0 4 1 1
4 Given A = B 3
1 2 2
4 1S 2
3 2S 4
5 2 R, B = C 0 1 3
58. B
6 2
61. B
2 3
1 64. C 0 0
67. B
p r
3 9
2 5 R C 0S 1 3 0 1 R C 1 1 0
2 0 2 3 1
0 2 2S C0 4 3
q a R B s b
1 5 4  2 S, and C = B 0 3 7
2 4S 2 1 2 0
2 1S 1
59. B
5 1
2 3 R B 4 1
62. B
9 3
2 0
65. 3 2
a 68. C d g
c R d
4
2 R 0
2 1 R C 1S 0 4 3 14 C 2 0
2 1 1
4 0S 4
b c x e fS CyS h i z
1 R , find each product if possible. 6
69. BA
70. AC
71. BC
72. CB
73. AB
74. CA
75. A2
76. A3
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77. Concept Check Compare the answers to Exercises 69 and 73, 71 and 72, and 70 and 74. Is matrix multiplication commutative?
(c) Use matrix multiplication to find a matrix giving the costs under the two purchase options to fill the day’s orders.
78. Concept Check For any matrices P and Q, what must be true for both PQ and QP to exist?
81. (Modeling) Predator–Prey Relationship In certain parts of the Rocky Mountains, deer are the main food source for mountain lions. When the deer population d is large, the mountain lions (m) thrive. However, a large mountain lion population drives down the size of the deer population. Suppose the fluctuations of the two populations from year to year can be modeled with the matrix equation
Solve each problem. 79. Income from Yogurt Yagel’s Yogurt sells three types of yogurt: nonfat, regular, and supercreamy, at three locations. Location I sells 50 gallons of nonfat, 100 gallons of regular, and 30 gallons of supercreamy each day. Location II sells 10 gallons of nonfat, 90 gallons of regular, and 50 gallons of supercreamy each day. Location III sells 60 gallons of nonfat, 120 gallons of regular, and 40 gallons of supercreamy each day. (a) Write a 3 * 3 matrix that shows sales for the three locations, with the rows representing the locations. (b) The incomes per gallon for nonfat, regular, and supercreamy are $12, $10, and $15, respectively. Write a 3 * 1 matrix displaying the incomes per gallon. (c) Find a matrix product that gives the daily income at each of the three locations. (d) What is Yagel’s Yogurt’s total daily income from the three locations? 80. Purchasing Costs The Bread Box, a small neighborhood bakery, sells four main items: sweet rolls, bread, cakes, and pies. The amount of each ingredient (in cups, except for eggs) required for these items is given by matrix A. Eggs Flour Sugar Shortening Milk Rolls (doz) Bread (loaves) Cakes Pies (crust)
1
4
1 4
1 4
1
0
3
0
1 4
0
4
3
2
1
1
0
1
0
1 3
0
E
U=A
The cost (in cents) for each ingredient when purchased in either large lots or small lots is given in matrix B. Cost Large lot Eggs Flour Sugar Shortening Milk
5 8 D 10 12 5
Small lot 5 10 12 T = B 15 6
(a) Use matrix multiplication to find a matrix giving the cost per item for the two purchase options. (b) Suppose a day’s orders consist of 20 dozen sweet rolls, 200 loaves of bread, 50 cakes, and 60 pies. Write the orders as a 1 * 4 matrix, and using matrix multiplication, write as a matrix the amount of each ingredient needed to fill the day’s orders.
B
mn + 1 0.51 R = B dn + 1  0.05
0.4 m R B nR. 1.05 dn
The numbers in the column matrices give the numbers of animals in the two populations after n years and n + 1 years, where the number of deer is measured in hundreds. (a) Give the equation for dn + 1 obtained from the second row of the square matrix. Use this equation to determine the rate the deer population will grow from year to year if there are no mountain lions. (b) Suppose we start with a mountain lion population of 2000 and a deer population of 500,000 (that is, 5000 hundred deer). How large would each population be after 1 year? 2 years? (c) Consider part (b), but change the initial mountain lion population to 4000. Show that the populations would both grow at a steady annual rate of 1%. 82. (Modeling) Northern Spotted Owl Population To analyze population dynamics of the northern spotted owl, mathematical ecologists divided the female owl population into three categories: juvenile (up to 1 year old), subadult (1 to 2 years old), and adult (over 2 years old). They concluded that the change in the makeup of the northern spotted owl population in successive years could be described by the following matrix equation: jn + 1 0 C sn + 1 S = C 0.18 an + 1 0
0 0 0.71
0.33 jn 0 S C sn S. an 0.94
The numbers in the column matrices give the numbers of females in the three age groups after n years and n + 1 years. Multiplying the matrices yields jn + 1 = 0.33an
Each year, 33 juvenile females are born for each 100 adult females.
sn + 1 = 0.18jn
Each year, 18% of the juvenile females survive to become subadults.
an + 1 = 0.71sn + 0.94an.
Each year, 71% of the subadults survive to become adults and 94% of the adults survive.
(Source: Lamberson, R. H., R. McKelvey, B. R. Noon, and C. Voss, “A Dynamic Analysis of Northern Spotted Owl Viability in a Fragmented Forest Landscape,” Conservation Biology, Vol. 6, No. 4.)
7.5
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Determinants and Cramer’s Rule
(c) In this model, the main impediment to the survival of the northern spotted owl is the number 0.18 in the second row of the 3 * 3 matrix. This number is low for two reasons: The first year of life is precarious for most animals living in the wild, and juvenile owls must eventually leave the nest and establish their own territory. If much of the forest near their original home has been cleared, then they are vulnerable to predators while searching for a new home. Suppose that, due to better forest management, the number 0.18 can be increased to 0.3. Rework part (a) under this new assumption. Let A = B
(a) Suppose there are currently 3000 female northern spotted owls: 690 juveniles, 210 subadults, and 2100 adults. Use the preceding matrix equation to determine the total number of female owls for each of the next 5 years. (b) Using advanced techniques from linear algebra, we can show that, in the long run, jn + 1 jn C sn + 1 S L 0.98359 C sn S. an + 1 an What can we conclude about the longterm fate of the northern spotted owl?
7.5
a11 a12 b b c R , B = B 11 12 R , and C = B 11 a21 a22 b21 b22 c21
c12 R, c22
where all the elements are real numbers. Use these matrices to show that each statement is true for 2 * 2 matrices. 83. A + B = B + A (commutative property) 84. A + (B + C) = (A + B) + C (associative property) 85. (AB)C = A(BC) (associative property) 86. A(B + C) = AB + AC (distributive property) 87. c(A + B) = cA + cB, for any real number c 88. (c + d) A = cA + dA, for any real numbers c and d 89. (cA)d = (cd) A 90. (cd) A = c(dA)
Determinants and Cramer’s Rule
Determinants of 2 : 2 Matrices • Determinants of Larger Matrices • Derivation of Cramer’s Rule • Using Cramer’s Rule to Solve Systems
For convenience, we use subscript notation to name the elements of a matrix, as in the following m by n matrix A: a11 a21 A = E a31 o am1
a12 a22 a32 o am2
a13 a23 a33 o am3
Á Á Á Á
a1n a2n a3n U o amn
With this notation, the row 1, column 1 element is a11; the row 2, column 3 element is a23; and, in general, the row i, column j element is aij. We use subscript notation in this section to define determinants.
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Determinants of 2 : 2 Matrices Associated with every square matrix A is a real number called the determinant of A. The symbols  A , d(A), and det A all represent the determinant of A. In this text, we use det A.
Looking Ahead to Calculus
Determinant of a 2 : 2 Matrix
Determinants are utilized in calculus to find vector cross products, which are used to study the effect of forces in the plane or in space. They are also used to express the results of certain vector operations.
The determinant of a 2 : 2 matrix A, where A = B
a11 a21
a12 R, a22
is a real number defined as det A ⴝ a 11a 22 ⴚ a 21a 12.
NOTE You should be able to distinguish between a matrix and its determinant. A matrix is an array of numbers, while a determinant is a single real number associated with a square matrix.
EXAMPLE 1
GCM
Find det A if A = B Analytic Solution Use the definition just given. det A = det B
3 6
Evaluating the Determinant of a 2 : 2 Matrix 3 6
4 R. 8 Graphing Calculator Solution The TI83/84 Plus graphing calculators can compute determinants. FIGURE 36 shows both 3A4 and det3A4.
4 R 8
=  3(8)  6(4) a11 a22
a21 a12
=  48 FIGURE 36
EXAMPLE 2
If A = B
Solving an Equation Involving a Determinant
x 3 R and det A = 33, find the value of x. 1 5
Solution Since det A = 5x  ( 1)(3), or 5x + 3, we have 5x + 3 = 33 5x = 30 x = 6.
Definition Subtract 3. Divide by 5.
7.5
Determinants and Cramer’s Rule
483
Determinants of Larger Matrices Determinant of a 3 : 3 Matrix The determinant of a 3 : 3 matrix A, where a11 A = C a21 a31
a12 a22 a32
a13 a23 S , a33
is a real number defined as det A ⴝ (a 11a 22a 33 ⴙ a 12a 23a 31 ⴙ a 13a 21a 32) ⴚ (a 31a 22a 13 ⴙ a 32a 23a 11 ⴙ a 33a 21a 12).
NOTE The above definition is not usually memorized in its general form. The method of cofactors, which is described next, is used more often because it allows us to calculate the determinant of any square matrix.
A method for calculating 3 * 3 determinants is found by rearranging and factoring the terms given in the definition to get a11 detC a21 a31
a12 a22 a32
a13 a23 S = a (a a  a a )  a (a a  a a ) 11 22 33 32 23 21 12 33 32 13 a33 + a31(a12a23  a22a13).
Each quantity in parentheses represents the determinant of a 2 * 2 matrix that is the part of the 3 * 3 matrix remaining when the row and column of the multiplier are eliminated, as shown here.
a11(a22a33
a11  a32a23) C a21 a31
a12 a22 a32
a13 a23 S a33
a21(a12a33
a11  a32a13) C a21 a31
a12 a22 a32
a13 a23 S a33
a31(a12a23
a11  a22a13) C a21 a31
a12 a22 a32
a13 a23 S a33
The determinant of each (red) 2 * 2 matrix is called a minor of the associated (blue) element in the 3 * 3 matrix. The symbol Mij represents the minor that results when row i and column j are deleted. The list on the next page gives some of the minors for the matrix A in the margin.
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Element
a11 A = C a21 a31
a12 a22 a32
a13 a23 S a33
Minor
Element
Minor
a11
a22 M11 = det B a 32
a23 a33 R
a22
a11 M22 = det B a 31
a13 a33 R
a21
a12 M21 = det B a 32
a13 a33 R
a23
a11 M23 = det B a 31
a12 a32 R
a31
a12 M31 = det B a 22
a13 a23 R
a33
a11 M33 = det B a 21
a12 a22 R
In a 4 * 4 matrix, the minors are determinants of 3 * 3 matrices. Similarly, an n * n matrix has minors that are determinants of (n  1) * (n  1) matrices. To find the determinant of a 3 * 3 or larger square matrix, first choose any row or column. Then the minor of each element in that row or column must be multiplied by +1 or  1, depending on whether the sum of the row number and column number is even or odd. The product of a minor and the number +1 or  1 is called a cofactor.
Cofactor Let Mij be the minor for element aij in an n * n matrix. The cofactor of aij , written Aij , is
#
Aij ⴝ (ⴚ1)iⴙ j Mij .
EXAMPLE 3
Finding the Cofactor of an Element Find the cofactor of each given element of the matrix 6 C8 1 (a) 6
(b) 3
2 9 2
4 3 S. 0
(c) 8
Solution (a) Because 6 is located in the first row, first column of the matrix, i = 1 and j = 1. 9 3 M11 = det B R =  6. The cofactor is 2 0 ( 1) 1 + 1( 6) = 1( 6) =  6. (b) Here i = 2 and j = 3. M23 = det B
6 1
2 R = 10. The cofactor is 2
( 1) 2 + 3(10) =  1(10) =  10. (c) We have i = 2 and j = 1. M21 = det B
2 2
4 R =  8. The cofactor is 0
( 1) 2 + 1( 8) =  1( 8) = 8.
7.5
Determinants and Cramer’s Rule
485
Finding the Determinant of an n : n Matrix Multiply each element in any row or column of the matrix by its cofactor. The sum of these products gives the value of the determinant.
The process of forming this sum of products is called expansion by a row or column.
GCM
Evaluating the Determinant of a 3 : 3 Matrix
EXAMPLE 4
2 Evaluate detC  1 1
3 4 0
2  3 S , expanding by the second column. 2
Solution First find the minor of each element in the second column. M12 = det B M22
1 1
2 = det B 1
M32 = det B
2 1
3 R =  1(2)  ( 1)( 3) =  5 2 2 R = 2(2)  ( 1)( 2) = 2 2
Use parentheses and keep track of all negative signs to avoid errors.
2 R = 2( 3)  ( 1)( 2) =  8 3
Now find the cofactor of each of these minors. A12 = ( 1) 1 + 2 A22 = ( 1) 2 + 2 A32 = ( 1) 3 + 2
# # #
M12 = ( 1) 3( 5) =  1( 5) = 5 M22 = ( 1) 4 # 2 = 1 # 2 = 2
M32 = ( 1) 5( 8) =  1( 8) = 8
Find the determinant by multiplying each cofactor by its corresponding element in the matrix and then finding the sum of the products.
This screen shows how a TI83/84 Plus calculator evaluates the 3 * 3 determinant in Example 4. FIGURE 37
For 3 : 3 Matrices + +
+ 
+ +
2 detC  1 1
3 4 0
2 3S = 2 = = =
a12 # A12 + a22 # A22 + a32 # A32  3(5) + ( 4)2 + 0(8)  15 + ( 8) + 0  23
See FIGURE 37.
In Example 4, we would have found exactly the same answer by using any row or column of the matrix. One reason we used column 2 is that it contains a 0 element, so it was not really necessary to calculate M32 and A32. One learns quickly that 0s are “friendly” in work with determinants. Instead of calculating ( 1) i + j for a given element, the sign checkerboard in the margin can be used. The signs alternate for each row and column, beginning with + in the first row, first column, position. If we expand a 3 * 3 matrix about row 3, for example, the first minor would have a + sign associated with it, the second minor a  sign, and the third minor a + sign. This array of signs can be extended to determinants of 4 * 4 and larger matrices.
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GCM
Evaluating the Determinant of a 4 : 4 Matrix
EXAMPLE 5
Evaluate. 1 0 detD 3 2
FIGURE 38
2 1 1 1
3 4 4 0
2 2 T 0 3
Solution The determinant of a 4 * 4 matrix can be found by using cofactors, but the computation is tedious. Using a graphing calculator with matrix capabilities is much easier. FIGURE 38 indicates that the desired determinant is  185.
Derivation of Cramer’s Rule Determinants can be used to solve a linear system of equations in the form a1 x + b1 y = c1 a2 x + b2 y = c2
(1) (2)
by elimination as follows: a1b2 x + b1b2 y = c1b2
Multiply (1) by b2.
 a2b1x  b1b2 y =  c2b1
Multiply (2) by  b1.
(a1b2  a2 b1)x = c1b2  c2 b1 x =
c1b2  c2 b1 a1b2  a2 b1
Add.
if a1b2  a2 b1 Z 0.
Similarly,  a1a2 x  a2 b1 y =  a2 c1
Multiply (1) by  a2.
a1a2 x + a1b2 y = a1c2
Multiply (2) by a1.
(a1b2  a2 b1)y = a1c2  a2 c1
Add.
y =
a1c2  a2 c1 a1b2  a2 b1
if a1b2  a2 b1 Z 0.
Both numerators and the common denominator of these values for x and y can be written as determinants, since
c1b2  c2 b1 = det B
c1 c2
b1 , b2 R
a1c2  a2 c1 = det B
a1 c1 , and a2 c2 R
a1b2  a2 b1 = det B
a1 b1 . a2 b2 R
With these determinants, the solutions for x and y become det B x = det B
c1 b1 c2 b2 R a1 b1 a2 b2 R
det B and y = det B
a1 c1 a2 c2 R a1 b1 a2 b2 R
if det B
a1 a2
b1 Z 0. b2 R
7.5
Determinants and Cramer’s Rule
487
These results are summarized as Cramer’s rule.
Cramer’s Rule for 2 : 2 Systems The solution of the system a1 x + b1 y = c1 a2 x + b2 y = c2 xⴝ
is given by
Dx D
and
yⴝ
Dy D
,
where Dx = det B
c1 c2
b1 a1 , Dy = det B R b2 a2
c1 , and c2 R
D = det B
a1 a2
b1 Z 0. b2 R
CAUTION Cramer’s rule does not apply if D ⴝ 0. When D = 0, the system is inconsistent or has dependent equations. For this reason, evaluate D first.
Using Cramer’s Rule to Solve Systems EXAMPLE 6
Applying Cramer’s Rule to a System with Two Equations Use Cramer’s rule to solve the system. 5x + 7y =  1 6x + 8y = 1 Analytic Solution Dy
Dx
and y = . Find D first, D D because if D = 0, Cramer’s rule does not apply. If D Z 0, then find Dx and Dy. By Cramer’s rule, x =
D = det B
5 6
Dx = det B
1 1
Dy = det B
5 6
Graphing Calculator Solution See FIGURE 39. Enter D, Dx , and Dy as 3A4, 3B4, and 3C4, respectively. Then find the desired quotients x =
det(3B4) det(3A4)
and
y =
det(3C4) det(3A4)
.
7 R = 5(8)  6(7) =  2 8 7 R =  1(8)  1(7) =  15 8 1 R = 5(1)  6( 1) = 11 1
From Cramer’s rule, x =
Dx D
=
 15 15 = 2 2
and
y =
Dy D
=
11 11 =  . 2 2
11 The solution set is E A 15 2 ,  2 B F , as can be verified by substituting into the given system.
FIGURE 39
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Cramer’s Rule for 3 : 3 Systems The solution of the system a1 x + b1 y + c1 z = d1 a2 x + b2 y + c2 z = d2 a3 x + b3 y + c3 z = d3 xⴝ
is given by
where
Dx D
yⴝ
,
Dy D
, and
zⴝ
Dz D
,
d1 b1 Dx = detC d2 b2 d3 b3
c1 c2 S , c3
a1 d1 Dy = detC a2 d2 a3 d3
a1 b1 Dz = detC a2 b2 a3 b3
d1 d2 S , d3
a1 b1 D = detC a2 b2 a3 b3
and
c1 c2 S , c3 c1 c2 S Z 0. c3
EXAMPLE 7
Applying Cramer’s Rule to a System with Three Equations Use Cramer’s rule to solve the system. x + y  z = 2 2x  y + z =  5 x  2y + 3z = 4 Solution Verify that the required determinants are
Thus,
1 D = detC 2 1
1 1 2
1 1 S =  3, 3
2 Dx = detC  5 4
1 Dy = detC 2 1
2 5 4
1 1 S =  22, 3
1 Dz = detC 2 1
x =
Dx D
=
7 7 =  , 3 3
so the solution set is E A
 73, 22 3,
y = 7B F.
Dy D
=
 22 22 = , 3 3
1 1 S = 7, 3
1 1 2
2  5 S =  21. 4
1 1 2 z =
Dz D
=
 21 = 7, 3
EXAMPLE 8
Verifying That Cramer’s Rule Does Not Apply Show why Cramer’s rule does not apply to the system. 2x  3y + 4z = 10 6x  9y + 12z = 24 x + 2y  3z = 5
FIGURE 40
Solution We must show that D = 0. FIGURE 40 confirms this fact, where D = det(3A4). When D = 0, the system either is inconsistent or contains dependent equations. We could use the elimination or row echelon method to tell which is the case. Verify that this system is inconsistent.
7.5
7.5
Determinants and Cramer’s Rule
3. det B
1 5
2 R 3
7. det B
3 5
489
EXERCISES
Checking Analytic Skills Find each determinant. Do not use a calculator. 1. det B
5 4
9 R 1
2. det B
1 2
5. det B
9 3
3 R 1
6. det B
0 1
3 R 9 2 R 5
4 R 2
4 R 1
4. det B
6 0
8. det B
9 2
7 R 6
Find the cofactor of each element in the second row for each matrix. 2 9. C 1 4
0 2 2
1 0S 1
1 10. C 1 0
1 0 3
2 2S 1
1 11. C 2 1
2 3 4
1 2S 1
2 12. C 3 2
1 0 1
4 1S 4
2 2 1
2 16. detC 4 2
1 7 4
1 2S 0
Find each determinant. 4 13. detC 2 6
7 1 3
10 17. detC  1 3
2 4 8
2 20. detC 1 1
3 9 2
0.4 23. detC 0.3 3.1
 0.8 0.9 4.1
8 3S 0
8 14. detC 7 5
1 3S 10 0 0S 0
2 0 1
4 3S 2
1 15. detC  1 0
7 18. detC 1 2
1 7 1
1 2S 1
1 19. detC 0 1
3 21. detC 2 6
3 6 6
1 0S 2
5 22. detC  5 1
 0.3 24. detC 2.5  0.1
0.6 0.7 S  2.8
0 1S 4
 0.1 4.9 0.4
0.9  3.2 S 0.8
2 0 10
17 25. detC 11 7
3 0S  12
3 3 0 4 5 9
2 2S 1 3  15 S 23
Solve each determinant equation for x. x R = 6 2
26. det B
5 3
29. det B
2x x R = 6 11 x
5 32. detC 0 4
3x 2 1
27. det B
 0.5 2 R = 0 x x
2 30. detC  1 5
3 1S = 7 x
0 3 2
2x 1 33. detC 0 4 3 0
28. det B
1 xS = 3 0
x 3 R = 4 x x
4 31. detC 2 3
1 xS = x 2
3 0 x
0 1S = 5 1
x x 2 34. detC 0 2 2 S = 96 0 0 3x
Evaluate each determinant. 3 0 35. detD 6 7
6 2 4 3
5 1 2 1
1 3 T 0 1
4 2 36. detD 5 0
5 3 1 2
1 1 3 1
1 0 T 9 5
4 1 37. detD 2 0
0 0 4 0
0 3 0 1
2 0 T 1 2
2 3 38. detD 0 9
0 6 0 0
4 0 0 2
2 4 T 3 1
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Area of a Triangle A triangle with vertices at (x1, y1), (x2, y2), and (x3, y3), as shown in the figure, has area equal to the absolute value of D, where x1 1 D = detC x2 2 x3
R(x3, y3) Q(x2, y2 )
y1 1 y2 1 S. y3 1
P(x1, y1)
0
x
Use D to find the area of each triangle with coordinates as given. 39. P(0, 0), Q(0, 2), R(1, 4)
40. P(0, 1), Q(2, 0), R(1, 5)
41. P(2, 5), Q( 1, 3), R(4, 0)
42. P(2,  2), Q(0, 0), R( 3,  4)
43. P(1, 2), Q(4, 3), R(3, 5)
44. P( 1, 0), Q( 3, 5), R(5, 2)
Use the concept of the area of a triangle discussed in Exercises 39–44 to determine whether the three points are collinear. 45. (1, 3), ( 3, 11), (2, 1)
46. (3, 6), ( 1,  6), (5, 11)
47. ( 2,  5), (4, 4), (2, 3)
48. (4,  5), ( 2, 10), (6,  10)
49. (4,  1), (6, 0), (12, 4)
50. ( 1,  4), (3, 8), (6, 17)
Several theorems are useful for calculating determinants. These theorems are true for square matrices of any dimension.
Determinant Theorems 1. If every element in a row (or column) of matrix A is 0, then det A = 0. 2. If the rows of matrix A are the corresponding columns of matrix B, then det B = det A. 3. If any two rows (or columns) of matrix A are interchanged to form matrix B, then det B =  det A. 4. Suppose matrix B is formed by multiplying every element of a row (or column) of matrix A by the real number k. Then det B = k # det A. 5. If two rows (or columns) of a matrix A are identical, then det A = 0. 6. Changing a row (or column) of a matrix by adding a constant times another row (or column) to it does not change the determinant of the matrix.
Use the determinant theorems to find each determinant. 1 51. detC 1 3 4 54. detC  1 2
0 0 0
0 1S 0 8 2 4
0 1S 3
1 52. detC 4 3
2 8 0
4 55. detC 2 0
1 0 2
4  16 S 5 4 1S 4
6 53. detC  1 4 6 56. detC 1 5
8 0 0 3 0 7
 12 2S 8 2 2S 3
Use Cramer’s rule to solve each system of equations. If D = 0, use another method to complete the solution. 57. x + y = 4 2x  y = 2
58. 3x + 2y =  4 2x  y =  5
59. 4x + 3y =  7 2x + 3y =  11
60. 4x  y = 0 2x + 3y = 14
61. 3x + 2y = 4 6x + 4y = 8
62. 1.5x + 3y = 5 2x + 4y = 3
63. 2x  3y =  5 x + 5y = 17
64. x + 9y =  15 3x + 2y = 5
7.6
65. 4x  y + 3z =  3 3x + y + z = 0 2x  y + 4z = 0
70. 3x + 5y =  7 2x + 7z = 2 4y + 3z =  8
72.  2x  2y + 3z = 4 5x + 7y  z = 2 2x + 2y  3z =  4
73.
9 75.
3x 2x x 2x  y
+ 2y  w + z + 2w + 2y  z + z + w
In your own words, explain what happens when you apply Cramer’s rule if D = 0.
7.6
71.
= = = =
9 76.
2x  y + 3z = 1  2x + y  3z = 2 5x  y + z = 2
74. x + 2y  z 2x  y y x
0 5 2 2
491
68. x + y + z = 4 2x  y + 3z = 4 4x + 2y  z =  15
67. 2x  y + 4z =  2 3x + 2y  z =  3 x + 4y + 2z = 17
66. 5x + 2y + z = 15 2x  y + z = 9 4x + 3y + 2z = 13
69. 5x  y =  4 3x + 2z = 4 4y + 3z = 22
Solution of Linear Systems by Matrix Inverses
+ w = 8 + 2w = 8 + 3z = 5  z = 4
Describe Dx , Dy , and Dz in terms of the coefficients and constants in a given system of equations.
Solution of Linear Systems by Matrix Inverses
Identity Matrices • Multiplicative Inverses of Square Matrices • Using Determinants to Find Inverses • Solving Linear Systems Using Inverse Matrices • Curve Fitting Using a System
In Section 7.4, we saw several similarities between the set of real numbers and the set of matrices. Another similarity is that both sets have identity and inverse elements for multiplication.
Identity Matrices By the identity property for real numbers, for any real number a, a
#
1 = a
and
1
#
a = a.
If there is to be a multiplicative identity matrix I, such that AI = A FIGURE 41 shows how the
and
IA = A,
for any matrix A, then A and I must be square matrices with the same dimension.
identity matrix I2 is displayed on a graphing calculator with matrix capabilities.
2 : 2 Identity Matrix If I2 represents the 2 * 2 identity matrix, then I2 ⴝ c
1 0
0 d. 1
FIGURE 41
492
CHAPTER 7
Systems of Equations and Inequalities; Matrices
GCM
EXAMPLE 1
Let A = c
2 3
Using the 2 : 2 Identity Matrix
6 d . Show that AI 2 = A and I2 A = A. 5
Analytic Solution
Graphing Calculator Solution Define 3A4 as shown in FIGURE AI2 = A and I2 A = A.
2 6 1 0 d c d 3 5 0 1  2(1) + 6(0)  2(0) + 6(1) = c d 3(1) + 5(0) 3(0) + 5(1) 2 6 = c d = A 3 5
AI 2 = c
1 0 2 6 d c d 0 1 3 5 1( 2) + 0(3) 1(6) + 0(5) = c d 0( 2) + 1(3) 0(6) + 1(5) 2 6 = c d = A 3 5
42.
Then show that
I2 A = c
FIGURE 42
FIGURE 43 shows how the
TI83/84 Plus graphing calculator displays I3 and I4.
n : n Identity Matrix
0 1
Á Á
0 0
Á Á
a ij
Á
For any value of n, there is an n * n identity matrix having 1s down the main diagonal and 0s elsewhere. The n : n identity matrix is given by
In ⴝ D
1 0 0
0
Á
T.
1
Here, aij = 1 when i = j (the diagonal elements), and aij = 0 otherwise. FIGURE 43
EXAMPLE 2
2 Let A = C 3 0
4 5 8
Using the 3 : 3 Identity Matrix 0 9 S . Show that AI 3 = A. 6
Analytic Solution Using the 3 * 3 identity matrix and the definition of matrix multiplication gives 2 AI 3 = C 3 0
4 5 8
0 1 9S C0 6 0
2 = C 3 0
4 5 8
0 9 S = A. 6
0 1 0
Graphing Calculator Solution See FIGURE 44.
0 0S 1
FIGURE 44
7.6
NOTE
Solution of Linear Systems by Matrix Inverses
493
It can also be shown that I3 A = A in Example 2.
Multiplicative Inverses of Square Matrices For every nonzero real number a, there is a multiplicative inverse 1a such that a
#
1 = 1 a
1 a
and
#
a = 1.
(Recall: 1a is also written a 1.) In a similar way, if A is an n * n matrix, then its multiplicative inverse, written A1, must satisfy both AA1 = In and
A1 A = In.
This result means that only a square matrix can have a multiplicative inverse. CAUTION
Although a 1 =
1 a
for any nonzero real number a, if A is a matrix, then Aⴚ1 ⴝ
In fact,
1 A
1 . A
has no meaning, since 1 is a number and A is a matrix.
EXAMPLE 3
Verifying an Inverse Determine if B is the inverse of A, where A = c
5 3
3 d 2
and
B = c
3 d. 5
2 3
Solution For B to be the inverse of A, it must satisfy AB = I2 and BA = I2. AB = c
5 3
3 2 d c 2 3
3 1 d = c 0 5
0 d = I2 1
BA = c
2 3
3 5 d c 5 3
3 1 d = c 0 2
0 d = I2 1
Thus, B is the inverse of A. That is, B = A1. The inverse matrix of an n * n matrix A (if it exists) can be found analytically by first forming the augmented matrix 3A  In4 and then performing matrix row operations, until the left side of the augmented matrix becomes the identity matrix. The resulting augmented matrix can be written as 3In  A14, where the right side of the matrix is A1. EXAMPLE 4
Finding a 2 : 2 Inverse Matrix
Find A1, if A = B
2 1
4 R. 1
Solution First form the 2 * 4 augmented matrix 3A  I24. Then use row transformations to obtain the 2 * 4 augmented matrix 3I2  A14. (continued)
494
CHAPTER 7
Systems of Equations and Inequalities; Matrices
c
2 1
4 1 ` 1 0
0 d 1
Augmented matrix 3A  I2 4
c
1 2
1 0 ` 4 1
1 d 0
Interchange R1 and R2 to get 1 in the upper left corner.
c
1 0
1 0 ` 6 1
1 d 2
 2R1 + R2
1 0
1 0 ` 1 1 6
1
1
0
0
1
B
C
Thus,
A1
=
1 B 16 6

`
1 6 1 6
2 3 1 R . To 3
 13

R
1 6 R2
R2 + R1; augmented matrix 3I2  A14
2 3 S 1 3
check, show that AA1 = I2 and A1A = I2.
If A1 exists, then it is unique. If A1 does not exist, then A is a singular matrix. The process for finding the multiplicative inverse A1 for any n * n matrix A that has an inverse is summarized as follows.
Finding an Inverse Matrix To obtain A1 for any n * n matrix A for which A1 exists, follow these steps:
Step 1 Form the augmented matrix 3A  In 4, where In is the n * n identity matrix. Step 2 Perform row transformations on 3A  In 4, to obtain a matrix of the form 3In  B4. Step 3 Matrix B is A1.
NOTE To confirm that two n * n matrices A and B are inverses of each other, it is sufficient to show that AB = In. It is not necessary to show also that BA = In. GCM
EXAMPLE 5
Find
A1
1 if A = C 2 3
Finding the Inverse of a 3 : 3 Matrix 0 2 0
1 1 S. 0
Solution Use row transformations as follows. Step 1
Write the augmented matrix 3A  I3 4. 1 C2 3
0 2 0
1 1 1 3 0 0 0
0 1 0
0 0S 1
7.6
Step 2
Solution of Linear Systems by Matrix Inverses
495
Since 1 is already in the upper lefthand corner, use the row transformation that will result in 0 for the first element in the second row. Multiply the elements of the first row by  2, and add the result to the second row. 1 C0 3
0 2 0
1 1 3 3 2 0 0
0 1 0
0 0S 1
 2R1 + R2
To get 0 for the first element in the third row, multiply the elements of the first row by  3 and add the result to the third row. 1 C0 0
0 2 0
1 1 3 3 2 3 3
0 1 0
0 0S 1
 3R1 + R3
Continuing this process, we transform the matrix as follows.
A graphing calculator can be used to find the inverse of a matrix as shown above. The screens support the result in Example 5. The elements of the inverse are expressed as fractions, so it is easier to compare with the inverse matrix found in the example.
Step 3
1 C0 0
0 1 0
1
1 C0 0
0 1 1 1 32 3 1 0 1 1
1 C0 0
0 1 0
0 0 3 3 2 1 1 1
1
0
0
0
0
D0
1
0 4  12
 12
0
0
1
3 2
3
3
1 1 3
0 
0
0  12 0
 13
0
1 3

 21 R2
0 0S  13 R3  1R3 + R1
0S  13
1 2
0
1
0 0S 1
1 2
0

1 3 1 2T 1 3
 32 R3 + R2
The last transformation shows that the inverse is 0
0
A1 = D  12
 12
1
0

1 3 1 2T 1 3
Confirm this result by forming the product A1 A or AA1, each of which equals the matrix I3.
Using Determinants to Find Inverses Determinants can be used to find the inverses of matrices.
Inverse of a 2 : 2 Matrix If A = c
a c
b 1 d d and det A Z 0, then A ⴚ 1 ⴝ c d det A ⴚc
ⴚb d. a
If det A ⴝ 0, then Aⴚ1 does not exist and A is a singular matrix.
496
Systems of Equations and Inequalities; Matrices
CHAPTER 7
GCM
Finding the Inverse of a 2 : 2 Matrix Use a determinant to find A1, if it exists, for each matrix. 2 3 3 6 (a) A = c d (b) A = c d 1 1 2 4 EXAMPLE 6
Analytic Solution
Graphing Calculator Solution
(a) First we find det A and determine whether ally exists.
A1
actu
(a) See FIGURE 45, which illustrates both 3A4 and 3A41.
det A = 2( 1)  1(3) = 5 Since  5 Z 0, A1 exists. Thus, from the box on the preceding page, A1
1 1 = c 5 1 1
=
B 51 5
3 5  25
3 d 2
FIGURE 45
(b) The calculator returns a singular matrix error message when it is directed to find the inverse of a matrix whose determinant is 0. See FIGURE 46.
R.
(b) Here, A1 does not exist, since det A = 0: det A = 3( 4)  2( 6) = 0.
A is singular. FIGURE 46
Solving Linear Systems Using Inverse Matrices We used matrices to solve systems of linear equations by the row echelon method in Section 7.3. Another way to use matrices to solve linear systems is to write the system as a matrix equation AX = B, where A is the matrix of the coefficients of the variables of the system, X is the column matrix of the variables, and B is the column matrix of the constants. Matrix A is called the coefficient matrix. To solve the matrix equation AX = B, first see if A1 exists. Assuming that it does, use the facts that A1 A = I and IX = X to get AX = = 1 (A A)X = IX = X =
A1(AX )
B A1B A1B A1B A1B.
Multiply each side by A1. Associative property Multiplicative inverse property Identity property
CAUTION When multiplying by matrices on each side of a matrix equation, be careful to multiply in the same order on each side. Multiplication of matrices is not commutative. EXAMPLE 7
Using a Matrix Inverse to Solve a System of Equations Use the inverse of the coefficient matrix to solve the system. 2x + y + 3z = 1 x  2y + z =  3  3x + y  2z =  4
7.6
Solution of Linear Systems by Matrix Inverses
Analytic Solution We find the determinant of the coefficient matrix to be sure that it is not 0. If 2 A = C 1 3
1 2 1
497
Graphing Calculator Solution Entering the coefficient matrix as 3A4, we find that det(3A4) =  10 and 3A41 exists. See FIGURE 47. Then we enter the column of constants as 3B4 and find the matrix product 3A413B4. The display shows the column matrix representing x = 4, y = 2, and z =  3.
3 1S , 2
then det A =  10 Z 0 and A1 exists. We evaluate A1B with x X = CyS z
and
2 = C 1 3
A1B
1 B = C 3 S. 4
1 2 1
3  10
 12
1 10 1 2
1 2 1 2
= D

3 1 1 1S C 3S 2 4 7  10

1 10 T 1 2
1
4 D 3T = C 2 S 3 4
FIGURE 47
The solution set is 5(4, 2,  3)6.
Since X = A1B, we have x = 4, y = 2, and z =  3. The solution set is 5(4, 2,  3)6. The elements of A1 can be found by hand, as demonstrated in Example 5, or by using a calculator.
NOTE If the determinant of the coefficient matrix A is 0, the system either is inconsistent or has dependent equations and A1 does not exist. FOR DISCUSSION
1. Using the window 3 5, 54 by 3 50, 504, reproduce the graph in FIGURE 48, based on the result of Example 8. 2. If your calculator has cubic regression, use it to find the same results as in Example 8.
Curve Fitting Using a System EXAMPLE 8
Using a System to Find the Equation of a Cubic Polynomial FIGURE 48 shows four views of the graph of a polynomial function of the form P(x) = ax 3 + bx 2 + cx + d. Use the points indicated to write a system of four equations in the variables a, b, c, and d, and then use the inverse matrix method to solve the system. What is the equation that defines this graph?
50
–5
5
–5
5
–50
–50
First view
Second view
50
50
50
–5
5
–5
5
–50
–50
Third view
Fourth view
FIGURE 48
(continued)
498
CHAPTER 7
Systems of Equations and Inequalities; Matrices
Solution We see from the graph in FIGURE 48 on the preceding page that P( 2) =  28, P( 1) =  10, P(1) =  4, and P(3) = 2. From the first of these, we get P( 2) = a( 2) 3 + b( 2) 2 + c( 2) + d =  28  8a + 4b  2c + d =  28. or, equivalently, Similarly, from the others, we find the following equations: From P( 1) =  10:  a + b  c + d =  10 a + b + c + d = 4 From P(1) =  4: 27a + 9b + 3c + d = 2. From P(3) = 2: Now we must solve the system formed by these four equations:  8a a a 27a
+ + + +
4b b b 9b
+ +
2c c c 3c
+ + + +
d d d d
= = = =
 28  10 4 2.
We use the inverse matrix method to solve the system, letting 8 1 A = D 1 27
4 1 1 9
2 1 1 3
 28  10 B = D T. 4 2
1 a 1 b T , X = D T , and 1 c 1 d
Because det A = 240 Z 0, a unique solution exists. On the basis of our earlier discussion, X = A1B. FIGURE 49 shows how the matrix inverse method yields the values a = 1, b =  3, c = 2, and d =  4. The polynomial is defined by P(x) = x 3  3x 2 + 2x  4. FIGURE 49
7.6
EXERCISES
Checking Analytic Skills Determine whether A and B are inverses by calculating AB and BA. Do not use a calculator. 1. A = c
5 2
3. A = c
1 3
7 3 d; B = c 3 2
7 d 5
2 5 d; B = c 5 3
0 5. A = C 0 1
1 0 1
1 7. A = C 4 0
1 5 1
2 d 1
0 1  2 S; B = C 1 0 0
0 1 0 0S 1 0
1 15 0 S; B = C  12 3 4
4 3 1
5 4S 1
2. A = c
2 1
3 1 d; B = c 1 1
3 d 2
4. A = c
2 3
1 2 d; B = c 2 3
1 d 2
1 6. A = C 0 0
2 1 1
0 1 0 S; B = C 0 0 0
1 8. A = C 1 1
3 4 3
3 7 3 S; B = C  1 4 1
2 1 1 3 1 0
0 0S 1 3 0S 1
Checking Analytic Skills For each matrix, find A1 if it exists Do not use a calculator. 9. A = c
3 2
7 d 5
10. A = c
5 8
3 d 5
11. A = c
1 3
2 d 4
7.6
12. A = c
1 2
2 d 1
13. A = c
6 3
4 d 2
14. A = c
15. A = c
0.6 0.5
0.2 d 0.1
16. A = c
0.8 0.5
 0.3 d  0.2
0 17. A = C 1 0
0 1 1
2 20. A = C 1 1
1 18. A = C 1 0
9 21.
0 1 1
0 0S 1
1 19. A = C 2 1
Under what condition will the inverse of a square matrix not exist?
499
Solution of Linear Systems by Matrix Inverses
1 3S 1
9 22.
10 d 6
5 3 0 0 1
1 0S 0 1 0 1
0 1S 0
Explain why a 2 * 2 matrix will not have an inverse if either a column or a row contains all 0s.
For each matrix, find A1 if it exists. 1 23. A = C 0 1
0 1 0
1 25. A = C 4 0
1 5 1
 0.4 27. A = C 0 0.3 2 29. A = C 5 3 31. A = B
0 0S 1
22  17
1.4 33. A = C 0.84 0.56
1 0S 3
0.1 0.6 0
1 10 6
1 24. A = C 1 1
1 2
0 1 1
0.8 28. A = C  0.2 0
0.2 0.8 S  0.2
32. A = B
2 3
34. A = D 0 1 2
0.2 0 0
0.1 0.3 S 0.5 2 2S 1
0.7 R 23
22 1 2
0.59 0.62 S 1.3
4 5S 2
3 3 0
5 30. A = C  5 1
R
0.5 1.36 0.47
3 3S 4
2 26. A = C 3 1
2 5S 3 0.5
3 4 3
1 4 1 4 1 2
1 3 1 3T 1 3
Solve each system by using the matrix inverse method. 35. 2x  y =  8 3x + y =  2
36. x + 3y =  12 2x  y = 11
37. 2x + 3y =  10 3x + 4y =  12
39. 2x  5y = 10 4x  5y = 15
40. 2x  3y = 2 4x  6y = 1
41.
43. x + 3y + z = 2 x  2y + 3z =  3 2x  3y  z = 34
46.
3x 2x x 2x  y
+ 2y  w + z + 2w + 2y  z + z + w
= = = =
0 5 2 2
38. 2x  3y = 10 2x + 2y = 5
2x + 4z = 14 3x + y + 5z = 19  x + y  2z =  7
42. 3x + 6y + 3z = 12 6x + 4y  2z =  4 y  z = 3 + + 
+ + 
+ 
44.
x + y  z = 6 2x  y + z =  9 x  2y + 3z = 1
45.
x 4x  3x x
47.
x  22 y = 2.6 0.75x + y = 7
48.
2.1x + y = 25 22 x  2y = 5
3y y y y
2z z z 3z
w 2w w 2w
= = = =
9 2 5 2
500
49.
Systems of Equations and Inequalities; Matrices
CHAPTER 7
50.
px + ey + 22 z = 1 ex + py + 22 z = 2 22 x + ey + pz = 3
(log 2)x + (ln 3)y + (ln 4)z = 1 (ln 3)x + (log 2)y + (ln 8)z = 5 (log 12)x + (ln 4)y + (ln 8)z = 9
Curve Fitting In Exercises 51 and 52, use the method of Example 8 to find the cubic polynomial P(x) that defines the curve shown in the four figures. 51.
20
20
–5
–5
5
5
–5
20
5
–5
5
–20
–20
–20
–20
First view
Second view
Third view
Fourth view
50
50
50
50
52.
20
–5
5
–5
5
–5
5
–5
5
–50
–50
–50
–50
First view
Second view
Third view
Fourth view
53. Curve Fitting Find the fourthdegree polynomial P(x) satisfying the following conditions: P( 2) = 13, P( 1) = 2, P(0) =  1, P(1) = 4, and P(2) = 41.
56. Determining Prices A group of students bought 3 soft drinks and 2 boxes of popcorn at a movie for $18.50. A second group bought 4 soft drinks and 3 boxes of popcorn for $26.
54. Curve Fitting Find the fifthdegree polynomial P(x) satisfying the following conditions: P( 2) =  8, P( 1) =  1, P(0) =  4, P(1) =  5, P(2) = 8, and P(3) = 167. (Modeling) Solve each problem. 55. Cost of CDs A music store has compact discs that sell for three prices, marked A, B, and C. The last column in the table shows the total cost of a purchase. Use this information to determine the cost of one CD of each type by setting up a matrix equation and solving it with an inverse.
A
B
C
Total
2 1 2
3 4 1
4 0 3
$120.91 $ 62.95 $ 79.94
(a) Find a matrix equation AX = B whose solution gives the individual prices of a soft drink and a box of popcorn. Solve this matrix equation by using A1. 9 (b) Could these prices be determined if both groups had bought 3 soft drinks and 2 boxes of popcorn for $18.50? Try to calculate A1 and explain your results.
7.6
57. Tire Sales In a study, the relationship among annual tire sales T in thousands, automobile registrations A in millions, and personal disposable income I in millions of dollars was investigated. Representative data for three different years are shown in the table. T
A
I
10,170 15,305 21,289
113 133 155
308 622 1937
Source: Jarrett, J., Business Forecasting Methods, Basil Blackwell.
The data were modeled by T = aA + bI + c, where a, b, and c are constants. (a) Use the data to write a system of linear equations whose solution gives a, b, and c. (b) Solve this linear system. Write a formula for T. (c) If A = 118 and I = 311, predict T. (The actual value for T was 11,314.)
P
S
C
122 130 158
1500 2000 2200
8 5 10
The data were related by P = a + bS + cC, where a, b, and c are constants. (a) Use the data to write a system of linear equations whose solution gives a, b, and c. (b) Estimate the selling price of a home with 1800 square feet and in condition 7. 60. Traffic Flow Refer to Exercises 73 and 74 in Section 7.3. The figure shows four oneway streets with intersections A, B, C, and D. Numbers indicate the average traffic flow in vehicles per minute. The variables x1, x2, x3, and x4 denote unknown traffic flows.
58. Plate Glass Sales The amount of plate glass G sold can be affected by the number of new building contracts B issued and automobiles A produced, since plate glass is used in buildings and cars. A plate glass company in California wanted to forecast sales. The table contains sales data for three consecutive years. All units are in millions.
4 cars/min
5
3
cars/min
cars/min
A
A
B
603 657 779
5.54 6.93 7.64
37.1 41.3 45.6
Source: Makridakis, S. and S. Wheelwright, Forecasting Methods for Management, John Wiley and Sons.
x1
6 5 cars/min
D
B
6 cars/min
x2
cars/min
x4
G
501
Solution of Linear Systems by Matrix Inverses
x3
C
7 cars/min
4 cars/min
(a) The number of vehicles per minute entering an intersection equals the number exiting an intersection. Verify that the given system of linear equations describes the traffic flow. A: x1 + 5 = 4 + 6 B: x2 + 6 = x1 + 3
The data were modeled by G = aA + bB + c,
C: x3 + 4 = x2 + 7
where a, b, and c are constants. (a) Write a system of linear equations whose solution gives a, b, and c. (b) Solve this linear system. Write a formula for G. (c) For the following year, it was estimated that A = 7.75 and B = 47.4. Predict G. (The actual value for G was $878 million.)
D: 6 + 5 = x3 + x4
59. Home Prices Realestate companies sometimes study how the selling price of a home is related to its size and condition. The table contains sales data for three homes. Price P is measured in thousands of dollars, home size S is in square feet, and condition C is rated on a scale from 1 to 10, where 10 represents excellent condition.
(b) Write the system as AX = B, and solve by using A1. 9 (c) Interpret your results. In Exercises 61–64, given a square matrix A1, find matrix A. 61. A1 = c
5 1
62. A1 = C
3 20 1  20
9 d 2 1 4 S 1 4
502
Systems of Equations and Inequalities; Matrices
CHAPTER 7
63. A1 =
2 3 D 13 1 3
 13
0
 53
1T
1 3
a 65. Let A = C 0 0
0 b 0
0 0 S, where a, b, and c are nonzero real c
0 numbers. Find A1.
0 64. A1 = C 0 1
0 1 0
1 0S 0
1 66. Let A = C 0 0
0 0 1
0  1 S. Show that A3 = I3 , and use this 1
result to find the inverse of A.
SECTIONS 7.4–7.6
REVIEWING BASIC CONCEPTS
Find the following, if possible, given matrices A = c
5 2
4 d, 1
B = c
0 3
2 d, 4
2 C = C 2 0
3 1 1
1 0 S, 4
0 D = C0 1
4 2 1
1 2 S. 1
1. A  B
2.  3B
3. A2
4. CD
5. det A
6. det C
7. A1
8. C 1
9. Roof Trusses Linear systems occur in the design of roof trusses for new homes and buildings. The simplest type of roof truss is a triangle. The truss shown in the figure is used to frame roofs of small buildings. If a 100pound force is applied at the peak of the truss, then the forces or weights W1 and W2 exerted parallel to each rafter of the truss are determined by the following linear system of equations: 100 lb
23 (W1 + W2) = 100 2 W1  W2 = 0. Solve the system, using Cramer’s rule to find W1 and W2. (Source: Hibbeler, R., Structural Analysis, PrenticeHall.)
W1 60°
10. Solve the system, using the inverse matrix method. 2x + y + 2z = 10 y + 2z = 4 x  2y + 2z = 1
W2 60°
7.7
7.7
Systems of Inequalities and Linear Programming
503
Systems of Inequalities and Linear Programming
Solving Linear Inequalities • Solving Systems of Inequalities • Linear Programming
y
Many mathematical descriptions of real situations are best expressed as inequalities rather than equations. For example, a firm might be able to use a machine no more than 12 hours a day, while the production of at least 500 cases of a certain product might be required to meet a contract. The simplest way to see the solution of an inequality in two variables is to draw its graph. A line divides a plane into three sets of points: the points of the line itself and the points belonging to the two regions determined by the line. Each of these two regions is called a half plane. The line is the boundary of each half plane. See FIGURE 50.
r P x
0
Q
Line r divides the plane into three different sets of points: line r, half plane P, and half plane Q. The points on r belong neither to P nor to Q.
Solving Linear Inequalities
FIGURE 50
Linear Inequality in Two Variables A linear inequality in two variables is an inequality of the form Ax ⴙ By ◊ C,
y
where A, B, and C are real numbers with A and B not both equal to 0. (The symbol … could be replaced with Ú , 6 , or 7 .) x
0
2
The graph of a linear inequality is a half plane, perhaps with its boundary. For example, to graph the linear inequality
3x – 2y = 6 –3
3x  2y … 6, first graph the boundary, 3x  2y = 6, as shown in FIGURE 51(a). Since the points of the line 3x  2y = 6 satisfy 3x  2y … 6, this line is part of the solution. To decide which half plane is part of the solution, solve the original inequality for y.
(a) y
3x – 2y ≤ 6
x
0
2
3x  2y … 6 Reverse the inequality symbol  2y …  3x + 6 when multiplying by 3 a negative number. y Ú x  3 2
Subtract 3x. Multiply by  12; change … to Ú .
–3
(b) FIGURE 51
For a particular value of x, the inequality will be satisfied by all values of y that are greater than or equal to 32 x  3. Thus, the solution contains the half plane above the line, as shown in FIGURE 51(b). CAUTION A linear inequality must be in slopeintercept form (solved for y) to determine, from the presence of asymbol, whether to shade the lower or upper halfplane. In FIGURE 51(b), the upper halfplane is shaded, even though the inequality is 3x  2y … 6 (with a 6 symbol) in standard form. Only when we write the inequality as y Ú 32 x  3 (slopeintercept form) does the 7 symbol indicate to shade the upper halfplane.
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Systems of Equations and Inequalities; Matrices
CHAPTER 7
EXAMPLE 1
GCM
Graphing a Linear Inequality
Graph x + 4y 7 4. Analytic Solution The boundary here is the line x + 4y = 4. Since the points on this line do not satisfy x + 4y 7 4, make the line dashed, as in FIGURE 52. To decide which half plane represents the solution, solve for y. x + 4y 7 4 4y 7  x + 4 1 y 7  x + 1 4
Subtract x. Divide by 4.
Since y is greater than  14 x + 1, the graph of the solution set is the half plane above the boundary, as shown in FIGURE 52.
x 0 4
y
y
x + 4y > 4
1 0 1 0
FIGURE 52
Alternatively, or as a check, choose a test point not on the boundary line and substitute into the inequality. Using (0, 0) makes the substitution easy.
x + 4y 7 4 0 + 4(0) 7 4 0 7 4
y = 
1 x + 1. 4
Graph this boundary, and use the appropriate commands for your calculator to shade the region above the boundary. See FIGURE 53. (Notice that the calculator does not tell you which region to shade.) 4
–2
x
4
Graphing Calculator Solution Solve the corresponding linear equation (the boundary) for y to get
Original inequality Substitute a test point.
6
–3 FIGURE 53
The calculator graph does not distinguish between solid boundary lines and dashed boundary lines. Whether the boundary is solid or dashed must be determined by looking at the inequality symbol. We must understand the mathematics to correctly interpret a calculator graph of an inequality.
False
Since the point (0, 0) is below the boundary, the points that satisfy the inequality must be above the boundary.
Graphing an Inequality 1. For a function ƒ, the graph of y 6 ƒ(x) consists of all the points that are below the graph of y = ƒ(x); the graph of y 7 ƒ(x) consists of all the points that are above the graph of y = ƒ(x). (Similar statements can be made for … and Ú , with boundaries included.) 2. If the inequality is not or cannot be solved for y, choose a test point not on the boundary. If the test point satisfies the inequality, the graph includes all points on the same side of the boundary as the test point. Otherwise, the graph includes all points on the other side of the boundary.
7.7
Systems of Inequalities and Linear Programming
505
Solving Systems of Inequalities The solution set of a system of inequalities is the intersection of the solution sets of its members. x 7 6  2y x 2 6 2y GCM
System of inequalities
EXAMPLE 2
Graphing a System of Two Inequalities Graph the solution set of the system. x 7 6  2y x 2 6 2y
Analytic Solution FIGURES 54 and 55 show the graphs of x 7 6  2y and x 2 6 2y. The methods of Section 7.1 can be used to show that the boundaries intersect at the points (2, 2) and A  3, 92 B . The solution set of the system is shown in FIGURE 56. Since the points on the boundaries of x 7 6  2y and x 2 6 2y do not belong to the graph of the solution, the boundaries are dashed.
Graphing Calculator Solution As usual, solve each inequality for y first. Enter the boundary equations, y1 =
y1 7
x > 6 – 2y
–2
0
x2 = 2y
2
2 x 2
4
–2
6
FIGURE 54
x
0
2
4
y2 =
x2 . 2
6  x 2
and
y2 7
x2 , 2
y is greater than an expression involving x. Therefore, use the capability of your calculator to shade above each boundary. See FIGURE 57.
x2 < 2y
x = 6 – 2y
and
In both inequalities,
y
y
6  x 2
10
6
FIGURE 55 y
x > 6 – 2y x2 < 2y
10
–10 –2
= 2y
FIGURE 57
(2, 2) x = 6 – 2y
The crosshatched region in FIGURE 57 supports the solution shown in FIGURE 56. The boundary lines are not included in the solution set.
x2
(–3, 92 )
2 –2
0
x 2
4
6
FIGURE 56
y
EXAMPLE 3
Graphing a System of Three Inequalities Graph the solution set of the system.
y ≥ 2x 2 9x + 4y 2 ≤ 36 2x + y < 1 0
x 3
FIGURE 58
y Ú 2x + 4y 2 … 36 2x + y 6 1
9x 2
Solution Graph the three inequalities on the same axes and shade the region common to all three. See FIGURE 58. Two boundary lines are solid and one is dashed.
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Linear Programming An important application of mathematics to business and social science is called linear programming. We use linear programming to find an optimum value—for example, minimum cost or maximum profit. Linear programming was first developed to solve problems in allocating supplies for the U.S. Air Force during World War II. EXAMPLE 4
Finding a Maximum Profit Model The Charlson Company makes two products: MP3 players and DVD players. Each MP3 player gives a profit of $30, while each DVD player produces a $70 profit. The company must manufacture at least 10 MP3 players per day to satisfy one of its customers, but no more than 50, because of production restrictions. The number of DVD players produced cannot exceed 60 per day, and the number of MP3 players cannot exceed the number of DVD players. How many of each should the company manufacture to obtain maximum profit? Solution First, we translate the statement of the problem into symbols. Let
x = number of MP3 players to be produced daily,
and
y = number of DVD players to be produced daily.
The company must produce at least 10 MP3 players (10 or more), so x Ú 10. Since no more than 50 MP3 players may be produced, x … 50. No more than 60 DVD players may be made in one day, so y … 60. The number of MP3 players may not exceed the number of DVD players, which translates as x … y. The numbers of MP3 players and of DVD players cannot be negative, so x Ú 0 and
y Ú 0.
These restrictions, or constraints, form the system of inequalities x Ú 10,
x … 50,
y … 60,
x … y,
x Ú 0, y Ú 0.
Each MP3 player gives a profit of $30, so the daily profit from production of x MP3 players is 30x dollars. Also, the profit from production of y DVD players will be 70y dollars per day. The total daily profit is thus y
profit = 30x + 70y.
(10, 60)
(50, 60) (50, 50)
Region of feasible solutions (10, 10) 0
FIGURE 59
x
This equation defines the function to be maximized, called the objective function. To find the maximum possible profit that the company can make, subject to these constraints, we sketch the graph of each constraint. The only feasible values of x and y are those which satisfy all constraints—that is, the values that lie in the intersection of the graphs of the constraints. The intersection is shown in FIGURE 59. Any point lying inside the shaded region or on the boundary in the figure satisfies the restrictions as to the number of MP3 players and DVD players that may be produced. (For practical purposes, however, only points with integer coefficients are useful.) This region is called the region of feasible solutions.
7.7
Systems of Inequalities and Linear Programming
507
The vertices (singular: vertex), or corner points, of the region of feasible solutions have coordinates (10, 10),
(10, 60),
(50, 50),
and (50, 60).
We must find the value of the objective function 30x + 70y for each vertex. We want the vertex that produces the maximum possible value of 30x + 70y. (10, 10): 30(10) (10, 60): 30(10) (50, 50): 30(50) (50, 60): 30(50)
+ + + +
70(10) 70(60) 70(50) 70(60)
= = = =
1000 4500 5000 5700
Maximum
The maximum profit is obtained when 50 MP3 players and 60 DVD players are produced each day. This maximum profit will be 30(50) + 70(60) = $5700 per day. To justify the procedure used in Example 4, consider the following scenario. The Charlson Company needed to find values of x and y in the shaded region of FIGURE 59 that produce the maximum profit—that is, the maximum value of 30x + 70y. To locate the point (x, y) that gives the maximum profit, add to the graph of that region lines corresponding to arbitrarily chosen profits of $0, $1000, $3000, and $7000: 30x + 70y = 0, 30x + 70y = 1000, 30x + 70y = 3000, 30x + 70y = 7000. For instance, each point on the line 30x + 70y = 3000 corresponds to production values that yield a profit of $3000. FIGURE 60 shows the region of feasible solutions, together with these lines. The lines are parallel, and the higher the line, the greater the profit. The line 30x + 70y = 7000 yields the greatest profit, but does not contain any points of the region of feasible solutions. To find the feasible solution of greatest profit, lower the line 30x + 70y = 7000 until it contains a feasible solution—that is, until it touches the region of feasible solutions at point A(50, 60), a vertex of the region. See FIGURE 61. y
y
A
10
30x + 70y = 7000
30x + 70y = 3000 0
x
10
A
30x + 70y = 7000 30x + 70y = 6300 30x + 70y = 5700
10 0
x 10
30x + 70y = 1000 30x + 70y = 0 FIGURE 60
FIGURE 61
The result observed in FIGURE 61 holds for every linear programming problem.
Fundamental Theorem of Linear Programming If the optimal value for a linear programming problem exists, it occurs at a vertex of the region of feasible solutions.
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Solving a Linear Programming Problem Step 1 Step 2 Step 3 Step 4 Step 5
Write the objective function and all necessary constraints. Graph the region of feasible solutions. Identify all vertices (corner points). Find the value of the objective function at each vertex. The solution is given by the vertex producing the optimal value of the objective function.
EXAMPLE 5
Finding a Minimum Cost Model Robin takes vitamin pills each day. She wants at least 16 units of Vitamin A, at least 5 units of Vitamin B1, and at least 20 units of Vitamin C. She can choose between red pills, costing 10¢ each, that contain 8 units of A, 1 of B1, and 2 of C; and blue pills, costing 20¢ each, that contain 2 units of A, 1 of B1, and 7 of C. How many of each pill should she buy to minimize her cost and yet fulfill her daily requirements? Solution Step 1 Let x represent the number of red pills to buy, and let y represent the number of blue pills to buy. Then the cost in pennies per day is given by cost = 10x + 20y. Robin buys x of the 10¢ pills and y of the 20¢ pills, and she gets 8 units of Vitamin A from each red pill and 2 units of Vitamin A from each blue pill. Thus, she gets 8x + 2y units of A per day. Since she wants at least 16 units, 8x + 2y Ú 16. Each red pill and each blue pill supplies 1 unit of Vitamin B1. Robin wants at least 5 units per day, so
y
x + y Ú 5. (0, 8)
For Vitamin C, the inequality is 2x + 7y Ú 20.
(1, 4) (3, 2)
Step 2
(10, 0) x
0
FIGURE 62
Since Robin cannot buy negative numbers of pills, x Ú 0 and y Ú 0. The intersection of the graphs of 8x + 2y Ú 16, x + y Ú 5, 2x + 7y Ú 20, x Ú 0, and y Ú 0
is given in FIGURE 62. Step 3 The vertices are (0, 8), (1, 4), (3, 2), and (10, 0). Steps 4 The minimum cost occurs at (3, 2). See the table. and 5 Point (0, 8) (1, 4) (3, 2) (10, 0)
Cost ⴝ 10x ⴙ 20y
10(0) 10(1) 10(3) 10(10)
+ + + +
20(8) 20(4) 20(2) 20(0)
= = = =
160 90 70 100
Minimum
Robin’s best choice is to buy 3 red pills and 2 blue pills, for a total cost of 70¢ per day. She receives just the minimum amounts of Vitamins B1 and C and an excess of Vitamin A.
7.7
7.7
Systems of Inequalities and Linear Programming
509
EXERCISES
Checking Analytic Skills Graph each inequality. Do not use a calculator. 1. x … 3
2. y …  2
3. x + 2y … 6
4. x  y Ú 2
5. 2x + 3y Ú 4
6. 4y  3x 6 5
7. 3x  5y 7 6
8. x 6 3 + 2y
9. 5x … 4y  2
10. 2x 7 3  4y
11. y 6 3x 2 + 2
12. y … x 2  4
13. y 7 (x  1) 2 + 2
14. y 7 2(x + 3) 2  1
15. x 2 + ( y + 3) 2 … 16
16. (x  4) 2 + ( y + 3) 2 … 9
9 17.
In your own words, explain how to determine whether the boundary of the graph of an inequality is solid or dashed.
Concept Check
9 18.
When graphing y 7 3x  6, would you shade above or below the line y = 3x  6? Explain your answer.
Use the concepts of this section to work Exercises 19–22.
19. For Ax + By Ú C, if B 7 0, would you shade above or below the boundary line?
20. For Ax + By Ú C, if B 6 0, would you shade above or below the boundary line?
21. Which one of the following is a description of the graph of the inequality (x  5) 2 + ( y  2) 2 6 4? A. The region inside a circle with center ( 5,  2) and radius 2 B. The region inside a circle with center (5, 2) and radius 2 C. The region inside a circle with center ( 5,  2) and radius 4 D. The region outside a circle with center (5, 2) and radius 4
22. Which one of the given inequalities satisfies the following description: the region outside an ellipse centered at the origin, with xintercepts 4 and  4 and yintercepts 9 and  9? y2 y2 x2 x2 A. B. + 7 1 7 1 4 9 16 81 y2 y2 x2 x2 C. D. + 7 0 + 7 1 16 81 16 81
Write an inequality that satisfies the description. 23. Inside the circle with radius 1 and center (0, 0)
24. Outside the circle with radius 3 and center (0, 0)
25. Outside the ellipse with xintercepts ⫾6 and yintercepts ⫾3
26. Inside the ellipse with xintercepts ⫾2 and yintercepts ⫾5
Concept Check In Exercises 27–30, match the inequality with the appropriate calculator graph. Do not use your calculator; instead, use your knowledge of the concepts involved in graphing inequalities. 27. y … 3x  6 A.
28. y Ú 3x  6 10
–10
B.
10
10 –10
–10
10
–10
29. y …  3x  6
30. y Ú  3x  6
C.
D.
10
–10
10
10 –10
–10
10
–10
Checking Analytic Skills Graph the solution set of each system of inequalities by hand. Do not use a calculator. 31.
x + y Ú 0 2x  y Ú 3
32. x + y … 4 x  2y Ú 6
33. 2x + y 7 2 x  3y 6 6
34. 4x + 3y 6 12 y + 4x 7  4
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37. 4x  3y … 12 y … x2
38. y …  x 2 y Ú x2  6
40. x + 2y … 4 y Ú x2  1
41. y … (x + 2) 2 y Ú  2x 2
42.
x  y 6 1 1 6 y 6 1
44. y 7 x 2 + 4x + 4 y 6  x2
45. y Ú (x  2) 2 + 3 y …  (x  1) 2 + 6
46.
x Ú 0 x + y … 4 2x + y … 5
48.  2 6 x 6 3 1 … y … 5 2x + y 6 6
49.  2 6 x 6 2 y 7 1 x  y 7 0
50. x + y … 4 x  y … 5 4x + y …  4
52. 2y + x y x y
53. 2x + 3y … 2x + 3y 7 3x + y 6 x Ú y Ú
54. y Ú 3x y Ú 2
35. 3x + 5y … 15 x  3y Ú 9
36.
39. x + y … 9 x … y2 43.
x + y … 36 4 … x … 4
47. 3x  2y Ú 6 x + y … 5 y … 4 51.
x x y x + 2y
… Ú Ú Ú
4 0 0 2
1 x 55. y … a b 2 y Ú 4
56.
y … x x2 + y2 6 1
Ú … … …
5 3 + x 0 0
ln x  y Ú 1 x 2  2x  y … 1
12 6 4 0 0
58. e x  y … 1 x  2y Ú 4
57. y … log x y Ú x  2
59. Concept Check Which one of the choices that follow is a description of the solution set of the following system?
60. Concept Check Fill in each blank with the appropriate response. The graph of the system
x 2 + 4y 2 6 36 y 6 x A. All points outside the ellipse above the line y = x B. All points outside the ellipse below the line y = x C. All points inside the ellipse x 2 + the line y = x D. All points inside the ellipse x 2 + the line y = x
y 7 x2 + 2 x 2 + y 2 6 16 y 6 7
x 2 + 4y 2 = 36 and x 2 + 4y 2 = 36 and
consists of all points
4y 2 = 36 and above
y = x 2 + 2,
4y 2 = 36 and below
(above/below)
(above/below)
(inside/outside)
the parabola given by
the circle x 2 + y 2 = 16, and
the line y = 7.
Concept Check In Exercises 61–64, match each system of inequalities with the appropriate calculator graph. Do not use your calculator; instead, use your knowledge of the concepts involved in graphing systems of inequalities. 61. y Ú x y … 2x  3 10 A.
–10
62. y Ú x 2 y 6 5 B.
10
–10
10
–15
15
–10
64. y … x y Ú 2x  3 D. 10
63. x 2 + y 2 … 16 y Ú 0 C. 10
–10
10
–10
10
–10
–10
Use the shading capabilities of your graphing calculator to graph each inequality or system of inequalities. 65. 3x + 2y Ú 6
66. y … x 2 + 5
67. x + y Ú 2 x + y … 6
68. y Ú  x + 2  y … 6
69. y Ú 2 x y … 8
70. y … x 3 + x 2  4x  4
7.7
71. Concept Check Find a system of linear inequalities for which the graph is the region in the first quadrant between and inclusive of the pair of lines x + 2y  8 = 0 and x + 2y = 12.
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Systems of Inequalities and Linear Programming
(Modeling) Solve each problem. 79. Inquiries about Displayed Products A wholesaler of party goods wishes to display her products at a convention of social secretaries in such a way that she gets the maximum number of inquiries about her whistles and hats. Her booth at the convention has 12 square meters of floor space to be used for display purposes. A display unit for hats requires 2 square meters, and one for whistles requires 4 square meters. Experience tells the wholesaler that she should never have more than a total of 5 units of whistles and hats on display at one time. If she receives three inquiries for each unit of hats and two inquiries for each unit of whistles on display, how many of each should she display in order to get the maximum number of inquiries? What is that maximum number?
72. Cost of Vitamins The figure shows the region of feasible solutions for the vitamin problem of Example 5 and the straightline graph of all combinations of red and blue pills for which the cost is 40 cents. y
A Region of feasible solutions B C
80. Profit from Farm Animals Farmer Jones raises only pigs and geese. She wants to raise no more than 16 animals, with no more than 12 geese. She spends $50 to raise a pig and $20 to raise a goose. She has $500 available for this purpose. Find the maximum profit she can make if she makes a profit of $80 per goose and $40 per pig. Indicate how many pigs and geese she should raise to achieve this maximum.
D x
0
y = –1x+2 2
(a) The cost function is c = 10x + 20y. Give the linear equation (in slope–intercept form) of the line of constant cost c. (b) As c increases, does the line of constant cost move up or down? (c) By inspection, find the vertex of the region of feasible solutions that gives the optimal solution.
81. Shipment Costs A manufacturer of refrigerators must ship at least 100 refrigerators to its two West Coast warehouses. Each warehouse holds a maximum of 100 refrigerators. Warehouse A holds 25 refrigerators already, while warehouse B has 20 on hand. It costs $12 to ship a refrigerator to warehouse A and $10 to ship one to warehouse B. How many refrigerators should be shipped to each warehouse to minimize cost? What is the minimum cost?
The graphs show regions of feasible solutions. Find the maximum and minimum values of each objective function. 74. 6x + y
73. 3x + 5y y
82. Diet Requirements Theo requires two food supplements: I and II. He can get these supplements from two different products A and B, as shown in the following table.
y
(5, 10) (2, 7)
(6, 8) (1, 5)
(6, 3) (1, 1) x
(9, 1)
(1, 2)
x
76. 5x + 5y
y
(1, 10)
77. 10y
(7, 9) (7, 6)
78. 3x  y
x
(1, 0)
I
II
Product A Product B
3 2
2 4
Theo’s physician recommends at least 15 grams of each supplement in Theo’s daily diet. If product A costs 25¢ per serving and product B costs 40¢ per serving, how can he satisfy his requirements most economically?
Find the maximum and minimum values of each objective function over the region of feasible solutions shown. 75. 3x + 5y
Supplement (grams/serving)
83. Gasoline Revenues A manufacturing process requires that oil refineries manufacture at least 2 gallons of gasoline for each gallon of fuel oil. To meet winter demand for fuel oil, at least 3 million gallons a day must be produced. The demand for gasoline is no more than 6.4 million gallons per day. If the price of gasoline is $1.90 per gallon and the price of fuel oil is $1.50 per gallon, how much of each should be produced to maximize revenue?
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84. Manufacturing Revenues A shop manufactures two types of bolts on three groups of machines. The time required on each group differs, as shown in the following table. Machine Group Bolt
I
II
III
Type A Type B
0.1 min 0.1 min
0.1 min 0.4 min
0.1 min 0.5 min
In a day, there are 240, 720, and 160 minutes available, respectively, on these machines. Type A bolts sells for 10¢ and Type B bolts for 12¢. How many of each type of bolt should be manufactured per day to maximize revenue? What is the maximum revenue?
7.8
85. Aid to Disaster Victims Earthquake victims in China need medical supplies and bottled water. Each medical kit measures 1 cubic foot and weighs 10 pounds. Each container of water is also 1 cubic foot, but weights 20 pounds. The plane can carry only 80,000 pounds, with total volume 6000 cubic feet. Each medical kit will aid 4 people, while each container of water will serve 10 people. How many of each should be sent in order to maximize the number of people aided? How many people will be aided? 86. Aid to Disaster Victims If each medical kit could aid 6 people instead of 4, how would the results in Exercise 85 change?
Partial Fractions
Decomposition of Rational Expressions • Distinct Linear Factors • Repeated Linear Factors • Distinct Linear and Quadratic Factors • Repeated Quadratic Factors
Decomposition of Rational Expressions Add rational expressions
2 3 5x + 3 + = x+1 x x(x + 1) Partial fraction decomposition
The sums of rational expressions are found by combining two or more such expressions into one rational expression. Here, the reverse process is considered: Given one rational expression, express it as the sum of two or more rational expressions. A special type of sum of rational expressions is called the partial fraction decomposition; each term in the sum is a partial fraction. The technique of decomposing a rational expression into partial fractions is useful in calculus and other areas of mathematics. To form a partial fraction decomposition of a rational expression, follow these steps.
ƒ( x ) g( x )
Looking Ahead to Calculus
Partial Fraction Decomposition of
In calculus, partial fraction decomposition provides a powerful technique for determining integrals of rational functions.
Step 1 If g(x) is not a proper fraction (a fraction whose numerator is of lesser degree than the denominator), divide ƒ(x) by g(x). For example,
ƒ(x)
14x + 6 x 4  3x 3 + x 2 + 5x = x 2  3x  2 + 2 . 2 x + 3 x + 3 Then apply the following steps to the remainder, which is a proper fraction. Step 2 Factor the denominator g(x) completely into factors of the form (ax + b) m or (cx 2 + dx + e) n, where cx 2 + dx + e is irreducible and m and n are integers. (continued)
7.8
Partial Fractions
513
Step 3 (a) For each distinct linear factor (ax + b), the decomposition must include the term ax A+ b. (b) For each repeated linear factor (ax + b) m, the decomposition must include the terms A1 ax + b
+
A2 (ax +
b) 2
+ Á +
Am (ax + b) m
.
Step 4 (a) For each distinct quadratic factor (cx 2 + dx + e), the decomposition must include the term
Bx + C . cx 2 + dx + e
(b) For each repeated quadratic factor (cx 2 + dx + e) n, the decomposition must include the terms B1x + C1 cx 2
+ dx + e
+
B2 x + C2 (cx 2
+ dx +
e) 2
+ Á +
Bn x + Cn (cx 2
+ dx + e) n
.
Step 5 Use algebraic techniques to solve for the constants in the numerators of the decomposition.
To find the constants in Step 5, the goal is to get a system with as many equations as there are unknowns in the numerators. One method for finding these equations is to substitute values for x in the rational equation formed in Steps 3 or 4.
Distinct Linear Factors EXAMPLE 1
Finding a Partial Fraction Decomposition 2x 4  8x 2 + 5x  2 . Find the partial fraction decomposition of x 3  4x Solution The numerator has greater degree than the denominator, so divide first. 2x x 3  4x冄2x 4  8x 2 + 5x  2 2x 4  8x 2 5x  2 2x 4  8x 2 + 5x  2 5x  2 = 2x + 3 . Now work with the remain3 x  4x x  4x der fraction. Factor the denominator as x 3  4x = x(x + 2)(x  2). Since the factors are distinct linear factors, use Step 3(a) to write the decomposition as The quotient is
B A C 5x  2 + = + , 3 x x  4x x + 2 x  2
(1)
where A, B, and C are constants that need to be found. Multiply each side of equation (1) by x(x + 2)(x  2) to get 5x  2 = A(x + 2)(x  2) + Bx(x  2) + Cx(x + 2).
(2)
(continued)
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Equation (1) is an identity, since both sides represent the same rational expression. Thus, equation (2) is also an identity. Equation (1) holds for all values of x except 0,  2, and 2. However, equation (2) holds for all values of x. In particular, substituting 0 for x in equation (2) gives  2 =  4A, so A = 12 . Similarly, choosing x =  2 gives  12 = 8B, so B =  32 . Finally, choosing x = 2 gives 8 = 8C, so C = 1. The remainder rational expression can be written as the following sum of partial fractions: 5x  2 1 3 1 = + + . x 3  4x 2x 2(x + 2) x  2 The given rational expression can be written as 1 3 1 2x 4  8x 2 + 5x  2 = 2x + + + . 3 x  4x 2x 2(x + 2) x  2 Check the work by combining the terms on the right. (a)
Y1 = Y2
We can use a graphing calculator to check the work in Example 1. We define Y1 using the original rational expression and Y2 using the partial fraction decomposition:
10.1
Y1 =
2X4  8X2 + 5X  2 1 3 1 , Y2 = 2X + + + . 3 X  4X 2X 2(X + 2) X  2
See FIGURE 63(a). When we graph both Y1 and Y2 on the same screen, the graphs should be indistinguishable. See FIGURE 63(b). –4.7
4.7 –1.1 (b) FIGURE 63
NOTE The method of graphical support just described is not foolproof; however, it gives a fairly accurate portrayal of whether the partial fraction decomposition is correct. A table can be used for support as well.
Repeated Linear Factors EXAMPLE 2
Finding a Partial Fraction Decomposition 2x Find the partial fraction decomposition of . (x  1) 3 Solution This is a proper fraction. The denominator is already factored with repeated linear factors. We write the decomposition as shown, using Step 3(b). 2x A B C = + + 3 2 (x  1) x  1 (x  1) (x  1) 3 We clear denominators by multiplying each side of this equation by (x  1) 3. 2x = A(x  1) 2 + B(x  1) + C Substituting 1 for x leads to C = 2, so 2x = A(x  1) 2 + B(x  1) + 2.
(1)
We substituted 1, the only root, and we still need to find values for A and B. However, any number can be substituted for x. For example, when we choose x =  1 (because it is easy to substitute), equation (1) becomes 2( 1) 2 4 2
= = = =
A( 1 4A 4A 2A 
 1) 2 + B( 1  1) + 2 2B + 2 2B B.
Let x =  1 in (1). Subtract 2. Divide by 2.
(2)
7.8
Partial Fractions
515
Substituting 0 for x in equation (1) gives 0 = A  B + 2 2 =  A + B.
(3)
Now we solve the system of equations (2) and (3) to get A = 0 and B = 2. The partial fraction decomposition is 2x 2 2 = + . (x  1) 3 (x  1) 2 (x  1) 3 We needed three substitutions because there were three constants to evaluate: A, B, and C. To check this result, we could combine the terms on the right.
Distinct Linear and Quadratic Factors EXAMPLE 3
Finding a Partial Fraction Decomposition x 2 + 3x  1 Find the partial fraction decomposition of . (x + 1)(x 2 + 2) Solution This denominator has distinct linear and quadratic factors, with neither repeated. Since x 2 + 2 cannot be factored, it is irreducible. The partial fraction decomposition by Steps 3(a) and 4(a) from the beginning of this section is x 2 + 3x  1 A Bx + C = + 2 . (x + 1)(x 2 + 2) x + 1 x + 2 Multiply each side by (x + 1)(x 2 + 2) to get x 2 + 3x  1 = A(x 2 + 2) + (Bx + C)(x + 1).
(1)
First, substitute  1 for x to get ( 1) 2 + 3( 1)  1 = A3( 1) 2 + 24 + 0 Use parentheses  3 = 3A around substituted A =  1. values to avoid errors. Replace A with  1 in equation (1) and substitute any value for x. If x = 0, then 0 2 + 3(0)  1 =  1(0 2 + 2) + (B # 0 + C)(0 + 1) 1 = 2 + C C = 1. Now, letting A =  1 and C = 1, substitute again in equation (1), using another number for x. For x = 1, 3 =  3 + (B + 1)(2) 6 = 2B + 2 B = 2. With A =  1, B = 2, and C = 1, the partial fraction decomposition is x 2 + 3x  1 1 2x + 1 = + 2 . (x + 1)(x 2 + 2) x + 1 x + 2 Again, this work can be checked by combining terms on the right.
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For fractions with denominators that have quadratic factors, another method is often more convenient. The system of equations is formed by equating coefficients of like terms on each side of the partial fraction decomposition. For instance, in Example 3, after each side was multiplied by the common denominator, the equation was x 2 + 3x  1 = A(x 2 + 2) + (Bx + C)(x + 1). Multiplying on the right and collecting like terms, we have x 2 + 3x  1 = Ax 2 + 2A + Bx 2 + Bx + Cx + C x 2 + 3x  1 = (A + B)x 2 + (B + C)x + (C + 2A). Now, equating the coefficients of like powers of x gives the three equations 1 = A + B 3 = B + C  1 = C + 2A. Solving this system for A, B, and C would give the partial fraction decomposition.
Repeated Quadratic Factors EXAMPLE 4
Finding a Partial Fraction Decomposition 2x . Find the partial fraction decomposition of 2 (x + 1) 2(x  1) Solution This expression has both a linear factor and a repeated quadratic factor. By Steps 3(a) and 4(b) from the beginning of this section, (x 2
2x E Ax + B Cx + D = 2 + 2 + . 2 2 + 1) (x  1) x + 1 (x + 1) x  1
Multiplying each side by (x 2 + 1) 2(x  1) leads to 2x = (Ax + B)(x 2 + 1)(x  1) + (Cx + D)(x  1) + E(x 2 + 1) 2.
(1)
If x = 1, equation (1) reduces to 2 = 4E, or E = 12. Substituting 12 for E in equation (1) and combining terms on the right gives 2x = aA +
1 4 b x + ( A + B)x 3 + (A  B + C + 1)x 2 2
+ ( A + B + D  C)x + a  B  D +
1 b. 2
(2)
To get additional equations involving the unknowns, equate the coefficients of like powers of x on each side of equation (2). Setting corresponding coefficients of x 4 equal yields 0 = A + 12, or A =  12 . From the corresponding coefficients of x 3, 0 =  A + B, which means that since A =  12 , B =  12 . Using the coefficients of x 2 gives 0 = A  B + C + 1. Since A =  12 and B =  12 , C =  1. Finally, from the coefficients of x, 2 =  A + B + D  C. Substituting for A, B, and C gives D = 1. With A =  12 , B =  12 , C =  1, D = 1, and E = 12 , the given fraction has the partial fraction decomposition 1  12 x  12 2x x + 1 2 = + + , (x 2 + 1) 2(x  1) x2 + 1 (x 2 + 1) 2 x  1
or
(x 2
 (x + 1) 1 x + 1 2x + = + 2 . 2 2 2 + 1) (x  1) 2(x + 1) (x + 1) 2(x  1)
Simplify complex fractions.
7.8
Partial Fractions
517
In summary, to solve for the constants in the numerators of a partial fraction decomposition, use either of the following methods or a combination of the two.
Techniques for Decomposition into Partial Fractions Method 1 For Linear Factors Step 1 Multiply each side of the resulting rational equation by the common denominator. Step 2 Substitute the zero of each factor into the resulting equation. For repeated linear factors, substitute as many other numbers as is necessary to find all the constants in the numerators. The number of substitutions required will equal the number of constants A, B, . . . . Method 2 For Quadratic Factors Step 1 Multiply each side of the resulting rational equation by the common denominator. Step 2 Collect terms on the right side of the equation. Step 3 Equate the coefficients of like terms to get a system of equations. Step 4 Solve the system to find the constants in the numerators.
7.8
EXERCISES
Find the partial fraction decomposition for each rational expression. 1.
5 3x(2x + 1)
2.
4.
x + 2 (x + 1)(x  1)
5.
7.
2x (x + 1)(x + 2) 2
8.
10.
13.
16.
19.
22.
x + 1  x)
x 2(1
x2
x2 + 2x + 1
6x 5 + 7x 4  x 2 + 2x 3x 2 + 2x  1 3 + 5)
x 2(x 2
3 x(x + 1)(x 2 + 1)
11.
14.
3x  1 x(x + 1)
3.
4x + 2 (x + 2)(2x  1)
6.
5x  3 (x + 1)(x  3)
9.
4 x(1  x)
4x 2  x  15 x(x + 1)(x  1)
12.
2x + 1 (x + 2) 3
3 + 4x + 3
15.
2x 5 + 3x 4  3x 3  2x 2 + x 2x 2 + 5x + 2
x2
x + 4x  5
x 2(x
x2
2 + 3)
17.
x3 + 4 9x 3  4x
18.
x3 + 2 x 3  3x 2 + 2x
20.
2x + 1 (x + 1)(x 2 + 2)
21.
3x  2 (x + 4)(3x 2 + 1)
23.
1 x(2x + 1)(3x 2 + 4)
24.
x4 + 1 x(x 2 + 1) 2
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25.
3x  1 x(2x 2 + 1) 2
26.
3x 4 + x 3 + 5x 2  x + 4 (x  1)(x 2 + 1) 2
27.
 x 4  8x 2 + 3x  10 (x + 2)(x 2 + 4) 2
28.
x2 x4  1
29.
5x 5 + 10x 4  15x 3 + 4x 2 + 13x  9 x 3 + 2x 2  3x
30.
3x 6 + 3x 4 + 3x x4 + x2
Determine whether each partial fraction decomposition is correct by graphing the left side and the right side of the equation on the same coordinate axes and observing whether the graphs coincide. 31.
33.
4x 2  3x  4 1 3 2 + = + x x  1 x + 2 x 3 + x 2  2x
(x 2
x 3  2x x  2 2 = 2 + 2 + 2x + 2) 2 x + 2x + 2 (x + 2x + 2) 2
SECTIONS 7.7 AND 7.8
32.
1 1 1 = (x  1)(x + 2) x  1 x + 2
34.
2x + 4 2 2 2 + 2 + = 2 x x  2 x (x  2) x
REVIEWING BASIC CONCEPTS
Graph the solution set of each inequality or system of inequalities. 1.  2x  3y … 6
2. x  y 6 5 x + y Ú 3
3. y Ú x 2  2 x + 2y Ú 4
4. x 2 + y 2 … 25 x2 + y2 Ú 9
7. The graph shows a region of feasible solutions. Find the maximum and minimum values of P = 3x + 5y over this region. y (5, 10)
5. Which one of the following systems is represented by the given graph?
(2, 7)
y (6, 3) 6
(1, 1) x
–6
x
0
6
–6
A. y 7 x  3 y 6  x2 + 4
B. y 6 x  3
C. y 7 x  3 y 7  x2 + 4
D. y 6 x  3
y 6  x2 + 4 y 7  x2 + 4
6. Find the minimum value of 2x + 3y, subject to the following constraints: x y x + y 2x + y
Ú Ú Ú …
0 0 4 8.
8. (Modeling) Minimizing Cost Two substances X and Y are found in pet food. Each substance contains the ingredients A and B. Substance X is 20% ingredient A and 50% ingredient B. Substance Y is 50% ingredient A and 30% ingredient B. The cost of substance X is $2 per pound and the cost of substance Y is $3 per pound. The pet store needs at least 251 pounds of ingredient A and at least 200 pounds of ingredient B. If cost is to be minimal, how many pounds of each substance should be ordered? Find the minimum cost. Decompose each rational expression into partial fractions. 9. 10.
10x + 13 x 2  x  20 3x 2  2x + 1 (x  1)(x 2 + 1)
CHAPTER 7
7
Summary
519
SUMMARY
KEY TERMS & SYMBOLS
KEY CONCEPTS
7.1 Systems of Equations
Systems of equations in two variables may be solved by the substitution method, the elimination method, or graphically, with the intersectionofgraphs method.
linear equation (or firstdegree equation) in n unknowns system of equations solutions system of linear equations (linear system) consistent system independent equations inconsistent system dependent equations substitution method elimination method equivalent systems nonlinear system of equations
7.2 Solution of Linear Systems in Three Variables ordered triple (x, y, z)
7.3 Solution of Linear Systems by Row Transformations matrix (matrices) element (entry) augmented matrix echelon (triangular) form main diagonal backsubstitution row echelon method
SUBSTITUTION METHOD (for a system of two equations in two variables) Solve one equation for one variable in terms of the other. Substitute for that variable in the other equation, and solve for its value. Then substitute that value into the other equation to find the value of the remaining variable. ELIMINATION METHOD (for a system of two equations in two variables) Multiply one or both equations by appropriate nonzero numbers so that the sum of the coefficients of one of the variables is 0. Add the equations, and solve for the value of the remaining variable. Then substitute that value into either of the given equations to solve for the value of the other variable.
SOLVING A SYSTEM OF LINEAR EQUATIONS IN THREE VARIABLES To solve a linear system with three unknowns, first eliminate a variable from any two of the equations. Then eliminate the same variable from a different pair of equations. Eliminate a second variable by using the resulting two equations in two variables to get an equation with just one variable whose value you can now determine. Find the values of the remaining variables by substitution.
MATRIX ROW TRANSFORMATIONS For any augmented matrix of a system of linear equations, the following row transformations will result in the matrix of an equivalent system: 1. Any two rows may be interchanged. 2. The elements of any row may be multiplied by a nonzero real number. 3. Any row may be changed by adding to its elements a multiple of the corresponding elements of another row.
reduced row echelon method
7.4 Matrix Properties and Operations
MATRIX ADDITION The sum of two m * n matrices A and B is the m * n matrix A + B in which each element is the sum of the corresponding elements of A and B.
dimension square matrix row matrix
MATRIX SUBTRACTION If A and B are matrices with the same dimension, then A ⴚ B ⴝ A ⴙ (ⴚB). (continued)
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Systems of Equations and Inequalities; Matrices
KEY TERMS & SYMBOLS
KEY CONCEPTS
column matrix zero matrix additive inverse (negative)
SCALAR MULTIPLICATION If k is a scalar and A is a matrix, then kA is the matrix of the same dimension where each entry of A is multiplied by k.
scalar
MATRIX MULTIPLICATION The product AB of an m * n matrix A and an n * k matrix B is an m * k matrix found as follows: To find the ith row, jth column element of AB, multiply each element in the ith row of A by the corresponding element in the jth column of B. The sum of these products will give the element of row i, column j of AB.
7.5 Determinants and Cramer’s Rule determinant det A minor cofactor expansion by a row or column
DETERMINANT OF A 2 : 2 MATRIX
a11 The determinant of a 2 * 2 matrix A, where A = c a 21 det A ⴝ a 11a 22 ⴚ a 21a 12.
a12 a22 d, is a real number defined as
COFACTOR Let Mi j be the minor for element ai j in an n * n matrix. The cofactor of ai j is given by Ai j ⴝ (ⴚ1)iⴙ j # Mi j . FINDING THE DETERMINANT OF AN n : n MATRIX Multiply each element in any row or column of the matrix by its cofactor. The sum of these products gives the value of the determinant. CRAMER’S RULE FOR 2 : 2 SYSTEMS The solution of the system a1 x + b1 y = c1 a2 x + b2 y = c2 is given by where
c1 Dx = detc c 2
xⴝ
Dx D
and
yⴝ
Dy D
b1 a1 c1 b2 d, Dy = detc a2 c2 d, and
, a1 D = detc a 2
b1 b2 d Z 0.
Cramer’s rule can be extended to 3 * 3 and larger systems.
7.6 Solution of Linear Systems by Matrix Inverses identity matrix In multiplicative inverse matrix singular matrix coefficient matrix
INVERSE OF A 2 : 2 MATRIX a b d and det A Z 0, then If A = c c d Aⴚ1 ⴝ
1 d ⴚb c d. det A ⴚc a
To find the multiplicative inverse of a 3 * 3 (or larger) matrix A, we often use a calculator.
CHAPTER 7
KEY TERMS & SYMBOLS
Summary
521
KEY CONCEPTS FINDING AN INVERSE MATRIX ANALYTICALLY To obtain A1 for any n * n matrix A for which A1 exists, follow these steps:
Step 1 Form the augmented matrix 3A  In4, where In is the n * n identity matrix. Step 2 Perform row transformations on 3A  In4 to obtain a matrix of the form 3In  B4. Step 3 Matrix B is A1.
7.7 Systems of Inequalities and Linear Programming half plane boundary linear inequality in two variables system of inequalities linear programming constraints objective function region of feasible solutions vertex (corner point)
7.8 Partial Fractions partial fraction decomposition partial fraction
GRAPHING AN INEQUALITY 1. For a function ƒ, the graph of y 6 ƒ(x) consists of all the points that are below the graph of y = ƒ(x); the graph of y 7 ƒ(x) consists of all the points that are above the graph of y = ƒ(x). (Similar statements can be made for … and Ú , with boundaries included.) 2. If the inequality is not or cannot be solved for y, choose a test point not on the boundary. If the test point satisfies the inequality, the graph includes all points on the same side of the boundary as the test point. Otherwise, the graph includes all points on the other side of the boundary. SOLVING A LINEAR PROGRAMMING PROBLEM Step 1 Write the objective function and all necessary constraints. Step 2 Graph the region of feasible solutions. Step 3 Identify all vertices (corner points). Step 4 Find the value of the objective function at each vertex. Step 5 The solution is given by the vertex producing the optimal value of the objective function.
To solve for the constants in the numerators of a partial fraction decomposition, use either of the following methods or a combination of the two. METHOD 1 FOR LINEAR FACTORS Step 1 Multiply each side of the resulting rational equation by the common denominator. Step 2 Substitute the zero of each factor into the resulting equation. For repeated linear factors, substitute as many other numbers as is necessary to find all the constants in the numerators. The number of substitutions required will equal the number of constants, A, B, Á . METHOD 2 FOR QUADRATIC FACTORS Step 1 Step 2 Step 3 Step 4
Multiply each side of the resulting rational equation by the common denominator. Collect terms on the right side of the equation. Equate the coefficients of like terms to get a system of equations. Solve the system to find the constants in the numerators.
522
Systems of Equations and Inequalities; Matrices
CHAPTER 7
7
REVIEW EXERCISES
Solve each system. Identify any systems with dependent equations and any inconsistent systems. 1. 4x  3y =  1 3x + 5y = 50
2. 0.5x  0.2y = 1.1 2x  0.8y = 4.4
3. 4x + 5y = 5 3x + 7y =  6
4. y = x 2  1 x + y = 1
5. x 2 + y 2 = 2 3x + y = 4
6. x 2 + 2y 2 = 22 2x 2  y 2 =  1
7. x 2  4y 2 = 19 x 2 + y 2 = 29
8. xy = 4 x  6y = 2
9. x 2  y 2 = 32 x = y + 4
9 10.
Use your calculator with window 3 18, 184 by 3 12, 124 to answer the following. (a) Do the circle x 2 + y 2 = 144 and the line x + 2y = 8 have any points in common? (b) Approximate any intersection points to the nearest tenth. (c) Find the exact solution set of the system.
9 12.
Can a system of two linear equations in two variables have exactly two solutions? Explain.
11. Consider the system in Exercise 5. (a) To graph the first equation, what two functions must you enter into your calculator? (b) To graph the second equation, what function must you enter? (c) What would be an appropriate window in which to graph this system?
9 13.
Can a system consisting of two equations in three variables have a unique solution? Explain.
Solve each system. Identify any systems with dependent equations and any inconsistent systems. 14. 2x  3y + z =  5 x + 4y + 2z = 13 5x + 5y + 3z = 14
15. x  3y = 12 2y + 5z = 1 4x + z = 25
16. x + y  z = 5 2x + y + 3z = 2 4x  y + 2z =  1
17.
5x  3y + 2z =  5 2x + y  z = 4  4x  2y + 2z =  1
21.
x + 2y + z = 0 3x + 2y  z = 4  x + 2y + 3z =  4
Use the reduced row echelon method to solve each system. Identify any systems with dependent equations and any inconsistent systems. 18.
2x + 3y = 10  3x + y = 18
19. 3x + y =  7 x  y = 5
20. x  z =  3 y + z = 6 2x  3z =  9
Perform each operation if possible. If not possible, say so. 22. c
5 2
24. c
2 1
26.  1c
4 1
9 1 d + c 2 4
5 9
8 3 d  c 2 7
3 1
5 7
7 d 5
4 d 1
2 0 d + 5c 4 1
28. Concept Check found by
2 5
3 8 1 23. C 2 S  C  4 S + C 0 S 5 6 2 25. 3c
2 d 3
2 1
27. 10c
4 5 d  2c 4 2
8 d 2
2x + 3y 4x + y  3x  y x + 6y d + 2c d x  5y 6x + 2y 4x + 2y 5x  y
Complete the following sentence: The sum of two m * n matrices A and B is .
523
Review Exercises
CHAPTER 7
Multiply if possible. If not possible, say so. 8 29. c 5
6 3 d c 2 7
1 32. C  3 0
2 4 2
1 d 2
5 4 7S c 10 1
3 30. c 4
2 5
4 33. c 10
3 d 6
2 1 d C 0 6 3
2 0
2 5
1 3 d C 3 6 0
0 2S 1 2 4 2
2 1
1 31. c 0 5 7S 1
4 1 1 04 C 2 5
34. 33  1
1 2 2 d D T 8 0 1 3 4 7
2 0S 3
Find AB and BA to determine whether A and B are inverses. 35. A = c
3 13
2 37. A = C 0 1
2 9 d; B = c 9  13 0 1 0
2 d 3
6 1 0 0 S; B = C 0 1 1 1 4 0
36. A = c 3 2
1 0 d; B = c 2 3 3
1 2
1 38. A = C 0 0
0S 1
0 2 0
0 d  13
2 1 4 S; B = C 0 1 0
0 1 2
0
2 2S 1
For each matrix, find A1 if it exists. 39. A = c
6 10
3 d 5
2 42. A = C 1 0
0 1 1
40. A = c
4 0
2 43. A = C 1 1
4 0S 2
2 d 3 1 0 2
2 1
0 R 5
2 44. A = C  2 1
3 3 4
41. A = B 0 1S 0
5 5S 2
Use the inverse matrix method to solve each system if possible. Identify any systems with dependent equations or any inconsistent systems. 45.
x + y = 4 2x + 3y = 10
46. 5x  3y =  2 2x + 7y =  9
48.
x  2y = 7 3x + y = 7
49.
x + 2y =  1 3y  z =  5 x + 2y  z =  3
50.
51. x + y + z = 1 2x  y =  2 3y + z = 2
52.
x = 3 y + z = 6 2x  3z =  9
53. 2x  4y + 4z = 0 x  3y + 2z =  3 x  y + 2z = 1
54. Concept Check
47. 2x + y = 5 3x  2y = 4 3x  2y + 4z = 1 4x + y  5z = 2  6x + 4y  8z =  2
If the solution set is given by 5(4  y, y)6, give one specific solution.
Evaluate each determinant. 55. detc
1 2
8 d 9
56. detc
2 0
4 d 3
2 4 1 57. detC 3 0 2 S 1 0 3
1 58. detC 4 5
2 0 1
3 3S 2
Solve each determinant equation for x. 59. detc
3 1
2 d = 5 x
60. detc
3x 7 d = 8 x 4
2 61. detC 1 0
5 3x 2
0 1S = 4 0
6x 2 62. detC 1 5 x 2
0 3 S = 2x 1
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Exercises 63 and 64 refer to the system
3x  y = 28 . 2x + y = 2
63. Suppose you are asked to solve this system by using Cramer’s rule. (a) What is the value of D? (b) What is the value of Dx ? (c) What is the value of Dy ? (d) Find x and y, using Cramer’s rule.
9 65.
64. Suppose you are asked to solve this system by applying the inverse matrix method to AX = B. (a) What is A? (b) What is B? 9 (c) Explain how you would use these matrices to go about solving the system.
Cramer’s rule has the condition D Z 0. Why is this necessary? What is true of the system when D = 0?
Solve each system, if possible, by using Cramer’s rule. Identify any systems with dependent equations or any inconsistent systems. 66. 3x + y =  1 5x + 4y = 10
67. 3x + 7y = 2 5x  y =  22
69. 3x + 2y + z = 2 4x  y + 3z =  16 x + 3y  z = 12
70.
68. 2x  5y = 8 3x + 4y = 10 71.  x + 3y  4z = 2 2x + 4y + z = 3 3x  z = 9
5x  2y  z = 8  5x + 2y + z =  8 x  4y  2z = 0
Solve each problem. 72. Determining the Contents of a Meal A cup of uncooked rice contains 15 grams of protein and 810 calories. A cup of uncooked soybeans contains 22.5 grams of protein and 270 calories. How many cups of each should be used for a meal containing 9.5 grams of protein and 324 calories?
A
W
113 138 152
39 53 65
142 181 191
Source: Brase, C.H. and C.P. Brase, Understandable Statistics, D.C. Heath and Company.
73. Determining Order Quantities A company sells CDs for $0.40 each and plastic holders for $0.30 each. The company receives $38 for an order of 100 CDs and holders. However, the customer neglected to specify how many of each to send. Determine the number of CDs and the number of holders that should be sent. 74. Mixing Teas Three kinds of tea worth $4.60, $5.75, and $6.50 per pound are to be mixed to get 20 pounds of tea worth $5.25 per pound. The amount of $4.60 tea used is to equal the amount of the other two kinds together. How many pounds of each tea should be used?
P
(a) Use the data to approximate values for the constants a, b, and c. (b) Estimate a typical systolic blood pressure for an individual who is 55 years old and weighs 175 pounds. 77. Curve Fitting Find the equation of the polynomial function of degree 3 whose graph passes through the points ( 2, 1), ( 1, 6), (2, 9), and (3, 26). 78. Curve Fitting Find the equation of the quadratic polynomial that defines the curve shown in the figure.
75. Mixing Solutions of a Drug A 5% solution of a drug is to be mixed with some 15% solution and some 10% solution to get 20 milliliters of 8% solution. The amount of 5% solution used must be 2 milliliters more than the sum of the other two solutions. How many milliliters of each solution should be used?
y
(–6, 4) –6
76. (Modeling) Blood Pressure In a study of adult males, it was believed that systolic blood pressure P was affected by both age A in years and weight W in pounds. This was modeled by P = a + bA + cW, where a, b, and c are constants. The table in the next column lists three individuals with representative blood pressures for the group.
(2, 4)
2 0
x 2 4
(–4, –2) –4
Graph the solution set of each system. 79.
x + y … 6 2x  y Ú 3
1 x  2 3 y 2 … 16  x 2
80. y …
CHAPTER 7
81. Find x Ú 0 and y Ú 0 such that
and 2x + 4y is maximized. 82. Find x Ú 0 and y Ú 0 such that
83. (Modeling) Company Profit A small company manufactures two products: radios and DVD players. Each radio results in a profit of $15, each DVD player a profit of $35.
7
Find the partial fraction decomposition of each rational expression. 84.
and 4x + 2y is minimized.
86.
5x  2 x2  4
x3
x + 2  x 2 + 4x
85.
x + 2 x 3 + 2x 2 + x
87.
6x 2  x  3 x3  x
TEST
1. Consider the system of equations. =  15 3x + y = 1
x2
5. Evaluate each determinant.
4y 2
(a) What type of graph does each equation have? (b) How many points of intersection of the two graphs are possible? (c) Solve the system. (d) Support the solutions, using a graphing calculator.
(a) detc
4 5
2x + y + z = 3 x + 2y  z = 3 3x  y + z = 5 3. Perform each matrix operation if possible. 3 2 4S  C 3 9 0
2 (a) 3 C 1 5
1 4 2 (b) c d + c d + c 2 6 7 1 0
1 3 d C2 5 3
6 1S 8 8 d 5
3 4S 2
4. Suppose A and B are both n * n matrices. (a) Can AB be found? (b) Can BA be found? 9 (c) Does AB = BA in general? Explain why or why not. (d) If A is n * n and C is m * n, can either AC or CA be found?
2 (b) detC  1 12
9 d  11
0 7 5
8 9S 3
6. Solve the system by using Cramer’s rule. 2x  3y =  33 4x + 5y = 11 7. Consider the system of equations. x + y  z = 4 2x  3y  z = 5 x + 2y + 2z = 3
2. Solve the system.
2 (c) c 4
525
Due to demand, the company must produce at least 5, and not more than, 25 radios per day. The number of radios cannot exceed the number of DVD players, and the number of DVD players cannot exceed 30. How many of each should the company manufacture to obtain maximum profit? What will that profit be?
3x + 2y … 12 5x + y Ú 5
x + y … 50 2x + y Ú 20 x + 2y Ú 30
Test
(a) Write the matrix of coefficients A, the matrix of variables X, and the matrix of constants B for this system. (b) Find A1. (c) Use the matrix inverse method to solve the system. 9 (d) If the first equation in the system is replaced by 0.5x + y + z = 1.5, the system cannot be solved with the matrix inverse method. Explain why. 8. World Population The table lists actual and projected world population in billions for three selected years. Year Population
1850
1950
2050
1.3
2.5
8.9
Source: United Nations.
(a) Find a function defined by ƒ(x) = ax 2 + bx + c that models the data, where x represents years after 1850. (b) Use ƒ to estimate when world population might reach 8 billion.
526
CHAPTER 7
Systems of Equations and Inequalities; Matrices
9. The solution set of a system of inequalities is shown. Which system is it? 10
–10
10
10. Storage Capacity An office manager wants to buy filing cabinets. Cabinet X costs $100, requires 6 square feet of floor space, and holds 8 cubic feet. Cabinet Y costs $200, requires 8 square feet of floor space, and holds 12 cubic feet. No more than $1400 can be spent, and the office has room for no more than 72 square feet of cabinets. The office manager wants the maximum storage capacity within the limits imposed by funds and space. How many of each type of cabinet should be bought?
–10
A. y 7 2  x y 7 x2  5
B. y 7 2  x
C. y 6 2  x y 6 x2  5
D. y 6 2  x y 7 x2  5
y 6 x2  5
Find the partial fraction decomposition for each rational expression. 11.
7x  1 2 x  x  6
12.
x 2  11x + 6 (x + 2)(x  2) 2
8 The Unit Circle and the Functions of Trigonometry MYSTERY SURROUNDS AN area of the Atlantic Ocean known as the Bermuda Triangle. According to the National Transportation Safety Board, more than 2000 boats and airplanes have disappeared while traversing this triangularshaped region with vertices at Bermuda, Puerto Rico, and Miami. What could explain the disappearance of so many boats and planes on otherwise routine journeys? One theory postulates that plumes of methane gas released from breaks in the ocean floor could displace water into the air and could force it over a boat, while simultaneously decreasing buoyancy under the boat. Clouds of the colorless, odorless gas could also rise above the water’s surface. Airplane mechanics and a team of scientists showed that with as little as a 1% concentration of methane gas in the air, a small airplane engine could stall, causing the plane to crash into the water. In this chapter, we use trigonometric functions to solve problems related to navigation, periodic phenomena, and rotation. (See Example 4 in Section 8.7.) Navigators rely on trigonometry to guide ships and airplanes, safely avoid mountains or submerged objects, and perhaps prevent unintended entry into the Bermuda Triangle. Source: www.bermudatriangle.org; www.discovery.com
CHAPTER OUTLINE 8.1 Angles, Arcs, and Their Measures 8.2 The Unit Circle and Its Functions 8.3 Graphs of the Sine and Cosine Functions 8.4 Graphs of the Other Circular Functions 8.5 Functions of Angles and Fundamental Identities 8.6 Evaluating Trigonometric Functions 8.7 Applications of Right Triangles 8.8 Harmonic Motion
From Chapter 8 of A Graphical Approach to Precalculus with Limits: A Unit Circle Approach, Fifth Edition. John Hornsby, Margaret Lial, Gark K. Rockswold. Copyright © 2011 by Pearson Education, Inc. All rights reserved.
528
The Unit Circle and the Functions of Trigonometry
CHAPTER 8
8.1
Angles, Arcs, and Their Measures
Basic Terminology • Degree Measure • Standard Position and Coterminal Angles • Radian Measure • Arc Lengths and Areas of Sectors • Linear and Angular Speed
Basic Terminology Line AB A
B
A
B
A
B
Segment AB
Ray AB FIGURE 1
Terminal side
Vertex A
Two distinct points A and B determine a line called line AB. The portion of the line between A and B, including points A and B, is segment AB. The portion of line AB that starts at A and continues through B and on past B is called ray AB. Point A is the endpoint of the ray. See FIGURE 1. In trigonometry, an angle consists of two rays in a plane with a common endpoint, or two line segments with a common endpoint. These two rays (or segments) are called the sides of the angle, and the common endpoint is called the vertex of the angle. Associated with an angle is its measure, generated by a rotation about the vertex. See FIGURE 2. This measure is determined by rotating a ray starting at one side of the angle, called the initial side, to the position of the other side, called the terminal side. A counterclockwise rotation generates a positive measure, while a clockwise rotation generates a negative measure. See FIGURE 3.
Initial side FIGURE 2
A
A
C
C
B B Positive angle
Negative angle FIGURE 3
An angle can be named by using the name of its vertex. For example, either angle in FIGURE 3 can be called angle C. Alternatively, an angle can be named using three letters, with the vertex letter in the middle. Thus, either angle also could be named angle ACB or angle BCA.
Degree Measure
A complete rotation of a ray gives an angle whose measure is 360°. 1 of a complete 360 rotation gives an angle whose measure is 1°. FIGURE 4
The most common unit used to measure the size of angles is the degree. To use degree measure, we assign 360 degrees to a complete rotation of a ray.* In FIGURE 4, notice that the terminal side of the angle corresponds to its initial side when it makes a complete 1 rotation. One degree, written 1ⴗ, represents 360 of a rotation. Therefore, 90° represents 1 180 1 90 360 = 4 of a complete rotation, and 180° represents 360 = 2 of a complete rotation. An angle measuring between 0° and 90° is an acute angle. An angle measuring exactly 90° is a right angle. An angle measuring more than 90° but less than 180° is an obtuse angle, and an angle of exactly 180° is a straight angle.
*The Babylonians were the first to subdivide the circumference of a circle into 360 parts, 4000 years ago. There are various theories as to why the number 360 was chosen. One is that it is approximately the number of days in a year, and it has many divisors, which makes it convenient to work with.
8.1
Angles, Arcs, and Their Measures
529
In FIGURE 5, we use the Greek letter u (theta)* to name each angle.
Acute angle 0° < < 90°
Right angle = 90°
Obtuse angle 90° < < 180°
Straight angle = 180°
FIGURE 5
If the sum of the measures of two positive angles is 90°, the angles are called complementary and the angles are complements. Two positive angles with measures whose sum is 180° are supplementary and the angles are supplements. EXAMPLE 1
Finding Measures of Angles Find the measure of each angle in FIGURE 6. (6m)° (3m)°
Solution (a) In FIGURE 6(a), the two angles form a right angle (as indicated by the they are complementary angles.
(a)
(4k)°
Don’t stop here.
(6k)° (b)
Combine like terms. Divide by 9.
Be sure that you determine the measure of each angle by substituting 10 for m. The two angles have measures of 6m = 6(10) = 60° and 3m = 3(10) = 30°.
FIGURE 6
(b) The angles in FIGURE 6(b) are supplementary.
y
A = 35°° O
x
Remember to substitute for k.
4k + 6k = 180 10k = 180 k = 18
Combine like terms. Divide by 10.
These angle measures are 4(18) = 72° and 6(18) = 108°.
FIGURE 7 GCM
6m + 3m = 90 9m = 90 m = 10
symbol), so
TECHNOLOGY NOTE
The measure of Angle A of FIGURE 7 is 35°. This measure can be expressed by saying that m(angle A) is 35°, where m(angle A) is read “the measure of angle A.” It is convenient, however, to abbreviate m(angle A) = 35° as simply angle A = 35°. Traditionally, portions of a degree have been measured with minutes and seconds. 1 One minute, written 1 œ , is 60 of a degree. 1œ ⴝ
1 ° or 60 œ ⴝ 1° 60
1 One second, 1 ﬂ , is 60 of a minute.
TI84
Graphing calculators can be set in either radian mode or degree mode.
1ﬂ ⴝ
1 œ 1 ° ⴝ or 60 ﬂ ⴝ 1 œ 60 3600
The measure 12° 42¿ 38– represents 12 degrees, 42 minutes, 38 seconds.
*In addition to u (theta), Greek letters such as A (alpha) and B (beta) are often used.
530
CHAPTER 8
The Unit Circle and the Functions of Trigonometry
GCM
EXAMPLE 2
Calculating with Degrees and Minutes Perform each calculation. (a) 51° 29¿ + 32° 46¿ (b) 90°  73° 12¿
Analytic Solution (a)
51° 29¿ + 32° 46¿ 83° 75¿
Add degrees and minutes separately.
Graphing Calculator Solution Your calculator can be in either degree or radian mode. (We discuss radian measure later in this section.) FIGURE 8 shows the calculations.
We can write the sum 83° 75¿ as 83° + 1° 15¿ 84° 15¿. (b)
89° 60¿  73° 12¿ 16° 48¿
75¿ = 60¿ + 15¿ = 1° 15¿
Write 90° as 89° 60¿. FIGURE 8
Angles are commonly measured in decimal degrees. For example, 12.4238° = 12
4238 ° . 10,000
Decimal degrees are often more convenient to use than degrees, minutes, and seconds. For this reason it is important to be able to convert between these two angle measures. These conversions can be done by hand or with a calculator.
GCM
EXAMPLE 3
Converting between Decimal Degrees and Degrees, Minutes, and Seconds
(a) Convert 74° 8¿ 14– to decimal degrees. Round to the nearest thousandth of a degree. (b) Convert 34.817° to degrees, minutes, and seconds. Analytic Solution 8 ° 14 ° 1° 1¿ = 60 and 1– = + 60 3600 L 74° + 0.1333° + 0.0039° L 74.137° Add; round.
(a) 74° 8¿ 14– = 74° +
(b) 34.817° = = = = = = =
34° + 0.817° 34° + 0.817(60¿) 34° + 49.02¿ 34° + 49¿ + 0.02¿ 34° + 49¿ + 0.02(60–) 34° + 49¿ + 1.2– 34° 49¿ 1.2–
1 ° 3600
Graphing Calculator Solution The first two results in FIGURE 9 show how the TI83/84 Plus converts 74° 8¿ 14– to decimal degrees. (The second result was obtained by fixing the display to three decimal places.) The final result shows how to convert decimal degrees to degrees, minutes, and seconds.
1° = 60¿
1¿ = 60– Degree mode FIGURE 9
8.1
Angles, Arcs, and Their Measures
531
Standard Position and Coterminal Angles An angle is in standard position if its vertex is at the origin and its initial side is along the positive xaxis. The angles in FIGURES 10(a) and 10(b) are in standard position. An angle in standard position is said to lie in the quadrant in which its terminal side lies. For example, an acute angle is in quadrant I (FIGURE 10(a)) and an obtuse angle is in quadrant II (FIGURE 10(b)). FIGURE 10(c) shows ranges of angle measures for each quadrant when 0° 6 u 6 360°. If the terminal side lies along an axis, then the angle does not lie in any quadrant. y
y
QI
90°
Q II Q II 90° < < 180°
Terminal side Vertex 0
Initial side
x
Acute angle in standard position
y
0
x
Obtuse angle in standard position
QI 0° < < 90° 0° 360°
180°
Q III Q IV 180° < < 270° 270° < < 360° 270°
420°° 60°°
(a)
(b) FIGURE 10
x
0
(c)
Coterminal angles
Quadrantal Angles
FIGURE 11
Angles in standard position having their terminal sides along the xaxis or yaxis, such as angles with measures 90°, 180°, 270°, and so on, are called quadrantal angles.
y
830° 110°° x
0
A complete rotation of a ray results in an angle measuring 360°. By continuing the rotation, angles of measure larger than 360° can be produced. The angles in FIGURE 11 with measures 60° and 420° have the same initial side and the same terminal side, but different amounts of rotation. Such angles are called coterminal angles—their measures differ by a multiple of 360°. The angles in FIGURE 12 are also coterminal.
Coterminal angles FIGURE 12 y
188°° 908°°
EXAMPLE 4
x
0
Finding Measures of Coterminal Angles Find the angle of smallest possible positive measure coterminal with each angle. (a) 908° (b)  75° (c)  800° Solution (a) Add or subtract 360° as many times as needed to obtain an angle with measure greater than 0° but less than 360°. Since
FIGURE 13 y
908°  2 # 360° = 908°  720° = 188°, an angle of 188° is coterminal with an angle of 908°. See FIGURE 13.
0
285°°
–75°°
FIGURE 14
x
(b) Use a rotation of 360° + ( 75°) = 285°. See FIGURE 14.
(c) The least integer multiple of 360° greater than 800° is 360° # 3 = 1080°. Add 1080° to  800° to obtain 1080° + ( 800°) = 280°.
532
CHAPTER 8
The Unit Circle and the Functions of Trigonometry
Sometimes we want to find an expression that will generate all angles coterminal with a given angle. For example, any angle coterminal with 60° can be obtained by adding an appropriate integer multiple of 360° to 60°. If we let n represent any integer, then the expression
#
60° + n 360°
Angles coterminal with 60°
represents all such coterminal angles. The following table shows a few possibilities. Examples of Angles Coterminal with 60° Value of n 2 1 0 1
Examples of Coterminal Quadrantal Angles Quadrantal Angle U
Coterminal with U
0° 90° 180° 270°
⫾360°, ⫾720°  630°,  270°, 450°  180°, 540°, 900°  450°,  90°, 630°
r 0
r
= 1 radian
60° 60° 60° 60°
+ + + +
2 # 360° = 780° 1 # 360° = 420° 0 # 360° = 60° (the angle itself) ( 1) # 360° =  300°
The table in the margin gives examples of coterminal quadrantal angles.
Radian Measure In work involving applications of trigonometry, angles are often measured in degrees. In more theoretical work in mathematics, radian measure of angles is preferred. FIGURE 15 shows an angle u in standard position along with a circle of radius r. The vertex of u is at the center of the circle. Because angle u intercepts an arc on the circle equal in length to the radius of the circle, we say that u has a measure of 1 radian.
y
Coterminal with 60°
x
Radian An angle with vertex at the center of a circle that intercepts an arc on the circle equal in length to the radius of the circle has measure 1 radian.
FIGURE 15
It follows that an angle of measure 2 radians intercepts an arc equal in length to twice the radius of the circle, an angle of measure 12 radian intercepts an arc equal in length to half the radius of the circle, and so on. The circumference of a circle—the distance around the circle—is given by C = 2pr, where r is the radius of the circle. The formula C = 2pr shows that the radius can be laid off 2p times around a circle. Therefore, an angle of 360°, which corresponds to a complete circle, intercepts an arc equal in length to 2p times the radius of the circle. Thus, an angle of 360° has measure 2p radians: 360° ⴝ 2P radians. An angle of 180° is half the degree measure of an angle of 360°, so an angle of 180° has half the radian measure of an angle of 360°. 180° ⴝ
1 (2P) radians ⴝ P radians 2
Degree–radian relationship
We can use the relationship 180° = p radians to develop a method for converting between degrees and radians as follows. 180° ⴝ P radians 1° ⴝ
P radian 180
Divide by 180.
or 1 radian ⴝ
180° P
Divide by p.
8.1
Angles, Arcs, and Their Measures
533
Converting between Degrees and Radians p 1. Multiply a degree measure by 180 radian and simplify to convert to radians. 2. Multiply a radian measure by 180° p and simplify to convert to degrees.
NOTE 1 radian is equivalent to approximately 57.3°. IF NO UNIT OF MEASURE IS SPECIFIED FOR AN ANGLE, RADIAN MEASURE IS UNDERSTOOD.
GCM
EXAMPLE 5
Converting Degrees to Radians Convert each degree measure to radian measure. (a) 45° (b)  270° (c) 249.8°
Analytic Solution (a) 45° = 45 a =
p radianb 180
Multiply by
p radian 4
(b)  270° =  270a
p 180
radian.
Simplify.
Graphing Calculator Solution For FIGURE 16, the calculator is in radian mode. When exact values involving P are required, such as P4 and ⴚ 3P 2 in parts (a) and (b), calculator approximations are not acceptable.
p 3p radianb = radians 180 2
(c) 249.8° = 249.8 a
p radianb 180 L 4.360 radians
Rounded to nearest thousandth
Radian mode FIGURE 16
GCM
EXAMPLE 6
Converting Radians to Degrees Convert each radian measure to degree measure. 9p 5p (a) (b) (c) 4.25 4 6
Analytic Solution (a)
9p 9p 180° b radians = a p 4 4 = 405°
(b) 
Multiply by
180° p .
Graphing Calculator Solution FIGURE 17 shows how a calculator in degree mode converts radian measures to degree measures.
Simplify.
5p 5p 180° b =  150° Multiply. radians = a p 6 6
(c) 4.25 radians = 4.25 a
180° b p L 243.5°
Rounded to nearest tenth
Degree mode FIGURE 17
NOTE Another way to convert a radian measure that is a rational multiple of p, such as 9p 4,
to degrees is to just substitute 180° for p. In Example 6(a), this would be 9(180°) = 405°. 4
534
The Unit Circle and the Functions of Trigonometry
CHAPTER 8 y
CAUTION
FIGURE 18
shows angles measuring 30 radians and 30°. Be careful not to con
fuse them.
The table that follows and FIGURE 19 give some equivalent angles measured in degrees and radians. Keep in mind that 180° ⴝ P radians.
x
0
30 radians
Equivalent Angle Measures in Degrees and Radians Radians
y
30°°
x
0
30 degrees
Radians
Degrees
Exact
Approximate
Degrees
Exact
Approximate
0° 30° 45° 60°
0 p 6 p 4 p 3
0 0.5236 0.7854 1.0472
90° 180° 270° 360°
p 2
1.5708 3.1416 4.7124 6.2832
FIGURE 18
p 3p 2
2p
y
90° = 2 60° = 120° = 2 3 3 45° = 135° = 3 4 4 30° = 150° = 5 6 6 180° =
0° = 0
x
210° = 7 330° = 11 6 6 225° = 5 315° = 7 4 4 240° = 4 300° = 5 3 270° = 3 3 2 Degree Measure
Radian Measure FIGURE 19
The angles identified in FIGURE 19 are important. You should learn these equivalences, as they will appear often in the chapters that follow.
Arc Lengths and Areas of Sectors y
T
Q
s R
r
radians 1 radian r O r P
FIGURE 20
x
The formula to find the length of an arc of a circle is derived from the fact (proved in geometry) that the length of an arc is proportional to the measure of its central angle. In FIGURE 20, angle QOP has measure 1 radian and intercepts an arc of length r on the circle. Angle ROT has measure u radians and intercepts an arc of length s on the circle. Since the lengths of the arcs are proportional to the measures of their central angles, s u = r 1 s = r u.
Multiply each side by r.
8.1
Angles, Arcs, and Their Measures
535
Arc Length The length s of the arc intercepted on a circle of radius r by a central angle of measure u radians is given by the product of the radius and the radian measure of the angle, or s ⴝ r U, U in radians.
CAUTION
In applying the formula s ⴝ r U, the value of U must be expressed in
radians.
Finding Arc Length by Using s ⴝ r U A circle has radius r = 18.20 centimeters. Find the length s of an arc intercepted by a central angle u having each of the following measures. 3p (a) radians (b) 144° 8 EXAMPLE 7
s
Solution (a) As shown in FIGURE 21, r = 18.20 centimeters and u =
3 8
s = r u = 18.20a
r = 18.20 cm
3p 8.
3p b L 21.44 centimeters 8
(b) The formula s = r u requires that u be measured in radians. First, convert u to p radians by multiplying 144° by 180 radian.
FIGURE 21
144° = 144 a
4p p b = radians 180 5
Convert from degrees to radians.
The length s is given by s = r u = 18.20 a Be sure to use radians for u in s = r u.
4p b L 45.74 centimeters. 5
EXAMPLE 8
Using Latitudes to Find Distance Reno, Nevada, is due north of Los Angeles. The latitude of Reno is 40° N, while that of Los Angeles is 34° N. (The N in 34° N means north of the equator.) If the radius of Earth is 6400 kilometers, find the north–south distance between the two cities.
Reno s 6° Los Angeles
Solution Latitude gives the measure of a central angle with vertex at Earth’s center whose initial side goes through the equator and whose terminal side goes through the given location. As shown in FIGURE 22, the central angle between Reno and Los Angeles is 40°  34° = 6°. The distance between the two cities can be found by the formula s = r u, after 6° is first converted to radians. 6° = 6 a
40° 34° 6400 km
Equator
p p b = radian 180 30
The distance between the two cities is FIGURE 22
s = r u = 6400 a
p b L 670 kilometers. 30
Let r = 6400 and u =
p 30 .
536
The Unit Circle and the Functions of Trigonometry
CHAPTER 8
The shaded region is a sector of the circle.
r
A sector of a circle is the portion of the interior of a circle intercepted by a central angle. Think of it as a “piece of pie.” See FIGURE 23. The interior of a circle can be thought of as a sector with measure 2p radians. If a central angle for a sector has meau sure u radians, then the sector makes up the fraction 2p of a complete circle. The area 2 inside a circle with radius r is A = pr . Therefore, the area of the sector is given by u the product of the fraction 2p and the total area, pr 2. area of sector =
FIGURE 23
u 1 (pr 2) = r 2 u, u in radians 2p 2
This discussion is summarized as follows.
Area of a Sector The area of a sector of a circle of radius r and central angle u is given by Aⴝ
1 2 r U, U in radians. 2
CAUTION As in the formula for arc length, the value of U must be in radians when the formula for the area of a sector is used.
EXAMPLE 9
Finding the Area of a SectorShaped Field FIGURE 24 shows a field in the shape of a sector of a circle. Find the area of the field.
15° = 15 a
15°
321
m
Solution First, convert 15° to radians. p p b = radian 180 12
Now use the formula for the area of a sector. A =
1 2 1 p r u = (321) 2 a b L 13,500 square meters 2 2 12
Be sure to use radians for u in A = 12 r 2u.
FIGURE 24
Linear and Angular Speed
y
P s θ
O
r
B
FIGURE 25
P moves at a constant speed along the circle. x
In many situations we need to know how fast a point on a circular disk is moving or how fast the central angle of such a disk is changing. Some examples occur with machinery involving gears or pulleys or the speed of a car around a curved portion of highway. Suppose that point P moves at a constant speed along a circle of radius r and center O. See FIGURE 25. The measure of how fast the position of P is changing is called linear speed. If v represents linear speed, then speed =
distance , time
s or v ⴝ , t
where s is the length of the arc traced by point P in time t. (This formula is just a restatement of d = rt with s as distance, v as rate (speed), and t as time.) As point P in FIGURE 25 moves along the circle, ray OP rotates around the origin. Since ray OP is the terminal side of angle POB, the measure of the angle changes as
8.1
Angles, Arcs, and Their Measures
537
P moves along the circle. The measure of how fast angle POB is changing is called angular speed. Angular speed, symbolized v (omega), is given as Vⴝ
U , U in radians, t
where u is the measure of angle POB. As with earlier formulas in this chapter, u must be measured in radians, with v expressed as radians per unit of time. Angular speed is used in physics and engineering, among other applications. The length s of the arc intercepted on a circle of radius r by a central angle of measure u radians is given by s = r u. Using this formula, the formula for linear speed, v = st, becomes v =
s ru = = r t t
#
u = rv. t
s = r u; v =
u t
The formula v ⴝ rV relates linear and angular speeds. As an example of linear and angular speeds, consider the following. The human joint that can be flexed the fastest is the wrist, which can rotate through 90°, or p2 radians, in 0.045 second while holding a tennis racket. The angular speed of a human wrist swinging a tennis racket is p
v =
u 2 = L 35 radians per second. t 0.045
If the radius (distance) from the tip of the racket to the wrist joint is 2 feet, then the speed at the tip of the racket is v = rv L 2(35) = 70 feet per second, or about 48 mph. In a tennis serve the arm rotates at the shoulder, so the final speed of the racket is considerably faster. (Source: Cooper, J. and R. Glassow, Kinesiology, Second Edition, C. V. Mosby.) Angular Speed Vⴝ
Linear Speed
U t
s t ru vⴝ t v ⴝ rV
vⴝ
(v in radians per unit time, u in radians)
EXAMPLE 10
Using Linear and Angular Speed Formulas Suppose that point P is on a circle with radius 10 centimeters, and ray OP is rotating p with angular speed 18 radian per second. (a) Find the angle generated by P in 6 seconds. (b) Find the distance traveled by P along the circle in 6 seconds. (c) Find the linear speed of P in centimeters per second. Solution (a) The angular speed of ray OP is v =
p 18
p u = 18 6 6p p u = = radians. 18 3
radian per second. Since v = ut , in 6 seconds p Let v = 18 and t = 6 in the angular speed formula.
Solve for u.
(continued)
538
CHAPTER 8
The Unit Circle and the Functions of Trigonometry
(b) From part (a), P generates an angle of eled by P along the circle is
p 3
radians in 6 seconds. The distance trav
p 10p s = r u = 10 a b = centimeters. 3 3 (c) From part (b), s =
10p 3
for 6 seconds, so for 1 second we divide by 6.
10p
s 10p 10p 3 v = = = , 6 = t 6 3 3 Be careful simplifying this complex fraction.
#
1 5p = centimeters per second 6 9
EXAMPLE 11
Finding Angular Speed of a Pulley and Linear Speed of a Belt A belt runs a pulley of radius 6 centimeters at 80 revolutions per minute. (a) Find the angular speed of the pulley in radians per second. (b) Find the linear speed of the belt in centimeters per second. Solution (a) In 1 minute, the pulley makes 80 revolutions. Each revolution is 2p radians, so 80(2p) = 160p radians per minute. Since there are 60 seconds in 1 minute, we find v, the angular speed in radians per second, by dividing 160p by 60. v =
160p 8p = radians per second 60 3
(b) The linear speed of the belt will be the same as that of a point on the circumference of the pulley. Thus, v = rv = 6 a
8.1
8p b = 16p L 50 centimeters per second 3
EXERCISES
Checking Analytic Skills Fill in the blanks with the appropriate short answers. Do not use a calculator. 1. An angle of 360° has an equivalent radian measure of .
2. An angle of p radians has an equivalent degree measure of .
3. The least positive angle coterminal with  180° has degree measure .
4. The complement of a 40° angle is plement of a 40° angle is
5. A formula relating r, u, and s is
6. A formula relating v, v, and r is
.
Checking Analytic Skills Find (a) the complement and (b) the supplement of each angle. Do not use a calculator. p p 7. 30° 8. 60° 9. 45° 10. 11. 3 4 13. Concept Check An angle measures x degrees. (a) What is the measure of its complement? (b) What is the measure of its supplement?
, and the sup. .
12.
p 12
14. Concept Check An angle measures x radians. (a) What is the measure of its complement? (b) What is the measure of its supplement?
8.1
539
Angles, Arcs, and Their Measures
Checking Analytic Skills Find the degree measure of the smaller angle formed by the hands of a clock at the following times. Do not use a calculator. 15.
16.
17. 3:15
18. 9:00
21.
22.
Find the measure of each angle in Exercises 19–24. 19.
20. (2y)°
(5k + 5)° (7x)°
(4y)°
(11x)°
23. Supplementary angles with measures 6x  4 degrees and 8x  12 degrees
(10m + 7)°
(3k + 5)°
(7m + 3)°
24. Complementary angles with measures 9z + 6 degrees and 3z degrees
Perform each calculation. 25. 62° 18¿ + 21° 41¿
26. 75° 15¿ + 83° 32¿
27. 47° 29¿  71° 18¿
28. 47° 23¿  73° 48¿
29. 90°  72° 58¿ 11–
30. 90°  36° 18¿ 47–
Convert each angle measure to decimal degrees. Use a calculator, and round to the nearest thousandth of a degree if necessary. 31. 20° 54¿
32. 38° 42¿
33. 91° 35¿ 54–
34. 34° 51¿ 35–
Convert each angle measure to degrees, minutes, and seconds. Use a calculator as necessary. Round to the nearest second. 35. 31.4296°
36. 59.0854°
37. 89.9004°
38. 102.3771°
Concept Check Sketch each angle in standard position. Draw an arrow representing the correct amount of rotation. Find the measure of two other angles, one positive and one negative, that are coterminal with the given angle. Give the quadrant of each angle. 39. 75°
40. 89°
41. 174°
42. 234°
43.  61°
44.  159°
Find the angle of least positive measure that is coterminal with the given angle. 45.  40° 49. 
46.  98°
p 4
50. 
47. 450°
p 3
51. 
48. 539°
3p 2
52.  p
Give an expression that generates all angles coterminal with each angle. Let n represent any integer. 53. 30° 57.
55.  90°
54. 45°
p 4
58.
p 6
59. 
56.  135°
3p 4
60. 
7p 6
Convert each degree measure to radians. Leave answers as rational multiples of p. 61. 60°
62. 90°
63. 150°
64. 270°
65.  45°
66.  210°
540
CHAPTER 8
The Unit Circle and the Functions of Trigonometry
Convert each radian measure to degrees. 67.
p 3
68.
8p 3
69.
7p 4
70.
2p 3
71.
11p 6
72.
15p 4
Convert each degree measure to radians. Round to the nearest hundredth. 73. 39°
74. 74°
75. 139° 10¿
76. 174° 50¿
77. 64.29°
78. 122.62°
83.  1.3
84.  4
Convert each radian measure to degrees. Round answers to the nearest minute. 79. 2
80. 5
81. 1.74
82. 0.3417
85. Concept Check The figure shows the same angles measured in both degrees and radians. Complete the missing measures. 2 3
90°;
____°; radians 3 ____°; 4 radians 150°; 180°;
2
radians 60°;
3 2
330°; ____ radians 315°; ____ radians ____°; 5 radians 3 radians
88. 2
4
3
93. r = 12.3 centimeters; u =
2p 3
94. r = 0.892 centimeter; u =
11p 10
radians radians
95. r = 4.82 meters; u = 60° 96. r = 71.9 centimeters; u = 135° Distance between Cities Find the distance in kilometers between the pair of cities whose latitudes are given. Assume that the cities are on a north–south line and that the radius of Earth is 6400 kilometers. Round answers to the nearest hundred kilometers. 97. Madison, South Dakota, 44° N, and Dallas, Texas, 33° N 98. Charleston, South Carolina, 33° N, and Toronto, Ontario, 43° N
Use the formula v = ut to find the value of the missing variable. In Exercise 104, round to the nearest thousandth.
Find the radius of each circle. 90. 2
Find the length of each arc intercepted by a central angle u in a circle of radius r. Round answers to the nearest hundredth.
100. Halifax, Nova Scotia, 45° N, and Buenos Aires, Argentina, 34° S
3 3 4
99. New York City, New York, 41° N, and Lima, Peru, 12° S
3 12
4
6
6
0°; 0 radians
Concept Check Find the exact length of each arc intercepted by the given central angle.
89.
92.
____ radian
86. Railroad Engineering The term grade has several different meanings in construction work. Some engineers use 1 the term to represent 100 of a right angle and express it as a percent. For instance, an angle of 0.9° would be referred to as a 1% grade. (Source: Hay, W., Railroad Engineering, John Wiley and Sons.) (a) By what number should you multiply a grade to convert it to radians? (b) In a rapidtransit rail system, the maximum grade allowed between two stations is 3.5%. Express this angle in degrees and in radians.
Concept Check
91.
____ radians ____°; 4 radian 30°;
____ radians
87.
Find the measure of each central angle (in
3
____ radians
210°; ____ radians 225°; ____ radians ____°; 4 radians 3 270°;
Concept Check radians).
101. u =
3p 4
radians, t = 8 seconds
102. u =
2p 5
radians, t = 10 seconds
103. u =
2p 9
radian, v =
5p 27
radian per minute
104. v = 0.90674 radian per minute, t = 11.876 minutes
8.1
The formula v = ut can be rewritten as u = vt. Substituting vt for u changes s = ru to s = rvt. Use the formula s = rvt to find the value of the missing variable. 105. r = 6 centimeters, v = 9 seconds 106. r = 9 yards, v =
2p 5
p 3
3p 4
11.7 cm
r
radians per second, t =
radians per second, t = 12 seconds
107. s = 6p centimeters, r = 2 centimeters, v = per second 108. s =
541
Angles, Arcs, and Their Measures
p 4
radian
kilometers, r = 2 kilometers, t = 4 seconds
113. Bicycle Chain Drive The figure shows the chain drive of a bicycle. To the nearest inch, how far will the bicycle move if the pedals are rotated through 180°? Assume the radius of the bicycle wheel is 13.6 inches.
Solve each problem.
1.38 in.
109. Pulley Raising a Weight Refer to the figure and answer each question. (a) How many inches will the weight rise if the pulley is rotated through an angle of 71° 50¿? (b) Through what angle, to the nearest minute, must the pulley be rotated to raise the weight 6 inches?
9.27 in.
110. Pulley Raising a Weight Approximate the radius of the pulley in the figure if a rotation of 51.6° raises the weight 11.4 centimeters.
4.72 in.
114. Pickup Truck Speedometer The speedometer of a pickup truck is designed to be accurate with tires of radius 14 inches. (a) Find the number of rotations, to the nearest whole number, of a tire in 1 hour if the truck is driven 55 mph. (b) Suppose that oversize tires of radius 16 inches are placed on the truck. If the truck is now driven for 1 hour with the speedometer reading 55 mph, how far has the truck gone? If the speed limit is 55 mph, does the driver deserve a speeding ticket? Approximate the area of a sector of a circle having radius r and central angle u. 115. r = 29.2 meters; u =
5p 6
116. r = 59.8 kilometers; u =
r
radians 2p 3
radians
117. r = 12.7 centimeters; u = 81° 118. r = 18.3 meters; u = 125°
111. Rotating Wheels The rotation of the smaller wheel in the figure causes the larger wheel to rotate. Through how many degrees, to the nearest tenth, will the larger wheel rotate if the smaller one rotates through 60.0°?
Concept Check Find the area of each sector. Express your answers in terms of p. 119.
120.
6 5.23 cm
4
2 8
8 . 16 cm
Solve each problem. 112. Rotating Wheels Refer to the figure at the top of the next column. Find the radius of the larger wheel, to the nearest tenth, in the figure if the smaller wheel rotates 80.0° when the larger wheel rotates 50.0°.
121. Doppler Radar Radar is used to identify severe weather. If Doppler radar can detect weather within a 240mile radius and creates a new image every 48 seconds, find the area scanned by the radar in 1 second.
542
The Unit Circle and the Functions of Trigonometry
CHAPTER 8
122. Land Required for a SolarPower Plant A 300megawatt solarpower plant needs approximately 950,000 square meters of land area to collect the required amount of energy from sunlight. (a) If the land area is circular, approximate its radius. (b) If the land area is a 35° sector of a circle, approximate its radius. 123. Area Cleaned by a Windshield Wiper The Ford Model A, built from 1928 to 1931, had a single windshield wiper on the driver’s side. The total arm and blade was 10 inches long and rotated back and forth through an angle of 95°. The shaded region in the figure is the portion of the windshield cleaned by the 7inch wiper blade. What is the area (to the nearest square inch) of the region cleaned?
126. Area of a Lot A frequent problem in surveying city lots and rural lands adjacent to curves of highways and railways is that of finding the area when one or more of the boundary lines is the arc of a circle. Find the area of the lot shown in the figure to the nearest hundred square yards. (Source: Anderson, J. and E. Michael, Introduction to Surveying, McGrawHill.)
10
in
.
95° 7i n.
40 yd
60°
124. Measures of a Structure The figure shows Medicine Wheel, a Native American structure in northern Wyoming. This circular structure is perhaps 2500 years old. There are 27 spokes in the wheel, all equally spaced.
30 yd
127. Speed of a Bicycle A bicycle has a tire 26 inches in diameter that is rotating at 15 radians per second. Approximate the speed of the bicycle in feet per second and in miles per hour.
26 in.
(a) Find the measure of each central angle in degrees and in radians. (b) If the radius of the wheel is 76 feet, approximate the circumference. (c) Approximate the length of each arc intercepted by consecutive pairs of spokes. (d) Find the area of each sector formed by consecutive spokes. 125. Size of a Corral The unusual corral in the figure at the top of the next column is separated into 26 areas, many of which approximate sectors of a circle. Assume that the corral has a diameter of 50 meters. (a) Approximate the central angle for each region, assuming that the 26 regions are all equal sectors with the fences meeting at the center. (b) What is the area of each sector (to the nearest square meter)?
128. Angular Speed of a Pulley The pulley shown has radius 12.96 centimeters. Suppose that it takes 18 seconds for 56 centimeters of belt to go around the pulley. Approximate the angular speed of the pulley in radians per second to the nearest hundredth.
12.96 cm
129. Angular Speed of Skateboard Wheels The wheels on a skateboard have diameter 2.25 inches. If a skateboarder is traveling downhill at 15.0 mph, approximate the angular speed of the wheels in radians per second.
8.1
130. Angular Speed of Pulleys The two pulleys in the figure have radii 15 centimeters and 8 centimeters, respectively. The larger pulley rotates 25 times in 36 seconds. Find the angular speed of each pulley in radians per second.
15 cm
8 cm
131. Orbit of Earth Earth travels about the sun in an orbit that is almost circular. Assume that the orbit is a circle, with radius 93,000,000 miles.
93,000,000 mi
Earth
Sun Not to scale
Angles, Arcs, and Their Measures
133. Speed of a Motor Propeller A 90horsepower outboard motor at full throttle rotates its propeller 5000 revolutions per minute. Find the angular speed of the propeller in radians per second. What is the linear speed in inches per second of a point at the tip of the propeller if its diameter is 10 inches? 134. Speed of a Golf Clubhead The shoulder joint can rotate at about 25 radians per second. Assuming that a golfer’s arm is straight and the distance from the shoulder to the clubhead is 5 feet, approximate the linear speed of the clubhead from the shoulder rotation. (Source: Cooper, J. and R. Glassow, Kinesiology, Second Edition, C.V. Mosby.) 135. Circumference of Earth The first accurate estimate of the distance around Earth was done by the Greek astronomer Eratosthenes (276–195 B.C.), who noted that the noontime position of the sun at the summer solstice differed by 7° 12¿ from the city of Syene to the city of Alexandria. (See the figure.) The distance between these two cities is 496 miles. Use the arc length formula to estimate the radius of Earth. Then approximate the circumference of Earth. (Source: Zeilik, M., Introductory Astronomy and Astrophysics, Third Edition, Saunders College Publishers.)
(a) Assume that a year is 365 days, and find u, the angle formed by Earth’s movement in 1 day. (b) Give the angular speed in radians per hour. (c) Approximate the linear speed of Earth in miles per hour. 132. Nautical Miles Nautical miles are used by ships and airplanes. They are different from statute miles, which equal 5280 feet. A nautical mile is defined to be the arc length along the equator intercepted by a central angle AOB of 1 minute, as illustrated in the figure. If the equatorial radius of Earth is 3963 miles, use the arc length formula to approximate the number of statute miles in 1 nautical mile. Round your answer to two decimal places.
543
Sun’s rays at noon
496 mi
7° 12' Shadow
Syene
Alexandria 7° 12'
136. Diameter of the Moon The distance to the moon is approximately 238,900 miles. Use the arc length formula to approximate the diameter d of the moon if angle u in the figure is measured to be 0.517°.
O B
1'
d
A Nautical mile
Not to scale
Not to scale
544
The Unit Circle and the Functions of Trigonometry
CHAPTER 8
8.2
The Unit Circle and Its Functions
Trigonometric (Circular) Functions • Using a Calculator to Find Function Values • Exact Function Values for
P P 4, 6,
and
P 3
Trigonometric (Circular) Functions y
x = cos s y = sin s
Arc of length s (0, 1)
(x, y) (–1, 0)
(1, 0) 0
(0, –1) Unit circle x2 + y2 = 1 (a)
1.5 – 4.7
4.7
x
As we saw in Chapter 6, the graph of the equation x 2 ⴙ y 2 ⴝ 1 is a circle with center at the origin and radius 1. We refer to its graph as the unit circle. Because the radius of the unit circle is 1, its circumference is C = 2p(1) = 2p. See FIGURE 26(a). Suppose that we consider the point P(1, 0) on the unit circle as a starting point. We can move either clockwise or counterclockwise along an arc of the unit circle any distance s we wish and stop at a point (x, y) on the unit circle. Let’s agree that if s represents a real number and s 7 0, then the movement is counterclockwise around the circle, while if s 6 0, the movement is clockwise around the circle. If s = 0, there is no movement. Thus, every real number s is paired with one and only one point on the graph of the unit circle. For example, if s = p2 , then we start at P(1, 0) and move p2 units counterclockwise, ending at the point (0, 1). (Note that the circumference of the circle is 2p, so a distance of p2 equals one fourth of the circumference.) We say that the point (0, 1) corresponds to p2 . See FIGURE 26(b). If (x, y) is the point on the unit circle that corresponds to the real number s, we can define the six circular functions of s as sine, cosine, tangent, cosecant, secant, and cotangent, abbreviated sin, cos, tan, csc, sec, and cot. (These are sometimes called trigonometric functions because of their relationship to right triangle side ratios.)
The unit circle
Circular (Trigonometric) Functions
(b) FIGURE 26
If (x, y) is the point on the unit circle that corresponds to the real number s, then sin s ⴝ y csc s ⴝ
1 ( y ⴝ 0) y
sec s ⴝ
y (x ⴝ 0) x x ( y ⴝ 0). cot s ⴝ y
tan s ⴝ
cos s ⴝ x 1 (x ⴝ 0) x
From the definitions, we see that there are three pairs of reciprocal functions:
y
Q(cos s, sin s)
sine and cosecant, cosine and secant,
(0, 1) s (–1, 0)
x
0
P(1, 0)
(0, –1) FIGURE 27
and tangent and cotangent.
FIGURE 27 indicates that, for the positive number s, the point Q corresponding to s has coordinates (cos s, sin s). FIGURE 28 on the next page justifies an important relationship between sin s and cos s for any real number s.* Since sin s = y and cos s = x, we can replace x and y in x 2 + y 2 = 1 and obtain the Pythagorean relationship (cos s) 2 + (sin s) 2 = 1, or, as it is usually written,
cos2 s ⴙ sin2 s ⴝ 1.
*The authors wish to thank Professor Linda K. Schmidt of Greenville Technical College for her suggestion to include FIGURE 28.
8.2
The Unit Circle and Its Functions
y
y 1
1
(x, y)
(cos s, sin s) s
y 0
–1
545
x
x
1
0
–1
–1
x
1
–1
x2 + y2 = 1
x = cos s, y = sin s
cos 2 s + sin 2 s = 1 FIGURE 28 GCM
TECHNOLOGY NOTE Powers of trigonometric functions are not usually entered on calculators as they are written mathematically. For example, we write sin2 s to represent the square of sin s, but this would be entered as (sin s)2 on many calculators.
Since the ordered pair (x, y) represents a point on the unit circle, and 1 … x … 1  1 … cos s … 1 and
so
 1 … y … 1,  1 … sin s … 1.
For any value of s, both sin s and cos s exist, so the domain of these functions is the y set of all real numbers. Note that tan s equals x , which is undefined when x = 0. Referring to the unit circle, we see that x = 0 when s = ⫾ p2 , ⫾ 3p 2 . Furthermore, if s 7p is more than one revolution, then x = 0 when s = ⫾ 5p 2 , ⫾ 2 , and so on. Therefore, the domain of the tangent function consists of real numbers s satisfying
s Z (2n + 1)
p , n any integer. 2
The definition of secant also has x in the denominator, so the domain of secant is the same as the domain of tangent. Both cotangent and cosecant are defined by expressions having denominator y. To guarantee that y Z 0, the domain for each of these functions must be the set of all values of s satisfying s Z np, n any integer. Looking Ahead to Calculus
The domains of the circular functions as defined in this section consist of sets of real numbers. Later in the chapter, we will see that the domains can also be sets of angles. Real number domains are used when studying these functions in calculus.
Domains The domains of the six circular functions are as follows. Assume that n is any integer and s is a real number. Sine and Cosine Functions:
(ⴚˆ, ˆ)
Tangent and Secant Functions:
e s ` s ⴝ (2n ⴙ 1)
Cotangent and Cosecant Functions:
P f 2
{s  s ⴝ nP}
From the definitions of the circular functions, the following statements are true for the real number s, provided that the denominator is not 0: tan s ⴝ
sin s , cos s
cot s ⴝ
cos s , sin s
sec s ⴝ
1 , cos s
If cos s Z 0, then
1 ⴙ tan2 s ⴝ sec 2 s,
and if sin s Z 0, then
1 ⴙ cot 2 s ⴝ csc 2 s.
csc s ⴝ
1 . sin s
Statements such as these are called identities. An identity is a statement that is true for all values in the domain of the variable.
546
CHAPTER 8
The Unit Circle and the Functions of Trigonometry
EXAMPLE 1
Finding Function Values by Using the Unit Circle For each number s, determine the point on the unit circle to which it corresponds and the circular function values for s. State which functions are undefined for s. 3p p (a) 0 (b) p (c) (d) 2 2 Solution (a) The starting point P(1, 0), seen in FIGURE 29, corresponds to s = 0. Thus,
y
P(1, 0)
x
0
(–1, 0)
(b) x2 + y2 = 1
cos 0 = 1
csc 0 is undefined
sec 0 =
y
(0, –1)
cos p =  1
csc p is undefined
sec p =
3p = 1 2 3p 1 csc = = 1 2 1
sin
x
P(1, 0) – 2
x2 + y2 = 1 FIGURE 30
1 = 1 1
sin p = 0
(c) The point (0,  1) corresponds to s =
0
tan 0 =
0 = 0 1
cot 0 is undefined.
FIGURE 29 shows that the point ( 1, 0) corresponds to s = p. (Since the circumference of the circle is 2p, a distance of p equals one half of the circumference.) Thus,
FIGURE 29
3 2
sin 0 = 0
tan p =
1 = 1 1 3p 2 ,
0 = 0 1
cot p is undefined.
as seen in FIGURE 30, leading to
3p = 0 2 3p is undefined sec 2 cos
3p is undefined 2 3p 0 cot = = 0. 2 1
tan
(d) The real number s =  p2 also corresponds to (0,  1), as seen in FIGURE 30. Thus, its circular function values are the same as those of 3p 2 in part (c). The following table gives function values of multiples of p2 from 0 to 2p. Exact Function Values for 0, P2 , P , 3P 2 , and 2P
TI83 Plus
TI84 Plus Radian mode FIGURE 31
s
sin s
cos s
tan s
cot s
sec s
csc s
0
0
1
0
Undefined
P 2
1
0
Undefined
0
Undefined
0
Undefined
1
Undefined
Undefined
1
P
0
1
3P 2
1
0
Undefined
0
2P
0
1
0
Undefined
1
1
Undefined 1
Undefined
Using a Calculator to Find Function Values When finding function values based on the unit circle, THE CALCULATOR MUST BE IN RADIAN MODE (as opposed to degree mode). See FIGURE 31.
8.2
The Unit Circle and Its Functions
547
EXAMPLE 2
Using a Calculator to Find Function Values Use a calculator in radian mode to find the six trigonometric function values for p. Solution FIGURE 32,
These values were found in Example 1(b) by using the unit circle. In we show how a calculator displays cos p, sin p, and tan p.
TECHNOLOGY NOTE In general, calculators do not have keys designated for secant, cosecant, and cotangent. To find these values, we use the reciprocal relationships.
Radian mode
Radian mode
Radian mode
FIGURE 32
FIGURE 33
FIGURE 34
To find sec p, we use the fact that the secant function is the reciprocal of the cosine function, and we enter 1/cos(p). The result is shown in FIGURE 33. The other two function values are undefined, and if we attempt to find csc p by entering the reciprocal of sin p, an error message occurs, as seen in FIGURE 34.
CAUTION The calculator functions cos1, sin1, and tan1 DO NOT represent reciprocals of the cosine, sine, and tangent. These are the inverse circular functions, introduced later in the text. For example, to find csc 3p 2 with a calculator, enter 1/sin(3p/2), as shown in FIGURE 35. Compare with the result of Example 1(c). csc 3 = –1 2 Radian mode FIGURE 35
Because the circumference of the unit circle is 2p, any integer multiple of 2p added to s corresponds to the same point as s on the unit circle. Thus, cos(s + 2np) = cos s
Radian mode FIGURE 36
FOR DISCUSSION Duplicate FIGURE 36, but use different integer values for n to further support the equations. Then show that similar statements, such as cos(p + 4p) = cos p, are true for the cosine function.
and
sin(s + 2np) = sin s
for any integer n. (The circular functions are examples of a broader class of functions called periodic functions, whose values repeat in a fixed manner.) FIGURE 36 illustrates these facts for the sine function, with s = p2 and n = 3. The definitions of the functions can be used to determine the signs of their values in each of the four quadrants. For example, if s corresponds to a point in quadrant I, then both x and y are positive, and all the functions have positive values. In quadrant II, x is negative and y is positive, so cos s = x is negative, sin s = y is posiy itive, tan s = x is negative, and so on.
Signs of Values of Circular Functions y
x < 0, y > 0 II Sine and cosecant have positive values.
x > 0, y > 0 I All functions have positive values. x
x < 0, y < 0 III Tangent and cotangent have positive values.
x > 0, y < 0 IV Cosine and secant have positive values.
548
CHAPTER 8
The Unit Circle and the Functions of Trigonometry
EXAMPLE 3
Finding the Cosine, Sine, and Quadrant Use a calculator to find approximations for cos 4.5 and sin 4.5. Then determine the quadrant in which the point on the unit circle corresponding to 4.5 lies. Solution
Use a calculator as in FIGURE 37 to find the approximations: cos 4.5 L  0.2107957994 and
sin 4.5 L  0.9775301177.
Since cos 4.5 6 0 and sin 4.5 6 0, the point corresponding to 4.5 lies in quadrant III, where both x and y are negative. This is also reasonable, since p 6 4.5 6 3p 2 . Radian mode FIGURE 37
Exact Function Values for x2 + y2 = 1
y=x 1.55
–2.35
2.35
and
P 3
We can find the exact sine and cosine values (and thus the other function values as well) for the real numbers p4 , p6 , and p3 . Because the unit circle is symmetric with respect to both axes and the origin, these results enable us to find all six circular function values of integer multiples of these numbers as well. The line y = x intersects the quadrant I portion of the unit circle, and the point of intersection will give us exact values of cos p4 and sin p4 . See FIGURE 38. We must solve the system x2 + y2 = 1 y = x.
–1.55 The display shows approximations of x = cos p4 and y = sin p4 . FIGURE 38
P P 4, 6,
(1) (2)
Substituting y for x in equation (1) gives y2 + y2 = 1 2y 2 = 1 1 y2 = 2
Since y = x, x =
Let y = x. Combine like terms. Divide by 2.
y =
1 B2
y =
1
Take square roots; y 7 0.
#
22
22 22
22 . Thus, 2
=
the point
22 . 2 a
Rationalize the denominator.
22 22 , b 2 2
corresponds to p4 , and
p 22 sin = 4 2
p 22 cos = 4 2
p tan = 4
p csc = 22 4
p sec = 22 4
p cot = 4
22 2 22 2 22 2 22 2
= 1
= 1.
FOR DISCUSSION 22
Use a calculator to find approximations for cos p4 , sin p4 , and . Show that all 2 three expressions give the same approximation. Compare them with the x and yvalues shown at the bottom of FIGURE 38. They should be the same (except for perhaps the final digits, depending on how many digits the calculator displays).
8.2
R(0, 1)
0
P(1, 0) Q' (x, –y)
549
To find the point on the unit circle that corresponds to p6 , see FIGURE 39. Point Q(x, y) corresponds to the real number p6 , and Q¿(x,  y) is its reflection across the xaxis, corresponding to  p6 . The chord QQ¿ cuts off an arc of length p6 +   p6  = p3 , as does the chord QR, since the length of arc QR is p2  p6 = p3 . Thus,
y
Q(x, y) 6
The Unit Circle and Its Functions
d(Q, Q¿) = d(Q, R)
x
2(x  x) 2 + 3y  ( y)42 = 2(x  0) 2 + ( y  1) 2 2(2y) 2 = 2x 2 + y 2  2y + 1
x2 + y2 = 1 FIGURE 39
Distance formula
Remember the middle term when squaring a binomial.
4y 2 = x 2 + y 2  2y + 1 4y 2
Square each side.
= 1  2y + 1
x2 + y2 = 1
4y 2 = 2  2y 2y 2
y =
b ⫾
2b 2
Add.
+ y  1 = 0  4ac
2a
Divide by 2; standard form.
y =
 1 ⫾ 21 + 8 4
Let a = 2, b = 1, c =  1 in the quadratic formula.
y =
1 . 2
Simplify; y 7 0, so reject y =  1.
To find x, we substitute 12 for y in x 2 + y 2 = 1. 1 2 x2 + a b = 1 2 x2 =
Let y = 12 . 2 A 12 B = 14 ; subtract 14 .
3 4
x =
3 B4
Take square roots; x 7 0.
x =
23 2
Simplify.
Therefore, the real number p6 corresponds to the point a
23 1 , b. 2 2
A similar argument shows that the real number p3 corresponds to a 12 ,
23 b . These 2
results lead to exact values of the circular function values for p6 and p3 , as summarized in the following table, along with the values for p4 found earlier.
Exact Function Values for
P P 6, 4,
and
P 3
s
sin s
cos s
tan s
cot s
sec s
csc s
P 6
1 2
23 2
23 3
23
223 3
2
P 4
22 2
22 2
1
1
22
22
P 3
23 2
1 2
23
23 3
2
223 3
550
CHAPTER 8
The Unit Circle and the Functions of Trigonometry
The symmetry of the unit circle and the values just found lead to other function values, as seen in FIGURE 40. y
(– 12 , √23 ) (– √22 , √22 ) 3 23
(0, 1) 2
3
4
(– √23 , 12 )
5 6
( 12 , √23 ) ( √22 , √22 ) 4 (√23 , 12 ) 6
0 2
(–1, 0)
(
– √3 , 2
–
1 2
0
11 6
7 6
)
5 4
7 4
4
3 (– √22 , – √22 ) (– 12 , – √23 )
(1, 0)
x
( √23 , – 12 )
5 3
3 2
(0, –1)
( √22 , – √22 ) ( 12 , – √23 )
Unit circle x2 + y2 = 1 FIGURE 40
EXAMPLE 4
Finding Exact Function Values Use FIGURE 40 to find each value. 4p 5p 2p (a) sin (b) cos (c) tan 6 3 3 2p p b (d) tan a(e) sec 3 3 Solution (a) The point a 
23 1 , b 2 2
corresponds to 5p 6 on the unit circle. Since the function is sine,
1 we use the yvalue of the point to obtain sin 5p 6 = 2.
(b) The point a  12 ,
23 b 2
corresponds to 2p 3 . The cosine function requires that we use
1 the xvalue, so cos 2p 3 =  2. y
(c) By definition, tan s = x . When s = 23
4p 2 tan = 1 3 
= 
4p 3 ,
23
(x, y) = a  12 ,  2 b , so
23 1 23 # 2 , a b = a  b = 23. 2 2 2 1
2
4p (d) The point on the unit circle for s =  2p 3 is the same as that for 3 . By the result
of part (c), tan A  2p 3 B = 23.
(e) sec
p 1 1 = p = 1 = 2 3 cos 3 2
8.2
8.2
551
The Unit Circle and Its Functions
EXERCISES
Checking Analytic Skills Determine what fraction of the circumference of the unit circle each value of s represents. For example, s = p represents 12 of the circumference of the unit circle. Do not use a calculator. 1. s =
p 3
2. s =
2p 3
3. s =
5. s =
p 6
6. s =
5p 6
7. s = 3p
9.
3p 2
p 4
4. s =
3p 4
8. s =
5p 2
10. p
Each figure shows an arc with length s along the circumference of a unit circle. Evaluate sin s, cos s, and tan s. y
11.
12. – 8 , 15 17 17 y
(
1
s –1
0
13.
)
( 45 , 35 )
s
x
1
y
0
x
0
s
(1213 , – 135 )
–1
y
14.
x
15. – 20 , 21
( 29 29 )
16.
y
y
(√22 , √22 ) s
0
x
x
0
x
0
s s
(
– 7 , – 24 25 25
)
Refer to FIGURE 26(a). For each real number s, determine the point on the unit circle to which it corresponds. Determine cos s and sin s. State which functions are undefined for s. 17.  p
18. p
3p 2
22. 
21.
19. 2p 3p 2
23.
p 2
20.  2p 24. 
p 2
Checking Analytic Skills Refer to FIGURE 26(a). Find the circular function values for each number s. State which functions are undefined for s. Do not use a calculator. 25.  5p 29. 
5p 2
26.  3p 30. 
7p 2
27. 6p
31.
5p 2
28. 8p
32.
7p 2
552
CHAPTER 8
The Unit Circle and the Functions of Trigonometry
Use a calculator in radian mode to find approximations for cos s and sin s for each number s. Give as many decimal places as your calculator displays. (These are NOT exact values—they are only approximations.) Then determine the quadrant in which the point on the unit circle corresponding to s lies. Finally, find approximations for tan s, cot s, sec s, and csc s. 33. 0.75
34. 0.95
35.  4.25
36.  3.75
37.  2.25
38.  2.75
39. 5.5
40. 5.75
RELATING CONCEPTS
For individual or group investigation (Exercises 41–46)
With your calculator in radian mode, work Exercises 41–46 in order.
9
41. Let s represent the number of letters in your first name. Find an approximation for cos s.
42. Let n represent the number of letters in your last name. Find an approximation for cos(s + 2np).
43. Compare your results from Exercises 41 and 42. Explain how and why they relate to each other as they do.
44. Let s represent the number of letters in your last name. Find an approximation for sin s.
45. Let n represent the number of letters in your first name. Find an approximation for sin(s + 2np).
9 46.
Compare your results from Exercises 44 and 45. Explain how and why they relate to each other as they do.
Use the figure and any appropriate relationship(s) among circular functions to find each value. 47. sin
7p 6
48. cos
5p 3
7p 50. cos 6
2p 51. sec 3
5p 6
54. cosa
53. cot
17p 56. tan 3 59. cosa
13p b 6
49. tan
y
3p 4
11p 52. csc 6 4p b 3
55. sin a
5p b 6
23p 57. sec 6
13p 58. csc 3
60. sin a 
61. tan a
9p b 4
13p b 4
(– 12 , √32 ) (– √22 , √22 ) 2 3 (– √32 , 12 ) 5 34 (–1, 0)
( 12 , √32 ) ( √22 , √22 ) 3 4 (√32 , 12 ) 6
(0, 1) 2
6
0 (1, 0) 2
7
(– √32 , – 12 ) 6 5 4 (– √22 , – √22 ) 4 3 (– 12 , – √32 )
11 √3 , 1 –2 7 6 2 5 4 √2 , √2 3 – 3 2 2 2 1 , √3 – (0, –1) 2 2
(
(
(
x2 + y2 = 1
9 62.
In the screen shown, the value 3 is stored in S. Then the value of (cos(S)) 2 + (sin(S)) 2 is shown to be 1. Duplicate this screen on your own calculator, but use several different values of S. Is the result always 1? Explain.
Use the identity cos2 s + sin2 s = 1 to find the value of x or y, as appropriate. Then, assuming that s corresponds to the given point on the unit circle, find the six circular function values for s. 3 63. a , yb, y 7 0 5
66. ax,
8 b, x 6 0 17
64. a
7 , yb, y 7 0 25
1 67. a  , yb , y 6 0 3
65. ax,
24 b, x 6 0 25
1 68. a , yb , y 6 0 4
)
)
x
)
Reviewing Basic Concepts (Sections 8.1 and 8.2)
Checking Analytic Skills calculator. 69. sin s =
553
For each of the following, find tan s, cot s, sec s, and csc s. Do not use a
23 1 , cos s = 2 2
70. sin s =
1 23 72. sin s =  , cos s = 2 2
3 27 , cos s = 4 4
73. sin s = 
23 1 , cos s = 2 2
71. sin s =
3 4 , cos s = 5 5
74. sin s =
12 5 , cos s = 13 13
Decide in what quadrant the point corresponding to s must lie to satisfy the following conditions for s. 75. sin s 7 0, cos s 6 0
76. cos s 7 0, tan s 7 0
77. sec s 6 0, csc s 6 0
78. tan s 7 0, cos s 6 0
79. cos s 7 0, sin s 6 0
80. tan s 6 0, sin s 7 0
81. Prove that if cos s Z 0, then 1 + tan2 s = sec2 s.
82. Prove that if sin s Z 0, then 1 + cot 2 s = csc2 s.
Let s be a real number corresponding to the point (a, b) on the unit circle. Use this information to determine an expression representing the sine and cosine of each real number. 83. s
84. s + 2p
85. s  6p
87.  s
88.  s + p
89. s 
SECTIONS 8.1 AND 8.2
p 2
86. s + p 90. s +
p 2
REVIEWING BASIC CONCEPTS
1. Give the complement and the supplement of each angle. p (b) (a) 35° 4 2. Convert 32.25° to degrees, minutes, and seconds. 3. Convert 59° 35¿ 30– to decimal degrees. 4. Find the angle of least positive measure coterminal with each angle. 2p (a) 560° (b) 3 5. Convert each angle measure as directed. 3p (a) 240° to radians (b) radians to degrees 4
7. Give the exact coordinates of the point on the unit circle that corresponds to the given value of s. 5p 5p (a)  2p (b) (c) 4 2 8. Give the values of the circular functions of  5p 2 . State which functions are undefined. 9. Give the exact values of the six trigonometric functions of each number. 7p 2p (a) (b) 6 3 10. Give calculator approximations for the six circular function values of 2.25.
6. Suppose a circle has radius 3 centimeters. (a) What is the exact length of an arc intercepted by a central angle of 120°? (b) What is the exact area of the sector formed in part (a)?
554
CHAPTER 8
8.3
The Unit Circle and the Functions of Trigonometry
Graphs of the Sine and Cosine Functions
Periodic Functions • Graph of the Sine Function • Graph of the Cosine Function • Graphing Techniques, Amplitude, and Period • Translations • Determining a Trigonometric Model Using Curve Fitting
Periodic Functions Many things in daily life repeat with a predictable pattern, such as weather, tides, and hours of daylight. Because the sine and cosine functions repeat their values in a regular pattern, they are examples of periodic functions. FIGURE 41 shows a periodic graph that represents a normal heartbeat.
FIGURE 41
Looking Ahead to Calculus
Periodic functions are used throughout calculus, so you will need to know their characteristics. One use of these functions is to describe the location of a point in the plane by polar coordinates, an alternative to rectangular coordinates. (See Chapter 10.)
Periodic Function A periodic function is a function ƒ such that ƒ(x) ⴝ ƒ(x ⴙ np), for every real number x in the domain of f, every integer n, and some positive real number p. The least possible positive value of p is the period of the function.
The circumference of the unit circle is 2p, so the least value of p for which the sine and cosine functions repeat is 2p. Therefore, the sine and cosine functions are periodic functions with period 2P .
Graph of the Sine Function In the previous section, we saw that for a real number s, the point on the unit circle corresponding to s has coordinates (cos s, sin s). See FIGURE 42, and trace along the circle to verify the results shown in the table.
y
(0, 1)
( x, y) = (cos s, sin s)
As s Increases from
s (–1, 0)
(1, 0) 0
(0, –1)
The unit circle x 2 + y2 = 1
FIGURE 42
x
sin s
cos s
0 to p2
Increases from 0 to 1
Decreases from 1 to 0
p 2
Decreases from 1 to 0
Decreases from 0 to  1
3p 2
Decreases from 0 to  1
Increases from  1 to 0
to 2p
Increases from  1 to 0
Increases from 0 to 1
to p
p to 3p 2
8.3
555
Graphs of the Sine and Cosine Functions
To avoid confusion when graphing the sine function, we use x rather than s. This corresponds to the letters in the xycoordinate system. Selecting key values of x and finding the corresponding values of sin x leads to the tables in FIGURE 43. To obtain the traditional graph of a portion of the sine function shown in FIGURE 43, we plot the points from the table of values and join them with a smooth curve. Since y = sin x is periodic and has ( q, q) as its domain, the graph continues in the same pattern in both directions. This graph is called a sine wave or sinusoid. A comprehensive graph of a sinusoid consists of at least one period of the graph and shows the extreme points.
FUNCTION CAPSULE ƒ(x) ⴝ sin x
SINE FUNCTION
Range: 3 1, 14
Domain: ( q, q) GCM
y
y ⴝ sin x
TECHNOLOGY NOTE
x
Graphing calculators often have a window designated for graphing circular functions. We refer to the window 3  2p, 2p4 by 3 4, 44 with Xscl = p2 and Yscl = 1 as the trig viewing window. Your model may use a different “standard” viewing window.
0
0
p 6
1 2
p 4
22 2
2 – 2 –2 – 3 – 2
23 2
p 3 p 2 3p 2
2p
3 2 0
–1
2
x 2
–2
1 0 1 0
p
f(x) = sin x, –2 ≤ x ≤ 2
1
f (x) = sin x 4
– 3 2 –2
2
–4
ΔTbl = 2 FIGURE 43 ■ ■ ■ ■
The graph is continuous over its entire domain, ( q, q). Its xintercepts are of the form np, where n is an integer. Its period is 2p. The graph is symmetric with respect to the origin. It is an odd function. For all x in the domain, sin(  x) =  sin x.
NOTE Recall that a function ƒ is an odd function if for all x in the domain of ƒ, ƒ( x ) =  ƒ( x ). The graph of an odd function is symmetric with respect to the origin (Section 2.1): If ( x, y) is on the graph of the function, then so is (  x,  y ). For example, A p2 , 1 B and A  p2 ,  1 B are points on the graph of y = sin x, illustrating sin (  x ) =  sin x.
Graph of the Cosine Function We find the graph of y = cos x in much the same way as the graph of y = sin x. In the tables for y = cos x shown with FIGURE 44 on the next page, we use the same values for x as before.
556
CHAPTER 8
The Unit Circle and the Functions of Trigonometry
FOR DISCUSSION The graph of the cosine function is a shift of the graph of the sine function. How many units and in which direction is this shift?
FUNCTION CAPSULE COSINE FUNCTION Domain: ( q, q)
The discussion of the derivative of a function in calculus shows that for the sine function, the slope of the tangent line at any point x is given by cos x. For example, look at the graph of y = sin x and notice that a tangent line at 5p x = ⫾ p2 , ⫾ 3p 2 , ⫾ 2 , Á will be horizontal and thus have slope 0. Now look at the graph of y = cos x and see that for these values, cos x = 0.
Range: 3 1, 14
y ⴝ cos x
x
Looking Ahead to Calculus
ƒ(x) ⴝ cos x
0
1
p 6
23 2 22 2
p 4 p 3 p 2
y 2
–2 – 3 – 2
1 2
3p 2
2p
– 2 –1
0
2
x 3 2
2
–2
0 1 0 1
p
f(x) = cos x, –2 ≤ x ≤ 2
1
f (x) = cos x 4
– 3 2 –2
2
–4
ΔTbl = Q 2 FIGURE 44 ■ ■ ■ ■
The graph is continuous over its entire domain, ( q, q). Its xintercepts are of the form (2n + 1) p2 , where n is an integer. Its period is 2p. The graph is symmetric with respect to the yaxis. It is an even function. For all x in the domain, cos(  x) = cos x.
NOTE Recall that a function f is an even function if for all x in the domain of ƒ, ƒ( x ) = ƒ( x ). The graph of an even function is symmetric with respect to the yaxis (Section 2.1): If ( x, y) is on the graph of the function, then so is ( x, y ). For example, A p2 , 0 B and A  p2 , 0 B are points on the graph of y = cos x, illustrating cos ( x ) = cos x.
Graphing Techniques, Amplitude, and Period The examples that follow show graphs that are “stretched” or “shrunk” either vertically, horizontally, or both, compared with the graphs of y = sin x or y = cos x. Graphing y ⴝ a sin x Graph y = 2 sin x, and compare with the graph of y = sin x. EXAMPLE 1
Solution For a given value of x, the value of y is twice as large as it would be for y = sin x, as shown in the table of values. The only change in the graph is the range, which becomes 3 2, 24. See FIGURE 45 on the next page, which includes a graph of y = sin x for comparison.
8.3
x 0
sin x 2 sin x 0 0
2 3 2 2
Graphs of the Sine and Cosine Functions
1
2
0
0
–1
–2
0
0
y
Period: 2
y 2 sin x
2
y sin x
1
2 2  3 2


557
0 2 1
x
3 2
2
2
FIGURE 45
We can think of the graph of y ⴝ a sin x as a vertical stretching of the graph of y ⴝ sin x when a>1 and a vertical shrinking when 02)X,
x=2 y = –1 r = √5
Divide by 2.
The point (2,  1) lies on the terminal side, and the corresponding value of r is
Graphing Calculator Solution FIGURE 80 shows the graph of
(2, –1) x + 2y = 0, x ≥ 0
or Y1 =  0.5X
with the restriction X Ú 0. We used the capability of the calculator to find the Yvalue for X = 2, which produced the point and values of X and Y shown on the screen.
r = 22 2 + ( 1) 2 = 25.
2 FIGURE 79
Thus, sin u =
y 1 1 = = r 25 25
# #
25 25 25
= 
25 5
2 25 5
Multiply by
x 2 2 = = r 25 25
tan u =
y 1 = x 2
csc u =
r 25 = =  25 y 1
sec u =
25 r = x 2
cot u =
x 2 = =  2. y 1
25
15
–3 ,
3
which is
equal to 1, to rationalize denominators.
cos u =
=
25
–2 FIGURE 80
We can now use these values of X and Y and the definitions of the trigonometric functions to find the six function values of u, as shown in the analytic solution.
584
The Unit Circle and the Functions of Trigonometry
CHAPTER 8
Recall that when the equation of a line is written using slope–intercept form y = mx + b, the coefficient of x is the slope of the line. In Example 2, x + 2y = 0 can be written as y =  12 x, so the slope is  12. Notice that tan u =  12. In general, it is true that m ⴝ tan U.
FOR DISCUSSION Either individually or with a group of students, rework Example 2, using a different value for x. Find the corresponding yvalue, and then show that the six trigonometric function values you obtain are the same as the previous ones.
NOTE The trigonometric function values we found in Example 2 are exact. If we were to use a calculator to approximate these values, the decimal results would not be acceptable if exact values were required.
Function Values of Quadrantal Angles If the terminal side of an angle in standard position lies along the yaxis, any point on this terminal side has xcoordinate 0. Similarly, an angle with terminal side on the xaxis has ycoordinate 0 for any point on the terminal side. Since the values of x and y appear in the denominators of some trigonometric functions, some trigonometric function values of quadrantal angles (i.e., those with terminal side on an axis) are undefined.
GCM
EXAMPLE 3
Finding Function Values of Quadrantal Angles Find the values of the six trigonometric functions for each angle, if possible. (a) An angle of 90° (b) An angle u in standard position with terminal side through (  3, 0)
Analytic Solution
Graphing Calculator Solution
(a) We select any point on the terminal side of a 90° angle, such as (0, 1) shown in FIGURE 81. Here, x = 0 and y = 1, so r = 1.
(a) A calculator set in degree mode returns the correct values for sin 90° and cos 90°. See FIGURE 83. The bottom screen shows an ERROR message for tan 90°, because 90° is not in the domain of the tangent function.
y 1 = = 1 r 1 0 x cos 90° = = = 0 r 1 y 1 tan 90° = = (undefined) x 0 sin 90° =
r 1 = = 1 y 1 r 1 sec 90° = = (undefined) x 0 0 x cot 90° = = = 0 y 1 csc 90° =
y
y
(0, 1)
90° 0
FIGURE 81
(b)
FIGURE 82
(–3, 0)
0
x
FIGURE 82
shows the angle. Here, x =  3, y = 0, and r = 3.
0 sin u = = 0 3 3 cos u = = 1 3 0 tan u = = 0 3
x
3 csc u = (undefined) 0 3 sec u = = 1 3 3 cot u = (undefined) 0
Degree mode FIGURE 83
On the TI83/84 Plus graphing calculator, there are no calculator keys for finding the function values of cotangent, secant, or cosecant. Later in this section, we discuss how to find these function values with a calculator.
8.5
585
Functions of Angles and Fundamental Identities
Undefined Function Values If the terminal side of a quadrantal angle lies along the 1. yaxis, the tangent and secant functions are undefined. 2. xaxis, the cotangent and cosecant functions are undefined. Function values of the most commonly used degreemeasured quadrantal angles, 0°, 90°, 180°, 270°, and 360°, are summarized in the following table, which is similar to the one in Section 8.2. (FIGURE 84 can be helpful in determining these values.) y
Function Values of Quadrantal Angles U
x=0 y=1 r=1
(0, 1) x = –1 y=0 r=1
(1, 0) 0
(–1, 0) x=0 y = –1 r=1
sin U
cos U
tan U
cot U
sec U
csc U
0°
0
1
0
Undefined
90°
1
0
Undefined
0
Undefined
180°
0
1
0
Undefined
1
Undefined
270°
1
0
Undefined
0
Undefined
1
360°
0
1
0
Undefined
1
Undefined 1
x
x=1 y=0 r=1 (0, –1)
1
Undefined
FIGURE 84
Values given in this table can also be found with a calculator that has trigonometric function keys. Make sure the calculator is set in degree mode. CAUTION One of the most common errors involving calculators in trigonometry occurs when the calculator is set for the wrong mode. Be sure that you know how to set your calculator in either radian or degree mode.
Reciprocal Identities In Section 8.2, we introduced several statements that are true for all real numbers for which the functions are defined. For example, sin2 s + cos2 s = 1
and
tan s =
1 cot s
Identities
are two of these equations, called identities. They hold for angles as well. The definitions of the trigonometric functions on page 582 were written so that funcy tions in the same column are reciprocals of each other. Since sin u = r and csc u = yr , sin u =
1 csc u
and
csc u =
1 , provided sin u Z 0. sin u
Also, cos u and sec u are reciprocals, as are tan u and cot u. The reciprocal identities hold for any angle u that does not lead to a 0 denominator.
Reciprocal Identities 1 csc U 1 csc U ⴝ sin U sin U ⴝ
1 sec U 1 sec U ⴝ cos U
cos U ⴝ
1 cot U 1 cot U ⴝ tan U
tan U ⴝ
586
CHAPTER 8
The Unit Circle and the Functions of Trigonometry
FIGURE 85(a) shows how csc 90°, sec 180°, and csc( 270°) are found by using the reciprocal identities and the reciprocal key of a graphing calculator in degree mode. Be sure not to use the inverse trigonometric function keys to find the reciprocal function values. Attempting to find sec 90° by entering cos190° produces an ERROR message, indicating that the reciprocal is undefined. See FIGURE 85(b). Compare these results with those in the table of quadrantal angle function values.
Degree mode (a)
NOTE Identities can be written in different forms. For example, sin U ⴝ
1 csc U
EXAMPLE 4 (b)
1 sin U
(b) Find sin u, if csc u = 
Solution (a) Since cos u is the reciprocal of sec u, cos u =
3
(b) Since sin u =

#
3 3 = . 5 5
1 csc u ,
1
sin u =
212 . 2
Be careful when simplifying a complex fraction.
1 5 1 = 5 = 1 , = 1 sec u 3
FOR DISCUSSION The signs of the functions in the four quadrants are determined mathematically by considering the signs of x, y, and r. A mnemonic device is the statement “All Students Take Calculus,” which indicates that all are positive in quadrant I, sine is positive in quadrant II, tangent is positive in quadrant III, and cosine is positive in quadrant IV. How can you remember the signs of secant, cosecant, and cotangent using this device? (Always remember that there are sound mathematical principles that justify mnemonic devices like this one.)
and (sin U)(csc U) ⴝ 1.
Using the Reciprocal Identities
(a) Find cos u, if sec u = 53.
FIGURE 85
csc U ⴝ
can also be written
212 2
=
2 212
=
2 2 23
212 = 24 # 3 = 223
=
1 23
=
1 23
#
23 23
23 . 3
Here we rationalize the denominator.
Signs and Ranges of Function Values In the definitions of the trigonometric functions, r is the distance from the origin to the point (x, y), so r 7 0. If we choose a point (x, y) in quadrant I, then both x and y will be positive, so the values of all six functions will be positive in quadrant I. A point (x, y) in quadrant II has x 6 0 and y 7 0. This makes the values of sine and cosecant positive for quadrant II angles, while the other four functions take on negative values. Similar results can be obtained for the other quadrants, as summarized here. (Compare this to the summary in Section 8.2.)
Signs of Function Values
= 
y
U in Quadrant
sin U
cos U
tan U
cot U
sec U
csc U
I
+
+
+
+
+
+
II
+




+
III


+
+


IV

+


+

x < 0, y > 0, r > 0 II Sine and cosecant positive x < 0, y < 0, r > 0 III Tangent and cotangent positive
x > 0, y > 0, r > 0 I All functions positive 0
x > 0, y < 0, r > 0 IV Cosine and secant positive
x
8.5
y
Functions of Angles and Fundamental Identities
587
EXAMPLE 5
r x
0
y x
Solution (a) Since sin u 7 0 in quadrants I and II and tan u 6 0 in quadrants II and IV, both conditions are met only in quadrant II.
y
r x
0
Identifying the Quadrant of an Angle Identify the quadrant (or possible quadrants) of an angle u that satisfies the given conditions. (a) sin u 7 0, tan u 6 0 (b) cos u 6 0, sec u 6 0
y x
(b) The cosine and secant functions are both negative in quadrants II and III, so in this
case u could be in either of these two quadrants.
y
r
y x
x
0
y
r
y x
x FIGURE 86
ⴚ1 ◊ cos U ◊ 1.
Similarly,
0
In Sections 8.3 and 8.4, we saw how the ranges of the circular functions were determined using the unit circle and graphs. Those ranges can also be justified by using anglebased definitions. FIGURE 86 shows an angle u as it increases in measure from near 0° toward 90°. In each case, the value of r is the same. As the measure of the angle increases, y increases, but never exceeds r, so y … r. Dividing each side by y the positive number r gives r … 1. In a similar way, angles in quadrant IV suggest that y  1 … r , so y 1 … … 1 r y ⴚ1 ◊ sin U ◊ 1. r = sin u for any angle u. y
The tangent of an angle is defined as x. It is possible that x 6 y, x = y, or x 7 y. y For this reason, x can take any value, so tan U can be any real number, as can cot U. The functions sec u and csc u are reciprocals of the functions cos u and sin u, respectively, making sec U ◊ ⴚ1 or sec U » 1
and
csc U ◊ ⴚ1 or csc U » 1.
In summary, the ranges of the trigonometric functions are as follows.
Ranges of Trigonometric Functions Trigonometric Function of U
Range (SetBuilder Notation)
Range (Interval Notation)
sin U, cos U tan U, cot U sec U, csc U
5y   y  … 16 5y  y is a real number6 5y   y  Ú 16
3 1, 14 ( q, q) ( q,  14 ´ 31, q)
EXAMPLE 6
Deciding whether a Value Is in the Range of a Function Decide whether each statement is possible for some angle u, or impossible. (a) sin u = 23 (b) tan u = 110.47 (c) sec u = 0.6 Solution (a) For any value of u,  1 … sin u … 1. Since 23 7 1, it is impossible to find a value of u for which sin u = 23. (b) The tangent function can take on any real number, so tan u = 110.47 is possible. (c) Since sec u …  1 or sec u Ú 1, the statement sec u = 0.6 is impossible.
588
CHAPTER 8
The Unit Circle and the Functions of Trigonometry
Pythagorean Identities We derive three new identities from the relationship x 2 + y 2 = r 2. y2 x2 r2 + = r2 r2 r2 y 2 x 2 a b + a b = 1 r r 2 (cos u) + (sin u) 2 = 1
Divide by r 2. Power rule for exponents cos u = xr , sin u =
sin2 U ⴙ cos2 U ⴝ 1
or
y r
Rewrite.
Starting again with x 2 + y 2 = r 2 and dividing through by x 2 gives y2 r2 x2 + = 2 2 2 x x x 2 y r 2 1 + a b = a b x x 2 1 + (tan u) = (sec u) 2
Divide by x 2. Power rule for exponents y
tan u = x , sec u =
1 ⴙ tan2 U ⴝ sec 2 U.
or
r x
Rewrite.
On the other hand, dividing through by
y2
(see Exercise 111) leads to
1 ⴙ cot 2 U ⴝ csc 2 U. These three identities are called the Pythagorean identities, since the original equation that led to them, x 2 + y 2 = r 2, comes from the Pythagorean theorem. (In the identities, we assume that all functions are defined for u.)
Pythagorean Identities sin2 U ⴙ cos2 U ⴝ 1
1 ⴙ tan2 U ⴝ sec 2 U
1 ⴙ cot 2 U ⴝ csc 2 U
As before, we have given only one form of each identity. However, algebraic transformations produce equivalent identities. For example, by subtracting sin2 u from each side of sin2 u + cos2 u = 1, we give the equivalent identity cos2 u = 1  sin2 u.
Alternative form
You should be able to transform these identities and recognize equivalent forms.
Quotient Identities Recall that sin u =
y r
and cos u = xr. Consider the quotient of sin u and cos u.
sin u = cos u
y r x r
=
y y x , = r r r
#
y r = = tan u, x x
cos u Z 0
u Similarly, cos sin u = cot u, for sin u Z 0. Thus, we have the quotient identities.
Quotient Identities sin U ⴝ tan U cos U
cos U ⴝ cot U sin U
8.5
Looking Ahead to Calculus
The reciprocal, Pythagorean, and quotient identities are used repeatedly in calculus to find limits, derivatives, and integrals of trigonometric functions. These identities are also used to rewrite expressions in a form that permits simplifying a square root. For example, if a Ú 0 and x = a sin u, =

sin2
u)
= 2a 2 cos2 u = a  cos u .
589
EXAMPLE 7
Finding Function Values, Given One Value and the Quadrant Find cos u and sin u if tan u = 43 and u is in quadrant III. Solution Since u is in quadrant III, sin u and cos u will both be negative. It is temptsin u 4 ing to say that since tan u = cos u and tan u = 3 , then sin u =  4 and cos u =  3. This is incorrect—both sin U and cos U must be in the interval 3ⴚ1, 14. We use the Pythagorean identity 1 + tan2 u = sec2 u to find sec u and then the reciprocal identity cos u = sec1 u to find cos u. 1 + tan2 u 4 2 1 + a b 3 16 1 + 9 25 Be careful to 9 choose the 5 correct sign 3 here. 3 5
2a 2  x 2 = 2a 2  a 2 sin2 u 2a 2(1
Functions of Angles and Fundamental Identities
= sec2 u
Pythagorean identity
= sec2 u
tan u =
= sec2 u
Apply the exponent.
= sec2 u
1 = 99 ; 99 +
= sec u
Choose the negative square root, since sec u is negative when u is in quadrant III.
= cos u
Secant and cosine are reciprocals.
4 3
16 9
=
25 9
Since sin2 u = 1  cos2 u, 3 2 sin2 u = 1  a  b 5 9 sin2 u = 1 25 16 sin2 u = 25 4 sin u =  . 5
y
x = –3 y = –4 r=5
3
0
–3
x 3
5 (–3, –4)
–4
FIGURE 87
cos u =  35 Apply the exponent. 1 =
25 25 25 ; 25

9 25
=
16 25
Choose the negative square root.
NOTE Example 7 can also be worked by drawing one possibility for u in standard position in quadrant III, finding r to be 5, and then using the definitions of sin u and cos u in terms of x, y, and r. See FIGURE 87. Be sure to use the correct signs for x and y. EXAMPLE 8
Using Identities to Find a Trigonometric Expression If u is in quadrant IV, find an expression for sec u in terms of sin u. Solution
sin2 u + cos2 u = 1 cos2 u = 1  sin2 u cos u = ⫾21  sin2 u 1 sec u = ⫾ 21  sin2 u
Pythagorean identity Subtract sin2 u. Take square roots. sec u =
1 cos u
Angle u is in quadrant IV, so sec u 7 0. Thus, we choose the positive expression. sec u =
1 21  sin2 u
=
21  sin2 u 1  sin2 u
Rationalize the denominator.
590
CHAPTER 8
The Unit Circle and the Functions of Trigonometry
An Application of Trigonometric Functions Grade, or slope, is a measure of steepness and indicates whether a highway is uphill or downhill. A 5% grade indicates that a road is increasing 5 vertical feet for each 100foot increase in horizontal distance. Grade resistance is the gravitational force acting on a vehicle and is given by R = W sin u, where W is the weight of the vehicle and u is the angle associated with the grade. See FIGURE 88. For an uphill grade, u 7 0 and for a downhill grade, u 6 0. (Source: Mannering, F. and W. Kilareski, Principles of Highway Engineering and Traffic Analysis, Second Edition, John Wiley and Sons.)
u0
FIGURE 88
EXAMPLE 9
Calculating Grade Resistance A downhill highway grade is modeled by the line y =  0.06x in quadrant IV. (a) Find the grade of the road. (b) Approximate the grade resistance for a 3000pound car. Interpret the result. y Not to scale 100 x
u –6
r
Solution (a) The slope of the line is  0.06, so when x increases by 100 feet, y decreases by 6 feet. See FIGURE 89. Thus, this road has a grade of  6%. (b) First we must find sin u. From FIGURE 89, we see that the point (100,  6) lies on the terminal side of u. Thus, r = 2x 2 + y 2 = 2100 2 + ( 6) 2 = 210,036,
(100, –6) FIGURE 89
and sin u =
y 6 = . r 210,036
The grade resistance is R = W sin u = 3000 a
6 210,036
b L  179.7 pounds.
On this stretch of highway, gravity would pull a 3000pound vehicle downhill with a force of about 180 pounds. Note that a downhill grade results in a negative grade resistance.
8.5
8.5
Functions of Angles and Fundamental Identities
591
EXERCISES
Checking Analytic Skills Sketch an angle u in standard position such that u has the least possible positive measure, and the given point is on the terminal side of u. Find the values of the six trigonometric functions for each angle. Rationalize denominators when applicable. Do not use a calculator. 1. (5,  12)
2. ( 12,  5)
3. ( 3, 4)
4. ( 4,  3)
5. ( 8, 15)
6. (15,  8)
7. (7,  24)
8. ( 24,  7)
9. (0, 2)
10. (0, 5)
11. ( 4, 0)
12. ( 5, 0)
13. (0,  4)
14. (0,  3)
15. (1, 23)
16. ( 1, 23)
17. (22, 22)
18. ( 22,  22)
19. ( 2 23,  2)
20. ( 223, 2)
In Exercises 21–30, an equation of the terminal side of an angle u in standard position is given with a restriction on x. Sketch the least positive angle u, and find the values of the six trigonometric functions of u. 21. 2x + y = 0, x Ú 0
22. 3x + 5y = 0, x Ú 0
23.  6x  y = 0, x … 0
24.  5x  3y = 0, x … 0
25.  4x + 7y = 0, x … 0
26. 6x  5y = 0, x Ú 0
27. x + y = 0, x Ú 0
28. x  y = 0, x Ú 0
29.  23x + y = 0, x … 0
30. 23x + y = 0, x … 0
To work Exercises 31–48, begin by reproducing the graph in FIGURE 84 on page 585. Keep in mind that for each of the four points labeled in the figure, r = 1. For each quadrantal angle, identify the appropriate values of x, y, and r to find the indicated function value. If it is undefined, say so. Check your answers with a calculator in degree mode. 31. cos 90°
32. sin 90°
33. tan 180°
34. cot 90°
35. sec 180°
36. csc 270°
37. sin( 270°)
38. cos( 90°)
39. cot 540°
40. tan 450°
41. csc( 450°)
42. sec( 540°)
43. sin 1800°
44. cos 1800°
45. csc 1800°
46. cot 1800°
47. sec 1800°
48. tan 1800°
Use the appropriate reciprocal identity to find each function value. Rationalize denominators when applicable. 49. sec u, given that cos u =
2 3
50. sec u, given that cos u =
5 8
51. csc u, given that sin u =  37
8 52. csc u, given that sin u =  43
53. cot u, given that tan u = 5
54. cot u, given that tan u = 18
55. cos u, given that sec u =  52
56. cos u, given that sec u =  11 7
57. sin u, given that csc u = 22
59. tan u, given that cot u =  2.5
60. tan u, given that cot u =  0.01
58. sin u, given that csc u =
226 3
592
CHAPTER 8
The Unit Circle and the Functions of Trigonometry
Concept Check If n is an integer, n # 180° represents an integer multiple of 180°, (2n + 1) # 90° represents an odd integer multiple of 90°, and so on. Decide whether each expression is equal to 0, 1, or  1 or is undefined. 61. cos[(2n + 1) # 90°]
62. cot(n
#
180°)
63. cos[(2n + 1)
#
180°]
64. cos(n
#
360°)
Identify the quadrant (or possible quadrants) of an angle u that satisfies the given conditions. 65. sin u 7 0, csc u 7 0
66. cos u 7 0, sec u 7 0
67. cos u 7 0, sin u 7 0
68. sin u 7 0, tan u 7 0
69. tan u 6 0, cos u 6 0
70. cos u 6 0, sin u 6 0
71. sec u 7 0, csc u 7 0
72. csc u 7 0, cot u 7 0
73. sec u 6 0, csc u 6 0
74. cot u 6 0, sec u 6 0
75. sin u 6 0, csc u 6 0
76. tan u 6 0, cot u 6 0
9
77. Explain why the answers to Exercises 67 and 71 are the same.
9
78. Explain why there is no angle u that satisfies tan u 7 0, cot u 6 0. Decide whether each statement is possible for some angle u, or impossible. 79. sin u = 2
80. sin u = 3
81. cos u =  0.96
82. cos u =  0.56
83. tan u = 0.93
84. cot u = 0.93
85. sec u =  0.3
86. sec u =  0.9
87. csc u = 100
88. csc u =  100 1 91. sin u = , csc u = 2 2
89. cot u =  4
90. cot u =  6
92. tan u = 2, cot u =  2
93. cos u =  2, sec u =
94. Concept Check
1 2
Is there an angle u for which tan u and cot u are both undefined?
Find all trigonometric function values for each angle u. 95. tan u =  15 8 , given that u is in quadrant II
96. cos u =  35, given that u is in quadrant III
97. sin u =
25 , 7
given that u is in quadrant I
99. cot u =
23 , 8
given that u is in quadrant I
100. csc u = 2, given that u is in quadrant II
101. sin u =
22 , 6
given that cos u 6 0
102. cos u =
103. sec u =  4, given that sin u 7 0
98. tan u = 23, given that u is in quadrant III
25 , 8
given that tan u 6 0
104. csc u =  3, given that cos u 7 0
Use fundamental identities to find each expression. 105. Write cos u in terms of sin u if u is acute.
106. Write sec u in terms of cos u.
107. Write sin u in terms of cot u if u is in quadrant III.
108. Write tan u in terms of cos u if u is in quadrant IV.
109. Write tan u in terms of sin u if u is in quadrant I or IV.
110. Write sin u in terms of sec u if u is in quadrant I or II.
8.6
Work each problem.
Evaluating Trigonometric Functions
117. (Modeling) Distance of a Rotating Beacon A rotating beacon is located at point A next to a long wall. See the figure. The beacon is 4 meters from the wall. The distance d is given by
111. Derive the identity 1 + cot 2 u = csc2 u by dividing x 2 + y 2 = r 2 by y 2.
d = 4 tan 2pt,
112. Using a method similar to the one given in this section sin u cos u showing that cos u = tan u, show that sin u = cot u. True or false? For all angles u,
113. Concept Check
593
sin u + cos u = 1.
where t is time measured in seconds since the beacon started rotating. (When t = 0, the beacon is aimed at point R. When the beacon is aimed to the right of R, the value of d is positive; d is negative if the beacon is aimed to the left of R.) d
If false, give an example showing why. R u 114. Concept Check True or false? Since cot u = cos sin u , 1 if cot u = 2 with u in quadrant I, then cos u = 1 and sin u = 2. If false, explain why.
115. Highway Grade Suppose the uphill grade of a highway can be modeled by the line y = 0.03x in quadrant I. (a) Find the grade of the hill. (b) Approximate the grade resistance for a gravel truck weighing 25,000 pounds. 116. Highway Grade Suppose the downhill grade of a highway can be modeled by the line y =  0.09x in quadrant IV. (a) Find the grade of the hill. (b) Approximate the grade resistance for a car weighing 3800 pounds.
8.6
4m
a
A
Find d to the nearest tenth for each time. (a) t = 0 (b) t = 0.2 (c) t = 0.8 (d) t = 1.2 9 (e) Why is 0.25 a meaningless value for t ? 118. (Modeling) Distance of a Rotating Beacon In the figure for Exercise 117, the distance a is given by a = 4  sec 2pt . Find a to the nearest tenth for each time. (a) t = 0 (b) t = 0.86 (c) t = 1.24
Evaluating Trigonometric Functions
Definitions of the Trigonometric Functions • Trigonometric Function Values of Special Angles • Cofunction Identities • Reference Angles • Special Angles as Reference Angles • Finding Function Values with a Calculator • Finding Angle Measures
Definitions of the Trigonometric Functions
y
B
r
A
In Section 8.5, we used angles in standard position to define the trigonometric functions. There is another way to approach them:
(x, y)
as ratios of the lengths of the sides of right triangles.
y
x
C
FIGURE 90
x
FIGURE 90 shows an acute angle A in standard position. The definitions of the trigonometric function values of angle A require x, y, and r. In FIGURE 90, x and y are the lengths of the two legs of the right triangle ABC; r is the length of the hypotenuse.
594
CHAPTER 8
The Unit Circle and the Functions of Trigonometry
The side of length y in FIGURE 90 is called the side opposite angle A, and the side of length x is called the side adjacent to angle A. We use the lengths of these sides to replace x and y in the definitions of the trigonometric functions, and the length of the hypotenuse to replace r, to get the following righttrianglebased definitions.
y
B
(x, y)
r
y
RightTriangleBased Definitions of Trigonometric Functions A
x
C
x
side opposite y ⴝ r hypotenuse side adjacent x cos A ⴝ ⴝ r hypotenuse side opposite y tan A ⴝ ⴝ x side adjacent
FIGURE 90 (repeated)
sin A ⴝ
A
24
Finding Trigonometric Function Values of an Acute Angle Find the values of sin A, cos A, and tan A in the right triangle in FIGURE 91. Solution The length of the side opposite angle A is 7, the length of the side adjacent to angle A is 24, and the length of the hypotenuse is 25. Thus,
B
7
hypotenuse r ⴝ y side opposite hypotenuse r sec A ⴝ ⴝ x side adjacent side adjacent x cot A ⴝ ⴝ . y side opposite
csc A ⴝ
EXAMPLE 1
25
C
For any acute angle A in standard position,
FIGURE 91
sin A =
side opposite 7 = hypotenuse 25
cos A =
side adjacent 24 = hypotenuse 25
tan A =
side opposite 7 = . side adjacent 24
NOTE Because the cosecant, secant, and cotangent ratios are the reciprocals of the sine, cosine, and tangent values, csc A =
25 7,
sec A =
25 24 ,
and cot A =
24 7
in Example 1.
Trigonometric Function Values of Special Angles 60° 2
2
60°
60° 2 Equilateral triangle (a)
30°
30°
2
2 x
x
60°
90° 90° 60° 1 1 30°– 60° Right triangles
Previously, we found exact function values for the real numbers p4 , p6 , and p3 . These real numbers correspond to radianmeasured angles, which in turn measure 45°, 30°, and 60°. The same values can be found by using angles in triangles. We start with an equilateral triangle—a triangle with all sides of equal length. Each angle of such a triangle measures 60°. While the results we will obtain are independent of the lengths of its sides, for convenience we choose the length of each side to be 2 units. See FIGURE 92(a). Bisecting one angle of this equilateral triangle leads to two right triangles, each of which has angles of 30°, 60°, and 90°, as shown in FIGURE 92(b). Since the hypotenuse of one of these right triangles has length 2, the shortest side will have length 1. (Why?) If x represents the length of the medium side, then 2 2 = 12 + x 2 4 = 1 + x2 3 = x2 23 = x.
Pythagorean theorem Apply the exponents. Subtract 1. Square root property; choose the positive root.
FIGURE 93 on the next page summarizes our results, using a 30°– 60° right triangle. In the figure, the side opposite the 30° angle has length 1; that is, for the 30° angle,
(b) FIGURE 92
hypotenuse = 2, side opposite = 1, side adjacent = 23.
8.6
595
Evaluating Trigonometric Functions
Now we use the definitions of the trigonometric functions. 30° √3
sin 30° ⴝ
side opposite 1 ⴝ hypotenuse 2
csc 30° ⴝ
cos 30° ⴝ
side adjacent 23 ⴝ hypotenuse 2
sec 30° ⴝ
tan 30° ⴝ
side opposite 1 23 ⴝ ⴝ side adjacent 3 23
cot 30° ⴝ
2
60° 1 30°–60° Right triangle FIGURE 93
2 = 2 1 2 23
ⴝ
2 23 3
23 ⴝ 23 1
EXAMPLE 2
Finding Trigonometric Function Values for 60° Find the six trigonometric function values for a 60° angle. Solution Refer to FIGURE 93 to find the following ratios. 23 2 223 csc 60° ⴝ 3 sin 60° ⴝ
1 2
tan 60° ⴝ 23
sec 60° ⴝ 2
23 3
cot 60° ⴝ
We find the values of the trigonometric functions for 45° by starting with a 45°–45° right triangle, as shown in FIGURE 94. This triangle is isosceles, and for convenience, we choose the lengths of the equal sides to be 1 unit. (As before, the results are independent of the length of the equal sides of the right triangle.) Since the shorter sides each have length 1, if r represents the length of the hypotenuse, then we have the following.
45° 1
cos 60° ⴝ
r = √2
12 + 12 = r 2 2 = r2
45° 1 45°– 45° Right triangle
22 = r
FIGURE 94
Pythagorean theorem Add. Choose the positive root.
Now we use the measures indicated on the 45°–45° right triangle in FIGURE 94. sin 45° ⴝ csc 45° ⴝ
GCM
Using the TI83/84 Plus or a scientific calculator to find the decimal value for cos 30° gives a 23 . 2
The
decimal is an approximation, while 23 2
cos 45° ⴝ
22 ⴝ 22 1
sec 45° ⴝ
22
ⴝ
22 2
tan 45° ⴝ
1 ⴝ1 1
22 ⴝ 22 1
cot 45° ⴝ
1 ⴝ1 1
1 22
ⴝ
Function values for 30° or p6 , 45° or p4 , and 60° or that follows.
TECHNOLOGY NOTE
decimal approximation for
22 2
1
is exact. In general, unless a
trigonometric function value is a rational number, a basic graphing calculator will provide only a decimal approximation for the irrational function value. (More advanced models, such as the TI89, do give exact values.)
p 3
are summarized in the table
Function Values for Special Angles U
sin U
cos U
tan U
cot U
sec U
csc U
30°, or P6
1 2
23 2
23 3
23
223 3
2
45°, or P4
22 2
22 2
1
1
22
22
60°, or P3
23 2
1 2
23
23 3
2
2 23 3
Trigonometry students should memorize the function values for the first two columns of this table. Identities will yield the remaining values.
596
CHAPTER 8
The Unit Circle and the Functions of Trigonometry
Cofunction Identities B a C
c
b
A
In a right triangle ABC with right angle C, the acute angles A and B are complementary. See FIGURE 95. The length of the side opposite angle A is a, and the length of the side opposite angle B is b. The length of the hypotenuse is c. In this triangle, sin A = ac. Since cos B is also equal to ac,
FIGURE 95
sin A ⴝ tan A ⴝ
Similarly,
a ⴝ cos B. c
a ⴝ cot B and b
sec A ⴝ
c ⴝ csc B. b
Since angles A and B are complementary and sin A = cos B, the functions sine and cosine are called cofunctions. Also, tangent and cotangent are cofunctions, as are secant and cosecant. Since angles A and B are complementary, A + B = 90°, or B = 90°  A, giving sin A = cos B = cos(90°  A). Similar results, called the cofunction identities, are true for the other trigonometric functions.
Cofunction Identities If A is an acute angle measured in degrees, then the following are true. sin A ⴝ cos(90° ⴚ A) cos A ⴝ sin(90° ⴚ A)
csc A ⴝ sec(90° ⴚ A) sec A ⴝ csc(90° ⴚ A)
tan A ⴝ cot(90° ⴚ A) cot A ⴝ tan(90° ⴚ A)
If A is an acute angle measured in radians, then the following are true. sin A ⴝ cos a
P ⴚ Ab 2 P cos A ⴝ sina ⴚ Ab 2 P tan A ⴝ cota ⴚ A b 2
P ⴚ Ab 2 P sec A ⴝ csca ⴚ A b 2 P cot A ⴝ tan a ⴚ A b 2 csc A ⴝ seca
NOTE These identities actually apply to all angles for which the functions are defined (not just acute angles). However, for our present discussion, we will need them only for acute angles.
EXAMPLE 3
Writing Functions in Terms of Cofunctions Write each expression in terms of its cofunction. p (a) cos 52° 16¿ (b) tan (c) sec 1 6 Solution (a) Since cos A = sin(90°  A), cos 52° 16¿ = sin(90°  52° 16¿) = sin 37° 44¿.
(b) Since tan A = cot A p2  A B , tan (c) sec 1 = csc A p2  1 B
p p p p = cota  b = cot . 6 2 6 3
8.6
Evaluating Trigonometric Functions
597
Reference Angles Associated with every nonquadrantal angle in standard position is a positive acute angle called its reference angle. A reference angle for an angle u, written U œ , is the positive acute angle made by the terminal side of angle u and the xaxis. The chart below shows several angles u (each less than one complete counterclockwise revolution) in quadrants II, III, and IV, with the reference angle u¿ also shown. In quadrant I, if u is acute, then u and u¿ are the same. If an angle u is negative or has measure greater than 360°, its reference angle is found by first finding its coterminal angle that is between 0° and 360° and then using the diagrams shown here. Reference Angle U œ for U, where 0°0
0
x
x
The polar coordinate system is based on a point, called the pole, and a ray, called the polar axis. The polar axis is usually drawn in the direction of the positive xaxis. See FIGURE 52. In FIGURE 53, the pole has been placed at the origin of a rectangular coordinate system, so that the polar axis coincides with the positive xaxis. Point P has rectangular coordinates (x, y). Point P can also be located by giving the directed angle u from the positive xaxis to ray OP and the directed distance r from the pole to point P. The ordered pair (r, u) gives polar coordinates of point P. If r 7 0, point P lies on the terminal side of u and if r 6 0, point P lies on the ray pointing in the opposite direction of the terminal side of u, a distance  r  from the origin. See FIGURE 54, which shows rectangular axes superimposed on a polar coordinate grid.
FIGURE 53 y
Relationships between Rectangular and Polar Coordinates
(r, )
O
(–r, )
r>0 x
If a point has rectangular coordinates (x, y) and polar coordinates (r, u), then these coordinates are related as follows. x ⴝ r cos U
y ⴝ r sin U
r 2 ⴝ x2 ⴙ y2
y tan U ⴝ , if x Z 0 x
FIGURE 54
742
CHAPTER 10
Applications of Trigonometry and Vectors
EXAMPLE 1
Plotting Points with Polar Coordinates Plot each point by hand in the polar coordinate system. Then determine the rectangular coordinates of each point. (a) P(2, 30°) (b) Q A  4, 2p (c) R A 5,  p4 B 3 B Solution (a) Since r = 2 and u = 30°, P is 2 units from the origin in the positive direction on a ray making a 30° angle with the polar axis. See FIGURE 55, which shows a polar grid superimposed onto a rectangular grid. Using the conversion equations, we can find the rectangular coordinates.
y
P(2, 30°) 1
30° x
0
1
x = r cos u
y = r sin u
x = 2 cos 30°
y = 2 sin 30°
x = 2a
23 b = 23 2
The rectangular coordinates are A 23, 1 B .
FIGURE 55
1 y = 2a b = 1 2
(b) Since r is negative, Q is 4 units in the opposite direction from the pole on an extension of the 2p 3 ray. See FIGURE 56. The rectangular coordinates are x =  4 cos
2p 2p 1 23 =  4 a  b = 2 and y =  4 sin = 4a b =  223. 3 2 3 2 y
y
2 3
Looking Ahead to Calculus
Techniques studied in calculus and associated with derivatives and integrals provide methods of finding slopes of tangent lines to polar curves, areas bounded by such curves, and lengths of their arcs.
1 –1
1
x
0
–1
(
Q –4,
2 3
0
)
– 4
x
R 5, – 4
(
)
FIGURE 57
FIGURE 56
(c) Point R is shown in FIGURE 57. Since u is negative, the angle is measured in the clockwise direction. Furthermore, we have x = 5 cos a 
p 5 22 b = 4 2
and
y = 5 sin a 
p 5 22 b = . 4 2
While a given point in the plane can have only one pair of rectangular coordinates, this same point can have infinitely many pairs of polar coordinates. For example, (2, 30°) locates the same point as (2, 390°), or (2,  330°), or (  2, 210°). y
P
EXAMPLE 2
140°
Giving Alternative Forms for Coordinates of a Point
(a) Give three other pairs of polar coordinates for the point P(3, 140°). 0
–220° 320°
FIGURE 58
x
– 40°
(b) Determine two pairs of polar coordinates, with r 7 0, for the point with rectangular coordinates ( 1, 1). Solution (a) Three such pairs are (3,  220°), ( 3, 320°), and ( 3,  40°). See FIGURE 58.
10.6
y
Polar Equations and Graphs
743
(b) As shown in FIGURE 59, the point ( 1, 1) lies in the second quadrant. Since tan u = 11 =  1, one possible value for u is 135°. Also, r = 2x 2 + y 2 = 2( 1) 2 + 12 = 22.
(–1, 1) x
0
Therefore, two pairs of polar coordinates are A 22, 135° B and A 22,  225° B . (Any angle coterminal with 135° could have been used for the second angle.)
Graphs of Polar Equations
FIGURE 59
Equations in x and y are called rectangular (or Cartesian) equations. An equation in which r and u are the variables is a polar equation. r = 3 sin u,
r = 2 + cos u,
r = u
Polar equations
The rectangular forms of lines and circles can also be defined in terms of polar coordinates. We typically solve a polar equation for r. ax + by = c Line: a(r cos u) + b(r sin u) = c r(a cos u + b sin u) = c This is the polar equation of ax + by = c.
Circle:
rⴝ
Rectangular equation of a line Convert to polar coordinates. Factor out r.
c a cos U ⴙ b sin U
x2 + y2 = a2
A 2x 2 +
These are the polar equations of x 2 + y 2 = a 2.
Polar equation of a line
Rectangular equation of a circle
B = a2
2 A 2k B = k
2 y2
r2 = a2 r ⴝ ⴞa
2x 2 + y 2 = r Polar equation of a circle; r can be negative in polar coordinates.
y
EXAMPLE 3
3
x
0
–3
3
Finding Polar Equations of Lines and Circles For each rectangular equation, give its equivalent polar equation and sketch its graph. (a) y = x  3 (b) x 2 + y 2 = 4 Solution (a) This is the equation of a line. Rewrite y = x  3 in standard form.
–3
x  y = 3 r cos u  r sin u = 3 r(cos u  sin u) = 3
y = x – 3 (rectangular) 3 r= (polar) cos – sin FIGURE 60
r =
y
Standard form: a x + by = c Substitute for x and y. Factor out r.
3 cos u  sin u
Divide by cos u  sin u.
Its graph is shown in FIGURE 60.
2
(b) The graph of x 2 + y 2 = 4 is a circle with center at the origin and radius 2. –2
0
x 2
–2
x2
y2
+ = 4 (rectangular) r = 2 or r = –2 (polar) FIGURE 61
x2 + y2 = 4
A 2x 2 + y 2 B 2 = (⫾2) 2 2x 2 + y 2 = 2 r = 2
or or
In polar coordinates, we may have r 6 0.
2x 2 + y 2 =  2 r = 2
The graphs of r = 2 and r =  2 coincide. See FIGURE 61.
744
CHAPTER 10
Applications of Trigonometry and Vectors
GCM
EXAMPLE 4
Graphing a Polar Equation (Cardioid) Graph r = 1 + cos u.
Analytic Solution We first find some ordered pairs, as in the table.* Once the pattern of values of r becomes clear, it is not necessary to find more ordered pairs. U
cos U
r ⴝ 1 ⴙ cos U
U
cos U
0° 30° 60° 90° 105° 135°
1 0.9 0.5 0  0.3  0.7
2 1.9 1.5 1 0.7 0.3
180° 225° 255° 270° 300° 330°
1  0.7  0.3 0 0.5 0.9
Graphing Calculator Solution We choose degree mode and graph values of u in the interval 30°, 360°4. The settings in FIGURE 63 and 64 will generate the graph in FIGURE 65.
r ⴝ 1 ⴙ cos U 0 0.3 0.7 1 1.5 1.9
FIGURE 63
Connect the points in the order in the order they appear in the table. See FIGURE 62. This curve is called a cardioid because of its heart shape.
r 1 + cos
This is a continuation of the previous screen.
900
FIGURE 64
1800
1
2
1.55
00 –2.35
2.35
2700 –1.55 FIGURE 65
FIGURE 62
GCM
90° r = 3 cos 2 7
Solution Because of the argument 2u, the graph requires a larger number of points than when the argument is u. A few ordered pairs are given in the table. Complete the table similarly through the first 180°, so that 2u has values up to 360°.
6
4
EXAMPLE 5 Graphing a Polar Equation (Rose) Graph r = 3 cos 2u.
1
Start 3 0° Finish
180° 5
U
0°
15°
30°
45°
60°
75°
90°
2U
0°
30°
60°
90°
120°
150°
180°
cos 2U
1
0.9
0.5
0
 0.5
 0.9
1
r ⴝ 3 cos 2U
3
2.6
1.5
0
 1.5
 2.6
3
8 2
3
270° FIGURE 66
*The tables in Examples 4–7 include approximations for cosine values and r.
10.6
Plotting these points gives the graph, called a fourleaved rose and is shown in FIGURE 66 on the preceding page. Notice how the graph is developed with a continuous curve, beginning with the upper half of the right horizontal leaf and ending with the lower half of that leaf. As the graph is traced, the curve goes through the pole four times.
3.1
–4.7
4.7
FOR DISCUSSION
–3.1 To graph the equation r = 3 cos 2 of Example 5 on a calculator, use polar graphing mode, degree mode, min = 0º, max = 360º, step = 15º, and window settings as shown.
Trace the rose curve of Example 5 and watch the cursor to verify that the numbering that appears in FIGURE 66 is indeed correct. The equation r = 3 cos 2u in Example 5 has a graph that belongs to a family of curves called roses. The graphs of r ⴝ a sin nU and r ⴝ a cos nU are roses, with n leaves if n is odd and 2n leaves if n is even. The value of a determines the length of the leaves.
EXAMPLE 6
GCM
Graph
r2
Graphing a Polar Equation (Lemniscate)
= cos 2u.
Analytic Solution Complete a table of ordered pairs, and sketch the graph, as in FIGURE 67. The point ( 1, 0°), with r negative, may be plotted as (1, 180°). Also, ( 0.7, 30°) may be plotted as (0.7, 210°), and so on. U
0°
30°
45°
135°
150°
180°
2U
0°
60°
90°
270°
300°
360°
cos 2U
1
0.5
0
0
0.5
1
⫾1
⫾0.7
0
0
⫾0.7
⫾1
r ⴝ ⴞ2cos 2U
745
Polar Equations and Graphs
Values of u for 45° 6 u 6 135° are not included in the table, because the corresponding values of cos 2u are negative (quadrants II and III) and so do not have real square roots. Values of u larger than 180° give 2u larger than 360° and would repeat the points already found. This curve is called a lemniscate.
GCM
Graphing Calculator Solution To graph r 2 = cos 2u with a graphing calculator, let r1 = 2cos 2u r2 =  2cos 2u.
and See
for the settings and for the graph.
FIGURE 68
FIGURE 69
90° FIGURE 68
rr22 == cos cos22 180°
1 1
0° –1.5
270° FIGURE 67
1.5
–1 FIGURE 69
EXAMPLE 7
Graphing a Polar Equation (Spiral of Archimedes) Graph r = 2u (u measured in radians). Solution Some ordered pairs are shown in the table on the next page. Since r = 2u, rather than a trigonometric function of u, we must also consider negative values of u. Radian measures have been rounded. The graph, shown in FIGURE 70 on the next page, is called a spiral of Archimedes. (continued)
746
CHAPTER 10
Applications of Trigonometry and Vectors 2
10
–10
U (radians)
r ⴝ 2U
U (radians)
r ⴝ 2U
p  p2  p4 0
 6.3  3.1  1.6 0 1
p 3 p 2
2.1 3.1 6.3 9.4 12.6
10
–10 –2 ⱕ ⱕ 2 More of the spiral can be seen in this calculator graph of the spiral in Example 7 than is shown in FIGURE 70.
p 6
p 3p 2
2p
r = 2
4
8
0
3 2 FIGURE 70
FOR DISCUSSION
Refer to the calculator graph of r = 2u in the margin. Experiment with various minimum and maximum values of u. What behaviors do you observe? (As you experiment, you will need to enlarge the window to accommodate the increased graph size.)
Classifying Polar Equations The table summarizes common polar graphs and forms of their equations. We also include limaçons. Cardioids are a special case of limaçons, where  ab  = 1. Circles and Lemniscates Circles
r ⴝ a cos U
Lemniscates
r 2 ⴝ a2 sin 2U
r ⴝ a sin U
r 2 ⴝ a2 cos 2U
Limaçons r ⴝ a ⴞ b sin U or r ⴝ a ⴞ b cos U
a b
6 1
a b
= 1
1 6
a b
a b
6 2
Ú 2
Rose Curves 2n leaves if n is even, n Ú 2 n = 2 n = 4
r ⴝ a sin nU
r ⴝ a cos nU
n leaves if n is odd n = 3
n = 5
r ⴝ a cos nU
r ⴝ a sin nU
10.6
Polar Equations and Graphs
747
Converting Equations In Example 3 we converted rectangular equations to polar equations. We conclude with an example that converts a polar equation to a rectangular one. EXAMPLE 8
Converting a Polar Equation to a Rectangular Equation 4 Consider the polar equation r = . 1 + sin u (a) Convert it to a rectangular equation. (b) Use a graphing calculator to graph the polar equation for 0 … u … 2p. 4 r= 1 + sin
(c) Use a graphing calculator to graph the rectangular equation. 10
–10
10
Solution (a) Multiply each side of the equation by the denominator on the right, to clear the fraction. 4 1 + sin u r + r sin u = 4 r =
–10 (a)
y = 2 – 18 x 2
2x 2
10
–10 (b) FIGURE 71
10.6
+ y = 4
2x 2 + y 2 = x2 + y2 = x2 + y2 = x2 = x2 =
10
–10
+
y2
(b)
FIGURE 71(a)
Polar equation Multiply by 1 + sin u. Let r = 2x 2 + y 2 and y = r sin u.
4  y (4  y) 2 16  8y + y 2  8y + 16  8( y  2)
Subtract y. Square each side. Expand the right side. Subtract y 2. Rectangular equation (a parabola)
shows a graph with polar coordinates.
(c) Solving x 2 =  8(y  2) for y, we obtain y = 2  18 x 2. Its graph is shown in FIGURE 71(b). Notice that the two graphs are the same parabola.
EXERCISES
1. Concept Check For each point given in polar coordinates, state the quadrant in which the point lies if it is graphed in a rectangular coordinate system. (a) (5, 135°) (b) (2, 60°) (c) (6,  30°) (d) (4.6, 213°)
2. Concept Check For each point given in polar coordinates, state the axis on which the point lies if it is graphed in a rectangular coordinate system. Also, state whether it is on the positive portion or the negative portion of the axis. (For example, (5, 0°) lies on the positive xaxis.) (a) (7, 360°) (b) (4, 180°) (c) (2,  90°) (d) (8, 450°)
Checking Analytic Skills Plot each point, given its polar coordinates. Give two other pairs of polar coordinates for each point. Do not use a calculator. 3. (1, 45°)
4. (3, 120°)
5. ( 2, 135°)
6. ( 4, 27°)
7. (5,  60°)
8. (2,  45°)
9. ( 3,  210°) 12. (4, 270°)
10. ( 1,  120°)
11. (3, 300°)
13. (0, 0°)
14. (0, 360°)
748
CHAPTER 10
Applications of Trigonometry and Vectors
Checking Analytic Skills Plot the point whose rectangular coordinates are given. Then determine two pairs of polar coordinates for the point with 0° … u 6 360°. Do not use a calculator. 15. ( 1, 1)
20.
16. (1, 1)
A  22, 22 B
21. a
17. (0, 3)
23 3 , b 2 2
22. a 
1 23 , b 2 2
A 22, 22 B
18. (0,  3)
19.
23. (3, 0)
24. ( 2, 0)
For each rectangular equation, give its equivalent polar equation and sketch its graph. 25. x  y = 4
26. x + y =  7
27. x 2 + y 2 = 16
28. x 2 + y 2 = 9
29. 2x + y = 5
30. 3x  2y = 6
Concept Check choices A–D.
In Exercises 31–34, match each polar graph to its corresponding equation from
31. r = 3 A.
90°
32. r = cos 3u
33. r = cos 2u
34. r =
B.
C.
D.
90°
90°
2 cos u + sin u 90°
2 180°
1
0°
180°
270°
0°
180°
270°
3
0°
180°
270°
1
0°
270°
Graph each polar equation for u in 30°, 360°). In Exercises 35–44, identify the type of polar graph. 35. r = 2 + 2 cos u
36. r = 8 + 6 cos u
37. r = 3 + cos u
38. r = 2  cos u
39. r = 4 cos 2u
40. r = 3 cos 5u
41. r 2 = 4 cos 2u
42. r 2 = 4 sin 2u
43. r = 4  4 cos u
44. r = 6  3 cos u
45. r = 2 sin u tan u (This is a cissoid.)
46. r =
9 47.
Explain the method you would use to graph (r, u) by hand if r 6 0.
cos 2u cos u (This is a cissoid with a loop.)
9 48.
Explain why, if r 7 0, the points (r, u) and ( r, u + 180°) have the same graph.
Concept Check Answer each question. 49. If a point lies on an axis in the rectangular plane, then what kind of angle must u be if (r, u) represents the point in polar coordinates?
50. What will the graph of r = k be, for k 7 0?
51. How would the graph of FIGURE 62 change if the equation were r = 1  cos u?
52. How would the graph of FIGURE 62 change if the equation were r = 1 + sin u?
53. The graphs of rose curves have equations of the form r = a sin nu or r = a cos nu. What does the value of a determine? What does the value of n determine?
54. In Exercise 53, if n = 1, what will the graph be? What will the value of a determine?
For each equation, find an equivalent equation in rectangular coordinates. Then graph the equation. 55. r = 2 sin u 57. r =
2 1  cos u
56. r = 2 cos u 58. r =
3 1  sin u
10.6
3 4 cos u  sin u
59. r =  2 cos u  2 sin u
60. r =
61. r = 2 sec u
62. r =  5 csc u
63. r =
2 cos u + sin u
64. r =
749
Polar Equations and Graphs
2 2 cos u + sin u
The graph of r = au is an example of the spiral of Archimedes. With your calculator set to radian mode, use the given value of a and interval of u to graph the spiral in the window specified. 65. a = 1, 0 … u … 4p, 3 15, 154 by 3 15, 154
67. a = 1.5,  4p … u … 4p, 3 20, 204 by 3 20, 204
66. a = 2,  4p … u … 4p, 3 30, 304 by 3 30, 304 68. a =  1, 0 … u … 12p, 3 40, 404 by 3 40, 404
Find the polar coordinates of the points of intersection of the given curves for the specified interval of u. 69. r = 4 sin u, r = 1 + 2 sin u; 0 … u 6 2p
70. r = 3, r = 2 + 2 cos u; 0° … u 6 360°
71. r = 2 + sin u, r = 2 + cos u; 0 … u 6 2p
72. r = sin 2u, r = 22 cos u; 0 … u 6 p
(c) Use graphing to determine whether Pluto is always the farthest of these from the sun.
(Modeling) Solve each problem. 73. Satellite Orbits The polar equation r =
9 74.
a(1  e 2) , 1 + e cos u
where a is the average distance in astronomical units from our sun and e is a constant called the eccentricity, can be used to graph the orbits of satellites of the sun. (See Section 6.3.) The sun will be located at the pole. The table lists a and e for the satellites.
Radio Towers and Broadcasting Patterns Many times, radio stations do not broadcast in all directions with the same intensity. To avoid interference with an existing station to the north, a new station may be licensed to broadcast only east and west. To create an east–west signal, two radio towers are sometimes used, as illustrated in the figure. Locations where the radio signal is received correspond to the interior of the curve defined by r 2 = 40,000 cos 2u,
Satellite
a
e
Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune Pluto
0.39 0.78 1.00 1.52 5.20 9.54 19.20 30.10 39.40
0.206 0.007 0.017 0.093 0.048 0.056 0.047 0.009 0.249
Source: Zeilik, M., S. Gregory, and E. Smith, Introductory Astronomy and Astrophysics, Saunders College Publishers.
(a) Graph the orbits of the four closest satellites on the same polar grid. Choose a viewing window that results in a graph with nearly circular orbits. (b) Plot the orbits of Earth, Jupiter, Uranus, and Pluto on the same polar grid. How does Earth’s distance from the sun compare with the distance from the sun to these satellites?
where the polar axis (or positive xaxis) points east.
N
E
(a) Graph r 2 = 40,000 cos 2u for 0° … u … 360°, with units in miles. Assuming that the radio towers are located near the pole, use the graph to describe the regions where the signal can be received and where the signal cannot be received. (b) Suppose a radio signal pattern is given by r 2 = 22,500 sin 2u. Graph this pattern and interpret the results.
750
CHAPTER 10
10.7
Applications of Trigonometry and Vectors
More Parametric Equations
Parametric Graphing Revisited • Parametric Equations with Trigonometric Functions • The Cycloid • Applications of Parametric Equations
Parametric Graphing Revisited In Example 1 we review parametric graphing, first introduced in Section 6.4. EXAMPLE 1
Let x =
t2
Graphing a Plane Curve Defined Parametrically and y = 2t + 3, for t in 3 3, 34. Graph the set of ordered pairs (x, y).
Analytic Solution Make a table of corresponding values of t, x, and y over the domain of t. Then plot the points as shown in FIGURE 72. The graph is a portion of a parabola with horizontal axis y = 3. The arrowheads indicate the direction the curve traces as t increases.
t
x
y
ñ3 ñ2 ñ1 0 1 2 3
9 4 1 0 1 4 9
ñ3 ñ1 1 3 5 7 9
y
Graphing Calculator Solution The parameters for the TI83/84 Plus and the graph are shown in FIGURE 73.
x t2 for t in [3, 3] y 2t + 3
9
(9, 9)
6 3 3
3
6
9
This is a continuation of the previous screen.
x
0
10
(9, ñ3)
3 FIGURE 72
–2
Note that more of the parabola would be traced if the domain of t were larger.
10 –4 FIGURE 73
Parametric Equations with Trigonometric Functions If we use trigonometric functions in parametric equations, many interesting curves can be drawn, as shown in FIGURE 74. 4
–6
4
6
–4
–6
6
–4 FIGURE 74
4
–6
6
–4
10.7
EXAMPLE 2
More Parametric Equations
751
Graphing a Circle with Parametric Equations
Graph x = 2 cos t, y = 2 sin t, for 0 … t … 2p. Find an equivalent equation by using rectangular coordinates. Solution Let X1T = 2 cos(T) and Y1T = 2 sin(T), and graph these parametric equations, as shown in FIGURE 75. To verify that the graph is a circle, consider the following.
2
–3
3
–2
When graphing parametric equations, be sure that your calculator is set in parametric mode. A square window is necessary for the curve in Example 2 to appear circular rather than elliptical.
x = 2 cos t, y = 2 sin t
x 2 + y 2 = 4 cos2 t + 4 sin2 t
Properties of exponents
x2
FIGURE 75
TECHNOLOGY NOTE
x 2 + y 2 = (2 cos t) 2 + (2 sin t) 2 +
y2
=
4(cos2
t +
sin2
t)
Distributive property
x2 + y2 = 4
cos2 t + sin2 t = 1
The parametric equations are equivalent to x 2 + y 2 = 4, which is the equation of a circle with center (0, 0) and radius 2.
EXAMPLE 3 Graphing an Ellipse with Parametric Equations Graph the plane curve defined by x = 2 sin t, y = 3 cos t, for t in 30, 2p4.
Solution x = 2 sin t
y = 3 cos t
x 2 = 4 sin2 t
Square each side.
y 2 = 9 cos2 t
Square each side.
x2 = sin2 t 4
Solve for sin2 t.
y2 = cos2 t 9
Solve for cos2 t.
Now add corresponding sides of the two equations. y2 x2 + = sin2 t + cos2 t 4 9 y2 x2 + = 1 4 9
sin2 t + cos2 t = 1
This is the equation of an ellipse, as shown in FIGURE 76. The ellipse can be graphed directly with a calculator in parametric mode. See FIGURE 77.
x = 2 sin t for y = 3 cos t t in [0, 2] y 3
4
–2
0
x 2
–5
5
–3
x 2 y2 + =1 4 9
–4
FIGURE 76
FIGURE 77
752
Applications of Trigonometry and Vectors
CHAPTER 10
Looking Ahead to Calculus
The Cycloid
The cycloid is a special case of a The path traced by a fixed point on the circumference of a circle rolling along a line curve traced out by a point at a is called a cycloid. A cycloid is defined by given distance from the center of a circle as the circle rolls along a x ⴝ at ⴚ a sin t, y ⴝ a ⴚ a cos t, for t in (ⴚˆ, ˆ). straight line. See FIGURE 78. Such a curve is called a trochoid. Parametrically defined curves are GCM EXAMPLE 4 Graphing a Cycloid studied in calculus.
Graph the cycloid with a = 1 for t in 30, 2p4.
Solution There is no simple way to find a rectangular equation for the cycloid from its parametric equations. Instead, begin with a table of values. (Some are approximations.) Plotting the ordered pairs (x, y) leads to the portion of the graph from 0 to 2p in FIGURE 78.
4
0
t
0
p 4
p 2
p
3p 2
2p
x
0
0.08
0.6
p
5.7
2p
2
y
0
0.3
1
2
1
0
1
2
P(x, y)
0 It is easier to graph the cycloid of Example 4 with a graphing calculator in parametric mode than with traditional methods.
P
y
1 x
2
FIGURE 78
Q
The cycloid has an interesting physical property. If a flexible cord or wire goes through points P and Q as in FIGURE 79, and, due to the force of gravity, a bead is allowed to slide without friction along this path from P to Q, the path that requires the least time takes the shape of an inverted cycloid.
FIGURE 79
Applications of Parametric Equations Parametric equations are used to simulate motion. If a ball is thrown with a velocity of v feet per second at an angle u with the horizontal, its flight can be modeled by the parametric equations
y
v0
θ
v0 sin θ
x ⴝ (v cos U)t and
y ⴝ (v sin U)t ⴚ 16t 2 ⴙ h,
where t is in seconds and h is the ball’s initial height in feet above the ground. The term  16t 2 occurs because gravity is pulling downward. See FIGURE 80. These equations ignore air resistance.
x
v0 cos θ FIGURE 80
GCM
EXAMPLE 5
Simulating Motion with Parametric Equations Three golf balls are hit simultaneously into the air at 132 feet per second (90 mph) at angles of 30°, 50°, and 70° with the horizontal. (a) Assuming that the ground is level, determine graphically which ball travels the farthest. Estimate this distance. (b) Which ball reaches the greatest height? Estimate this height. Solution (a) The three sets of parametric equations determined by the three golf balls are as follows, since h = 0. x1 = (132 cos 30°)t, x2 = (132 cos 50°)t, x3 = (132 cos 70°)t,
y1 = (132 sin 30°)t  16t 2 y2 = (132 sin 50°)t  16t 2 y3 = (132 sin 70°)t  16t 2
10.7
(x3, y3) (x2, y2)
(x1, y1)
400
753
More Parametric Equations
Graphs are shown in FIGURE 81(a), where 0 … t … 9. We used a graphing calculator in simultaneous mode so that we simulate all three balls in flight at the same time. From the graph in FIGURE 81(b), we can see that the ball hit at 50° travels the farthest distance. Using the trace feature, we estimate this distance to be about 540 feet. (Verify this.) (b) The ball hit at 70° reaches the greatest height, about 240 feet. (Verify this.)
0
600 0 (a)
(x3, y3) (x2, y2) (x1, y1) 400
0
EXAMPLE 6
Examining Parametric Equations of Flight A small rocket is launched from a table that is 3.36 feet above the ground. Its initial velocity is 64 feet per second, and it is launched at an angle of 30° with respect to the ground. Find the rectangular equation that models this path. What type of path does the rocket follow? Solution Its path is defined by the parametric equations
600 0
x = (64 cos 30°)t and y = (64 sin 30°)t  16t 2 + 3.36
(b)
or, equivalently,
FIGURE 81
x = 32 23t and
y =  16t 2 + 32t + 3.36.
Solving x = 32 23t for t, we obtain t = ric equation yields the following.
FOR DISCUSSION If a golf ball is hit at 88 feet per second (60 mph), use trial and error to find the angle u that results in a maximum distance for the ball.
y =  16 ¢ y = 
x 32 23
2
x 32 13
≤ + 32 ¢
. Substituting for t in the other paramet
x 32 23
≤ + 3.36
1 2 23 x + x + 3.36 192 3
Simplify.
Because this equation defines a parabola, the rocket follows a parabolic path.
GCM
EXAMPLE 7
Analyzing the Path of a Projectile Determine the total flight time and the horizontal distance traveled by the rocket in Example 6.
Analytic Solution The equation y =  16t 2 + 32t + 3.36 gives the vertical position of the rocket at time t. We need to determine those values of t for which y = 0, since these values correspond to the rocket at ground level. This yields
Graphing Calculator Solution FIGURE 82 shows that when t = 2.1, the horizontal distance covered is approximately 116.4 feet, which agrees with the analytic solution.
0 =  16t 2 + 32t + 3.36. From the quadratic formula, the solutions are t =  0.1 or t = 2.1. Since t represents time, t =  0.1 is an unacceptable answer. Therefore, the flight time is 2.1 seconds. The rocket was in the air for 2.1 seconds, so we can use t = 2.1 and the parametric equation that models the horizontal position, x = 3223t, to get x = 32 23 (2.1) L 116.4 feet.
50
–30
150 –10 FIGURE 82
754
CHAPTER 10
Applications of Trigonometry and Vectors
Parametric equations are used frequently in computer graphics to design a variety of figures, letters, and shapes. GCM
2
–3
3
(a)
2
3
–2 (b)
Creating a Drawing with Parametric Equations Graph a smiley face using parametric equations. Solution Head We use a circle centered at the origin for the head. If the radius is 2, then we let x = 2 cos t and y = 2 sin t for 0 … t … 2p. See FIGURE 83(a). Eyes For the eyes, we use two small circles. The eye in the first quadrant can be modeled by x = 1 + 0.3 cos t and y = 1 + 0.3 sin t for 0 … t … 2p, representing a circle centered at (1, 1) with radius 0.3. The eye in the second quadrant can be modeled by x =  1 + 0.3 cos t and y = 1 + 0.3 sin t for 0 … t … 2p, representing a circle centered at ( 1, 1) with radius 0.3. See FIGURE 83(b).
–2
–3
EXAMPLE 8
Mouth For the smile, we can use the lower half of a circle. Using trial and error, we arrive at x = 0.5 cos 12 t and y =  0.5  0.5 sin 12 t, a semicircle centered at (0,  0.5) with radius 0.5. Since we are letting 0 … t … 2p, the term 12 t ensures that only half a circle (semicircle) is drawn. The minus sign before 0.5 sin 12 t in the yequation results in the lower half of the semicircle being drawn rather than the upper half. The final result is shown in FIGURE 83(c). We can add the pupils by plotting the points (1, 1) and ( 1, 1). Turning off the coordinate axes rids the face of lines. FOR DISCUSSION
Modify the face in Example 8 so that it is frowning. Try to find a way to make the right eye be shut rather than open. (c) FIGURE 83
WHAT WENT WRONG? A student graphed x = cos 12 t, y = sin 12 t, which should be a circle with radius 1 because x 2 + y 2 = cos2 12 t + sin2 12 t = 1. However, the following screen shows the graph obtained by the student. 2
3
–3
–2
What Went Wrong? to a circle?
What should the student do to change the graph from a semicircle
Answer to What Went Wrong? The periods for y = cos 12 t and y = sin 12 t are both 4p. The interval for t should be 0 … t … 4p, rather than 0 … t … 2p, which is the interval the student used.
10.7
10.7
More Parametric Equations
755
EXERCISES
Checking Analytic Skills Find a rectangular equation for each curve and describe the curve. Do not use a calculator. 1. x = 3 sin t, y = 3 cos t ; for t in 3 p, p4
2. x = 2 sin t, y = 2 cos t ; for t in 30, 2p4
3. x = 2 cos2 t, y = 2 sin2 t ; for t in c0,
4. x = 25 sin t, y = 23 cos t ; for t in 30, 2p4
5. x = 3 tan t, y = 2 sec t ; for t in a 
p d 2
p p , b 2 2
6. x = cot t, y = csc t ; for t in (0, p)
Graph each pair of parametric equations for 0 … t … 2p. Describe any differences in the two graphs. 7. (a) x = 3 cos t, y = 3 sin t (b) x = 3 cos 2t, y = 3 sin 2t 9. (a) x = 3 cos t, y = 3 sin t (b) x = 3 sin t, y = 3 cos t
8. (a) x = 2 cos t, y = 2 sin t (b) x = 2 cos t, y =  2 sin t 10. (a) x =  1 + cos t, y = 2 + sin t (b) x = 1 + cos t, y = 2 + sin t
Find a rectangular equation for each curve and graph the curve. 11. x = sin t, y = csc t ; for t in (0, p)
12. x = tan t, y = cot t ; for t in a0,
13. x = 2 + sin t, y = 1 + cos t ; for t in 30, 2p4
14. x = 1 + 2 sin t, y = 2 + 3 cos t ; for t in 30, 2p4
p b 2
Graph each pair of parametric equations. 15. x = 2 + cos t, y = sin t  1; 0 … t … 2p
16. x =  2 + cos t, y = sin t + 1; 0 … t … 2p
17. x =
18. x = cos5 t, y = sin5 t ; 0 … t … 2p
cos3
t, y =
sin3
t ; 0 … t … 2p
19. x =  3 sin t , y =  3 cos t ; 0 … t … p
20. x = 3 sin 2t, y = 3 cos t ; 0 … t … 2p
Graph each cycloid for t in the specified interval. 21. x = t  sin t, y = 1  cos t ; for t in 30, 4p4
22. x = 2t  2 sin t, y = 2  2 cos t ; for t in 30, 8p4
Graph each pair of parametric equations for 0 … t … 2p in the window 30, 64 by 30, 44. Identify the letter of the alphabet that is being graphed. 23. x1 = 1, x2 = 1 + t>(3p), x3 = 1 + t>(2p),
y1 = 1 + t>p y2 = 2 y3 = 3
25. x1 = 1, x2 = 1 + 1.3 sin 0.5t,
y1 = 1 + t>p y2 = 2 + cos 0.5t
24. x1 x2 x3 x4
= = = =
y1 1, 1 + t>(3p), y2 1 + t>(2p), y3 1 + t>(2p), y4
= = = =
26. x1 = 2 + 0.8 cos 0.85t, x2 = 1.2 + t>(1.3p),
1 + t>p 2 3 1 y1 = 2 + sin 0.85t y2 = 2
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CHAPTER 10
Applications of Trigonometry and Vectors
(Modeling) Designing Letters Find a set of parametric equations that results in a letter similar to the one shown in each figure. Use the window 3 4.7, 4.74 by 3 3.1, 3.14, and turn off the coordinate axes. Answers may vary. 27.
28.
29.
30.
31. (Modeling) Designing a Face Refer to Example 8. Use parametric equations to create your own smiley face. This face should have a head, a mouth, and eyes.
32. (Modeling) Designing a Face Add a nose to the face that you designed in Exercise 31.
Lissajous Figures The middle screen in FIGURE 74 is an example of a Lissajous figure. Lissajous figures occur in electronics and may be used to find the frequency of an unknown voltage. Graph each Lissajous figure for 0 … t … 6.5 in the window 3 6, 64 by 3 4, 44. 33. x = 2 cos t, y = 3 sin 2t
34. x = 3 cos 2t, y = 3 sin 3t
35. x = 3 sin 4t, y = 3 cos 3t
36. x = 4 sin 4t, y = 3 sin 5t
(Modeling) In Exercises 37–40, do the following. (a) Determine the parametric equations that model the path of the projectile. (b) Determine the rectangular equation that models the path of the projectile. (c) Determine the time the projectile is in flight and the horizontal distance covered.
39. Flight of a Softball Sally hits a softball when it is 2 feet above the ground. The ball leaves her bat at an angle of 20° with respect to the ground at a velocity of 88 feet per second.
37. Flight of a Model Rocket A model rocket is launched from the ground with a velocity of 48 feet per second at an angle of 60° with respect to the ground.
40. Flight of a Baseball Pronk hits a baseball when it is 2.5 feet above the ground. The ball leaves his bat at an angle of 29° from the horizontal with a velocity of 136 feet per second.
T
38. Flight of a Golf Ball Tyler McGinnis is playing golf. He hits a golf ball from the ground at an angle of 60° with respect to the ground at a velocity of 150 feet per second.
29° 2.5 ft
41. (Modeling) Simulating Gravity on the Moon If an object is thrown on the moon, then the parametric equations of flight are 60°
x = (v cos u)t
and y = (v sin u)t  2.66t 2 + h.
Estimate the distance that a golf ball hit at 88 feet per second (60 mph) at an angle of 45° with the horizontal travels on the moon if the moon’s surface is level.
CHAPTER 10
42. (Modeling) Flight of a Baseball A baseball is hit from a height of 3 feet at a 60° angle above the horizontal. Its initial velocity is 64 feet per second. (a) Write parametric equations that model the flight of the baseball. (b) Determine the horizontal distance traveled by the ball in the air. Assume that the ground is level. (c) What is the maximum height of the baseball? At that time, how far has the ball traveled horizontally? (d) Would the ball clear a 5foothigh fence that is 100 feet from the batter? (Modeling) Path of a Projectile In Exercises 43 and 44, a projectile has been launched from the ground with an initial velocity of 88 feet per second. You are given parametric equations that model the path of the projectile. (a) Graph the parametric equations. (b) Approximate u, the angle the projectile makes with the horizontal at launch, to the nearest tenth of a degree.
SECTIONS 10.6 AND 10.7
757
(c) On the basis of your answer to part (b), write parametric equations for the projectile, using the cosine and sine functions. 43. x = 82.69265063t, y =  16t 2 + 30.09777261t 44. x = 56.56530965t, y =  16t 2 + 67.41191099t 45. The spiral of Archimedes has polar equation r = au, where r 2 = x 2 + y 2. Show that a parametric representation of the spiral of Archimedes is x = a u cos u,
y = a u sin u,
for u in ( q, q).
46. Show that the hyperbolic spiral given by ru = a, where r 2 = x 2 + y 2, is given parametrically by x =
a cos u , u
y =
a sin u , u
for u in (  q, 0) ´ (0, q).
REVIEWING BASIC CONCEPTS
1. For the point with polar coordinates ( 2, 130°), state the quadrant in which the point lies if it is graphed in a rectangular coordinate system. 2. Let the point P have rectangular coordinates ( 2, 2). Determine two pairs of equivalent polar coordinates for point P.
5. Find an equivalent equation in polar coordinates for the equation x + y = 6. 6. Find a rectangular equation for the curve described by x = 2 cos t, y = 4 sin t for 0 … t 6 2p. 7. Graph the parametric equations x = 2  sin t,
3. Graph the polar equation r = 2  2 cos u. Identify the type of polar graph. 4. Find an equivalent equation in rectangular coordinates for the polar equation r = 2 cos u.
10
Summary
y = cos t  1,
for 0 … t 6 2p.
8. (Modeling) Flight of a Golf Ball Jeffery hits a golf ball with a 45° angle of elevation from the top of a ridge that is 50 feet above an area of level ground. The initial velocity of the ball is 88 feet per second, or 60 mph. Find the horizontal distance traveled by the ball in the air.
SUMMARY
KEY TERMS & SYMBOLS
KEY CONCEPTS
10.1 The Law of Sines
LAW OF SINES In any triangle ABC with sides a, b, and c,
side–angle–side (SAS) angle–side–angle (ASA) side–side–side (SSS) side–angle–angle (SAA) side–side–angle (SSA) oblique triangle ambiguous case
a b ⴝ , sin A sin B or, in compact form
a c ⴝ , sin A sin C
and
c b ⴝ , sin B sin C
a b c ⴝ ⴝ . sin A sin B sin C
(continued)
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CHAPTER 10
Applications of Trigonometry and Vectors
KEY TERMS & SYMBOLS
KEY CONCEPTS
10.2 The Law of Cosines and Area Formulas
LAW OF COSINES In any triangle ABC with sides a, b, and c,
semiperimeter
a 2 ⴝ b 2 ⴙ c 2 ⴚ 2bc cos A, b 2 ⴝ a 2 ⴙ c 2 ⴚ 2ac cos B, c 2 ⴝ a 2 ⴙ b 2 ⴚ 2ab cos C.
and
HERON’S AREA FORMULA (SSS) If a triangle has sides of lengths a, b, and c, and if the semiperimeter is s ⴝ 12 (a ⴙ b ⴙ c), then the area of the triangle is Ꮽ ⴝ 2s(s ⴚ a)(s ⴚ b)(s ⴚ c). AREA OF A TRIANGLE (SAS) In any triangle ABC, the area Ꮽ is given by the following formulas: Ꮽⴝ
10.3 Vectors and Their Applications scalars vector quantities ! vector, v, OP, or OP initial point terminal point magnitude,  OP  parallelogram rule resultant vector opposite of v zero vector scalar multiplication position vector, 8a, b9 horizontal component vertical component direction angle unit vectors, i, j dot product angle between two vectors orthogonal vectors airspeed ground speed
1 1 bc sin A, Ꮽ ⴝ ab sin C, 2 2
and Ꮽ ⴝ
1 ac sin B. 2
RESULTANT VECTOR The resultant (or sum) A ⴙ B of vectors A and B is shown in the figure.
A+B
B
A
MAGNITUDE AND DIRECTION ANGLE OF A VECTOR The magnitude (length) of vector u = 8a, b9 is given by 円 u円 ⴝ 2a 2 ⴙ b 2. The direction angle u satisfies tan u = ba , where a Z 0. VECTOR OPERATIONS For any real numbers a, b, c, d, and k,
8a, b9 ⴙ 8c, d9 ⴝ 8a ⴙ c, b ⴙ d9, k # 8a, b9 ⴝ 8ka, kb9,
if a ⴝ 8a 1, a 29, then ⴚa ⴝ 8ⴚa 1, ⴚa 29,
8a, b9 ⴚ 8c, d9 ⴝ 8a, b9 ⴙ ⴚ8c, d9 ⴝ 8a ⴚ c, b ⴚ d9.
If u = 8x, y9 has direction angle u, then u = 8 u  cos u,  u  sin u9. i, j FORM FOR VECTORS If v = 8a, b9, then v = ai + bj, where i = 81, 09 and j = 80, 19.
DOT PRODUCT The dot product of the two vectors u = 8a, b9 and v = 8c, d9, denoted u # v, is given by u # v ⴝ ac ⴙ bd.
If u is the angle between u and v, where 0° … u … 180°, then u # v ⴝ 円u円 円v円 cos U.
CHAPTER 10
KEY TERMS & SYMBOLS
Summary
759
KEY CONCEPTS PROPERTIES OF THE DOT PRODUCT For all vectors u, v, and w and real numbers k, (a) u # v ⴝ v # u (b) u # (v ⴙ w) ⴝ u # v ⴙ u # w (c) (u ⴙ v) # w ⴝ u # w ⴙ v # w (d) (ku) # v ⴝ k(u # v) ⴝ u # (kv) # (e) 0 u ⴝ 0 (f) u # u ⴝ 円 u円 2.
10.4 Trigonometric (Polar) Form of Complex Numbers
TRIGONOMETRIC (POLAR) FORM OF A COMPLEX NUMBER The expression r (cos U ⴙ i sin U) is called the trigonometric form (or polar form) of the complex number x + yi. The expression cos u + i sin u is sometimes abbreviated cis u. In this notation, r (cos U ⴙ i sin U) is written r cis U.
real axis imaginary axis complex plane rectangular form trigonometric (polar) form modulus (absolute value) argument
PRODUCT THEOREM If r1(cos u1 + i sin u1) and r2(cos u2 + i sin u2) are any two complex numbers, then
10.5 Powers and Roots of Complex Numbers
3r1 (cos U1 ⴙ i sin U1 )4 # 3r2 (cos U2 ⴙ i sin U2 )4 ⴝ r1 r23cos(U1 ⴙ U2 ) ⴙ i sin(U1 ⴙ U2)4,
or, in compact form,
(r1 cis U1 )(r2 cis U2 ) ⴝ r1 r2 cis(U1 ⴙ U2 ).
QUOTIENT THEOREM If r1 (cos u1 + i sin u1 ) and r2 (cos u2 + i sin u2 ) are any two complex numbers, where r2(cos u2 + i sin u2) Z 0, then r1 r1(cos U1 ⴙ i sin U1) ⴝ 3cos(U1 ⴚ U2) ⴙ i sin(U1 ⴚ U2)4, r2(cos U2 ⴙ i sin U2) r2 r1 cis U1 r1 or, in compact form, ⴝ cis(U1 ⴚ U2). r2 cis U2 r2 DE MOIVRE’S THEOREM If r(cos u + i sin u) is a complex number, and if n is any real number, then 3r(cos U ⴙ i sin U)4n ⴝ r n(cos nU ⴙ i sin nU), 3r cis U4n ⴝ r n(cis nU).
or, in compact form,
nth root of a complex number
nTH ROOT For a positive integer n, the complex number a + bi is an nth root of the complex number x + yi if (a ⴙ bi )n ⴝ x ⴙ yi. nTH ROOT THEOREM If n is any positive integer, r is a positive real number, and u is in degrees, then the nonzero complex number r(cos u + i sin u) has exactly n distinct nth roots, given by n
where
10.6 Polar Equations and Graphs polar coordinate system pole polar axis polar coordinates
Aⴝ
U ⴙ 360° # k , n
2r (cos A ⴙ i sin A), or ␣ ⴝ
U 360° # k ⴙ , k = 0, 1, 2, Á , n  1. n n
RELATIONSHIPS BETWEEN RECTANGULAR AND POLAR COORDINATES The following relationships hold between the point (x, y) in the rectangular coordinate plane and the point (r, u) in the polar coordinate plane. x ⴝ r cos U,
y ⴝ r sin U,
r 2 ⴝ x 2 ⴙ y 2,
tan U ⴝ
y , if x Z 0 x (continued)
760
CHAPTER 10
Applications of Trigonometry and Vectors
KEY TERMS & SYMBOLS
KEY CONCEPTS
polar equation cardioid polar grid rose curve (fourleaved rose) lemniscate spiral of Archimedes limaçon
POLAR EQUATIONS AND GRAPHS
10.7 More Parametric Equations
HEIGHT OF AN OBJECT If an object has an initial velocity v and initial height h, and travels so that its initial angle of elevation is u, then its flight after t seconds is modeled by the parametric equations
r ⴝ a cos U f r ⴝ a sin U
r ⴝ a ⴞ b sin U f r ⴝ a ⴞ b cos U
Limaçons
x ⴝ (v cos U)t and
cycloid
10
r 2 ⴝ a 2 sin 2U f r 2 ⴝ a 2 cos 2U
Circles
r ⴝ a sin nU f r ⴝ a cos nU
Lemniscates
Rose curves
y ⴝ (v sin U)t ⴚ 16t 2 ⴙ h.
REVIEW EXERCISES
Use the law of sines or the law of cosines to find the indicated part of each triangle ABC.
Approximate the height of the balloon above the ground to the nearest tenth of a mile.
1. C = 74° 10¿, c = 96.3 meters, B = 39° 30¿; find b 2. a = 165 meters, A = 100.2°, B = 25.0°; find b 3. a = 86.14 inches, b = 253.2 inches, c = 241.9 inches; find A
C
4. a = 14.8 feet, b = 19.7 feet, c = 31.8 feet; find B 5. A = 129° 40¿, a = 127 feet, b = 69.8 feet; find B 6. B = 39° 50¿, b = 268 centimeters, a = 340 centimeters; find A 7. B = 120.7°, a = 127 feet, c = 69.8 feet; find b 8. A = 46.2°, b = 184 centimeters, c = 192 centimeters; find a Find the area of each triangle ABC with the given information. 9. b = 840.6 meters, c = 715.9 meters, A = 149.3° 10. a = 6.90 feet, b = 10.2 feet, C = 35° 10¿ 11. a = 0.913 kilometer, b = 0.816 kilometer, c = 0.582 kilometer 12. a = 43 meters, b = 32 meters, c = 51 meters Solve each problem. 13. Height of a Balloon The angles of elevation of a balloon from two points A and B on level ground are 24° 50¿ and 47° 20¿, respectively. As shown in the figure, points A and B are in the same vertical plane and are 8.4 miles apart.
B
47° 20´ 24° 50´ 8.4 mi
A
14. Distance in a Boat Race The course for a boat race starts at point X and goes in the direction S 48° W to point Y. It then turns and goes S 36° E to point Z and finally returns back to point X. If point Z lies 10 kilometers directly south of point X, find the distance from Y to Z, to the nearest kilometer.
CHAPTER 10
15. Radio Direction Finders Radio direction finders are placed at points A and B, which are 3.46 miles apart on an east–west line, with A west of B. From A, the bearing of an illegal radio transmitter is 48.0°; from B, the bearing is 302°. Find the distance between the transmitter and A.
Review Exercises
761
24. Concept Check When applying the law of cosines to find an angle, how can we tell if the angle is acute or obtuse before evaluating the inverse cosine value? 25. Use the vectors shown here to sketch a + 3c.
16. Distance across a Canyon To measure the distance AB across a canyon for a power line, a surveyor measures angles B and C and the distance BC. (See the figure.) What is the distance from A to B?
a c
Find the magnitude and direction angle for u, rounded to the nearest tenth of a degree.
A
26. u = 821,  209
27. u = 8 9, 129
Canyon
B
58.4°
27.9° 125 ft
Vector v has the given magnitude and direction angle. Write v in the form 8a, b9. C
28.  v  = 50, u = 45° (give exact values) 29.  v  = 69.2, u = 75°
17. Length of a Brace A banner on an 8.0foot pole is to be mounted on a building at an angle of 115°, as shown in the figure. Find the length of the brace.
0
8.
Find (a) the dot product and (b) the angle between the pairs of vectors. 31. u = 86, 29, v = 83,  29
ft
115°
30.  v  = 964, u = 154° 20¿
22°
32. u = H 223, 2 I , v = H 5, 523 I
9 33. Brace
Are the vectors u = 85,  19 and v = 8 2,  109 orthogonal? Explain.
Solve each problem. 18. Hanging Sculpture A hanging sculpture in an art gallery is to be hung with two wires of lengths 15.0 feet and 12.2 feet so that the angle between them is 70.3°. How far apart should the ends of the wire be placed on the ceiling? 19. Pipeline Position A pipeline is to run between points A and B, which are separated by a protected wetlands area. To avoid the wetlands, the pipe will run from point A to C and then to B. The distances involved are AB = 150 kilometers, AC = 102 kilometers, and BC = 135 kilometers. What angle should be used at point C ?
9 20. 9 21.
If we are given a, A, and C in a triangle ABC, does the possibility of the ambiguous case exist? If not, explain why. Can a triangle ABC exist if a = 4.7, b = 2.3, and c = 7.0? If not, explain why. Answer this question without using trigonometry.
22. Concept Check Given that a = 10 and B = 30° in triangle ABC, determine the values of b for which A has (a) exactly one value, (b) two values, (c) no value. 23. Concept Check If angle C of a triangle ABC measures 90°, what does the law of cosines become?
34. Weight of a Sled and Child Paula and Steve are pulling their daughter Jessie on a sled as shown in the figure. Steve pulls with a force of 18 pounds at an angle of 10°. Paula pulls with a force of 12 pounds at an angle of 15°. Find the magnitude of the resultant force. 12 lb 15°
Sled
10° 18 lb
35. Barge Movement One rope pulls a barge directly east with a force of 1000 newtons. Another rope pulls the barge to the northeast with a force of 2000 newtons. Find the resultant force acting on the barge, and find the angle between the resultant and the first rope. 36. Direction and Airspeed A plane has an airspeed of 520 mph. The pilot wishes to fly on a course of 310°. A wind of 37 mph is blowing from a bearing of 212°. On what bearing should the pilot fly, and what will be her ground speed? 37. Bearing and Speed A longdistance swimmer starts out swimming a steady 3.2 mph due north. A 5.1mph current is flowing on a bearing of 12°. What is the swimmer’s resulting bearing and speed?
762
9 38.
CHAPTER 10
Applications of Trigonometry and Vectors
(Modeling) Wind and Vectors A wind can be described by v = 6i + 8j, where vector j points north and represents a south wind of 1 mph. (a) What is the speed of the wind? (b) Find 3v. Interpret the result. (c) Interpret the wind if it switches to u =  8i + 8j.
Graph each complex number as a vector in the complex plane. 40.  4 + 2i
39. 5i
Find the resultant of each pair of complex numbers. 41. 7 + 3i and  2 + i
42. 2  4i and 5 + i
58. Solve the equation x 4 + i = 0. Leave solutions in polar form. Convert to rectangular coordinates. Give exact values. 60. a  8, 
59. (12, 225°)
p b 3
Convert to polar coordinates, with 0° … u 6 360°. Give exact values. 62. (0,  5)
61. ( 6, 6)
Use a graphing calculator to graph each polar equation for 0° … u … 360°. Use a square window. 63. r = 4 cos u (circle) 64. r = 1  2 sin u (limaçon with a loop)
Perform each conversion in Exercises 43–46. Rectangular Form
43.
Trigonometric Form
r = 1 + 2 sin u 3(cos 90° + i sin 90°)
45.
2 cis 225°  4 + 4i23
Perform each indicated operation. Give answers in rectangular form. 47. 35(cos 90° + i sin 90°)4 36(cos 180° + i sin 180°)4 48. 33 cis 135°4 32 cis 105°4
2(cos 60° + i sin 60°) 49. 8(cos 300° + i sin 300°) 50. 51.
4 cis 270° 2 cis 90°
A 23 + i B
3
52. (2  2i ) 5 53. (cos 100° + i sin 100°) 6 54. (cis 20°) 3
Find the indicated roots and graph as vectors in the complex plane. Leave answers in polar form. 55. The cube roots of  27i 56. The fourth roots of 16i 57. The fifth roots of 32
66. Sketch a traditional graph of
 2 + 2i
44.
46.
65. r = 2 sin 4u (eightleaved rose) (limaçon with a loop).
Find an equivalent equation in rectangular coordinates. 67. r =
3 1 + cos u
68. r =
4 2 sin u  cos u
69. r = sin u + cos u 70. r = 2 Find an equivalent equation in polar coordinates. 71. x =  3
72. y = x
73. y = x 2
74. x = y 2
Find a rectangular equation for each plane curve with the given parametric equations. 75. x = cos 2t, y = sin t, for t in ( p, p) 76. x = 5 tan t, y = 3 sec t, for t in A  p2 , p2 B 77. Graph the curve defined by the following parametric equations: x = t + cos t, y = sin t, for t in 30, 2p4. 78. (Modeling) Flight of a Baseball A baseball is hit when it is 3.2 feet above the ground. It leaves the bat with a velocity of 118 feet per second at an angle of 27° with respect to the ground. Find the horizontal distance the baseball travels in the air.
CHAPTER 10
10
Test
763
TEST
1. Find the indicated part of each triangle. (a) A = 25.20°, a = 6.920 yards, b = 4.820 yards; find C (b) C = 118°, b = 132 kilometers, a = 75.1 kilometers; find c (c) a = 17.3 feet, b = 22.6 feet, c = 29.8 feet; find B 2. Given that a = 10 and B = 150° in triangle ABC, determine the values of b for which A has (a) exactly one value (b) two values (c) no value.
9 3. What conditions determine whether three positive numbers can represent the sides of a triangle? 4. Solve each problem. (a) Distance between Boats Two boats leave a dock together. Each travels in a straight line. The angle between their courses measures 54.2°. One boat travels 36.2 kilometers per hour, and the other travels 45.6 kilometers per hour. How far apart will they be after 3 hours? (b) Distance between Points on a Softball Field The pitcher’s mound on a regulation softball field is 46 feet from home plate. The distance between the bases is 60 feet, as shown in the figure. How far from third base (point T ) is the pitcher’s mound (point M )? Give your answer to the nearest foot. 60 ft
60 ft M
T
46 ft 60 ft
60 ft
5. For the vectors u = 8 1, 39 and v = 82,  69 find the following. (a) u + v (b)  3v (c) u # v (d) The angle between u and v 6. Find the following for the complex numbers 4 cis 240°
and
 4 + 4i23.
(a) The rectangular form of 4 cis 240° (b) The trigonometric form of  4 + 4i23 (c) Their resultant in the form a + bi 7. Perform each indicated operation. Give the answer in rectangular form. (a) 3(cos 30° + i sin 30°) # 5(cos 90° + i sin 90°) 2 cis 315° (b) 4 cis 45° (c)
A 1  i23 B
5
(d) Find the fourth roots of 23 + i. Leave the answers in trigonometric form. 8. Do the following for the polar equation r = 4 cos u. (a) Graph the equation for 0° … u … 360°. (b) Find an equation equivalent to the given one, using rectangular coordinates. 9 (c) Is the graph from part (a) what you would expect for the graph of the equation from part (b)? Explain. 9. Find an equivalent equation in polar coordinates for
(c) Horizontal and Vertical Components Find the horizontal and vertical components of the vector with magnitude 569 that is inclined 127.5° from the horizontal. (d) Magnitude of a Resultant Find the magnitude of the resultant of forces of 475 pounds and 586 pounds that form an angle of 78.2°.
 x + 2y = 4 in the form r = ƒ(u). 10. Graph the curve defined by the following parametric equations: x = 2 cos 2t, y = 2 sin 2t,
for t in 30, 2p4.
11 Further Topics in Algebra EARTH HAS A natural circulatory system called the Great Ocean Conveyor Belt. This system circulates warm water from the Gulf Stream to northern latitudes and cool water from the North Atlantic to southern latitudes, greatly influencing and moderating temperatures throughout the world. Global warming disrupts the normal flow of the Conveyor Belt. When greenhouse gases raise the average yearly temperature of the atmosphere, heavierthannormal precipitation falls over the North Atlantic. Runoff from the land and polar ice melt add more fresh water, which lowers the ocean’s salinity level and decreases its density. With lowered density, the Conveyor Belt slows or stops. As circulation slows, climate changes occur around the globe, including drought in some regions and torrential rainfall in others. When the Conveyor Belt stops, winter temperatures in northern latitudes get progressively colder without the warming waters of the Gulf Stream to mod813.094 erate them, and a pt new Ice Age begins. Topics in this chapter such as sequences and probability allow scientists to model the effect of global warming on Earth’s climate, plants, and animals. (See Exercises 72 and 73 of Section 11.1.) Source: Woods Hole Oceanographic Institute; United Nations Environment Programme.
CHAPTER OUTLINE 11.1 Sequences and Series 11.2 Arithmetic Sequences and Series 11.3 Geometric Sequences and Series 11.4 Counting Theory 11.5 The Binomial Theorem 11.6 Mathematical Induction 11.7 Probability
From Chapter 11 of A Graphical Approach to Precalculus with Limits: A Unit Circle Approach, Fifth Edition. John Hornsby, Margaret Lial, Gark K. Rockswold. Copyright © 2011 by Pearson Education, Inc. All rights reserved.
11.1
11.1
Sequences and Series
765
Sequences and Series
Sequences • Series and Summation Notation • Summation Properties
Sequences A sequence is a function that computes an ordered list. For example, the average person in the United States uses 100 gallons of water each day. The function defined by ƒ(n) = 100n generates the terms of the sequence 100, 200, 300, 400, 500, 600, 700, Á , when n = 1, 2, 3, 4, 5, 6, 7, Á . This function represents the number of gallons of water used by the average person after n days. As another example, if $100 is deposited into a savings account paying 5% interest compounded annually, then the function defined by g (n) = 100(1.05) n calculates the account balance after n years. The terms of the sequence are g (1), g (2), g (3), g (4), g (5), g (6), g (7), Á . , and can be approximated as
y 10
105, 110.25, 115.76, 121.55, 127.63, 134.01, 140.71, Á .
6
f (x) = 2x
2 x
0
–5
Sequence
5
A sequence is a function that has a set of natural numbers (positive integers) as its domain. FIGURE 1 an 10 6
an = 2n
2 –5
0
FIGURE 2
n 5
Instead of using ƒ(x) notation to indicate a sequence, it is customary to use an, where a n ⴝ ƒ(n). The letter n is used instead of x as a reminder that n represents a natural number. The elements in the range of a sequence, called the terms of the sequence, are a1, a2, a3, . . . . The elements of both the domain and the range of a sequence are ordered. The first term is found by letting n = 1, the second term by letting n = 2, and so on. The general term, or nth term, of the sequence is an. FIGURE 1 shows the graph of ƒ(x) = 2x, and FIGURE 2 shows that of an = 2n. Notice that ƒ(x) is a continuous function and an is discontinuous. To graph an, we plot points of the form (n, 2n) for n = 1, 2, 3, . . . . EXAMPLE 1
Finding Terms of Sequences Write the first five terms of each sequence. ( 1) n n + 1 (a) an = (b) an = ( 1) n # n (c) an = n + 2 2n Solution (a) Replacing n in an =
n + 1 n + 2
1 1 3 n = 3: a3 = 3 5 n = 5: a5 = 5 n = 1: a1 =
with 1, 2, 3, 4, and 5 gives + + + + + +
1 2 = ; 2 3 1 4 = ; 2 5 1 6 = . 2 7
2 2 4 n = 4: a4 = 4 n = 2: a2 =
+ + + +
1 3 = ; 2 4 1 5 = ; 2 6
(continued)
766
CHAPTER 11
Further Topics in Algebra
(b) Replace n in an = ( 1) n
#
n with 1, 2, 3, 4, and 5 to obtain
n = 1: a1 = ( 1) 1 # 1 =  1; n = 3: a3 = ( 1) 3 # 3 =  3; n = 5: a5 = ( 1) 5 # 5 =  5.
(c) For an =
an = 1n
2, 4, 6, 8, 10, 12, 14, Á ,
1 16 ,
1 . and a5 =  32
is infinite,
but the sequence of days in June, 0 1 2 3 4 5 6 7 8 9 10
TECHNOLOGY NOTE Some graphing calculators have a designated sequence mode to investigate and graph sequences defined in terms of n, where n is a natural number. Using sequence mode on the TI83/84 Plus to list the first 10 terms of the sequence with general term an = n + 1n produces the result shown in the top figure. Additional terms can be seen by scrolling to the right. The bottom figure shows a calculator graph of an = n + 1n . Notice that for n = 5, the term is 5 +
1 5
1, 2, 3, 4, Á , 29, 30,
n
FIGURE 3
GCM
we have a1 =  12 , a2 = 14 , a3 =  18 , a4 =
A sequence is a finite sequence if the domain is the set 51, 2, 3, 4, Á , n6, where n is a natural number. An infinite sequence has the set of all natural numbers as its domain. For example, the sequence of even natural numbers,
an 1.6 1.4 1.2 1.0 0.8 0.6 0.4 0.2
(  1) n 2n ,
n = 2: a2 = ( 1) 2 # 2 = 2; n = 4: a4 = ( 1) 4 # 4 = 4;
= 5.2.
is finite.
If the terms of an infinite sequence get closer and closer to some real number, the sequence is said to be convergent and to converge to that real number. For example, the sequence defined by an = 1n approaches 0 as n becomes large. Thus, an is a convergent sequence that converges to 0. A graph of this sequence for n = 1, 2, 3, Á , 10 is shown in FIGURE 3. The terms of an approach the horizontal axis. A sequence that does not converge to some number is divergent. The terms of the sequence an = n 2 are 1, 4, 9, 16, 25, 36, 49, 64, 81, Á . This sequence is divergent because as n becomes large, the values for an do not approach a fixed number; rather, they increase without bound. Some sequences are defined by a recursive definition, a definition in which each term is defined as an expression involving the previous term or terms. By contrast, the sequences in Example 1 were defined explicitly, with a formula for an that does not depend on a previous term.
EXAMPLE 2
Using a Recursion Formula Find the first four terms of each sequence. (a) a1 = 4; for n 7 1, an = 2 # an  1 + 1 (b) a1 = 2; for n 7 1, an = an  1 + n  1 Solution (a) This is a recursive definition. We know that a1 = 4. Since an = 2 # an  1 + 1, a2 = 2 # a1 + 1 = 2 # 4 + 1 = 9
a3 = 2 # a2 + 1 = 2 # 9 + 1 = 19
11
a4 = 2 # a3 + 1 = 2 # 19 + 1 = 39.
(b) a1 = 2 0
11 0 The fifth term is 5.2.
a2 = a1 + 2  1 = 2 + 1 = 3 a3 = a2 + 3  1 = 3 + 2 = 5 a4 = a3 + 4  1 = 5 + 3 = 8
11.1
767
Sequences and Series
FOR DISCUSSION
One of the most famous sequences in mathematics is the Fibonacci sequence, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, Á , named for the Italian mathematician Leonardo of Pisa, who was also known as Fibonacci. The Fibonacci sequence is found in numerous places in nature. For example, male honeybees hatch from eggs that have not been fertilized, so a male bee has only one parent, a female. On the other hand, female honeybees hatch from fertilized eggs, so a female has two parents, one male and one female. The number of ancestors in consecutive generations of bees follows the Fibonacci sequence. Successive terms in the sequence also appear in plants, such as the daisy head, pineapple, and pinecone. Leonardo of Pisa (Fibonacci) (1170–1250)
1. Try to discover the pattern in the Fibonacci sequence. 2. Using the given description, write a recursive definition that calculates the number of ancestors of a male bee in each generation.
EXAMPLE 3
Modeling Insect Population Growth Frequently, the population of a particular insect grows rapidly at first and then levels off because of competition for limited resources. In one study, the behavior of the winter moth was modeled with a sequence similar to the following, where an represents the population density in thousands per acre during year n. (Source: Varley, G. and G. Gradwell, “Population models for the winter moth,” Symposium of the Royal Entomological Society of London 4.) a1 = 1 an = 2.85an  1  0.19an  1 2, for n Ú 2 (a) Give a table of values for n = 1, 2, 3, Á , 10. (b) Graph the sequence. Describe what happens to the population density of the winter moth. Solution (a) Evaluate a1, a2, a3, Á , a10 recursively. Since a1 = 1, a2 = 2.85a1  0.19a1 2 = 2.85(1)  0.19(1) 2 = 2.66, a3 = 2.85a2  0.19a2 2 = 2.85(2.66)  0.19(2.66) 2 L 6.24.
and FIGURE 4
FOR DISCUSSION In Example 3, the insect population stabilizes near the value k = 9.7 thousand. This value of k can be found by solving the quadratic equation k = 2.85k  0.19k 2. Explain why.
Approximate values for other terms are given in the table. FIGURE 4 shows the computation of the sequence, denoted by u(n) rather than an, with a calculator. n
1
2
3
4
an
1
2.66
6.24
10.4
5 9.11
6
7
10.2
8
9.31
10.1
9 9.43
10 9.98
(b) The graph of a sequence is a set of discrete points. Plot the points (1, 1),
(2, 2.66),
(3, 6.24),
Á,
(10, 9.98),
as shown in FIGURE 5(a) on the next page. At first, the insect population increases rapidly and then oscillates about the line y = 9.7. (See the “For Discussion” in the margin.) The oscillations become smaller as n increases, indicating that the population density may stabilize near 9.7 thousand per acre. In FIGURE 5(b), the first 20 terms have been plotted with a calculator. (continued)
768
CHAPTER 11
Further Topics in Algebra
an
Insect Density (in thousands/acre)
12
14
10 8 6 4 2 2
4
6
8
21
0
n 0
0
10 12
Year (a)
(b) FIGURE 5
Series and Summation Notation Suppose a person has a starting salary of $30,000 and receives a $2000 raise each year. Then 30,000 , 32,000 , 34,000 , 36,000 , 38,000 are terms of the sequence that describe this person’s salaries over a 5year period. The total earned is given by the finite series 30,000 + 32,000 + 34,000 + 36,000 + 38,000, Looking Ahead to Calculus
Sigma (summation) notation, © , is introduced in a first calculus course in conjunction with the definite integral, symbolized with an elongated S A 1 B to which limits of integration are applied. The definition of a definite integral is b
La
n
ƒ( x ) dx = limq a ƒ( xi ) ¢xi . n: i=1
The definite integral can be used to calculate the area between two curves.
whose sum is $170,000. Any sequence can be used to define a series. For example, the infinite sequence 1,
1 1 1 1 1 , , , , ,Á 3 9 27 81 243
defines the terms of the infinite series 1 +
1 1 1 1 1 + + + + + Á. 3 9 27 81 243
If a sequence has terms a1, a2, a3, Á , then Sn is defined as the sum of the first n terms. That is, Sn = a1 + a2 + a3 + Á + an. The sum of the terms of a sequence, called a series, is written using summation notation. The symbol π, the Greek capital letter sigma, is used to indicate a sum.
Series A finite series is an expression of the form Sn ⴝ a 1 ⴙ a 2 ⴙ a 3 ⴙ
Á
n
ⴙ an ⴝ a ai , iⴝ 1
and an infinite series is an expression of the form Sˆ ⴝ a 1 ⴙ a 2 ⴙ a 3 ⴙ
Á
ⴙ an ⴙ
Á
ˆ
ⴝ a ai . iⴝ 1
The letter i is called the index of summation.
11.1
Sequences and Series
769
CAUTION Do not confuse this use of i with the use of i to represent the imaginary unit. Other letters, such as k and j, also may be used for the index of summation. GCM
EXAMPLE 4
Using Summation Notation 6
Evaluate the series a (2 k + 1). k=1
Analytic Solution Write each of the six terms. Then evaluate the sum. 6
k 1 2 3 a (2 + 1) = (2 + 1) + (2 + 1) + (2 + 1)
k=1
+ (2 4 + 1) + (2 5 + 1) + (2 6 + 1)
Graphing Calculator Solution A graphing calculator can store the sequence into a list L 1 and then compute the sum of the six terms in the list. The screen in FIGURE 6 confirms the analytic result.
= (2 + 1) + (4 + 1) + (8 + 1) + (16 + 1) + (32 + 1) + (64 + 1) = 3 + 5 + 9 + 17 + 33 + 65 = 132 FIGURE 6
EXAMPLE 5
Using Summation Notation with Subscripts Write the terms for each series. Evaluate each sum if possible. 6
Looking Ahead to Calculus
Sums like the one in Example 5(c) are frequently used in calculus.
(a) a aj j=3
3
(b) a (6xi  2) if x1 = 2, x2 = 4, and x3 = 6 i=1
4
(c) a ƒ(xi) ¢x if ƒ(x) = x 2, x1 = 0, x2 = 2, x3 = 4, x4 = 6, and ¢x = 2 i=1
Solution 6
(a) a aj = a3 + a4 + a5 + a6 j=3
(b) Let i = 1, 2, and 3, respectively, with x1 = 2, x2 = 4, and x3 = 6. 3
a (6 xi  2) = (6 x1  2) + (6 x2  2) + (6 x3  2)
i=1
Use the order of operations.
= (6 # 2  2) + (6 # 4  2) + (6 # 6  2) Substitute the given values for x1, x2, and x3.
= 10 + 22 + 34 = 66
Simplify. Add.
(c) Let i = 1, 2, 3, and 4, respectively, with ƒ(x) = x 2, x1 = 0, x2 = 2, x3 = 4, and x4 = 6. 4
a ƒ(xi) ¢x = ƒ(x1) ¢x + ƒ(x2) ¢x + ƒ(x3) ¢x + ƒ(x4) ¢x
i=1
= x1 2 ¢x + x2 2 ¢x + x3 2 ¢x + x4 2 ¢x = 0 2(2) + 2 2(2) + 42(2) + 62(2)
ƒ( x ) = x 2; ¢x = 2
= 0 + 8 + 32 + 72
Simplify.
= 112
Add.
770
CHAPTER 11
Further Topics in Algebra
Estimating P with a Series The infinite series given by EXAMPLE 6
1 1 1 1 1 1 p4 = 4 + 4 + 4 + 4 + 4 + Á + 4 + Á 90 1 2 3 4 5 n can be used to estimate p. (a) Approximate p by finding the sum of the first four terms. (b) Use a calculator to approximate p by summing the first 50 terms. Compare the result with the decimal value of p. Solution (a) Summing the first four terms gives p4 1 1 1 1 L 4 + 4 + 4 + 4 L 1.078751929. 90 1 2 3 4 This approximation can be solved for p by multiplying by 90 and then taking the fourth root. Thus, p L 2 4 90(1.078751929) L 3.139.
FIGURE 7
(b) As shown in FIGURE 7, the first 50 terms of the series provide the approximation p L 3.141590776. This computation matches the value of p for the first five decimal places.
Summation Properties Properties of summation provide useful shortcuts for evaluating series.
Summation Properties If a1, a2, a3, Á , an and b1, b2, b3, Á , bn are two sequences and c is a constant, then, for every positive integer n, n
(a) a c ⴝ nc iⴝ 1 n
n
iⴝ 1
iⴝ 1
(b) a ca i ⴝ c a a i n
n
n
iⴝ 1
iⴝ 1
iⴝ 1
n
n
n
iⴝ 1
iⴝ 1
iⴝ 1
(c) a (a i ⴙ bi) ⴝ a a i ⴙ a bi (d) a (a i ⴚ bi) ⴝ a a i ⴚ a bi .
To prove Property (a), expand the series to get c + c + c + c + Á + c, where there are n terms of c, so the sum is nc.
11.1
Sequences and Series
771
Property (c) also can be proved by first expanding the series: n
Á + (a + b ) a (ai + bi ) = (a1 + b1) + (a2 + b2) + n n
i=1
= (a1 + a2 + Á + an) + (b1 + b2 + Á + bn ) Commutative and associative properties n
n
i=1
i=1
= a ai + a bi . Proofs of the other two properties are similar. The following results about summations can be proved by mathematical induction. (See Section 11.6.)
Summation Rules
Á
n
aiⴝ1ⴙ2ⴙ
ⴙnⴝ
iⴝ 1 n
2 2 2 ai ⴝ1 ⴙ2 ⴙ
n(n ⴙ 1) 2
Á
ⴙ n2 ⴝ
n(n ⴙ 1)(2n ⴙ 1) 6
Á
ⴙ n3 ⴝ
n 2(n ⴙ 1)2 4
iⴝ 1 n
3 3 3 ai ⴝ1 ⴙ2 ⴙ
iⴝ 1
EXAMPLE 7
Using the Summation Properties Use the summation properties to find each sum. 40
22
(a) a 5
14
(b) a 2i
i=1
i=1
(c) a (2i 2  3) i=1
Solution 40
(a) a 5 = 40(5) = 200
Property (a) with n = 40 and c = 5
i=1 22
22
i=1
i=1
(b) a 2i = 2 a i = 2
#
Property (b) with c = 2 and ai = i
22(22 + 1) 2
= 506
Summation rules Simplify.
14
14
14
i=1
i=1
(c) a (2i 2  3) = a 2i 2  a 3 i=1
14
14
i=1
i=1
= 2 a i2  a 3 14(14 + 1)(2 # 14 + 1)  14(3) 6 = 1988 = 2
#
Property (d) with ai = 2i 2 and bi = 3 Property (b) with c = 2 and ai = i 2 Summation rules; Property (a) Simplify.
772
Further Topics in Algebra
CHAPTER 11
EXAMPLE 8
Using the Summation Properties
6
Evaluate a (i 2 + 3i + 5). i=1
Analytic Solution 6
6
6
6
i=1
i=1
i=1
i=1
2 2 a (i + 3i + 5) = a i + a 3i + a 5 6
6
6
i=1
i=1
i=1
6
6
i=1
i=1
Property (c)
= a i2 + 3 a i + a 5
Property (b)
= a i 2 + 3 a i + 6(5) =
Property (a)
Graphing Calculator Solution FIGURE 8 confirms the analytic result. We use two commands on one line, to define the sequence and then compute its sum. (This approach is more efficient than using two separate commands, as seen in FIGURE 6 of Example 4.)
6(6 + 1)(2 # 6 + 1) 6(6 + 1) Summation + 3c d + 6(5) rules 6 2
= 91 + 3(21) + 6(5) = 184
Simplify. FIGURE 8
11.1
EXERCISES
Checking Analytic Skills Write the first five terms of each sequence. Do not use a calculator. 1. an = 4n + 10
2. an = 6n  3
1 n 5. an = a b (n  1) 3
6. an = ( 2) n(n)
8. an = ( 1) n  1(n + 1)
9. an =
9 11.
7. an = ( 1) n(2n)
4n  1 n2 + 2
Your friend does not understand what is meant by the nth term, or general term, of a sequence. How would you explain this idea?
Concept Check
4. an =  3n
3. an = 2 n  1
10. an =
9 12.
n2  1 n2 + 1
How are sequences related to functions?
Decide whether each sequence is finite or infinite.
13. The sequence of days of the week
14. The sequence of dates in the month of November
15. 1, 2, 3, 4
16.  1,  2,  3,  4
17. 1, 2, 3, 4, Á
18.  1,  2,  3,  4, Á
19. a1 = 3; for 2 … n … 10, an = 3 # an  1
20. a1 = 1; a2 = 3; for n Ú 3, an = an  1 + an  2
Find the first four terms of each sequence. 21. a1 =  2, an = an  1 + 3, for n 7 1 23. a1 = 1, a2 = 1, an = an  1 + an  2 , for n Ú 3 (the Fibonacci sequence)
22. a1 =  1, an = an  1  4, for n 7 1
24. a1 = 2, an = n # an  1, for n 7 1
11.1
Sequences and Series
773
Find the sum for each series. 5
6
25. a (2i + 1)
4 1 27. a j=1 j
26. a (3i  2)
i=1
i=1
4
4
i=1
i = 1
40. a 3( 2) i  34
5
38. a (2i 2 + 1)
i = 1
36. a 5(2) i
i = 2
6
37. a (i 2  2i)
2
35. a 2(3) i
i=3
5
i=1
3
34. a (5i + 2)
i=2
32. a ( 1) i + 1 # i 2
k=1
7
33. a (6  3i)
7
31. a ( 1) k # k
k=1
5
i=1
6
30. a (k + 1) 2
29. a i i
5
28. a (i + 1) 1
4
39. a (3i  4)
i=3
i=1
i=1
Evaluate the terms of each sum, where x1 =  2, x2 =  1, x3 = 0, x4 = 1, and x5 = 2. 5
5
5
42. a  xi
41. a xi i=1
i=1
3
5
46. a (xi 2 + 1)
i=1
44. a x i2
i=1
3
45. a (3xi  x i2)
4
43. a (2xi + 3)
i=1
xi + 1
5 xi 48. a i = 1 xi + 3
47. a i = 2 xi + 2
i=1
4
Evaluate the terms of a ƒ(xi) ¢ x, with x1 = 0, x2 = 2, x3 = 4, x4 = 6, and ¢ x = 0.5, for each function. i=1
49. ƒ(x) = 4x  7
50. ƒ(x) = 6 + 2x
52. ƒ(x) = x 2  1
53. ƒ(x) =
51. ƒ(x) = 2x 2
2 x + 1
54. ƒ(x) =
5 2x  1
Find the sum for each series. 100
20
55. a 6
15
i=1
i=1
5
60. a (8i  1) i=1
50
57. a i 2
56. a 5i
i=1
5
61. a (4i 2  2i + 6) i=1
6
i=1
62. a (2 + i  i 2) i=1
5
59. a (5i + 3)
58. a 2i 3
i=1
4
63. a (3i 3 + 2i  4) i=1
6
64. a (i 2 + 2i 3) i=1
Use a graphing calculator to graph the first 10 terms of each sequence. Make a conjecture as to whether the sequence converges or diverges. If you think it converges, determine the number to which it converges. 65. an =
n + 4 2n
68. an = n(n + 2)
66. an =
1 + 4n 2n
69. an = a1 +
Solve each problem. 71. Estimating p Find the sum of the first six terms of the series p4 1 1 1 1 1 1 = 4 + 4 + 4 + 4 + 4 + Á + 4 + Á 90 1 2 3 4 5 n presented in Example 6. Use your result to estimate p. Compare your answer with the actual value of p. 72. (Modeling) Insect Population Suppose an insect population density in thousands per acre during year n can be modeled by the recursively defined sequence a1 = 8 an = 2.9an  1  0.2an  1 2, for n 7 1.
1 n b n
67. an = 2e n 70. an = 5 
1 n
(a) Find the population for n = 1, 2, 3. 9 (b) Graph the sequence for n = 1, 2, 3, Á , 20. Use the window 30, 214 by 30, 144. Interpret the graph. 73. (Modeling) Bacteria Growth If certain bacteria are cultured in a medium with sufficient nutrients, they will double in size and then divide every 40 minutes. Let N1 be the initial number of bacteria cells, N2 the number after 40 minutes, N3 the number after 80 minutes, and Nj the number after 40( j  1) minutes. (Source: Hoppensteadt, F. and C. Peskin, Mathematics in Medicine and the Life Sciences, SpringerVerlag.) (a) Write Nj + 1 in terms of Nj for j Ú 1. (b) Determine the number of bacteria after two hours if N1 = 230.
774
CHAPTER 11
Further Topics in Algebra
(b) Graph the sequence Nj for j = 1, 2, 3, Á , 20. Use the window 30, 204 by 30, 60004. 9 (c) Describe the growth of these bacteria when there are limited nutrients. (d) 9 Make a conjecture as to why K is called the saturation constant. Test your conjecture by changing the value of K in the given formula.
(c) Graph the sequence Nj for j = 1, 2, 3, Á , 7. Use the window 30, 104 by 30, 15,0004. 9 (d) Describe the growth of these bacteria when there are unlimited nutrients. 74. (Modeling) Verhulst’s Model for Bacteria Growth Refer to Exercise 73. If the bacteria are not cultured in a medium with sufficient nutrients, competition will ensue and the growth will slow. According to Verhulst’s model, the number of bacteria Nj at time 40( j  1) minutes can be determined by the sequence Nj + 1 = c
75. Estimating Powers of e ea L 1 + a +
2 dN , 1 + (Nj >K) j
where
where K is a constant and j Ú 1. (Source: Hoppensteadt, F. and C. Peskin, Mathematics in Medicine and the Life Sciences, SpringerVerlag.) (a) If N1 = 230 and K = 5000, make a table of Nj for j = 1, 2, 3, Á , 20. Round values in the table to the nearest integer.
11.2
The series a3 an a2 + + Á + , 2! 3! n!
n! = 1 # 2 # 3 # 4
#
Á
#
n,
ea
can be used to estimate the value of for any real number a. Use the first eight terms of this series to approximate each expression. Compare this estimate with the actual value. Give values to six decimal places. (a) e (b) e 1 (c) 2e
Arithmetic Sequences and Series
Arithmetic Sequences • Arithmetic Series
Arithmetic Sequences A sequence in which each term after the first is obtained by adding a fixed number to the previous term is an arithmetic sequence (or arithmetic progression). The fixed number that is added is the common difference. The sequence 5, 9, 13, 17, 21, Á is an arithmetic sequence because each term after the first is obtained by adding 4, the common difference, to the previous term. That is, 9 = 5 + 4,
13 = 9 + 4,
17 = 13 + 4,
21 = 17 + 4,
and so on. The common difference is 4. If the common difference of an arithmetic sequence is d, then, for every positive integer n in its domain, d ⴝ a nⴙ 1 ⴚ a n.
Common difference d
EXAMPLE 1
Finding the Common Difference Find the common difference d of the arithmetic sequence  9,  7,  5,  3,  1, Á . Solution We find d by choosing any two consecutive terms and subtracting the first from the second. Choosing  7 and  5 gives Be careful when subtracting a negative number. d =  5  ( 7) = 2. Choosing  9 and  7 gives d =  7  ( 9) = 2, the same result.
11.2
Arithmetic Sequences and Series
775
EXAMPLE 2
Finding Terms Given a1 and d Write the first five terms of each arithmetic sequence. (a) The first term is 7 and the common difference is  3. (b) a1 =  12, d = 5 Solution (a) a1 = a2 = a3 = a4 = a5 = (b) a1 a2 a3 a4 a5
= = = = =
7 7 + ( 3) = 4 4 + ( 3) = 1 1 + ( 3) =  2  2 + ( 3) =  5
a1 = 7, d =  3
 12  12 + 5 =  7 7 + 5 = 2 2 + 5 = 3 3 + 5 = 8
Starting with a1, add d to each term to get the next term.
If a1 is the first term of an arithmetic sequence and d is the common difference, then the terms of the sequence are given by a1 a2 a3 a4 a5 a6 and, in general,
= = = = = =
a1 a1 a2 a3 a4 a5
+ + + + +
d d d d d
= = = =
a1 a1 a1 a1
+ + + +
d + d = a1 + 2d 2d + d = a1 + 3d 3d + d = a1 + 4d 4d + d = a1 + 5d,
an = a1 + (n  1)d.
nth Term of an Arithmetic Sequence In an arithmetic sequence with first term a1 and common difference d, the nth term is given by a n ⴝ a 1 ⴙ (n ⴚ 1)d.
EXAMPLE 3
Finding Terms of an Arithmetic Sequence Find a13 and an for the arithmetic sequence  3, 1, 5, 9, Á . Solution Here, a1 =  3 and d = 1  ( 3) = 4. First find a13.
Work inside the parentheses first.
an = a1 + (n  1)d a13 = a1 + (13  1)d
Formula for the nth term
a13 =  3 + (12)4 a13 = 45
a1 =  3, d = 4
n = 13
Simplify.
Find an by substituting values for a1 and d in the formula for an. an =  3 + (n  1) # 4 an =  3 + 4n  4 an = 4n  7
a1 =  3, d = 4 Distributive property Simplify.
776
Further Topics in Algebra
CHAPTER 11
EXAMPLE 4
Finding Terms of an Arithmetic Sequence Find a18 and an for the arithmetic sequence having a2 = 9 and a3 = 15. Solution Find d first; d = a3  a2 = 15  9 = 6. a2 = a1 + d,
Since
9 = a1 + 6
a2 = 9, d = 6
a1 = 3. Then,
and
Simplify.
a18 = 3 + (18  1)6
Formula for an ; a1 = 3, n = 18, d = 6
a18 = 105,
Simplify.
an = 3 + (n  1)6
Formula for an
an = 3 + 6n  6
Distributive property
an = 6n  3.
Simplify.
EXAMPLE 5
Finding the First Term of an Arithmetic Sequence Suppose that an arithmetic sequence has a8 =  16 and a16 =  40. Find a1. Solution Since a16 = a8 + 8d, 8d = a16  a8 =  40  ( 16) =  24, so d =  3. To find a1, use the equation a8 = a1 + 7d.  16 = a1 + 7d
a 8 =  16
 16 = a1 + 7( 3)
d = 3
a1 = 5
an 14
Simplify.
The graph of a sequence consists of discrete points. To determine the characteristics of the graph of an arithmetic sequence, start by rewriting the formula for the nth term.
12 10 8
an = a1 + (n  1)d
6 4 2 0
1
2
3
4
5
6
7
n
(a)
Formula for the nth term
= a1 + nd  d
Distributive property
= dn + (a1  d)
Commutative and associative properties
= dn + c
c = a1  d
The points on the graph of an arithmetic sequence are determined by ƒ(n) = dn + c, where n is a natural number. Thus, the points on the graph of ƒ must lie on the line
bn 14
y = dx + c.
12 10 8
Slope yintercept
6 4 2 n 0
1
2
3
4
5
(b) FIGURE 9
6
7
For example, the sequence an shown in FIGURE 9(a) is an arithmetic sequence because the points that make up its graph are collinear (lie on a line). The slope determined by these points is 2, so the common difference d equals 2. The sequence bn shown in FIGURE 9(b) is not an arithmetic sequence, however, because the points are not collinear.
11.2
Arithmetic Sequences and Series
777
EXAMPLE 6
Finding the nth Term from a Graph Find a formula for the nth term of the sequence an shown in FIGURE 10. What are the domain and range of this sequence? an 4.0 3.5 3.0 2.5 2.0 1.5 1.0 0.5 n 0
an
1
2
3
4
5
6
7
FIGURE 10
4.0 3.5 3.0 2.5 2.0 1.5 1.0 0.5
Solution The points in FIGURE 10 lie on a line, so the sequence is arithmetic. The equation of the dashed line shown in FIGURE 11 is y =  0.5x + 4, so the nth term of this sequence is determined by an =  0.5n + 4. n 0
1
2
3
4
5
FIGURE 11
6
7
The sequence is composed of the points (1, 3.5), (2, 3), (3, 2.5), (4, 2), (5, 1.5), and (6, 1). Therefore, the domain of the sequence is 51, 2, 3, 4, 5, 66, and the range is 51, 1.5, 2, 2.5, 3, 3.56.
Arithmetic Series The sum of the terms of an arithmetic sequence is an arithmetic series. To illustrate, suppose that a person borrows $3000 and agrees to pay $100 per month plus interest of 1% per month on the unpaid balance until the loan is paid off. The first month, $100 is paid to reduce the loan, plus interest of (0.01)3000 = 30 dollars. The second month, another $100 is paid toward the loan, and (0.01)2900 = 29 dollars is paid for interest. Since the loan is reduced by $100 each month, interest payments decrease by (0.01)100 = 1 dollar each month, forming the arithmetic sequence 30, 29, 28, Á , 3, 2, 1. The total interest paid is given by the sum of the terms of this sequence. Now we develop a formula to find that sum without adding all 30 numbers directly. Since the sequence is arithmetic, we can write the sum of the first n terms as S = a + 3a + d4 + 3a + 2d4 + Á + 3a + (n  1)d4. (1) n
1
1
1
1
We used the formula for the general term in the last expression. Now we write the same sum in reverse order, beginning with an and subtracting d. S = a + 3a  d4 + 3a  2d4 + Á + 3a  (n  1)d4 (2) n
n
n
n
n
Adding respective sides of equations (1) and (2) term by term, we obtain S + S = (a + a ) + (a + a ) + Á + (a + a ) n
n
1
n
1
n
2Sn = n(a1 + an) n Sn = (a1 + an ) 2 n Sn = 3a1 + a1 + (n  1)d4 2 n Sn = 32a1 + (n  1)d4. 2
1
n
There are n terms of ( a1 + an ). Divide by 2. an = a1 + ( n  1)d Alternative form
778
CHAPTER 11
Further Topics in Algebra
FOR DISCUSSION Explain why there is no formula for the sum of the terms of an infinite arithmetic sequence.
Sum of the First n Terms of an Arithmetic Sequence If an arithmetic sequence has first term a1 and common difference d, then the sum of the first n terms is given by Sn ⴝ
n n (a 1 ⴙ a n ) or Sn ⴝ 32a 1 ⴙ (n ⴚ 1)d4. 2 2
The first formula is used when the first and last terms are known; otherwise, the second formula is used. For example, in the sequence of interest payments given earlier, n (a + an) 2 1 30 (30 + 1) = 15(31) = 465, = 2
Sn = gives
S30
First formula for Sn n = 30, a1 = 30, an = 1
so a total of $465 interest will be paid over the 30 months. EXAMPLE 7
Using the Sum Formulas
(a) Evaluate S12 for the arithmetic sequence  9,  5,  1, 3, 7, Á . (b) Use a formula for Sn to evaluate the sum of the first 60 positive integers. Solution (a) We want the sum of the first 12 terms. n Sn = 32a1 + (n  1)d4 Second formula for Sn 2 12 32( 9) + 11(4)4 = 156 S12 = n = 12, a1 =  9, d = 4 2 n (b) Sn = (a1 + an) First formula for Sn 2 60 (1 + 60) = 1830 S60 = n = 60, a1 = 1, a60 = 60 2 EXAMPLE 8
Using the Sum Formulas The sum of the first 17 terms of an arithmetic sequence is 187. If a17 =  13, find a1 and d. Solution 17 (a + a17) 2 1 17 187 = (a  13) 2 1 22 = a1  13 S17 =
a1 = 35
Use the first formula for Sn , with n = 17. S17 = 187, a17 =  13 Multiply by
2 17 .
Simplify.
Since a17 = a1 + (17  1)d,  13 = 35 + 16d  48 = 16d d =  3.
a17 =  13, a1 = 35 Subtract 35. Divide by 16; rewrite.
11.2
Arithmetic Sequences and Series
779
Any sum of the form n
a (di ⴙ p),
iⴝ 1
where d and p are real numbers, represents the sum of the terms of an arithmetic sequence having first term a1 = d(1) + p = d + p and common difference d. These sums can be evaluated with the formulas in this section.
EXAMPLE 9
Using Summation Notation
Evaluate each sum. 10
(a) a (4i + 8) i=1
9
(b) a (4  3k) k=3
Analytic Solution (a) This sum contains the first 10 terms of an arithmetic sequence. a1 = 4 # 1 + 8 = 12 a10 = 4 # 10 + 8 = 48
First term Last term
10
Thus,
10 a (4i + 8) = S10 = 2 (12 + 48) = 300. i=1
Graphing Calculator Solution The screen in FIGURE 12 shows the sums for the series in parts (a) and (b). These results agree with the analytic solutions. Remember that a series is the sum of the terms of a sequence.
(b) We use the second formula instead of the first. The first few terms are 34  3(3)4 + 34  3(4)4 + 34  3(5)4 + Á =  5 + ( 8) + ( 11) + Á . Thus a1 =  5 and d =  3. If the sequence started with k = 1, there would be nine terms. Since it starts at 3, two of those terms are missing, so there are seven terms and n = 7. FIGURE 12
7 a (4  3k) = 2 32( 5) + 6( 3)4 =  98 k=3 9
11.2
EXERCISES
Checking Analytic Skills Find the common difference d for each arithmetic sequence. Do not use a calculator. 1. 2, 5, 8, 11, Á
2. 4, 10, 16, 22, Á
3. 3,  2,  7,  12, Á
4.  8,  12,  16,  20, Á
5. x + 3y, 2x + 5y, 3x + 7y, Á
6. t 2 + q,  4t 2 + 2q,  9t 2 + 3q, Á
Checking Analytic Skills Write the first five terms of each arithmetic sequence. Do not use a calculator. 7. The first term is 8, and the common difference is 6. 9. a1 = 5, d =  2
10. a1 = 4, d = 3
8. The first term is  2, and the common difference is 12. 11. a3 = 10, d =  2
12. a1 = 3  22, a2 = 3
Find a8 and an for each arithmetic sequence. 13. a1 = 5, d = 2
14. a1 =  3, d =  4
15. a3 = 2, d = 1
16. a4 = 5, d =  2
17. a1 = 8, a2 = 6
18. a1 = 6, a2 = 3
780
Further Topics in Algebra
CHAPTER 11
19. a10 = 6, a12 = 15
20. a15 = 8, a17 = 2
21. a1 = x, a2 = x + 3
22. a2 = y + 1, d =  3
23. a3 = p + 22e, a4 = p + 32e
24. a2 = 26 + 322, a6 = 26 + 1522
Find a1 for each arithmetic sequence. 25. a5 = 27, a15 = 87
26. a12 = 60, a20 = 84
27. S16 =  160, a16 =  25
28. S28 = 2926, a28 = 199
Find the sum of the first 10 terms of each arithmetic sequence. 29. a1 = 8, d = 3
30. a1 =  9, d = 4
31. a3 = 5, a4 = 8
32. a2 = 9, a4 = 13
33. 5, 9, 13, Á
34. 8, 6, 4, Á
35. a1 = 10, a10 = 5.5
36. a1 =  8, a10 =  1.25
39. S12 =  108, a12 =  19
40. S25 = 650, a25 = 62
Find a1 and d for each arithmetic sequence. 37. S20 = 1090, a20 = 102
38. S31 = 5580, a31 = 360
Find a formula for the nth term of the arithmetic sequence shown in each graph. Then state the domain and range of the sequence. an
41.
42. 2
2 4
3 n
0
n
0
2
6
4
2
6
–2
–2
an
43.
an
1 n 0 1 2 3 4 5 6 7
44.
45.
an
10
15 10 5 0 –5
46.
an
–10
n 1
3
5
n 3
1
–50
5
5 3
–30 1
an
–1
n 1
3
5
–70
Use a formula to find the sum of each arithmetic series. 47. 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17
48. 7.5 + 6 + 4.5 + 3 + 1.5 + 0 + ( 1.5)
49. 1 + 2 + 3 + 4 + Á + 50
50. 1 + 3 + 5 + 7 + Á + 97
51.  7 + ( 4) + ( 1) + 2 + 5 + Á + 98 + 101
52. 89 + 84 + 79 + 74 + Á + 9 + 4
53. The first 40 terms of the series given by an = 5n
54. The first 50 terms of the series given by an = 1  3n
Evaluate each sum. 3
55. a (i + 4) i=1 12
59. a ( 5  8i ) i=1
5
56. a (i  8) i=1 19
60. a ( 3  4k) k=1
10
57. a (2j + 3) j=1
15
58. a (5j  9) j=1
1000
61. a i i=1
2000
62. a k k=1
Use the sequence feature of a graphing calculator to evaluate the sum of the first 10 terms of the arithmetic sequence. In Exercises 65 and 66, round to the nearest thousandth. 63. an = 4.2n + 9.73
64. an = 8.42n + 36.18
65. an = 28n + 23
66. an =  2 3 4n + 27
11.3
Solve each problem. 67. Integer Sum to 71.
Find the sum of all the integers from 51
68. Integer Sum Find the sum of all the integers from  8 to 30.
Geometric Sequences and Series
781
of the sequence. A child who measures 96 centimeters at age 3 would have his expected height in subsequent years represented by the sequence 102, 108, 114, 120, 126, 132, 138, 144. Each term differs from the adjacent terms by the common difference, 6.
69. Clock Chimes If a clock strikes the proper number of chimes each hour on the hour, how many times will it chime in a month of 30 days? 70. Telephone Pole Stack A stack of telephone poles has 30 in the bottom row, 29 in the next, and so on, with one pole in the top row. How many poles are in the stack? 71. Population Growth Five years ago, the population of a city was 49,000. Each year, the zoning commission permits an increase of 580 in the population. What will the maximum population be 5 years from now? 72. Supports on a Slide A slide of uniform slope is to be built on a level piece of land. There are to be 20 equally spaced supports, with the longest support 15 meters long and the shortest 2 meters long. Find the total length of all the supports.
(a) If a child measures 98.2 centimeters at age 3 and 109.8 centimeters at age 5, what would be the common difference of the arithmetic sequence describing his yearly height? (b) What would we expect his height to be at age 8?
73. Rungs of a Ladder How much material would be needed for the rungs of a ladder of 31 rungs if the rungs taper uniformly from 18 inches to 28 inches?
75. Concept Check Find all arithmetic sequences a1, a2, a3, Á such that a1 2, a2 2, a3 2, Á is also an arithmetic sequence.
74. (Modeling) Growth Pattern for Children The normal growth pattern for children aged 3–11 follows an arithmetic sequence. An increase in height of about 6 centimeters per year is expected. Thus, 6 would be the common difference
76. Suppose that a1, a2, a3, Á and b1, b2, b3, Á are arithmetic sequences. Let dn = an + c # bn, for any real number c and every positive integer n. Show that d1, d2, d3, Á is an arithmetic sequence.
11.3
Geometric Sequences and Series
Geometric Sequences • Geometric Series • Infinite Geometric Series • Annuities
Geometric Sequences Suppose you agreed to work for 1¢ the first day, 2¢ the second day, 4¢ the third day, 8¢ the fourth day, and so on, with your wages doubling each day. How much will you earn on day 20? How much will you have earned altogether in 20 days? These questions will be answered in this section. A geometric sequence (or geometric progression) is a sequence in which each term after the first is obtained by multiplying the preceding term by a constant nonzero real number, called the common ratio. The sequence of wages 1, 2, 4, 8, 16, Á is an example of a geometric sequence in which the first term is 1 and the common ratio is 2. Notice that if we divide any term (except the first) by the preceding term, we obtain the common ratio r = 2. a2 a3 a4 a5 2 4 8 16 = = 2; = = 2; = = 2; = = 2 a1 a2 a3 a4 1 2 4 8
782
CHAPTER 11
Further Topics in Algebra
If the common ratio of a geometric sequence is r, then, by the definition of a geometric sequence, rⴝ
a nⴙ 1 an
Common ratio r
for every positive integer n. Therefore, the common ratio can be found by choosing any term except the first and dividing it by the preceding term. In the geometric sequence 2, 8, 32, 128, Á , we have r = 4. Notice that 8 = 2 # 4
32 = 8 # 4 = (2 # 4) # 4 = 2 # 42
128 = 32 # 4 = (2 # 42) # 4 = 2 # 43. To generalize this pattern, assume that a geometric sequence has first term a1 and common ratio r. The second term can be written as a2 = a1r, the third can be written as a3 = a2 r = (a1r)r = a1 r 2, and so on. Following this pattern, the nth term is an = a1r n  1.
nth Term of a Geometric Sequence In a geometric sequence with first term a1 and common ratio r, neither of which is zero, the nth term is given by a n ⴝ a 1 r nⴚ 1.
EXAMPLE 1
Finding the nth Term of a Geometric Sequence of Wages The formula for the nth term of a geometric sequence can be used to answer the first question posed at the beginning of this section. How much will be earned on day 20 if daily wages follow the sequence 1, 2, 4, 8, 16, Á ? Solution To answer the question, let a1 = 1 and r = 2, and find a20. a20 = a1 r 19 = 1(2) 19 = 524,288 cents, or $5242.88
EXAMPLE 2
Using the Formula for the nth Term Find a5 and an for the geometric sequence 4,  12, 36,  108, Á . Solution The first term, a1, is 4. Find r by choosing any term except the first and dividing it by the preceding term. For example, r = 3612 =  3. Since a4 =  108, a5 =  3( 108) = 324. The fifth term also could be found by using the formula an = a1 r n1 and replacing n with 5, r with  3, and a1 with 4. a5 = 4 # ( 3) 5  1 = 4 # ( 3) 4 = 324
By the formula,
an = 4 # ( 3) n  1.
11.3
Geometric Sequences and Series
783
EXAMPLE 3
Using the Formula for the nth Term Find a1 and r for the geometric sequence with third term 20 and sixth term 160. Solution Use the formula for the nth term of a geometric sequence. For n = 3, a3 = a1 r 2 = 20; for n = 6, a6 = a1 r 5 = 160. Because a1 r 2 = 20, a1 =
20 . r2
Substitute this value for a1 in the second equation. a1 r 5 = 160
a
20 5 br r2 20r 3 r3 r
= 160 = 160 = 8 = 2
Substitute. r5 r2
= r 52 = r3
Divide by 20. Take cube roots.
Since a1 r 2 = 20 and r = 2, a1(2) 2 = 20 4a1 = 20 a1 = 5.
Substitute. Apply the exponent. Divide by 4.
EXAMPLE 4
Modeling a Population of Fruit Flies A population of fruit flies is growing in such a way that each generation is 1.5 times as large as the last generation. Suppose there were 100 insects in the first generation. How many would there be in the fourth generation? Solution The population can be written as a geometric sequence with a1 as the firstgeneration population, a2 the secondgeneration population, and so on. Then the fourthgeneration population is a4. Use the formula for an. a4 = a1 r 3 = 100(1.5) 3 = 100(3.375) = 337.5
n = 4, a1 = 100, r = 1.5
In the fourth generation, the population numbers about 338 insects.
Geometric Series A geometric series is the sum of the terms of a geometric sequence. In applications, it may be necessary to find the sum of the terms of such a sequence. For example, a scientist might want to know the total number of insects in four generations of the population discussed in Example 4. This population would equal a1 + a2 + a3 + a4, or 100 + 100(1.5) + 100(1.5) 2 + 100(1.5) 3 = 812.5 L 813 insects. To find a formula for the sum Sn of the first n terms of a geometric sequence, first write the sum as S = a + a + a + Á + a n
or
1
2
n
3
Sn = a1 + a1 r + a1
r2
+ Á + a1 r n1.
(1)
If r = 1, Sn = na1, which is a correct formula for this case. If r Z 1, multiply each side of equation (1) by r to obtain rSn = a1 r + a1 r 2 + a1 r 3 + Á + a1 r n. (2)
784
CHAPTER 11
Further Topics in Algebra
If equation (2) is subtracted from equation (1), Sn = a1 + a1 r + a1 r 2 + Á + a1 r n  1 rSn = a1 r + a1 r 2 + Á + a1 r n  1 + a1 r n Pay attention to this step involving factoring.
 a1
Sn  rSn = a1 Sn (1  r) = a1 (1  r n) Sn =
a1(1 
r n)
1  r
rn
(1) (2) Subtract. Factor.
, where r Z 1.
Divide by 1  r.
Sum of the First n Terms of a Geometric Sequence If a geometric sequence has first term a1 and common ratio r, then the sum of the first n terms is given by Sn ⴝ
a 1(1 ⴚ r n) 1ⴚr
, where r Z 1.
We can use this formula to find the total fruit fly population in Example 4 over the fourgeneration period. With n = 4, a1 = 100, and r = 1.5, S4 =
100(1  5.0625) 100(1  1.54) = = 812.5 L 813 insects. 1  1.5  0.5
Same result
EXAMPLE 5
Applying the Sum of the First n Terms At the beginning of this section, we posed the following question: How much will you have earned altogether after 20 days? Analytic Solution We must find the total amount earned in 20 days with daily wages of 1, 2, 4, 8, Á cents. Since a1 = 1 and r = 2,
Graphing Calculator Solution See FIGURE 13, which confirms the analytic solution.
1(1  2 20) 1  2 = 1,048,575 cents, or $10,485.75.
S20 =
FIGURE 13
EXAMPLE 6
Finding the Sum of the First n Terms
6
Find a 2 # 3i. i=1
Solution This series is the sum of the first six terms of a geometric sequence having a1 = 2 # 31 = 6 and r = 3. From the formula for Sn, S6 =
6(1  36) 6(1  729) 6( 728) = = = 2184. 1  3 2 2
11.3
Geometric Sequences and Series
785
Infinite Geometric Series We can extend our discussion of sums of sequences to include infinite geometric sequences such as 2, 1,
1 1 1 1 , , , ,Á, 2 4 8 16
with first term 2 and common ratio 12. Using the formula for Sn gives the sequence of sums S1 = 2,
S2 = 3,
S3 = 3.5,
S4 = 3.75,
S5 = 3.875,
S6 = 3.9375.
With a calculator in function mode, we define Y1 as Sn. 2 A 1  A 12 B B X
Y1 =
FIGURE 14
1 
1 2
Using the table in FIGURE 14, we can observe the first few terms as X takes on the values 1, 2, 3, Á . These terms seem to be getting closer and closer to the number 4. For no value of n is Sn = 4. However, if n is large enough, then Sn is as close to 4 as desired. As mentioned earlier, we say that the sequence converges to 4, expressed as lim Sn = 4.
n:q
Looking Ahead to Calculus
In calculus, different types of infinite series are studied. One important question to answer for each series is whether lim Sn q n:
(Read: “The limit of Sn as n increases without bound is 4.”) Since limq Sn = 4, the n: number 4 is called the sum of the terms of the infinite geometric sequence 1 1 2, 1, , , Á , 2 4
and
2 + 1 +
1 1 + + Á = 4. 2 4
converges to a real number. In the discussion of lim Sn = 4,
n:q
we used the phrases “large enough” and “as close as desired.” This description is made more precise in a standard calculus course.
WHAT WENT WRONG? The preceding discussion justified the statement that the sum of the terms 2, 1, 12, 14, 1 1 8 , 16 , Á gets closer and closer to 4 by adding more and more terms of the sequence. However, this sum will always differ from 4 by some small amount. The following figure is an extension of the table in FIGURE 14. According to the calculator, the sum is 4 when X Ú 17.
What Went Wrong? this example.
Discuss the limitations of technology, basing your comments on
Answers to What Went Wrong? The calculator has room to display numbers to only four decimal places in the table. When the value of Y1 reaches 3.99995 or greater, the calculator rounds up to 4. Try positioning the cursor on the 4 next to the 17 in the table. At the bottom of the screen a more accurate value of 3.99996948242 will appear for Y1.
786
CHAPTER 11
Further Topics in Algebra
EXAMPLE 7
Summing the Terms of an Infinite Geometric Sequence 1 1 1 + Á. Find 1 + + + 3 9 27 Solution Use the formula for the sum of the first n terms of a geometric sequence to obtain S1 = 1,
4 , 3
S2 =
1 C 1  A 13 B D
13 , 9
S3 =
S4 =
40 , 27
n
Sn =
and, in general,
1 
1 3
a1 = 1, r =
.
1 3
The following table shows the approximate value of A 13 B for larger and larger values of n. n
n
1
10
100
200
1 n 3
1 3
0.0000169
1.94 * 10 48
3.76 * 10 96
A B
As n gets larger and larger, A 13 B gets closer and closer to 0. That is, n
1 n limq a b = 0, n: 3
2
making it reasonable that 1 C 1  A 13 B D n
0
limq Sn = limq
n:
6 0 This graph of the first six values of Sn in Example 7 shows its value approaching 3 1 2 . (The yscale here is 2 .)
1 
n:
1 3
=
1(1  0) 1 
1 3
=
1 2 3
=
3 . 2
Be careful simplifying a complex fraction.
Hence,
1 +
1 1 1 3 + + + Á = . 3 9 27 2
If a geometric sequence has first term a1 and common ratio r, then Sn =
a1(1  r n)
(r Z 1)
1  r
for every positive integer n. If  1 6 r 6 1, then limq r n = 0, and n:
lim Sn = q
n:
This quotient, 1
a1(1  0) 1  r
=
a1 1  r
a1  r , is called the sum of the terms of an infinite geometric sequence. ˆ
The limit limˆ Sn is often expressed as Sˆ or a a i . n: iⴝ 1
.
11.3
Geometric Sequences and Series
Looking Ahead to Calculus
Sum of the Terms of an Infinite Geometric Sequence
In calculus, you will find sums of the terms of infinite sequences that are not geometric.
The sum of the terms of an infinite geometric sequence with first term a1 and common ratio r, where  1 6 r 6 1, is given by Sˆ ⴝ
a1 1ⴚr
787
.
If  r  7 1, the terms get larger and larger in absolute value, so there is no limit as n : q. Hence, the terms of the sequence will not have a sum.
EXAMPLE 8
Finding Sums of the Terms of Infinite Geometric Sequences
Find each sum. q 3 1 i1 (a) a a b a b 4 2 i=1
q 3 i (b) a a b 5 i=1
Solution (a) Here, a1 =  34 and r =  12. Since  1 6 r 6 1, the preceding formula applies. Sq =
a1
=
1  r
 34
1  A
 12
B
=
 34 3 2
= 
3 3 3 , = 4 2 4
#
2 1 = 3 2
Be careful simplifying a complex fraction. 3
q 3 i 5 (b) a a b = 5 1 i=1
3 5
=
3 5 2 5
=
3 2 3 , = 5 5 5
#
5 3 = 2 2
a1 = 35 , r =
3 5
Annuities A sequence of equal payments made at equal intervals, such as car payments or house payments, is called an annuity. If the payments are accumulated in an account (with no withdrawals), the sum of the payments and interest on the payments is called the future value of the annuity. The formula that follows is derived from the formula for the sum of the terms of a geometric sequence.
Future Value of an Annuity The formula for the future value of an annuity is S ⴝ Rc
(1 ⴙ i) n ⴚ 1 d, i
where S is the future value, R is the payment at the end of each period, i is the interest rate in decimal form per period, and n is the number of periods.
788
CHAPTER 11
Further Topics in Algebra
EXAMPLE 9
Finding the Future Value of an Annuity To save money for a trip, Callie Daniels deposited $1000 at the end of each year for 4 years in an account paying 3% interest compounded annually. Find the future value of this annuity. Solution Use the formula for S, with R = 1000, i = 0.03, and n = 4. S = 1000 c
(1 + 0.03) 4  1 d L 4183.63 0.03
The future value of the annuity is $4183.63.
11.3
EXERCISES
Checking Analytic Skills Write the terms of the geometric sequence that satisfies the given conditions. Do not use a calculator. 1. a1 =
5 , r = 3, n = 4 3
2 3 2. a1 =  , r = , n = 4 4 3
3. a4 = 5, a5 = 10, n = 5
4. a3 = 16, a4 = 8, n = 5
Checking Analytic Skills Find a5 and an for each geometric sequence. Do not use a calculator. 5. a1 = 5, r =  2
6. a1 = 8, r =  5
8. a3 =  2, r = 4
9. a4 = 243, r =  3
11.  4,  12,  36,  108, Á
14.
1 2 8 32 , , , ,Á 2 3 9 27
7. a2 =  4, r = 3 10. a4 = 18, r = 2 25 4 , 2, 5, , Á 5 2
12.  2, 6,  18, 54, Á
13.
5 5 15. 10,  5, ,  , Á 2 4
9 27 81 16. 3,  , ,  , Á 4 16 64
Find a1 and r for each geometric sequence. 17. a3 = 5, a8 =
1 625
18. a2 =  6, a7 =  192
1 1 19. a4 =  , a9 = 4 128
20. a3 = 50, a7 = 0.005
Use the formula for Sn to find the sum of the first five terms for each geometric sequence. Round the answers for Exercises 25 and 26 to the nearest hundredth. 21. 2, 8, 32, 128, Á
22. 4, 16, 64, 256, Á
4 4 24. 12,  4, ,  , Á 3 9
9 9 23. 18,  9, ,  , Á 2 4
25. a1 = 8.423, r = 2.859
26. a1 =  3.772, r =  1.553
Use a formula to find each sum. 5
4
27. a 3i
28. a ( 2) i
2 j 30. a 243 a b 3 j=1
31. a 2 k
i=1 5
i=1 10
k=4
33. Concept Check Under what conditions does the sum of the terms of an infinite geometric sequence exist?
6 1 j 29. a 48a b 2 j=1 9
32. a ( 3) k k=3
34. Concept Check The number 0.999 Á can be written as the sum of the terms of an infinite geometric sequence: 0.9 + 0.09 + 0.009 + Á . Here we have a1 = 0.9 and r = 0.1. Use the formula for Sq to find this sum. Does your intuition indicate that your answer is correct?
11.3
Geometric Sequences and Series
789
Write the sum of each geometric series as a rational number. (See Exercise 34.) 35. 0.8 + 0.08 + 0.008 + 0.0008 + Á
36. 0.7 + 0.07 + 0.007 + 0.0007 + Á
37. 0.45 + 0.0045 + 0.000045 + Á
38. 0.36 + 0.0036 + 0.000036 + Á
Find r for each infinite geometric sequence. Identify any whose sum does not converge. 39. 12, 24, 48, 96, Á
40. 625, 125, 25, 5, Á
41.  48,  24,  12,  6, Á
42. 2,  10, 50,  250, Á
Find each sum that converges. 43. 16 + 2 +
1 1 + + Á 4 32
44. 18 + 6 + 2 +
46. 128 + 64 + 32 + Á q 1 i1 49. a 3 a b 4 i=1
47.
2 + Á 3
4 2 1 + + + Á 3 3 3
q 1 i1 50. a 5a  b 4 i=1
45. 100 + 10 + 1 + Á 48.
q
51. a (0.3) k k=1
1 1 1 2  + + Á 4 6 9 27 q
52. a 10 k k=1
Use a graphing calculator to evaluate each sum. Round to the nearest thousandth. 10
6
54. a  (3.6) j
53. a (1.4) i i=1
Annuity Values
j=1
Find the future value of each annuity.
57. Payments of $1000 at the end of each year for 9 years at 4% interest compounded annually 58. Payments of $800 at the end of each year for 12 years at 3% interest compounded annually 59. Payments of $2430 at the end of each year for 10 years at 2.5% interest compounded annually 60. Payments of $1500 at the end of each year for 6 years at 1.5% interest compounded annually Solve each problem. 61. (Modeling) Investment for Retirement According to T. Rowe Price Associates, a person who has a moderate investment strategy and n years to retirement should have accumulated savings of an percent of his or her annual salary. The geometric sequence defined by an = 1276(0.916) n gives the appropriate percent for each year n. (a) Find a1 and r. 9 (b) Find and interpret the terms a10 and a20. 62. (Modeling) Investment for Retirement Refer to Exercise 61. For someone who has a conservative investment strategy with n years to retirement, the geometric sequence is an = 1278(0.935) n. (Source: T. Rowe Price Associates.) (a) Repeat part (a) of Exercise 61. 9 (b) Repeat part (b) of Exercise 61. 9 (c) Why are the answers in parts (a) and (b) larger than in Exercise 61?
8
55. a 2(0.4) j j=3
9
56. a 3(0.25) i i=4
63. (Modeling) Bacteria Growth The strain of bacteria described in Exercise 73 in the first section of this chapter will double in size and then divide every 40 minutes. Let a1 be the initial number of bacteria cells, a2 the number after 40 minutes, and an the number after 40(n  1) minutes. (Source: Hoppensteadt, F. and C. Peskin, Mathematics in Medicine and the Life Sciences, SpringerVerlag.) (a) Write a formula for the nth term an of the geometric sequence a1, a2, a3, Á , an, Á . (b) Determine the first value for n where an 7 1,000,000 if a1 = 100. (c) How long does it take for the number of bacteria to exceed 1 million? 64. Photo Processing The final step in processing a blackandwhite photographic print is to immerse the print in a chemical fixer. The print is then washed in running water. Under certain conditions, 98% of the fixer will be removed with 15 minutes of washing. How much of the original fixer would be left after 1 hour of washing? 65. Chemical Mixture A scientist has a vat containing 100 liters of a pure chemical. Twenty liters are drained and replaced with water. After complete mixing, 20 liters of the mixture are again drained and replaced with water. What will be the strength of the mixture after nine such drainings? 66. HalfLife of a Radioactive Substance The halflife of a radioactive substance is the time it takes for half the substance to decay. Suppose the halflife of a substance is 3 years and 10 15 molecules of the substance are present initially. How many molecules will be present after 15 years?
790
CHAPTER 11
Further Topics in Algebra
67. Depreciation in Value Each year a machine loses 20% of the value it had at the beginning of the year. Find the value of the machine at the end of 6 years if it cost $100,000 new. 68. Sugar Processing A sugar factory receives an order for 1000 units of sugar. The production manager thus orders the production of 1000 units of sugar. He forgets, however, that the production of sugar requires some sugar (to prime the machines, for example), so he ends up with only 900 units of sugar. He then orders an additional 100 units, and receives only 90 units. A further order for 10 units produces 9 units. Finally seeing that he is wrong, the manager decides to try mathematics. He views the production process as an infinite geometric progression with a1 = 1000 and r = 0.1. Find the number of units of sugar that he should have ordered originally.
72. (Modeling) Drug Dosage Certain medical conditions are treated with a fixed dose of a drug administered at regular intervals. Suppose that a person is given 2 milligrams of a drug each day and that during each 24hour period the body utilizes 40% of the amount of drug that was present at the beginning of the period. (a) Show that the amount of the drug present in the body at the end of n days is n i a 2(0.6) .
i=1
(b) What will be the approximate quantity of the drug in the body at the end of each day after the treatment has been administered over a long period? 73. Salaries You are offered a 6week summer job and are asked to select one of the following salary options:
69. Swing of a Pendulum A pendulum bob swings through an arc 40 centimeters long on its first swing. Each swing thereafter, it swings only 80% as far as on the previous swing. How far will it swing altogether before coming to a complete stop?
Option 1: $5000 for the first day, with a $10,000 raise each day for the remaining 29 days (that is, $15,000 for day 2, $25,000 for day 3, and so on)
70. Height of a Dropped Ball Beth Schiffer drops a ball from a height of 10 meters and notices that on each bounce the ball returns to about 34 of its previous height. About how far will the ball travel before it comes to rest? (Hint: Consider the sum of two sequences.)
Which option would you choose?
Option 2: 1¢ for the first day, with the pay doubled each day thereafter (that is, 2¢ for day 2, 4¢ for day 3, and so on) 74. Number of Ancestors Suppose a genealogical Web site allows you to identify all your ancestors that lived during the last 300 years. Assuming that each generation spans about 25 years, guess the number of ancestors that would be found during the 12 generations. Then use the formula for a geometric series to find the actual value. 75. Side Length of a Triangle A sequence of equilateral triangles is constructed. The first triangle has sides 2 meters in length. To get the second triangle, midpoints of the sides of the original triangle are connected. What is the length of the side of the eighth such triangle? See the figure.
71. Number of Ancestors Each person has two parents, four grandparents, eight greatgrandparents, and so on. What is the total number of ancestors a person has, going back five generations? ten generations?
…
76. Perimeter and Area of Triangles (a) In Exercise 75, if the process could be continued indefinitely, what would be the total perimeter of all the triangles? (b) What would be the total area of all the triangles, disregarding the overlapping?
11.4
SECTIONS 11.1–11.3
REVIEWING BASIC CONCEPTS
1. Find the first five terms of the sequence an = ( 1) n  1(4n). 5
2. Evaluate the series a (3i + 1). i=1
3. Decide whether the sequence an = 1  2n converges or diverges. If it converges, determine the number to which it converges. 4. Write the first five terms of the arithmetic sequence with a1 = 8 and d =  2. 5. Find a1 for the arithmetic sequence with a5 = 5 and a8 = 17.
11.4
791
Counting Theory
6. Find the sum of the first 10 terms of the arithmetic sequence with a1 = 2 and d = 5. 7. Find a3 and an for the geometric sequence with a1 =  2 and r =  3. 8. Is the series 5 + 3 + 95 + Á + 243 625 arithmetic or geometric? Find the sum of the terms of this series. 9. Find the sum of the infinite geometric series a 3 A 23 B k. q
k=1
10. Annuity Value Find the future value of an annuity if payments of $500 are made at the end of each year for 13 years at 3.5% interest compounded annually.
Counting Theory
Fundamental Principle of Counting • n Factorial • Permutations • Combinations • Distinguishing between Permutations and Combinations
Fundamental Principle of Counting Baker
Creswich Creswich
Albany
Baker
Creswich Creswich
Baker FIGURE 15
Creswich Creswich
If there are 3 roads from Albany to Baker and 2 roads from Baker to Creswich, in how many ways can one travel from Albany to Creswich by way of Baker? For each of the 3 roads from Albany to Baker, there are 2 different roads from Baker to Creswich. Hence, there are 3 # 2 = 6 different ways to make the trip, as shown in the tree diagram in FIGURE 15. Each choice of a road is an example of an event. Two events are independent events if neither influences the outcome of the other. The opening example illustrates the fundamental principle of counting with independent events.
Fundamental Principle of Counting If n independent events occur, with
and then there are
m1 ways for event 1 to occur, m2 ways for event 2 to occur, . . . mn ways for event n to occur,
#
m1 m2
# Á # mn
different ways for all n events to occur.
792
CHAPTER 11
Further Topics in Algebra
EXAMPLE 1
Using the Fundamental Principle of Counting A restaurant offers a choice of 3 salads, 5 main dishes, and 2 desserts. Use the fundamental principle of counting to find the number of different 3course meals that can be selected. Solution Three events are involved: selecting a salad, selecting a main dish, and selecting a dessert. The first event can occur in 3 ways, the second in 5 ways, and the third in 2 ways; thus, there are 3 # 5 # 2 = 30 possible meals.
EXAMPLE 2
Using the Fundamental Principle of Counting A teacher has 5 different books that she wishes to arrange in a row. How many different arrangements are possible? Solution Five events are involved: selecting a book for the first spot, selecting a book for the second spot, and so on. For the first spot, the teacher has 5 choices. After a choice has been made, the teacher has 4 choices for the second spot. Continuing in this manner, we get 3 choices for the third spot, 2 for the fourth spot, and 1 for the fifth spot. By the fundamental principle of counting, there are 5 # 4 # 3 # 2 # 1 = 120 arrangements.
EXAMPLE 3
Arranging r of n Items ( r 3 4x – 5 if x 3
7
ƒ(x) approaches 7.
Use f(x) = 2x + 1 for xvalues less than 3, and use f(x) = 4x  5 for xvalues greater than 3.
x
0
3
FIGURE 4
From the table and the graph in FIGURE 4, we see that as x approaches 3 from either the left or the right, ƒ(x) approaches 7. Therefore, lim ƒ(x) = 7.
x:3
GCM
EXAMPLE 3
Find lim
x:0
Finding a Limit of a Trigonometric Function
sin x , where x is measured in radians. x
Solution In this example, 0, the number approached by x, is not in the domain of the function. However, since the limit as x approaches 0 is not affected by what happens at x = 0, we can continue as before. From the following table, the calculator table in FIGURE 5, and the calculator graph in FIGURE 6, we conclude that lim
x:0
sin x = 1. x x approaches 0.
x ƒ(x)
 0.1 0.9983342
 0.01 0.9999833
 0.0019 9 l 0.9999998 l
k 0.001
0.01
0.1
k 0.9999998
0.9999833
0.9983342
ƒ(x) approaches 1. Looking Ahead to Calculus
The concept of the derivative of a function is central to calculus. The limit evaluated in Example 3 is used in calculus to show that the derivative of the sine function is the cosine function. Therefore, the value of the cosine function at any xvalue gives a measure of the “steepness” of the graph of the sine function at that same xvalue.
3
y1 = sinx x
–2
2
Radian mode
–1 Radian mode
FIGURE 5
FIGURE 6
12.1
GCM
833
An Introduction to Limits
EXAMPLE 4
Finding a Limit as x Approaches a Negative Number + x + 1 . x + 2
ex+2
Find lim
x : 2
Solution Since x approaches a negative number, the table looks a little different from the previous tables in this section. The numbers approaching  2 from the left are greater in absolute value than 2, and the numbers approaching  2 from the right are smaller in absolute value than 2. The table and the graph in FIGURE 7 indicate that ex+2 + x + 1 = 2. x : 2 x + 2 lim
x approaches  2.  2.1
x ƒ(x)
1.9516
 2.01
 2.001
:
;  1.999
1.9995 :
1.9950
;
 1.99
2.0005
 1.9
2.0050
2.0517
ƒ( x) approaches 2. “Hole”
8
8 x+2+x+1 y1 = e x+2
x+2+x+1 y1 = e x+2
–4.7
4.7
–4.7
4.7
–2
–2 FIGURE 7
Limits That Do Not Exist If there is no single value that is approached by ƒ(x) as x approaches a, we say that ƒ(x) does not have a limit as x approaches a, or lim ƒ(x) does not exist. x:a
EXAMPLE 5
Determining whether a Limit Exists
Find lim ƒ(x), where ƒ(x) = b x:2
4x  5 3x  5
if x … 2 if x 7 2.
Solution From the table and the graph in FIGURE 8, we see that as x gets closer and closer to 2 from the left, the values of ƒ(x) get closer and closer to 3. However, as x gets closer and closer to 2 from the right, the values of ƒ(x) get closer and closer to 1. Since ƒ(x) approaches different values depending on whether x approaches 2 from the left or from the right, lim ƒ(x) does not exist. x:2
y
3
x approaches 2. x
1.9
1.99
1.999 :
; 2.001
2.01
2.1
ƒ(x)
2.6
2.96
2.996 :
; 1.003
1.03
1.3
ƒ(x) approaches 3.
1 0
< f (x) = 4x – 5 if x > 2 3x – 5 if x 2
ƒ(x) approaches 1.
Limit does not exist.
x 2
–5
FIGURE 8
834
CHAPTER 12
Limits, Derivatives, and Definite Integrals
EXAMPLE 6
Determining whether a Limit Exists 1 Find lim ƒ(x), where ƒ(x) = 2 . x:0 x y
Solution The following table and the graph in FIGURE 9 illustrate that as the values of x approach 0, the corresponding values of ƒ(x) grow arbitrarily large. A limit L must be a (finite) real number. Therefore, we conclude that lim x12 does not exist. x:0
300
f (x) = 1 x2
x approaches 0.
x –0.1
x
 0.1
 0.01
ƒ(x)
100
10,000
 0.001
:
;
1,000,000 :
0.001
; 1,000,000
0.01
0.1
10,000
100
0.1
ƒ( x) becomes arbitrarily large. FIGURE 9
EXAMPLE 7
Determining whether a Limit Exists
1 Find lim sin . x x:0 Solution A graph of ƒ(x) = sin 1x is shown in FIGURE 10. y
TECHNOLOGY NOTE
2
f (x) = sin 1 x
The graph in FIGURE 10 is difficult to obtain using the typical low resolution of graphing calculators. Be sure to use radian mode.
x –0.5
0.5
–2
FIGURE 10
The graph oscillates faster and faster as x approaches 0, which leads us to believe that the limit does not exist. To help confirm our suspicion, we vary the table so that we can look at more values of x to the right of 0 than before. (Since ƒ(x) is an odd function, the corresponding values to the left of 0 will just be the negatives of these.) x approaches 0. x
0.1
0.01
0.001
ƒ(x)
 0.544
 0.506
0.827
0.0001  0.306
0.00001 0.036
0.000001  0.350 ƒ(x) shows no limit.
To further substantiate the fact that the limit does not exist, notice that horizontal lines between y =  1 and y = 1 intersect the graph infinitely many times. Therefore, as x : 0, the graph of ƒ(x) does not approach a single unique value. This limit does not exist.
12.1
835
An Introduction to Limits
The following summarizes the conclusions drawn from Examples 5–7.
Some Conditions under Which lim ƒ( x ) Fails to Exist x:a
1. ƒ(x) approaches a number L as x approaches a from the left and approaches a different number M as x approaches a from the right. (See Example 5.) 2. ƒ(x) becomes infinitely large in absolute value as x approaches a from either side. (See Example 6.) 3. ƒ(x) oscillates infinitely many times between two fixed values as x approaches a. (See Example 7.)
12.1 Concept Check
EXERCISES
Tell whether each statement is true or false. If false, tell why.
1. If a function ƒ is defined at x = a, then lim ƒ(x) = ƒ(a).
2. If lim ƒ(x) does not exist, then ƒ(x) necessarily approaches x:a one value as x approaches a from the left and a different value as x approaches a from the right.
3. If lim ƒ(x) = 5, then 1 is in the domain of ƒ(x).
4. If lim ƒ(x) = 5, then 5 is in the range of ƒ(x).
5. If lim ƒ(x) =  5, then ƒ(x) is between  5.001 and  4.999
6. If lim ƒ(x) = b, then  ƒ(x)  b  6 0.0001 for some value
x:a
x:1
x:1
x:a
x:a
of x near a.
for some value of x near a.
Checking Analytic Skills Decide from the graph whether each limit exists. If a limit exists, find its value. Do not use a calculator. 7. lim ƒ(x)
8. lim F(x)
x:3
9. lim F(x)
x:2
y
x:2
y
y
y = F(x) 4
y = f (x)
10
2.5
x
0 –10 x
0
10. lim ƒ(x)
2
11. lim ƒ(x)
x:3
y = F(x)
x
0
3
2
12. lim h(x)
x:0
x:1
y
y
y
y = f (x)
2 0
x
2
3 1 0
y = f (x) –40
x
y = h(x)
0
x 1
836
CHAPTER 12
Limits, Derivatives, and Definite Integrals
13. lim ƒ(x)
14. lim ƒ(x)
x:1
15. lim g (x)
x:2
x:3
y
y
y
y = f (x)
3
y = f(x)
y = g(x) 2
2
1.5
0 0 x
0
2
1
16. lim g (x)
17. lim h (x)
x : 2
18. lim ƒ(x)
x : 0.5
x:2
y
y
y
y = h(x)
y = f (x)
x
0
–2
x 3
x
1.5
1
y = g(x) –1
x
0
0.5
0
Infinitely many oscillations near 0.5
19. Use the table of values to predict lim ƒ(x).
Infinitely many oscillations near 2
20. Use the table of values to predict lim ƒ(x).
x:1
x:2
x
0.9
0.99
0.999
1.001
1.01
1.1
x
ƒ(x)
1.9
1.99
1.999
2.001
2.01
2.1
ƒ(x)
1.9
1.99
1.999
2.001
 1.3
 1.05
 1.002
 0.997
Complete each table and use the results to predict the indicated limit, if it exists. 21. If ƒ(x) = 2x 2  4x + 3, find lim ƒ(x). x:1
x
0.9
0.99
ƒ(x)
22. If ƒ(x) = x ƒ(x)
23. If ƒ(x) =
ƒ(x)
 3.68
ƒ(x)
1.01
1.000002
1.000002
1.1
1.99
1.999
2.001
2.01
2.1
10.0601
2x 3 + 3x 2  4x  5 , find lim ƒ(x). x : 1 x + 1  1.1
x
1.001
9.9401
x
24. If ƒ(x) =
0.999
x 3  2x  4 , find lim ƒ(x). x:2 x  2 1.9
 1.01
 1.001
 0.999
 0.99
 0.9  4.28
2x  2 , find lim ƒ(x). x:1 x  1 0.9
0.99
0.999
x 2
1.001
1.01
1.1
2.01
2.1
 0.993
 0.985
12.1
An Introduction to Limits
837
2x  3 , find lim ƒ(x). x:3 x  3
25. If ƒ(x) = x
2.9
2.99
2.999
3.001
3.01
3.1
ƒ(x)
26. If ƒ(x) = x
x 3 + 3x 2 + x + 3 , find lim ƒ(x). x : 3 x + 3  3.1
 3.01
 3.001
 2.999
 2.99
 2.9
ƒ(x)
27. If ƒ(x) =
sin 2x , find lim ƒ(x). x:0 x  0.1
x
 0.01
 0.001
0.001
0.01
0.1
ƒ(x)
28. If ƒ(x) =
sin 5x , find lim ƒ(x). x:0 2x  0.1
x
 0.01
 0.001
0.001
0.01
0.1
ƒ(x)
Use a table and/or graph to decide whether each limit exists. If a limit exists, find its value. 29. lim  2x  4 
30. lim 25 + 3x
x + 1 32. lim x : 3 (x  3) 2
33. lim
x:5
x : 1
x : 2
x2  x  2 x:2 x  2
x2 + 2 x + 2
4x  x  1 x:1 x  1
36. lim
35. lim
x 2  3x  10 x:5 x  5 x 2 + 5x + 4 34. lim x : 4 x + 4 31. lim
37. lim ƒ(x), where x:3
x + 7 5x  5
ƒ(x) = b 39. lim ƒ(x), where
38. lim ƒ(x), where x:1
ƒ(x) = b
x:2
3x  5 6  2x
if x … 1 if x 7 1
41. lim ƒ(x), where
42. lim
x:1
x:1
ƒ(x) = b
ex 2x
ƒ(x) = b
ƒ(x) = b
23  x if x 6  1 4 if x 7  1 1  x
x3 x : 0 x  sin x
43. lim
46. lim
48. lim
ln x x:1 x  1
49. lim
51. lim (x sin x)
52. lim cos
e 2x  1 x:0 ex  1 e x  1 x:0 x
50. lim
x : 1
cos x  1 x:0 x
45. lim
47. lim
2x  1 x  1
40. lim ƒ(x), where
if x … 1 if x 7 1
sin x x : 0 sin 2x
44. lim
 3 if x 6 2 5  x 2 if x 7 2 x2
if x … 3 if x 7 3
x:0
tan2 x x : p 1 + sec x ln x 2 x : 1 ln x x:0
1 x
838
CHAPTER 12
12.2
Limits, Derivatives, and Definite Integrals
Techniques for Calculating Limits
Rules for Limits • Limits Involving Trigonometric Functions
Rules for Limits The following rules are often helpful when evaluating limits.
Rules for Limits 1. Constant rule If k is a constant real number, then lim k ⴝ k. 2. Limit of x rule
x:a
lim x ⴝ a
x:a
For the following rules, we assume that lim ƒ(x) and lim g (x) both exist. x:a
3. Sum and difference rules
x:a
x:a
5. Quotient rule
The discussions at the right are not proofs of these rules. Proofs require the “epsilondelta” definition of limit, discussed in the margin on page 831.
x:a
lim [ƒ(x) # g (x)] ⴝ lim ƒ (x) # lim g (x)
4. Product rule
Looking Ahead to Calculus
x:a
lim [ƒ(x) ⴞ g (x)] ⴝ lim ƒ(x) ⴞ lim g (x)
lim
x:a
x:a
x:a
x:a
lim ƒ(x) ƒ(x) x:a ⴝ , provided that lim g (x) Z 0. x:a g (x) lim g (x) x:a
1. In the constant rule, the constant function defined by ƒ(x) ⴝ k always has the same value, k. Therefore, as x gets closer and closer to a, the values of ƒ(x) are always equal to k. (How much closer to k can you get?) 2. In the limit of x rule, the value of the identity ƒ(x) at x is the same as x. Therefore, as x gets closer and closer to a, the value of ƒ(x), which is the same as the value of x, gets closer and closer to a. 3. The sum and difference rules say that the limit of a sum or difference of two functions is the sum or difference of the limits. Suppose lim ƒ(x) = A and
lim g (x) = B.
x:a
x:a
As x gets closer and closer to a, the values of ƒ(x) and g (x) get closer and closer to A and B, respectively. Thus, the sum ƒ(x) + g (x) gets closer and closer to the sum A + B; the difference ƒ(x)  g (x) gets closer and closer to the difference A  B. 4. and 5. The product and quotient rules follow, since reasoning similar to that for the sum or difference applies for the product or quotient of two functions. EXAMPLE 1
Finding a Limit of a Linear Function
Find lim (3 + 2x). x:4
Solution
lim (3 + 2x) = lim 3 + lim 2x
x:4
x:4
x:4
= 3 + lim 2 x:4
First multiply, then add.
= 3 + 2 # 4 = 11
#
lim x
x:4
Sum rule Constant and product rules Constant and limit of x rules Order of operations
12.2
Techniques for Calculating Limits
839
Letting ƒ(x) = 3 + 2x, we see from Example 1 that lim ƒ(x) = ƒ(4). In fact, for x:4 any linear function defined by ƒ(x) = mx + b,
#
lim ƒ(x) ⴝ m a ⴙ b ⴝ ƒ(a).
x:a
EXAMPLE 2
Finding a Limit of a Polynomial Function with One Term
Find lim 3x 2. x:5
Solution lim 3x 2 = lim 3
x:5
x:5
= 3
#
#
lim x 2
Product rule
x:5
lim x
x:5
#
= 3 # 5 # 5
lim x
x:5
Constant and product rules Limit of x rule
= 75
Multiply.
Letting ƒ(x) = 3x 2, we see from Example 2 that lim ƒ(x) = 3 # 52. Actually, for x:5 any polynomial function in the form ƒ(x) = kx n,
#
lim ƒ(x) ⴝ k an ⴝ ƒ(a).
x:a
EXAMPLE 3
Find
lim (4x 3 x:2
Finding a Limit of a Polynomial Function  6x + 1).
Solution lim (4x 3  6x + 1) = lim 4x 3  lim 6x + lim 1
x:2
x:2
x:2
= 4 # 23  6 # 2 + 1
Follow the order of operations.
x:2
Difference and sum rules Generalization following Example 2; constant rule
= 32  12 + 1
Apply exponent; multiply.
= 21
Subtract; add.
Example 3 shows that lim ƒ(x) = 4 # 2 3  6 # 2 + 1 when ƒ(x) = 4x 3  6x + 1. x:2
In fact, for any polynomial p(x), lim p(x) ⴝ p(a).
x:a
EXAMPLE 4
Find
p(x) lim q(x), x:a
Finding a Limit of a Rational Function
where p(x) and q(x) are polynomials and q(a) Z 0.
Solution Since q(x) is a polynomial, lim q(x) = q(a) Z 0. Therefore, x:a
lim
x:a
lim p(x) p(a) p(x) x:a = = . q(x) lim q(x) q(a)
Quotient rule; generalization following Example 3
x:a
840
CHAPTER 12
Limits, Derivatives, and Definite Integrals
Some additional rules for limits are summarized as follows.
Rules for Limits (Continued) For the following rules, we assume that lim ƒ(x) and lim g (x) both exist. x:a
x:a
If p(x) defines a polynomial function, then
6. Polynomial rule
lim p(x) ⴝ p(a).
x:a
p(x)
7. Rational function rule If ƒ(x) defines a rational function q(x) with q(a) Z 0, then p(a) lim ƒ(x) ⴝ . x:a q(a) 8. Equal functions rule If ƒ(x) = g (x) for all x Z a, then lim ƒ(x) ⴝ lim g (x).
x:a
9. Power rule
x:a
For any real number k,
lim [ƒ(x)]k ⴝ C lim ƒ(x) D
x:a
k
x:a
provided that this limit exists.* 10. Exponent rule For any real number b 7 0, lim ƒ(x)
lim b ƒ(x) ⴝ b x : a .
x:a
11. Logarithm rule
For any real number b 7 0 with b Z 1,
lim [log b ƒ(x)] ⴝ log b C lim ƒ(x) D
x:a
x:a
provided that lim ƒ(x) 7 0. x:a
The equal functions rule is valid since the limit of a function as x approaches a does not depend on the value of the function at a. In the power rule, k can be any real number for which the equation makes sense. For instance, if k = 12, then the power rule applies to the square root of a function. The exponent and logarithm rules allow us to compute limits of generalized exponential and logarithmic functions. EXAMPLE 5
Finding a Limit of a Square Root Function
Find lim 25x  1. x:2
Solution Apply the power rule with ƒ(x) = 5x  1 and k = 12. lim 25x  1 = lim (5x  1) 1/2
x:2
x:2
= C lim (5x  1) D =
x:2 9 1/2
= 3
1/2
n
2a = a1/n Power rule Polynomial rule Simplify.
*The limit does not exist, for example, when lim ƒ(x) 6 0 and k = 12 , or when lim ƒ(x) = 0 and k … 0. x:a
x:a
12.2
EXAMPLE 6
Find lim
x:2
841
Techniques for Calculating Limits
Finding a Limit of an Exponential Function
2 3(x + 1). 2 + 1)
lim 3(x
Solution
x:2
EXAMPLE 7
lim (x2 + 1)
= 3x : 2
Exponent rule
= 35
Polynomial rule
= 243
Apply exponent.
Finding a Limit of a Logarithmic Function
Find lim 3log(3x 5 + 4)4. x:2
The screens show how a calculator evaluates the functions (and thus computes the limits) in Examples 5–7.
Solution
lim 3log(3x 5 + 4)4 = log C lim (3x 5 + 4) D
x:2
Logarithm rule
x:2
= log 100
Polynomial rule
= 2
102 = 100
y
EXAMPLE 8
Finding a Limit of a Rational Function
x 2 + 2x  3 . x : 1 x 2  3x + 2
Find lim 1
x 1
–4
2 f(x) = x2 + 2x – 3 x – 3x + 2
Solution The rational function rule cannot be applied directly, since the value of the denominator is 0 when x = 1. We can simplify as follows. (x + 3)(x  1) x + 3 x 2 + 2x  3 = = 2 x  3x + 2 (x  2)(x  1) x  2
Factor, and then divide out the common factor.
x 2 + 2x  3
12.4
x + 3
2 y1 = x2 + 2x – 3 Now we apply the equal functions rule with ƒ(x) = x 2  3x + 2 and g (x) = x  2, two x – 3x + 2 functions that are identical when x Z 1.
lim
x:1
–2.7
6.7
–6.2 FIGURE 11
x 2 + 2x  3 x + 3 = lim x:1 x  2 x 2  3x + 2 1 + 3 = 1  2 = 4
Equal functions rule Rational function rule Simplify.
See the graphs in FIGURE 11, which support this result. NOTE In Example 8, we evaluated the limit by using fraction reduction. This technique will be used extensively in Section 12.4.
Limits Involving Trigonometric Functions Some limits, such as lim sinx x, can be obtained only with a geometric argument or by x:0 concepts covered in calculus. In Exercises 47–52, geometric arguments are used to derive lim sin x ⴝ 0,
x:0
lim cos x ⴝ 1,
x:0
Other limits, such as lim cos xx  1 x:0
and
sin x ⴝ 1. x x:0 lim
= 0, can be obtained algebraically from these limits.
842
CHAPTER 12
Limits, Derivatives, and Definite Integrals
EXAMPLE 9
Find 8
Finding a Limit of a Trigonometric Function
lim tan x. x:0 x
Solution The graph in FIGURE 12 suggests that this limit may equal 1; this is not, however, a proof. To show that 1 is the limit, we use trigonometric identities.
x y1 = tan x
Rewrite in terms of sine and cosine. –1.57
1.57
#
tan x =
1 x
#
#
sin x sin x = cos x x
1 . cos x
Therefore,
–2
lim
x:0
FIGURE 12
12.2
1 tan x = x x
sin x tan x = lim a x x x:0
#
1 sin x b = lim cos x x x:0
#
lim
x:0
1 = 1 cos x
x:4
1 = 1. 1
EXERCISES
Checking Analytic Skills Let lim ƒ(x) = 16 and lim g (x) = 8. Use the limit rules to find each limit. x:4 x:4 Do not use a calculator. ƒ(x) 1. lim 3ƒ(x)  g (x)4 2. lim 3g (x) # ƒ(x)4 3. lim 4. lim 3log 2 ƒ(x)4 x:4 x:4 x : 4 g (x) x:4 6. lim 31 + ƒ(x)42
#
8. lim 2 3 g (x)
7. lim 2 g(x) x:4
x:4
9. lim
x:4
ƒ(x) + g (x) 2g (x)
5. lim 2ƒ(x) x:4
10. lim
x:4
5g (x) + 2 1  ƒ(x)
Determine each limit, if it exists. 11. lim 7
12. lim ( 5)
13. lim x
15. lim 4x 2
16. lim ( 3x 5)
17. lim 4x 3
18. lim (5x 8  3x 2 + 2)
19. lim (x 3 + 4x 2  5)
x3  1 20. lim 2 x:3 x + 1
2x + 3 21. lim x : 1 3x + 4
22. lim
x2  4 x + 2
25. lim
x2  x  6 x + 2
28. lim
x 2  7x + 10 x 2  25
x : 3
14.
lim x
x : 22 x : 1
x:6 x:3 x:1
x:p
x : 2 x:2
x:0
x 2 + 2x x
x2  9 x:3 x  3
24. lim
26. lim
x 2  3x  10 x:5 x  5
27. lim
29. lim 26x  2
30. lim (1  6x) 3/2
31. lim 9 1/(x + 1)
33. lim 3log 3(2x  1)4
34. lim C log 2 A 14 + 2x B D
35. lim C 2x(1 + x) D
36. lim 32 3x  ln(x + 1)4
37. lim
38. lim 23  x
39. lim 2x
40. lim 2x
sin x  3x 41. lim x:0 x
sin x 42. lim x : 0 5x
43. lim (x cot x)
23. lim
x:3
32. lim
x:3
52x + 1
x:1
x:1
sin2 x x:0 x2
44. lim
x : 2
x:1
x2 + x  2 x  1
x : 4 x:5
x:0
x : 1
45. lim
x:0
cos x  1 3x
x : 2
x:5
x:1 x:4
x:3
2x + 1 log 2(5x + 1)
x:0
x:0
46. lim
x:0
cos x + 2 sin x  1 3x
12.3
RELATING CONCEPTS
OneSided Limits; Limits Involving Infinity
For individual or group investigation (Exercises 47–52)
The definitions of the trigonometric functions imply that AB = sin x, OB = cos x, and CD = tan x in the figure. Work Exercises 47–52 in order.
C
A
47. Use the figure to show that lim sin x = 0
x:0
and
tan x
lim cos x = 1.
1
x:0
sin x
48. Show that the area of triangle OAB is 12 sin x cos x.
x
49. Show that the area of triangle OCD is 12 tan x. 50. Use the fact that the area of triangle OAB 6 the area of sector OAD 6 the area of triangle OCD to obtain the inequality cos x 6
sin x 1 6 . x cos x
A Hint: If a 6 b 6 c, then 1c 6
1 b
6 1a. B
51. Why can we conclude that, for positive values of x, sinx x approaches 1 as x approaches 0?
12.3
843
O
B
D
cos x
52. Use the result of Exercise 51 and the rules for limits to show that lim
x:0
1  cos x = 0. x
A Hint: Multiply the numerator and denominator of
1  cos x x
by the conjugate of the numerator, 1 + cos x, and then factor out sinx x. B
OneSided Limits; Limits Involving Infinity
Right and LeftHand Limits • Infinity as a Limit • Limits as x Approaches ⴞ ˆ
Right and LeftHand Limits The limits we have discussed so far are called twosided limits, since the values of x get close to a from both the right and the left sides of a. Other types of limits, called onesided limits, involve only values of x on one side of a. The righthand limit, lim ƒ(x) ⴝ L,
x : aⴙ
is read “The limit of ƒ(x) as x approaches a from the right equals L.” It means that as x gets closer and closer to a from the right (that is, x 7 a), then the corresponding values of ƒ(x) get closer and closer to L. The value of ƒ(x) can be made as close to L as desired by taking values of x greater than a and close enough to a. The lefthand limit, lim ƒ(x) ⴝ L,
x : aⴚ
read “The limit of ƒ(x) as x approaches a from the left equals L,” is defined similarly, but with x restricted to the left of a (that is, x 6 a).*
*Some textbooks refer to these limits as the “limit from the right” and the “limit from the left.”
844
CHAPTER 12
Limits, Derivatives, and Definite Integrals
As with twosided limits, limits from the right or left are not affected by whether ƒ(a) is defined; if it is, the value of ƒ(a) has no influence on the limit. The rules for limits presented in Section 12.2 are also valid for onesided limits. EXAMPLE 1
y
Finding OneSided Limits Function x + 6 Find lim+ ƒ(x) and limⴚ ƒ(x), where ƒ(x) = c 5 x:2 x:2 1 2 2x
x + 6 if x < 2 5 if x = 2 1 2 x if x > 2 2
f(x) =
8
of a PiecewiseDefined if x 6 2 if x = 2 if x 7 2.
Solution Since x 7 2 in the first limit, use the formula ƒ(x) = 12 x 2. To find the second limit, where x 6 2, use ƒ(x) = x + 6.
5
1 2 1 x = (2 2) = 2 x:2 x:2 2 2 lim ƒ(x) = lim (x + 6) = 2 + 6 = 8 lim+ ƒ(x) = lim+
2 x 0
x:2
2
x:2
These results are supported by the graph of ƒ in FIGURE 13.
FIGURE 13
EXAMPLE 2
Finding OneSided Limits of the Square Root Function If possible, find lim+ 2x and lim 2x. x:0
y
x:0
Solution The value of 2x is defined whenever x Ú 0. Thus, by the power function rule, lim 2x = 2 lim+ x = 20 = 0.
x : 0+
y = √x
2
x:0
However, the value of 2x is undefined whenever x 6 0. Thus, lim 2x does not exist.
1
x:0
0
x 1
4
Notice in FIGURE 14 that the graph of ƒ(x) = 2x does not exist to the left of x = 0. Therefore, a lefthand limit cannot exist at x = 0.
FIGURE 14
If both the right and lefthand limits of ƒ(x) exist at x = a and are equal, then the twosided limit exists and has this common value. The converse is also true, so lim ƒ(x) ⴝ L if and only if
x:a
lim ƒ(x) ⴝ L and
x : aⴙ
lim ƒ(x) ⴝ L.
x : aⴚ
Infinity as a Limit shows the graph of a function with a vertical asymptote. Neither the righthand nor the lefthand limit exists for this function at x = a. However, as x approaches a from the right, the values of ƒ(x) increase without bound. We describe this behavior by writing FIGURE 15
y
lim ƒ(x) ⴝ ˆ.
x : aⴙ
a
FIGURE 15
x
The infinity symbol does not represent a real number. Even though we say “The limit of ƒ(x) as x approaches a from the right equals infinity,” the function does not have a righthand limit. The notation lim+ ƒ(x) = q merely expresses the particular x:a way in which the limit does not exist, namely, the function grows arbitrarily large as x approaches a from the right.
12.3
OneSided Limits; Limits Involving Infinity
845
Similarly, we write lim ƒ(x) ⴝ ⴚˆ
x : aⴙ
if the function values decrease without bound as x approaches a from the right. Analogous definitions can be made for lim ƒ(x) ⴝ ˆ
and
x : aⴚ
lim ƒ(x) ⴝ ⴚˆ.
x : aⴚ
If lim ƒ(x) = q and lim+ ƒ(x) = q, we write x:a
x:a
lim ƒ(x) ⴝ ˆ.
x:a
Similarly, if lim ƒ(x) =  q and lim+ ƒ(x) =  q, we write x:a
x:a
lim ƒ(x) ⴝ ⴚˆ.
x:a
The line given by x = a is a vertical asymptote of ƒ(x) if at least one of the infinite behaviors discussed occurs as x approaches a. FIGURE 16 shows the four different types of situations near vertical asymptotes that occur in the graphs of rational functions. y
y
y
a
y
x
x
a
a
x
a
x
lim
f (x) = ∞
lim
f (x) = – ∞
lim
f (x) = ∞
lim
f (x) = – ∞
lim
f (x) = ∞
lim
f (x) = – ∞
lim
f (x) = – ∞
lim
f (x) = ∞
x→ a – x→ a +
lim x→ a
f (x) = ∞
x→ a – x→ a +
x→ a – x→ a +
f (x) = – ∞
lim x→ a
x→ a – x→ a +
lim f (x) does not exist.
lim f (x) does not exist.
x→ a
x→ a
Vertical asymptotes FIGURE 16
Two familiar functions with vertical asymptotes are shown in FIGURE 17. For the graph of the natural logarithmic function, lim+ ƒ(x) =  q. For the graph of the x:0
tangent function,
lim
x : p/2
ƒ(x) =  q and lim  ƒ(x) = q. + x : p/2
y
y
f (x) = tan x
2
0
3
f(x) = ln x
1
x 1
2
– 2
2
3
x
–1 –3 –2
x = – and x = are 2 2 vertical asymptotes.
x = 0 is a vertical asymptote. FIGURE 17
846
CHAPTER 12
Limits, Derivatives, and Definite Integrals
EXAMPLE 3
Finding OneSided Limits
Find lim+ ƒ(x) and lim ƒ(x), where ƒ(x) = x:2
y
Solution If x is slightly to the right of 2, say, x = 2.001, then
5
f(x) =
x:2
1 x–2
ƒ(2.001) = x
2
1 . x  2
1 1 = = 1000. 2.001  2 0.001
As x moves closer and closer to 2, say, x = 2.0001, 2.00001, and 2.000001, the corresponding values of ƒ(x) are 10,000, 100,000, and 1,000,000. Continuing in this manner, we see that the values of ƒ(x) get arbitrarily large. Therefore,
–5
lim ƒ(x) = q,
x : 2+
and the line x = 2 is a vertical asymptote. A similar analysis from the left has x taking on values 1.999, 1.9999, 1.99999, and 1.999999 and ƒ(x) taking on values  1000,  10,000,  100,000, and  1,000,000. Since these values of ƒ(x) are negative and get arbitrarily large in absolute value, lim ƒ(x) =  q.
x : 2
FIGURE 18
See the graph in FIGURE 18. The table also supports our conclusion.
Limits as x Approaches ⴞ ˆ shows portions of the graphs of two functions with horizontal asymptotes. We describe these asymptotic behaviors by writing FIGURE 19
lim ƒ(x) = 2 and
x:q
lim g (x) = 1.
x:  q
y
y
3 2 f(x) = 4x 3 – 3x + 5 , x > 0 2x + 6x – 1
g(x) = 1 + e x
2
2 1
x 0
5
–1
0
x
FIGURE 19
In general, the notation lim ƒ(x) ⴝ L,
x:ˆ
read “The limit of ƒ(x) as x approaches infinity equals L,” means that ƒ(x) gets closer and closer to L as x gets larger and larger. Similarly, the notation lim ƒ(x) ⴝ L,
x : ⴚˆ
read “The limit of ƒ(x) as x approaches negative infinity equals L,” means that ƒ(x) gets closer and closer to L as x assumes negative values of larger and larger magnitude. Limit Rules 1, 3–5, and 9–11 from the previous section are valid when x : a is replaced with either x : q or x :  q.
12.3
GCM
EXAMPLE 4
Finding Limits at Infinity
Find limq ƒ(x) and limq ƒ(x), where ƒ(x) = 5 + x:
Analytic Solution The values of e  0.25x become arbitrarily close to 0 as x : q, so the 10 expression 5 + 1 + e  0.25x approaches 5 +
10 1 + 0
x: 
10 . 1 + e  0.25x
Graphing Calculator Solution There are two ways to determine these limits with a graphing calculator: Look at the graph or analyze a table of function values. The graph in FIGURE 20 appears to approach the horizontal asymptote y = 15 as x moves to the right and to approach the horizontal asymptote y = 5 as x moves to the left. The tables in FIGURE 21 support these conclusions.
= 15. Therefore, lim ƒ(x) = 15.
x:q
10 1 + e–0.25x
y1 = 5 +
For negative values of x, the expression 1 + e  0.25x gets arbitrarily large as x gets larger and larger in magnitude. Therefore, 5 +
847
OneSided Limits; Limits Involving Infinity
y1 = 5 + 20
20
10 1 + e  0.25x
–30
30
10 1 + e–0.25x
–30
30 –2
–2
approaches 5 + 0 = 5 as x :  q. That is,
FIGURE 20
lim ƒ(x) = 5.
x:  q
Thus, the graph of ƒ has two horizontal asymptotes: y = 15 and y = 5.
FIGURE 21
NOTE The graph of a function can have two horizontal asymptotes, as shown in Example 4.
EXAMPLE 5
Find
lim
x:  q
1 , x
lim
x:q
1 , and x2
lim
x:  q
1 . x2 1 x
Solution The following table, as well as the graphs of ƒ(x) = in FIGURE 22, indicate that
y
g(x) = 1 x2
lim
2
x:q
1 = 0, x
limq
x: 
1 = 0, x
limq
x:
1 = 0, x2
and
and g (x) =
limq
x: 
1 x2
shown
1 = 0. x2
x
0
f (x) = 1 x
1 lim , x
x:q
Finding Limits at Infinity
2 –2
FIGURE 22
4
x
 100
 10
1
1
10
100
1 ƒ(x) ⴝ x 1 g (x) ⴝ 2 x
 0.01
 0.1
1
1
0.1
0.01
0.0001
0.01
1
1
0.01
0.0001
The results of Example 5 suggest the following rule.
848
CHAPTER 12
Limits, Derivatives, and Definite Integrals
Looking Ahead to Calculus
The concept of series discussed in Chapter 11 can be extended to infinite series. Functions of the form 1 x n are often used to decide whether an infinite series is convergent. For example, if 0 … ƒ( x ) … x1n , where
1 xn
Limits at Infinity of
For any positive integer n, limˆ
x:
q
1 ⴝ 0 and xn
limˆ
x:ⴚ
1 ⴝ 0. xn
n 7 1, then a ƒ( k ) is convergent. k=0
NOTE The first limit can be extended to include n as a positive real number. If x is positive, x n exists for all positive real numbers n, so the first of the preceding two limits is valid.
EXAMPLE 6
Find limq x:
Finding a Limit at Infinity
5x 2  7x + 1 . 2x 2 + x + 5
Solution We can find limits for rational functions by dividing both numerator and denominator by the greatest power of x appearing in the denominator. (We used this technique to find asymptotes of rational functions in Chapter 4.) 1 7 + 2 x  7x + 1 x = limq lim x : q 2x 2 + x + 5 x: 5 1 2 + + 2 x x 5 
5x 2
y1 =
5x2 – 7x + 1 2x2 + x + 5
Divide each term in the numerator and denominator by x 2.
5
lim a 5 
= –10
10
–3
x:q
5 1 limq a 2 + + 2 b x x: x
#
1 1 + lim 2 q x x: x = 1 1 lim 2 + lim + 5 # lim 2 x:q x:q x x:q x lim 5  7
x:q
FIGURE 23
1 7 + 2b x x
=
5  0 + 0 2 + 0 + 0
=
5 2
lim
x:q
Quotient rule
Constant, sum, difference, and product rules
1
lim n x:q x
= 0
This result is supported by the table and the graph in FIGURE 23. One final limit behavior combines features of both limits at infinity and onesided limits that get arbitrarily large. Functions such as x n (where n 7 0) and e x do not have a limit as x : q. They get larger and larger, increasing without bound. We describe this situation by writing lim ƒ(x) ⴝ ˆ.
x:ˆ
The limits described by limˆ ƒ(x) ⴝ ⴚˆ, limˆ ƒ(x) ⴝ ˆ, and limˆ ƒ(x) ⴝ ⴚˆ have x: x:ⴚ x:ⴚ analogous meanings.
12.3
12.3
OneSided Limits; Limits Involving Infinity
849
EXERCISES
Checking Analytic Skills Determine each limit. Do not use a calculator. 1. (a) lim+ ƒ(x)
(b) lim ƒ(x)
x:2
2. (a) lim+ ƒ(x)
x:2
x:4
1 2
if x 6 2 if x = 2 if x 7 2
x where ƒ(x) = c 3 4
(b) lim ƒ(x)
x:4
if x 6 4 if x = 4 if x 7 4
x2
where ƒ(x) = c 5 2x
y
3. (a) lim+ ƒ(x)
x 5(3  x) 3
y 5
8
3
x:3
where ƒ(x) =
y
4
(b) lim ƒ(x)
x:3
5
2
x
0
1 0
3
2 x 1
2
3
4. (a) lim + ƒ(x) x : 2
where ƒ(x) =
4
x
0
5. (a) lim + ƒ(x)
(b) lim  ƒ(x) x : 2
(b) lim  ƒ(x)
x : 1
x2
y
x : 1
x (x + 1) 2
where ƒ(x) =
3(x + 2)
–5
4
6. (a) lim+ ƒ(x)
(b) lim ƒ(x)
x:1
x:1
2x (x  1) 2
where ƒ(x) =
y
y
0
x
–1 2 –2
x
0
–5
5
0
x 1
7. lim+ (3x  5)
11. lim  2x + 3
x:1
x:0
16. lim
x:  q
x:7
x : 4
10. lim+ 2x  1 13. lim
9. lim 100
8. lim  x 3
x:5
12. lim 22  x
x : 3
x
14. lim+
x
x:0
6x 2 + 1 2x 2 + 3
x
15. lim 
x
x : 3
17. limq (2 + e x)
x + 3 x + 3 19. limq ax +
18. lim (x sin x) x:q
x:
y
x:2
x:
y
y
1 b x
y
4 20 4 2
–5
0
0
2 x 5
0
x 5
x 20
0
x 4
850
Limits, Derivatives, and Definite Integrals
CHAPTER 12
Use a table and/or graph to determine each limit. 5 20. lim a x sin b x:q x
21. lim ax sin x:q
22. lim A 2x 2 + x  x B
1 b x2
23. lim A x  2x 2 + 5 B
x:q
x:q
Use a table and/or graph to find the asymptote(s) of each function. 24. ƒ(x) =
ex ex  1
25. ƒ(x) = tan 1
x x  1
x  cos x x + sin x
26. ƒ(x) =
27. ƒ(x) = 5  e x
Determine each limit. (b) lim ƒ(x)
28. (a) lim+ ƒ(x) x:1
29. (a) lim+ ƒ(x)
x:1
2x + 3 if x 6 1 where ƒ(x) = c 4 if x = 1 2 x if x 7 1 30. (a) lim + ƒ(x)
x:
(b) lim ƒ(x)
33. (a) lim + ƒ(x)
x:3
x: 
8x + 2 2x  5
38. limq
2x 3
39. limq
x 2 + 2x  5 3x 2 + 2
2x 2  1 3x 4 + 2
42. lim
2x 3  x  3 6x 2  x  1
45. lim
2x  3x 3 4x 3 + x
x:
43. lim q
x 4  x 3  3x 7x 2 + 9
44. lim q
x:
x: 
36. limq x: 
x 2 + 2x  2x + 1
41. lim q
x:
x:q
 x 3  3x + 1 4x 3 + 5x 2  x
x:  q
46. Evaluate limq
xn + 3 for n 6 m, n = m, and n 7 m. Assume m and n are positive integers. xm  3
Concept Check
Write an expression for a function defined by ƒ(x) with the given features.
x:
48. ƒ(x) is a quotient of two polynomials of degree greater than 2, limq ƒ(x) = 0
47. lim+ ƒ(x) = q, lim ƒ(x) =  q x:5
2x + 3 4x  7
5x 3x  1
x:
3x 3 + 2x  1 2x 4  3x 3  2
x:
x : 3
35. limq
40. lim q x:
(b) lim  ƒ(x)
x : 3
x where ƒ(x) = (x + 3) 3
3x 5x  1
37. limq
x:4
x where ƒ(x) = (4  x) 3
1 where ƒ(x) = (x  3) 2 34. limq
if x … 2 if x 7 2 (b) lim ƒ(x)
x:4
1 where ƒ(x) = (1 + x) 3 x:3
7x x  1
31. (a) lim+ ƒ(x)
x : 1
32. (a) lim+ ƒ(x)
x:2
where ƒ(x) = b
(b) lim  ƒ(x)
x : 1
(b) lim ƒ(x)
x:2
x:5
x:
49. ƒ(x) is a polynomial, limq ƒ(x) =  q x:
Concept Check
50. lim+ ƒ(x) = q, lim ƒ(x) = q x:5
x:5
Find two functions defined by ƒ(x) and g(x) with the given properties.
51. lim ƒ(x) = q, lim g (x) = q, and lim 3ƒ(x)  g (x)4 = q q q q
52. limq ƒ(x) = q, limq g (x) = q, and limq 3ƒ(x)  g (x)4 = 2
53. lim ƒ(x) = 0, lim g (x) = q, and lim 3ƒ(x) # g (x)4 = q q q q
54. limq ƒ(x) = 0, limq g (x) = q, and limq 3ƒ(x) # g (x)4 = 0
x:
x:
x:
x:
x:
x:
x:
x:
x:
x:
x:
x:
12.3
Concept Check domain 30, q).
In Exercises 55 and 56, assume that ƒ(x) has
9 60.
55. Find lim ƒ(x) if the graph of y = ƒ(x) has oblique asympq
851
OneSided Limits; Limits Involving Infinity
(Modeling) Speed of a Skydiver If air resistance is not ignored during free fall, then the speed (in feet per second) of a skydiver after t seconds is given by
x:
ƒ(t) = 176(1  e 0.2t ).
tote y = 12 x + 3.
Calculate lim ƒ(t) and give an interpretation of its value.
56. Find lim ƒ(x) if the graph of y = ƒ(x) has oblique asympq
t:q
x:
tote y =  2x + 3.
Solve each problem. 57. Use a calculator to answer each of the following. (a) From a graph of y = xe x, what do you think is the value of lim (xe x)? Support your answer by evaluatx:q
ing the function for several large values of x. (b) Repeat part (a), but this time use the graph of the function y = x 2e x. (c) On the basis of your results from parts (a) and (b), what do you think is the value of lim (x ne x) for other x:q positive integers n?
9 61.
58. Use a calculator to answer each of the following. (a) From a graph of y = lnx x, what do you think is the value of lim
ln x
x:q x
? Support your answer by evaluating
the function for several large values of x. (b) Repeat part (a), but this time use the graph of the 2 function y = (ln xx) . (c) On the basis of your results from parts (a) and (b), n what do you think is the value of lim (ln xx) , where n x:q is a positive integer? (Modeling) Quantity of a Drug Suppose a patient is given a 0.05mg injection of a drug daily and each day 60% of the drug in the body is eliminated. The function defined by y = ƒ(t) in the figure shows the quantity of the drug in the body during the first two days of treatment. Interpret lim ƒ(t) and
lim ƒ(t).
t : 1
t : 1+
p (t ) = 12  4e 0.5t, where p (t) is the price in dollars after t days. Calculate lim p (t) and give an interpretation of its value. t:q
9 62.
Explain why limq sin x does not exist. x:
RELATING CONCEPTS For individual or group investigation (Exercises 63–66)
Recall from Section 3.5 that the end behavior of the graph of a polynomial function is determined by the degree of the polynomial and the sign of the leading coefficient. Work Exercises 63–66 in order, to relate the concepts introduced in Section 3.5 with those of this chapter. 63. Rewrite the polynomial function defined by ƒ(x) = an x n + an  1 x n  1 + Á + a1 x + a0
Level of a Drug y
Quantity (in mg)
9 59.
(Modeling) Evan’s Price Adjustment Model If there is excess demand for a commodity, the price will rise rapidly at first and then more slowly according to what economists call the “Evan’s price adjustment model.” A typical function describing this behavior is given by
as a product by factoring out the leading term, an x n.
0.08
64. Use the rules for limits to show that lim ƒ(x) = an
0.05
x:q
#
lim x n and
x:q
lim ƒ(x) = an
x:  q
#
lim x n.
x:  q
65. Determine lim ƒ(x) and lim ƒ(x) in each case. q q x:
0
1
Time (in days)
2
(a) an positive, n even (c) an positive, n odd
x: 
(b) an negative, n even (d) an negative, n odd
66. Give the results of Exercise 65 in terms of endbehavior diagrams for graphs of polynomial functions.
852
Limits, Derivatives, and Definite Integrals
CHAPTER 12
SECTIONS 12.1–12.3
REVIEWING BASIC CONCEPTS
1. Find each limit if it exists. (a) lim ƒ(x)
(b) lim F(x)
x:3
(c) lim ƒ(x)
x:2
y
(d) lim g(x)
x:0
x:3
y
y
y
y = F(x) y = f (x)
3
–2
y = f (x)
–2 –1 0
x
0
y = g(x)
2
4 2
3 2 1
x 1 2
–4
1 2 3
0
–2
–2
x
x
0
3 –2
2
–3
In Exercises 2 and 3, use the graph to find lim ƒ(x), lim+ ƒ(x), and lim ƒ(x), if they exist. x:a
2. (a) a =  2
x:a
(b) a =  1
x:a
3. (a) a = 1
(b) a = 2
y
y
1 –2
2
–1
1 x
0
y = f (x)
1
y = f (x)
–1
0
–2
x 1
2
–2
4. Find each limit if it exists. (a) lim ƒ(x)
(b) lim g (x)
x:q
x:  q
y
y 6
3
3
y = g(x)
y = f (x) 0
–2
0
–2
x
x 4
4 –3
5. Find lim
x:4
2x  2 , using a table of values near x = 4. x  4
6. Let lim ƒ(x) = 32 and lim g(x) = 4. Use the limit rules to find each limit. x:8 x:8 ƒ(x) (a) lim 3ƒ(x)  g (x)4 (b) lim 3g (x) # ƒ(x)4 (c) lim x:8 x:8 x : 8 g (x) (e) lim 2ƒ(x)
3 g(x) (f ) lim 2
(g) lim 2
ƒ(x)  g(x) (i) limx:8 4g(x)
2g(x) + 3 ( j) lim+ x : 8 1 + ƒ(x)
(k) lim+ 3ƒ(x) + g(x)42
x:8
x:8
g(x)
x:8
(d) lim 3log 2 ƒ(x)4 x:8
(h) lim 31 + ƒ(x)42 x:8
x:8
(l) lim 3log 6 ( ƒ(x) + g(x))4 x:8
Find each limit if it exists. x2  9 x:3 x  3
7. lim
2x 3  x + 3 9. limq 3 x :  6x + 4x  9
8. limq x:
2x 2  1 3x 4 + 5
10. lim ƒ(x), where ƒ(x) = b x:4
x + 2 x2  6
if x 6 4 if x Ú 4
12.4
12.4
Tangent Lines and Derivatives
853
Tangent Lines and Derivatives
The Tangent Line as a Limit of Secant Lines • Derivative of a Function • Interpretation of the Derivative as a Rate of Change • Marginal Concept in Economics
The Tangent Line as a Limit of Secant Lines
y
Tangent line P
x
0
FIGURE 24 y
P1
P2 P3
P4
P5
x
0
FIGURE 25
In geometry, a tangent line to a circle is defined as a line that touches the circle at only one point, as at the point P in FIGURE 24 (which shows the top half of a circle). If you think of this half circle as part of a curving road on which you are driving at night, then the tangent line indicates the direction of the light beam from your headlights as you pass through the point P. The tangent line to an arbitrary curve at a point P on the curve should touch the curve at P, but not at any other nearby points, and should indicate the direction of the curve. In FIGURE 25, for example, the lines through P1 and P3 are tangent lines, while the lines through P2 and P5 are not. The tangent lines just touch the curve, while the other lines pass through it. To decide about the line at P4, we need to define the idea of a tangent line to the graph of a function more carefully. Let R be a fixed point with coordinates (a, ƒ(a)) on the graph of a function y = ƒ(x), as in FIGURE 26. Choose a different point S on the graph, and draw the line through R and S; this line is called a secant line. If S has coordinates (x, ƒ(x)), then, by the definition of slope, the slope of the secant line RS is given by ƒ(x) ⴚ ƒ(a) . Slope of secant line xⴚa y
y
y = f (x) (x, f (x))
S
f (x) – f (a) (a, f (a))
0
S2
Secant lines
Secant line
R
S1
(x, f (x))
S3
Points slide down graph.
S4 R(a, f (a))
x–a
a
x
FIGURE 26
x
Tangent line x
0
FIGURE 27
This slope equals the average rate of change of y with respect to x over the interval from a to x. As x approaches a, point S will slide along the curve, getting closer and closer to the fixed point R. FIGURE 27 shows successive positions S1, S2, S3, and S4 of the point S. Notice that the corresponding secant lines get closer and closer to the tangent line at R. If the slopes of the corresponding secant lines approach a limit as x approaches a, then the slope of the tangent line at point R is defined to be this limit.
Tangent Line The tangent line of the graph of y = ƒ(x) at the point (a, ƒ(a)) is the line through this point and having slope ƒ(x) ⴚ ƒ(a) m ⴝ lim , xⴚa x:a provided that this limit exists. If the limit does not exist, then there is either no tangent line or a vertical tangent line at the point.
854
CHAPTER 12
Limits, Derivatives, and Definite Integrals
The slope of the tangent line at a point is also called the slope of the curve at the point and corresponds to the “instantaneous” rate of change of y with respect to x at the point. In certain applications of mathematics, it is necessary to determine the equation of the line tangent to the graph of a function at a given point.
GCM
EXAMPLE 1
Finding the Equation of a Tangent Line Find the equation of the tangent line to the graph of the function defined by ƒ(x) = x 2 + 2 when x has the value  1.
Analytic Solution First find the slope of the tangent line at the given point by using the definition on the preceding page, with ƒ(x) = x 2 + 2 and a =  1. lim
x:a
3x 2 + 24  3( 1) 2 + 24 ƒ(x)  ƒ(a) = lim x  a x : 1 x  ( 1) 2 x  1 = lim x : 1 x + 1 (x + 1)(x  1) = lim x : 1 (x + 1) = lim (x  1) x : 1
= ⴚ2
Substitute. Simplify. Factor.
Graphing Calculator Solution Graphing calculators can draw a tangent line at a specified point on a graph. FIGURE 28 was generated with the Tangent command from the DRAW menu on the TI83/84 Plus. At the bottom of the screen, the first line gives the xcoordinate of the point of “tangency” and the second line gives the equation of the tangent line.
Simplify.
f (x) = x 2 + 2
Polynomial rule
6
When x =  1, ƒ(x) = ƒ( 1) = + 2 = 3. Now find the equation of the line through the point ( 1, 3) having slope ⴚ2. ( 1) 2
y  y1 y  3 y  3 y  3 y
= = = = =
m(x  x1) ⴚ23x  ( 1)4  2(x + 1)  2x  2  2x + 1
y = x2 + 2
y = –2x + 1
4
Work inside parentheses. Distributive property
–2 FIGURE 28
Add 3.
shows the result of using a calculator to zoom in on the point ( 1, 3) in Notice that in this closeup view, the graph and its tangent line appear nearly identical. This gives us another interpretation of the tangent line. Suppose, as we zoom in on a function, the graph appears to become a straight line. Then this line is the tangent line to the graph at that point. In other words, the tangent line captures the behavior of the function very close to the point of tangency. Consequently, another way to approximate the slope of the curve at a point is to zoom in on the curve near the point until it appears to be a straight line (the tangent line). Then we can estimate the slope by using any two points on that section of the curve. FIGURE 28.
–0.7 2.7
–4
y1 = 3, m =  2, x1 =  1
FIGURE 29
3.3
–1.3
Point–slope form
FIGURE 29
Derivative of a Function A function defined by ƒ(x) associates, with each number a in its domain, the height of the graph of y = ƒ(x) above the xaxis at the point where x = a. (Height is taken as negative for points below the xaxis.) Related to ƒ is the function denoted ƒ œ , read “ƒ prime,” whose value at a is the slope of the curve y = ƒ(x) at the point where x = a. The function ƒ¿ is called the derivative of ƒ.
12.4
Tangent Lines and Derivatives
855
Derivative If a is in the domain of ƒ, then the derivative of ƒ at a is defined by ƒ œ (a) ⴝ lim
x:a
ƒ(x) ⴚ ƒ(a) , xⴚa
provided that this limit exists.
NOTE The derivative of ƒ at a is equal to the slope of the tangent line to the graph of ƒ at ( a, ƒ (a)). They are one and the same concept.
EXAMPLE 2
GCM
Finding a Derivative
Find ƒ¿(4), where ƒ(x) = 2x. Analytic Solution Let a = 4, and thus ƒ(a) = 24 . ƒ¿(4) = lim
x:4
= lim
x:4
= lim
x:4
= lim
2x  24 x  4
Definition of the derivative
2x  2 x  4
24 = 2
2x  2
A 2x  2 B A 2x + 2 B 1
2x + 2 1 1 = = 4 24 + 2 x:4
Graphing Calculator Solution Calculators can approximate the value of ƒ¿(a) with a single instruction. In FIGURE 30, ƒ¿(4) is found with a TI83/84 Plus. nDeriv( denotes “numerical approximation to the derivative,” 2(X) is the function, X is the variable, and 4 is the value of a.
≈
Factor.
1 4
Simplify. FIGURE 30
Evaluate limit.
Therefore, ƒ¿(4) = 14.
Looking Ahead to Calculus
In calculus, we derive analytic techniques for converting the expression for ƒ( x ) into an expression for ƒ¿( x ). It can be shown that if ƒ( x ) = 2x, then ƒ¿(4)
ƒ¿( x ) = 1 . Therefore, 2 1x 1 = 14 , the same = 2 14
result obtained in Example 2. In Example 3, ƒ¿( x) = 2x. Verify that ƒ¿( 2) =  4.
EXAMPLE 3
Finding a Derivative Find ƒ¿( 2), where ƒ(x) = x 2. Solution ƒ(x)  ƒ( 2) x  ( 2)
Definition of the derivative
= lim
x2  4 x + 2
ƒ( x ) = x 2; ƒ( 2) = 4
= lim
(x  2)(x + 2) x + 2
Factor.
ƒ¿( 2) = lim
x:  2
x : 2
x : 2
= lim (x  2)
Simplify.
= 4
Evaluate limit:  2  2 =  4.
x : 2
856
CHAPTER 12
Limits, Derivatives, and Definite Integrals
Interpretation of the Derivative as a Rate of Change Consider a quantity that is changing with time. The derivative gives valuable information about the rate of the change. The value of ƒ œ (a) is the rate at which the quantity is changing at time t ⴝ a. For instance, let ƒ(t) be the population of the United States (in millions) t years after 1800. In FIGURE 31, the fact that ƒ¿(40) = 0.5 tells us that in the year 1840 the population was growing at the rate of 0.5 million (that is, 500,000) people per year. Note that the slope of the tangent line at t = 40 is equal to the derivative ƒ¿(40). U.S. Population
Distance Traveled by a Car y
y = f (t)
100
Distance (in feet)
Population (in millions)
y
80 60 40
slope = 0.5
20 0
100
t 20
40
60
80
t 2
4
6
8
Time (in seconds) FIGURE 32
FIGURE 31
The function s¿( t ) is also denoted v ( t ), since it gives the velocity of a car at any time. In calculus, we learn why v ¿( t ) tells how fast the car is accelerating at any time.
slope = 20 50
0
100
Years after 1800
Looking Ahead to Calculus
y = s(t)
150
As another example, let s(t) be the number of feet traveled by a car after t seconds. FIGURE 32 shows the distance traveled by a car on an 8second test run. The fact that s¿(2) = 20 tells us that after 2 seconds the car is traveling at a velocity of 20 feet per second. Note that the slope of the tangent line at t = 2 is equal to the derivative s¿(2). We can analyze the function defined by s(t) to see why s¿(2) is the car’s velocity at time t = 2. (This value is sometimes called the instantaneous velocity at time t = 2.) Consider the time interval from time 2 seconds to 4 seconds. The average velocity during that interval is average velocity =
s(4)  s(2) distance traveled = . time elapsed 4  2
This average velocity is an approximation to the (instantaneous) velocity at time 2 seconds. Actually, for any time t 7 2, the average velocity during the time interval from 2 to t is average velocity =
s(t)  s(2) distance traveled = . time elapsed t  2
The closer t is to 2, the better the approximation. As t gets closer and closer to 2, the average velocity gets closer and closer to the (instantaneous) velocity at time 2. That is, average velocity approaches instantaneous velocity as t : 2. Therefore, instantaneous velocity at time 2 = lim
t:2
s(t)  s(2) , t  2
or s¿(2).
A car traveling along a road is an example of motion along a straight line. The function defined by s(t) is often called a position function, since it gives the location or position of the car on the straight line at time t. The velocity of the car tells how fast the position of the car changes and the direction of the car. The velocity at time t will be positive when the car is moving forward and negative when the car is backing up.
12.4
Tangent Lines and Derivatives
857
This discussion is summarized as follows.
Velocity as a Derivative If an object is moving along a straight line and its position on the line at time t is s(t), then the velocity at time a is s œ (a) ⴝ lim t:a
s(t) ⴚ s(a) . tⴚa
Another example of an object moving along a straight line is a ball thrown straight up into the air. The ball initially rises (positive direction) and then falls (negative direction). In physics, it is shown that the height (in feet) of the ball above the ground after t seconds, ignoring air resistance, is s(t) ⴝ ⴚ16t 2 ⴙ v0t ⴙ h 0, where v0 is the initial velocity in feet per second and h0 is the initial height in feet.
EXAMPLE 4
Finding Information about a Ball Thrown into the Air Suppose a ball thrown straight up into the air has an initial height of 112 feet and an initial velocity of 96 feet per second. (a) Give the equation for s(t), the height of the ball after t seconds. (b) How high above the ground is the ball at 5 seconds? (c) What is the velocity of the ball at 5 seconds? Solution (a) Since v0 = 96 and h0 = 112, s(t) =  16t 2 + 96t + 112. (b) The height at 5 seconds is s(5) =  16(5) 2 + 96(5) + 112 = 192 feet. (c) The velocity of the ball at 5 seconds is s(t)  s(5) t:5 t  5 ( 16t 2 + 96t + 112)  192 lim t:5 t  5 2  16t + 96t  80 lim t:5 t  5  16(t 2  6t + 5) lim t:5 t  5  16(t  5)(t  1) lim t:5 t  5 lim3 16(t  1)4
s¿(5) = lim = = Be careful with signs.
= = =
t:5
=  64.
Definition of the derivative Substitute the results from parts (a) and (b). Simplify. Factor out  16. Factor the trinomial. Simplify. Evaluate the limit.
At 5 seconds, the ball is falling downward at a speed of 64 feet per second.
858
CHAPTER 12
Limits, Derivatives, and Definite Integrals
The interpretation of the derivative as an instantaneous rate of change applies to any quantity, even when the independent variable is not time. The derivative ƒ œ (a) measures the rate of change of ƒ(x) at x ⴝ a.
Marginal Concept in Economics In economics, derivatives are often described by the adjective “marginal.” For instance, if C(x) is a cost function (the cost of producing x units of a commodity), then the value of the derivative C¿(a) is called the marginal cost at the production level of a units. Since the marginal cost is a derivative, its value gives the rate at which costs are increasing (or decreasing) with respect to the level of production, assuming that production is at level a. For instance, if C(1000) = 20,000 and C¿(1000) = 15, then the cost of producing 1000 units of goods is $20,000 and the marginal cost when 1000 units of goods are produced is $15. The marginal cost can be interpreted as the approximate cost of producing one additional unit of goods. That is, when x = 1000, the cost is increasing at the rate of 15 dollars per unit, so the cost of producing 1001 units will be about $20,015. If R(x) and P(x) define revenue and profit functions, then the values of R œ (a) and P œ (a) give the marginal revenue and the marginal profit, respectively, at x ⴝ a.
Summary of Ideas Related to ƒ œ ( x ) 1. Geometric definition of ƒ œ (a) ƒ¿(a) is the slope of the tangent line to the graph of ƒ(x) at x = a. ƒ(x)  ƒ(a) 2. Algebraic definition of ƒ œ (a) ƒ¿(a) = lim x  a x:a œ (a) 3. Interpretation of ƒ ƒ¿(a) represents the rate of change of ƒ(x) when x = a.
12.4
Estimate the slope of the tangent line to each curve at the given point (x, y).
Concept Check 1.
EXERCISES 2.
y
3.
y
4.
y
y 4
6 2
4 4
(5, 3)
(2, 2)
2
(–2, 2)
4
0
–4
2
2 x 2
4
6
–2
0
x 2
–2
–4 –2
0
x 2
4
–2
–2
0
x 2
–2
(–3, –3) –4
Checking Analytic Skills Find the slope of the tangent line to each curve when x has the given value. Do not use a calculator. 5. ƒ(x) = x 2; x = 4 8. ƒ(x) = 6x 2  4x; x =  1
6. ƒ(x) = x 2 + 5; x = 2 2 9. ƒ(x) =  ; x = 4 x
7. ƒ(x) =  4x 2 + 11x; x =  2 6 10. ƒ(x) = ; x =  1 x
12.4
Tangent Lines and Derivatives
11. ƒ(x) =  3 2x; x = 1
12. ƒ(x) = x 3; x = 1
13. ƒ(x) = 2x 3; x = 1
14. ƒ(x) = 22x; x = 2
15. ƒ(x) = 4  x 2; x =  1
16. ƒ(x) =
859
1 + 1; x = 2 x
Find the equation of the tangent line to each curve when x has the given value. Verify your answer by graphing both ƒ(x) and the tangent line with a calculator. 17. ƒ(x) = x 2 + 2x; x = 3 20. ƒ(x) =
3 ;x = 1 x + 1
5 ;x = 2 x
18. ƒ(x) = 6  x 2; x =  1
19. ƒ(x) =
21. ƒ(x) = 42x; x = 9
22. ƒ(x) = 2x; x = 25
Concept Check By considering the graph of the function but not calculating any limits, give the value of ƒ¿(2) for each function. 23. ƒ(x) = 5
24. ƒ(x) = x
25. ƒ(x) =  x
26. ƒ(x) = 3x + 4
Use a calculator to estimate ƒ¿(a) for the given value of a. 27. ƒ(x) = e x; a = 0 30. ƒ(x) =
29. ƒ(x) =
28. ƒ(x) = sin x; a = 0
1 ;a = 0 1 + x2
31. ƒ(x) = x cos x; a =
Concept Check In Exercises 33–36, use the figure to determine the derivative.
p 4
10x ;a = 2 1 + 0.25x 2
32. ƒ(x) = xe x; a = 1
36. Consider the curve y = ƒ(x) in the figure. Estimate ƒ¿(1).
33. The figure shows the graph of ƒ(x) = 1 + x +x 3, along with its tangent line at the point (0, 1). What is ƒ¿(0)?
y
(1.4, 1.3) (1.6, 1.4) (1.2, 1.1) y = f (x)
y
2
y=
1 3
x+1
1
2
(1, 0.8) 0
0
x
x 1
2
1 2
–2
f(x) = 1 +
x x+3
34. The figure shows the graph of ƒ(x) = ln x, along with its tangent line at the point (1, 0). What is ƒ¿(1)? y
y= x–1 3 2 1 0
Concept Check The position in feet of a car along a straight racetrack after t seconds is approximated by s(t). Find the car’s velocity in feet per second after 3 seconds. 37. s(t) = 9t 2
38. s(t) = 50 2t
39. s(t) = 3t 3  t 2
40. s(t) = 4t 2 + 5t + 1
x 1 2 3
f(x) = ln x
For the given ƒ(x), find a formula for ƒ¿(a). 35. Consider the curve y = ƒ(x) in the figure, with the sequence of secant lines through the point (a, ƒ(a)). Find a and ƒ(a). Estimate ƒ¿(a). y = f (x) y = 2.03x – 0.53 y = 2.02x – 0.52 y = 2.01x – 0.51
a
41. ƒ(x) = x 2
42. ƒ(x) =  2x 2
43. ƒ(x) = 3x  1
44. ƒ(x) = 5  4x
45. ƒ(x) = 4x  x 2
46. ƒ(x) = x 2  5x
47. ƒ(x) = 2x
48. ƒ(x) = x 3
x
860
Limits, Derivatives, and Definite Integrals
CHAPTER 12
(Modeling) Solve each problem. 49. Interest Rates on Treasury Bonds In the figure, ƒ(t) is the interest rate (as a percent) on a 3month United States treasury bond t years after January 1, 1979. The straight line is the tangent line to the graph of y = ƒ(t) when t = 10. How fast were interest rates rising on January 1, 1989? y
s(t ) = t 2 + t. (a) How long will it take for the helicopter to rise 20 feet? (b) Find the vertical velocity of the helicopter when it is 20 feet above the ground. 54. Marginal Profit Suppose that the total profit in hundreds of dollars from selling x items is given by
15
Interest Rate (as a percent)
53. Velocity of a Helicopter A helicopter is gradually rising straight up in the air. Its distance from the ground t seconds after takeoff is s(t) feet, where
P(x) = 2x 2  5x + 6.
y = f (t)
Find the marginal profit at x = 2.
10
55. Marginal Revenue The revenue (in thousands of dollars) from producing x units of an item is modeled by
5
R(x) = 10x  0.002x 2. Find the marginal revenue at x = 1000.
t
0
2
4
6
8
10
12
Years after Jan. 1, 1979 Source: Statistical Abstract of the United States.
50. Growth Rate in a Savings Account One hundred dollars is deposited in a savings account at 5% interest compounded continuously. The function defined by ƒ(t) shown in the figure gives the balance in the account after t years. At what rate (in dollars per year) is the balance growing after 14 years? Balance in a Savings Account y
p(t ) = t 2 + 2t for 0 … t … 5. Find the instantaneous rate of change of the percent of the population infected at time t = 3. (Modeling) In Exercises 57 and 58, use a calculator to determine the derivative. 57. Spread of a Rumor A rumor is spreading through a city. The number of people who have heard the rumor after t days is modeled by
480
y = f(t)
420
56. Spread of a Disease Epidemiologists estimate that t days after the flu begins to spread in a small town, the percent of the population infected by the flu is approximated by
Dollars
360
ƒ(t ) =
300 240
Graph ƒ(t) in the window 30, 324 by 30, 100,0004. Determine how fast the rumor is spreading after 8 days.
180 120 60 0
6
12
18
24
30
t
Years
51. Weight of a Small Fish The weight of a small fish in grams after t weeks is modeled by W(t) =
0.1t 2.
At what rate is the fish growing at time t = 4? 52. Velocity of a Ball A ball thrown straight up into the air has an initial height of 5 feet and an initial velocity of 128 feet per second. What is the velocity of the ball after 2 seconds?
100,000 . 1 + 9.134(0.8) t
58. Level of a Drug When a drug is taken orally, the amount of the drug in the bloodstream after t hours is modeled by the function defined by ƒ(t ) = 120(e 0.2t  e t ) units. Graph the function in the window 30, 164 by 30, 704. How many units of the drug are in the bloodstream after 6 hours? At what rate is the level of the drug in the bloodstream decreasing after 6 hours?
12.5
12.5
861
Area and the Definite Integral
Area and the Definite Integral
Areas by Approximation • The Definite Integral
Areas by Approximation Under certain conditions, the area of a region can be thought of as a sum of parts. FIGURE 33 shows a region bounded by the positive yaxis, the positive xaxis, and the graph of ƒ(x) = 24  x 2. We can find a rough approximation of the area of this region by using two circumscribed rectangles, as in FIGURE 34. The height of the rectangle on the left is ƒ(0) = 2, and the height of the rectangle on the right is ƒ(1) = 23. The width of each rectangle is 1, making the total area of the two rectangles 1 # ƒ(0) + 1 # ƒ(1) = 2 + 23 L 3.7321 square units. y
y
y
2
2
2
f (x) =
√4 – x2
f (x) = f (0)
f (1)
f (1) x
x 0
2
FIGURE 33
f(x) =
√4 – x2
0.5
1
1.5
1
2
0
FIGURE 34
2 FIGURE 35
If the left endpoint gives an answer too large, and the right endpoint an answer too small, it seems reasonable to average the two answers. This produces the method called the trapezoidal rule. In this example, we get
2
FIGURE 36
3.7321 + 1.7321 = 2.7321. 2
y
2
Another way to get an improved answer would be to use the midpoint of each interval, rather than the left endpoint or the right endpoint. This is called the midpoint rule. See FIGURE 36. In this example, that procedure leads to
1
1 # ƒ(0.5) + 1 # ƒ(1.5) = 23.75 + 21.75 L 3.2594 square units. x
0
1
1 # ƒ(1) + 1 # ƒ(2) = 23 + 0 L 1.7321 square units.
x 0
1
f (2) = 0 x
Here, the function is decreasing, and we overestimate the area when we evaluate the function at the left endpoint to determine the height of the rectangle in that interval. If we use the right endpoint, however, the answer will be too small. For example, using the right endpoints 1 and 2, as shown in FIGURE 35, the area of the two rectangles is
y 2
0
√4 – x2
1 2
1
3 2
FIGURE 37
2
To improve the accuracy of all of the previous approximations, we could divide the interval from x = 0 to x = 2 into more parts. The result obtained by using the left endpoint again with four equal parts, each of width 12, is shown in FIGURE 37.
862
CHAPTER 12
Limits, Derivatives, and Definite Integrals
This approximation is greater than the actual area. As before, the height of each rectangle is given by the value of ƒ at the left side of the rectangle, and its area is the width, 12, multiplied by the height. The total area of the four rectangles is 1 2
#
=
1 1 215 1 1 27 (2) + a b + A 23 B + a b 2 2 2 2 2 2
ƒ(1) +
1 2
#
3 ƒa b 2
ƒ(0) +
1 2
#
1 1 ƒa b + 2 2
#
215 23 27 + + 4 2 4 L 3.4957 square units. = 1 +
y
This approximation looks better, but it is still greater than the actual area. To improve the approximation, we divide the interval from x = 0 to x = 2 into eight parts with equal widths of 14. See FIGURE 38. The total area of all these rectangles is
2
1 4
#
ƒ(0) +
1 4
#
1 1 ƒa b + 4 4
#
1 1 ƒa b + 2 4
#
3 1 ƒa b + 4 4
#
ƒ(1)
1 4
#
5 1 ƒa b + 4 4
#
3 1 ƒa b + 2 4
#
7 ƒa b 4
1
+
L 3.3398 square units.
x 0
1 2
1
3 2
FIGURE 38
2
The process of approximating the area under a curve by using more and more rectangles to get a better and better approximation can be generalized. We divide the interval from x = 0 to x = 2 into n equal parts. Each of these n intervals has width 2 2  0 = . n n Thus, each rectangle has width 2n and height determined by the function value at the left side of the rectangle, the right side, or the midpoint. We could also average the leftand rightside values as before. Using a computer or graphing calculator to find approximations to the area for several values of n gives the following results. n
Left Sum
Right Sum
Trapezoidal Rule
Midpoint Rule
2 4 8 10 20 50 100 500
3.7321 3.4957 3.3398 3.3045 3.2285 3.1783 3.1604 3.1455
1.7321 2.4957 2.8398 2.9045 3.0285 3.0983 3.1204 3.1375
2.7321 2.9957 3.0898 3.1045 3.1285 3.1383 3.1404 3.1415
3.2594 3.1839 3.1567 3.1524 3.1454 3.1426 3.1419 3.1416
The numbers in the last four columns of this table represent approximations to the area under the curve, above the xaxis, and between the lines x = 0 and x = 2. As n becomes larger and larger, all four approximations become better and better, getting closer to the actual area. In this example, the exact area can be found by a formula from plane geometry. We rewrite the given function as y = 24  x 2 y2 = 4  x2 2 x + y 2 = 4.
Let ƒ( x ) = y. Square each side. Add x 2.
12.5
Area and the Definite Integral
863
The resulting equation x 2 + y 2 = 4 is the equation of a circle centered at the origin with radius 2. The region in FIGURE 33 is the quarter of this circle that lies in quadrant I. The actual area of this region is onequarter of the area of the entire circle, or 1 p(2) 2 = p L 3.1416. 4
A = pr 2
As the number of rectangles increases without bound, the sum of the areas of these rectangles gets closer and closer to the actual area of the region, p. This statement can be written as lim (sum of areas of n rectangles) = p.
n:q
(The value of p was originally investigated by a process similar to this.) In this example, for a particular value of n, the midpoint rule gave the best answer (the one closest to the true value of p, or about 3.1416), followed by the trapezoidal rule, followed by the left and right sums. In fact, the midpoint rule with n = 20 gives a value (3.1454) that is slightly more accurate than the left sum with n = 500 (3.1455).
The Definite Integral To approximate the area bounded by the curve y = ƒ(x), the xaxis, and the vertical lines x = a and x = b, as shown in FIGURE 39, we could divide the region under the curve into 10 rectangles (FIGURE 39(a)) and then into 20 rectangles (FIGURE 39(b)). The sums of the areas of the rectangles give approximations to the area under the curve. y
y
y = f (x)
y = f(x)
x a
a
b
Ten rectangles of equal width
b
x
Twenty rectangles of equal width
(a)
(b) FIGURE 39
To develop a process that would yield the exact area, we divide the interval from a to b into n pieces of equal width, using each of these n pieces as the base of a rectangle. See FIGURE 40 on the next page. Let x1 be an arbitrary point in the first interval, x2 be an arbitrary point in the second interval, and so on, up to the nth interval. The symbol ≤x represents the width of each interval. The red rectangle is an arbitrary rectangle called the ith rectangle. Its area is the product of its length and width. Since the width of the ith rectangle is ¢x and the length of the ith rectangle is given by the height ƒ(xi), area of the ith rectangle = ƒ(xi)
#
¢x.
The total area under the curve is approximated by the sum of the areas of all n of the rectangles. With sigma notation, the approximation to the total area becomes n
area of all n rectangles = a ƒ(xi)
#
¢x.
i=1
864
CHAPTER 12
Limits, Derivatives, and Definite Integrals y
Δx
y = f (x)
f (x)
… x1 x2 x3
… xi
x
xn
n rectangles of equal width FIGURE 40
GCM
TECHNOLOGY NOTE Some calculators have a builtin function for evaluating the definite integral. The figure shows how a TI83/84 Plus calculator computes
n
exact area = limq a ƒ(xi)¢x. n: i=1
This limit, called the definite integral of ƒ(x) from a to b, is written as follows.
2
L0
The exact area is defined to be the limit of this sum (if the limit exists) as the number of rectangles increases without bound:
24  x 2 dx,
giving an answer of 3.141593074. This is an approximation with an error of about 0.0000004.
Definite Integral
If ƒ is defined on the interval 3a, b4, the definite integral of ƒ from a to b is given by b
La
n
ƒ(x) dx ⴝ lim a ƒ(xi)≤x, n:ˆ iⴝ 1
provided that the limit exists, where ¢x = the ith interval.
b  a n
and xi is any value of x in
The answer is an approximation for .
The b above the integral sign in the definition is called the upper limit of integration, and the a is the lower limit of integration. This use of the word limit has nothing to do with the limit of the sum; it refers to the limits, or boundaries, on x. Note that the definite integral of a continuous function ƒ will always exist. The definite integral can be approximated by n
a ƒ(xi)¢x.
i=1
If ƒ(x) Ú 0 on the interval 3a, b4, the definite integral gives the area under the curve (and above the xaxis) between x = a and x = b. In the midpoint rule, xi is the midpoint of the ith interval. We may also let xi be the left endpoint, the right endpoint, or any other point in the ith interval. In the example at the beginning of this section, the area bounded by the xaxis, the curve y = 24  x 2, and the lines x = 0 and x = 2 could be written as the definite integral 2
L0
24  x 2 dx = p.
12.5
Area and the Definite Integral
865
Finding the definite integral of a function can be thought of as a mathematical process that gives the sum of an infinite number of individual parts (within certain limits). The definite integral represents area only if the function is nonnegative (ƒ(x) Ú 0) at every xvalue in the interval 3a, b4. There are other interpretations of the definite integral, all of which involve approximation by appropriate sums.
EXAMPLE 1
Approximating Area
Approximate 4
L0
2x dx,
the area of the region under the graph of ƒ(x) = 2x, above the xaxis, and between x = 0 and x = 4, by using four rectangles of equal width whose heights are the values of the function at the midpoint of each rectangle. y 8
f (x) = 2x
Solution We want to find the area of the shaded region in FIGURE 41. The heights of the four rectangles given by ƒ(xi) for i = 1, 2, 3, and 4 are as follows.
6
i
4
1 2 3 4
2 x 0
2
4
FIGURE 41
ƒ(xi )
xi x1 x2 x3 x4
The width of each rectangle is ¢x = tangles is
= = = =
0.5 1.5 2.5 3.5
4  0 4
ƒ(0.5) ƒ(1.5) ƒ(2.5) ƒ(3.5)
= = = =
1 3 5 7
= 1. The sum of the areas of the four rec
4
a ƒ(xi)¢x = ƒ(x1)¢x + ƒ(x2)¢x + ƒ(x3)¢x + ƒ(x4)¢x
i=1
= ƒ(0.5)¢x + ƒ(1.5)¢x + ƒ(2.5)¢x + ƒ(3.5)¢x = 1(1) + 3(1) + 5(1) + 7(1) = 16.
Approximated area
Using the formula for the area of a triangle, A = 12bh, with b, the length of the base, equal to 4 and h, the height, equal to 8, gives A =
The figure shows how the TI83/84 Plus confirms our answer in Example 1.
1 1 bh = (4)(8) = 16. 2 2
Exact area
The approximation equals the exact area because using the midpoints of each subinterval distributed the error evenly above and below the graph of a linear function. If ƒ were a polynomial of degree greater than 1, then the midpoint method would not typically get the exact area.
866
CHAPTER 12
12.5
Limits, Derivatives, and Definite Integrals
EXERCISES
1. Let ƒ(x) = 2x + 1, x1 = 0, x2 = 2, x3 = 4, x4 = 6, and ¢x = 2. 4
(a) Find a ƒ(xi)¢x. i=1
5
14. Find 10 (5  x) dx by using the formula for the area of a triangle. 4
15. Find 10 ƒ(x) dx for each graph of y = ƒ(x).
(b) The sum in part (a) approximates a definite integral by using rectangles. The height of each rectangle is given by the value of the function at the left endpoint. Write the definite integral that the sum approximates.
y
(a)
1
2. Use the midpoint rule with n = 4 to approximate the area above the xaxis bounded by the graph of
y
(b)
1 x
x 0
ƒ(x) = 216  x 2
(c)
0
1
(d)
y
in the first quadrant.
1 y
4 2
In Exercises 3–12, approximate the area under the graph of ƒ(x) and above the xaxis, using each of the following methods with n = 4. (a) Use left endpoints. (b) Use right endpoints. (c) Average the answers in parts (a) and (b). (d) Use midpoints.
1 x 0
5. ƒ(x) = x + 2 from x = 0 to x = 4
3
16.
7. ƒ(x) =
from x = 1 to x = 5
8. ƒ(x) =  x 2 + 4 from x =  2 to x = 2 9. ƒ(x) = e x  1 from x = 0 to x = 4 10. ƒ(x) = e x + 1 from x =  2 to x = 2
4
L3
0
29  x 2 dx
17.
L4
3
18.
6. ƒ(x) = 3 + x from x = 1 to x = 3 x2
2
Checking Analytic Skills Find the exact value of each integral, using formulas from geometry. Do not use a calculator.
3. ƒ(x) = 3x + 2 from x = 1 to x = 5 4. ƒ(x) = x + 5 from x = 2 to x = 4
x 0
1
L1
5
(5  x) dx
19.
21  (x  1) 2 dx
21.
L0
(1 + 2x) dx
L2
2
20.
216  x 2 dx
4
L1
(2x  1) dx
22. (Modeling) Distance The curves at the top of the next page show the velocity at t seconds after a car has accelerated from a dead stop. (Source: Road & Track.) To find the total distance traveled by the car in reaching 100 mph, we must estimate the definite integral T
1 11. ƒ(x) = from x = 1 to x = 5 x 2 12. ƒ(x) = from x = 1 to x = 9 x 13. Consider the region below ƒ(x) = 2x , above the xaxis, and between x = 0 and x = 4. Let xi be the midpoint of the ith subinterval. (a) Approximate the area of the region, using four rectangles. 4 (b) Find 10 ƒ(x) dx by using the formula for the area of a triangle.
L0
v (t) dt,
where T represents the number of seconds it takes for the car to reach 100 mph. Use the graphs to estimate this distance by adding the areas of rectangles with widths representing 5 seconds. (The last rectangle has a width representing 4 seconds in part (a) and 3 seconds in part (b).) Use the midpoint rule. To adjust your answer to miles per hour, divide by 3600 (the number of seconds in an hour). You then have the number of miles that the car traveled in reaching 100 mph. Finally, multiply by 5280 feet per mile to convert the answer to feet.
Reviewing Basic Concepts (Sections 12.4 and 12.5)
(a) Estimate the distance traveled by the Porsche 928, using the graph.
(b) Estimate the distance traveled by the BMW 733i, using the graph. 100
100
3rd4th
80
2nd3rd
70 60
1st50 2nd 40 30 20
0
5
10
15
20
25
SECTIONS 12.4 AND 12.5
50
1st
40 2nd 30 20
5
10
15
20
25
30
35
4. If ƒ(x) = k, where k is a real number, what is the value of ƒ¿(a)? 5. (Modeling) Marginal Revenue The revenue in dollars from producing (and selling) x units of an item is modeled by
4
R(x) =  0.0012x 2 + 3x.
2 2 0
Find the marginal revenue at a production level of 1800 units.
x
4
(3, –1)
–2
6. Find the equation of the tangent line to the graph of ƒ(x) = 6  x 2 when x =  1.
–4
7. Find ƒ¿(9), where ƒ(x) = 2x + 2.
3.
y
y
8. Approximate the area under the graph of ƒ(x) = x 2 and above the xaxis from x = 0 to x = 5 with n = 5, using midpoints.
6 4 4 2
(4, 2)
2
(1, 2)
0
2nd3rd
REVIEWING BASIC CONCEPTS
y
2.
60
Elapsed Time (in seconds) BMW 733i
Estimate the slope of the tangent line to each curve at the given point (x, y).
–2
70
0
30 35
Elapsed Time (in seconds) Porsche 928
–4
80
10
10
1.
3rd4th
90
Speed (in miles per hour)
Speed (in miles per hour)
90
867
2
4
x 0
x 2
4
6
9. Find the area bounded by the xaxis, the yaxis, and the graph of y = 8  x, using the formula for the area of a triangle. 10. Use the formula for the area of a circle to find the exact 3 value of 10 29  x 2 dx.
868
CHAPTER 12
12
Limits, Derivatives, and Definite Integrals
SUMMARY
KEY TERMS & SYMBOLS
KEY CONCEPTS
12.1 An Introduction to Limits
LIMIT OF A FUNCTION Let ƒ be a function and let a and L be real numbers. L is the limit of ƒ(x) as x approaches a, written lim ƒ(x) ⴝ L, if the following conditions are met:
limit
x:a
1. As x takes values closer and closer (but not equal) to a on both sides of a, the corresponding values of ƒ(x) get closer and closer (and perhaps equal) to L. 2. The value of ƒ(x) can be made as close to L as desired by taking values of x arbitrarily close to a. SOME CONDITIONS UNDER WHICH lim ƒ(x) FAILS TO EXIST x:a
1. ƒ(x) approaches a number L as x approaches a from the left and approaches a different number M as x approaches a from the right. 2. ƒ(x) becomes infinitely large in absolute value as x approaches a from either side. 3. ƒ(x) oscillates infinitely many times between two fixed values as x approaches a.
12.2 Techniques for Calculating Limits
RULES FOR LIMITS 1. Constant rule If k is a constant real number, then lim k ⴝ k. x:a
lim x ⴝ a
2. Limit of x rule
x:a
For the following rules, we assume that lim ƒ(x) and lim g (x) both exist. x:a
3. Sum and difference rules 4. Product rule
x:a
lim 3ƒ(x) ⴞ g(x)4 ⴝ lim ƒ(x) ⴞ lim g(x)
x:a
x:a
x:a
lim 3ƒ(x) # g(x)4 ⴝ lim ƒ(x) # lim g(x)
x:a
x:a
x:a
lim ƒ(x)
5. Quotient rule 6. Polynomial rule
lim
x:a
ƒ(x) x:a ⴝ , provided that lim g(x) Z 0 x:a g(x) lim g(x) x:a
If p (x) defines a polynomial function, then lim p (x) ⴝ p (a). x:a
7. Rational function rule If ƒ(x) defines a rational function p (a) . lim ƒ(x) ⴝ x:a q (a)
p (x) q (x) ,
with q (a) Z 0, then
8. Equal functions rule If ƒ(x) = g(x) for all