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l parts are adequately supported. Problems like these can be modeled and solved by using vectors. See Exercise 50 in Section 10.6. 616 Chapter Outline 10.1 The Law of Cosines 10.2 The Law of Sines 10.3 The Complex Plane and Polar Form for Complex Numbers 10.4 DeMoivre’s Theorem and nth Roots of Complex Numbers 10.5 Vectors in the Plane 10.6 Applications of Vectors in the Plane 10.6.A Excursion: The Dot Product Chapter Review can do calculus Euler’s Formula Interdependence of Sections 10.1 10.2 > 10.3 10.5 > > 10.4 10.6 Trigonometry has a variety of useful applications in geometry, alge- bra, and the physical sciences. Several applications are discussed in this chapter. 10.1 The Law of Cosines Objectives • Solve oblique triangles by using the Law of Cosines. Sections 6.1 and 6.2 presented right triangle trigonometry and its applications. In this section, the solutions to oblique triangles, ones that do not contain a right angle, are considered. A c B b a Figure 10.1-1 Standard notation of triangles, which is used in this section and the next, is shown in Figure 10.1-1 and is described below. C Each vertex is labeled with a capital letter, and the length of the side opposite that vertex is denoted by the same letter in lower case. The letter A will also be used to label the angle at vertex A, and similarly for B and C. Thus, statements such as A 37° or cos B 0.326 will be made. The first fact needed to solve oblique triangles is the Law of Cosines, whose proof is given at the end of this section. Law of Cosines In any triangle ABC, with lengths a, b, c, as in Figure 10.1-1, a2 b2 c2 2bc cos A b2 a2 c2 2ac cos B c2 a2 b2 2ab cos C 617 618 Chapter 10 Trigonometric Applications You need only memorize one of these equations since each of them provides essentially the same information: the square of the length of one side of a triangle is given in terms of the angle opposite it and the other two sides. Solving the first equation in the Law of Cosines for cos A results in the alternate form of the Law of Cosines, given below. In any triangle ABC, with sides of lengths a, b, and c, as in Figure 10.1-1, cos A b2 c2 a2 2bc . The other two equations in the Law of Cosines can be similarly rewritten in an alternate form. In this form, the Law of Cosines provides a description of each angle of a triangle in terms of the three sides. Consequently, the Law of Cosines can be used to solve triangles in the following cases. 1. Two sides and the angle between them are known (SAS) 2. Three sides are known (SSS) Example 1 Solve a Triangle with SAS Information Solve triangle ABC shown in Figure 10.1-2. Solution Because c is the unknown quantity, use the third equation in the Law of Cosines. c2 a2 b2 2ab cos C c2 162 102 2 10 21 c2 356 320 cos 110° c 2356 320 cos 110° c 21.6 16 2 1 cos 110° Take the square root of each side. Use the alternate form of the Law of Cosines to find the measure of angle A. cos A b2 c2 a2 2bc cos A 102 21.62 162 2 1 cos A 0.7172 A cos A 44.2° 1 0.7172 21.6 10 21 2 1 COS Use the calculator in degree mode. key with the Alternate Form: Law of Cosines NOTE When C is a right angle, then c is the hypotenuse and cos C cos 90° 0, so that the third equation in the Law of Cosines becomes the Pythagorean theorem. c2 a2 b2 C 110° 10 16 A c B Figure 10.1-2 NOTE Throughout this chapter, no rounding is done in the actual computation until the final quantity is obtained. Section 10.1 The Law of Cosines 619 Because the sum of the angle measures in a triangle is 180°, B 180° Thus, c 21.6, A 44.2°, and 44.2° 110° 1 B 25.8°. 2 25.8°. ■ A 15 8.3 Example 2 Solve a Triangle with SSS Information Find the angles of triangle ABC shown in Figure 10.1-3. C 20 B Figure 10.1-3 Solution and c 8.3. To find angles, use a 20, b 15, The given information is the alternate form of the Law of Cosines. cos A b2 c2 a2 cos A 152 8.32 202 15 2 21 1 cos A 0.4261 A 115.2° 2bc 8.3 2 2ac cos B a2 c2 b2 cos B 202 8.32 152 2 1 cos B 0.7346 B 42.7° 8.3 20 21 2 Use the sum of angle measures in a triangle to find the third angle. Thus, A 115.2°, B 42.7°, C 180° 115.2° 42.7° C 22.1°. 1 and 22.1° 2 ■ Example 3 The Distance Between Two Vehicles 125° Two trains leave a station on different tracks. The tracks make an angle of with the station as the vertex. The first train travels at an average speed of 100 kilometers per hour, and the second train travels at an average speed of 65 kilometers per hour. How far apart are the trains after 2 hours? Solution The first train, A, traveling at 100 kilometers per hour for 2 hours, goes a kilometers. The second train, B, travels a disdistance of kilometers. The situation is shown in Figure 10.1-4. tance of 100 2 200 65 2 130 B 130 125° C Station c 200 A Figure 10.1-4 By the Law of Cosines: c2 a2 b2 2ab cos C c2 1302 2002 2 200 130 c2 56,900 52,000 cos 125° c 256,900 52,000 cos 125° c 294.5 21 2 1 cos 125° The trains are about 294.5 kilometers apart after 2 hours. ■ 620 Chapter 10 Trigonometric Applications Example 4 Find Angles with the Horizontal A 100-foot tall antenna tower is to be placed on a hillside that makes an with the horizontal. It is to be anchored by two cables from angle of the top of the tower to points 85 feet uphill and 95 feet downhill from the base. How much cable is needed? 12° Solution The situation is shown in Figure 10.1-5, where AB represents the tower and AC and AD represent the cables. A b 100 k 85 D C 95 12° B E Figure 10.1-5 If the hillside makes an angle of BEC, angle E is a right angle and angle C measures the angle measures in the triangle. with the horizontal, then in triangle Use the sum of 12°. 12° mCBE 180° 90° 12° 78° 1 As shown in Figure 10.1-5, adjacent angles ABC and CBE form a straight angle, which measures 180°. 2 mABC 180° 78° 102° Using the SAS information in triangle ABC, apply the Law of Cosines. b2 a2 c2 2ac cos B b2 952 1002 2 95 2 b2 19,025 19,000 cos 102° b 219,025 19,000 cos 102° b 151.58 100 21 1 cos 102° The length of the downhill cable is about 151.58 feet. To find the length of the uphill cable, notice that adjacent angles ABC and ABD form a straight angle. mABD 180° mABC 180° 102° 78° Section 10.1 The Law of Cosines 621 Using the SAS information in triangle ABD, apply the Law of Cosines. 100 k2 1002 852 2 85 2 k2 17,225 17,000 cos 78° k 217,225 17,000 cos 78° k 117.01 21 1 cos 78° The length of the uphill cable is about 117.01 feet. The amount of cable needed is the sum of the lengths of the uphill and downhill cables. 151.58 117.01 268.59 Therefore, the length of the cable needed is about 268.59 feet. ■ Proof of the Law of Cosines Given triangle ABC, position it on a coordinate plane so that angle A is in standard position with initial side c and terminal side b. Depending on the size of angle A, there are two possibilities, as shown in Figure 10.1-6. y b C(x, y) a B x A c y a C(x, y) b x A c B Figure 10.1-6 The coordinates of B are (c, 0). Let (x, y) be the coordinates of C. Now C is a point on the terminal side of angle A, and the distance from C to the origin A is b. Therefore, according to the definitions of sine and cosine, the following statements are true. x b y b cos A, or equivalently, x b cos A sin A, or equivalently, y b sin A 2 1 2 b cos cos A c 2 b sin A 0 2 Use the distance formula to find a, the distance from C to B. a 2 1 a 2 a2 1 a2 b2 cos2 A 2bc cos A c2 b2 sin2 A a2 b2 cos2 A b2 sin2 A c2 2bc cos A a2 b2 c2 2bc cos A cos2 A sin2 A 2 a2 b2 c2 2bc cos A Rearrange terms. Square each side. Factor out b sin A Simplify. 2 2 b2. 1 2 2 2 1 1 2 Pythagorean identity Substitute for x and y. 622 Chapter 10 Trigonometric Applications This proves the first equation in the Law of Cosines. Similar arguments beginning with angles B or C in standard position prove the other two equations. Exercises 10.1 Standard notation for triangle ABC is used throughout. Use a calculator and round your answers to one decimal place at the end of each computation. In Exercises 1–16, solve the triangle ABC under the given conditions. 1. A 20°, b 10, c 7 2. B 40°, a 12, c 20 3. C 118°, a 6, b 10 4. C 52.5°, a 6.5, b 9 5. A 140°, b 12, c 14 6. B 25.4°, a 6.8, c 10.5 7. C 78.6°, a 12.1, b 20.3 8. A 118.2°, b 16.5, c 10.7 9. a 7, b 3, c 5 10. a 8, b 5, c 10 11. a 16, b 20, c 32 12. a 5.3, b 7.2, c 10 13. a 7.2, b 6.5, c 11 14. a 6.8, b 12.4, c 15.1 15. a 12, b 16.5, c 21.3 16. a 5.7, b 20.4, c 16.8 17. Find the angles of the triangle whose vertices are 1, 4 5, 2 and 0 18. Find the angles of the triangle whose vertices are 2, 1 3, 4 , and 6, 1 , . 19. Two trains leave a station on different tracks. The tracks make a 112° angle with the station as vertex. The first train travels at an average speed of 90 kilometers per hour and the second at an average speed of 55 kilometers per hour. How far apart are the trains after 2 hours and 45 minutes? 20. One plane flies west from Cleveland at 350 miles per hour. A second plane leaves Cleveland at the same time and flies southeast at 200 miles per hour. How far apart are the planes after 1 hour and 36 minutes? 21. The pitcher’s mound on a standard baseball diamond (which is actually a square) is 60.5 feet from home plate. How far is the pitcher’s mound from first base? 2nd base 90 ft 90 ft Pitcher's mound 3rd base 1st base 90 ft 60.5 ft 90 ft Home plate 22. If the straight-line distance from home plate over second base to the center field wall in a baseball stadium is 400 feet, how far is it from first base to the same point in center field? Adapt the figure above. 23. A stake is located 10.8 feet from the end of a closed gate that is 8 feet long. The gate swings open, and its end hits the stake. Through what angle does the gate swing? 8 10.8 24. The distance from Chicago to St. Louis is 440 kilometers, from St. Louis to Atlanta 795 kilometers, and from Atlanta to Chicago 950 kilometers. What are the angles in the triangle with these three cities as vertices? 25. A boat runs in a straight line for 3 kilometers, 45° then makes a 6 kilometers. How
far is the boat from its starting point? turn and goes for another 45° 3 6 Start 26. A plane flies in a straight line at 400 miles per 15° hour for 1 hour and 12 minutes. It makes a turn and flies at 375 miles per hour for 2 hours and 27 minutes. How far is it from its starting point? 27. The side of a hill makes an angle of with the horizontal. A wire is to be run from the top of a 175-foot tower on the top of the hill to a stake located 120 feet down the hillside from the base of the tower. What length of wire is needed? 12° 28. Two ships leave port, one traveling in a straight course at 22 miles per hour and the other traveling a straight course at 31 miles per hour. Their courses diverge by How far apart are they after 3 hours? 38°. 29. An engineer wants to measure the width CD of a sinkhole. He places a stake B and determines the Section 10.1 The Law of Cosines 623 measurements shown in the figure below. How wide is the sinkhole? B 103° 1 2 0 ft 7 4 ft C D 30. A straight tunnel is to be dug through a hill. Two people stand on opposite sides of the hill where the tunnel entrances are to be located. Both can see a stake located 530 meters from the first person and 755 meters from the second. The angle determined by the two people and the stake (the vertex) is How long must the tunnel be? 77°. 31. One diagonal of a parallelogram is 6 centimeters long, and the other is 13 centimeters long. They form an angle of with each other. How long are the sides of the parallelogram? Hint: The diagonals of a parallelogram bisect each other. 42° 32. A parallelogram has diagonals of lengths 12 and How 15 inches that intersect at an angle of long are the sides of the parallelogram? 63.7°. 33. A ship is traveling at 18 miles per hour from 22° Corsica to Barcelona, a distance of 350 miles. To avoid bad weather, the ship leaves Corsica on a route below). After 7 hours the bad weather has been bypassed. Through what angle should the ship now turn to head directly to Barcelona? south of the direct route (see the figure Barcelona Angle of turn Corsica 22° 34. In aerial navigation, directions are given in degrees clockwise from north. Thus east is 180°, and so on, as shown in the following figure. A plane leaves South Bend for Buffalo, 400 miles away, intending to fly a straight course in the south is 90°, 624 Chapter 10 Trigonometric Applications 70°. After flying 180 miles, the pilot direction realizes that an error has been made and that he has actually been flying in the direction 55°. 0° North 270° West 90° East 180° South a. At that time, how far is the plane from Buffalo? b. In what direction should the plane now go to reach Buffalo? 35. Assume that the earth is a sphere of radius 3980 miles. A satellite travels in a circular orbit around the earth, 900 miles above the equator, making one full orbit every 6 hours. If it passes directly over a tracking station at 2 P.M., what is the distance from the satellite to the tracking station at 2:05 P.M.? 36. Two planes at the same altitude approach an airport. One plane is 16 miles from the control tower and the other is 22 miles from the tower. The angle determined by the planes and the tower, with the tower as the vertex, is far apart are the planes? 11°. How 37. Assuming that the circles in the following figure are mutually tangent, find the lengths of the sides and the measures of the angles in triangle ABC. 8.23 C B 11.27 13 A 38. Assuming that the circles in the following figure are mutually tangent, find the lengths of the sides and the measures of the angles in triangle ABC. A 20.62 7.35 C B 8.04 39. Critical Thinking A rope is attached at points A and B and taut around a pulley whose center is at C, as shown in the following figure. The rope lies on the pulley from D to E and the radius of the pulley is 1 meter. How long is the rope 40. Critical Thinking Use the Law of Cosines to prove that the sum of the squares of the lengths of the two diagonals of a parallelogram equals the sum of the squares of the lengths of the four sides. Section 10.2 The Law of Sines 625 10.2 The Law of Sines Objectives • Solve oblique triangles by using the Law of Sines • Use area formulas to find areas of triangles Law of Sines In Section 10.1, the Law of Cosines was used to solve oblique triangles when SAS or SSS information was given. When different information is given about the triangle, the Law of Cosines may not be sufficient to solve them. In this case, the Law of Sines may be used. In any triangle ABC (in standard notation) a sin A b sin B c sin C . Proof Position triangle ABC on a coordinate plane so that angle C is in standard position, with initial side b and terminal side a, as shown in Figure 10.2-1. 90° < C < 180° y 0° < C < 90 Figure 10.2-1 a B h b D c x A In each case, sin C can be computed by using the point B on the terminal side of angle C. The second coordinate of B is h, and the distance from B to the origin is a. Therefore, by the definition of sine, sin C h a , or equivalently, h a sin C. In each case, right triangle ADB shows that sin A opposite hypotenuse h c , or equivalently, h c sin A. Combine these two expressions for h. a sin C c sin A Because angles in a triangle are nonzero, sin each side of the last equation by (sin A)(sin C). A 0 and sin C 0. Divide a sin A c sin C 626 Chapter 10 Trigonometric Applications This proves one proportion in the Law of Sines. Similar arguments beginning with angles A or B in standard position prove the other proportions. The Law of Sines can be used to solve triangles in the following cases. 1. Two angles and one side are known (AAS) 2. Two sides and the angle opposite one of them are known (SSA) Example 1 Solve a Triangle with AAS Information B 20°, C 31°, If lengths. See Figure 10.2-2. and b 210 , find the other angle measure and side C 31° 210 a 20° A c B Figure 10.2-2 Solution Because the sum of the angle measures of a triangle is A 180° 20° 31° 129°. 180°, 1 2 In order to find a, notice that three of the four quantities in one of the Law of Sines proportions are known. a sin A a sin 129° b sin B 210 sin 20° a 210 sin 129° a 477.2 sin 20° Multiply each side by sin 129° . c is found in a similar manner. Use a Law of Sines proportion involving c and the three known quantities. c sin C c sin 31° b sin B 210 sin 20° c 210 sin 31° c 316.2 sin 20° Multiply each side by sin 31°. Therefore, A 129°, a 477.2, and c 316.2. ■ Section 10.2 The Law of Sines 627 The Ambiguous Case Given AAS information, such as in Example 1, there is exactly one triangle satisfying the given data (see note). But when two sides of a triangle and the angle opposite one of them are known, as in the SSA case, there may be one, two, or no triangles that satisfy the given data. This is called the ambiguous case. NOTE The four Triangle Congruence Theorems—AAS, SAS, SSS, and ASA—state that a unique triangle can be formed that is congruent to a given triangle. However, SSA is not a triangle congruence theorem because a unique triangle congruent to the given one is not guaranteed; zero, one, or two triangles can be formed from the side-side-angle information. To see why the ambiguous case occurs, suppose sides a and b and angle A are given. Place angle A in standard position with terminal side b. If angle A is less than then there are four possibilities for side a. 90°, Ambiguous Case: SSA Information with A 66 90 (i) a b, and side a is too short to reach the third side: no solution. (ii) a b, and side a just reaches the third side and is perpendicular to it: one solution Figure 10.2-3 Figure 10.2-4 (iii) a b, and an arc of radius a (iv) meets the third side at 2 points to the right of A: two solutions. a b, so that an arc of radius a meets the third side at just one point to the right of A: one solution Figure 10.2-5 A B Figure 10.2-6 628 Chapter 10 Trigonometric Applications If angle A is greater than 90°, then there are only two possibilities. Ambiguous Case: SSA Information with A 7 90° (i) so that side a is too a b, short to reach the third side: no solution. so that an arc of radius a 7 b, a meets the third side at just one point to the right of A: one solution. (ii Figure 10.2-7 Figure 10.2-8 Recall from Section 8.3 that when finding all solutions of a trigonometric equation involving the sine function, the identity is used. This same identity, stated in terms of degrees rather than radians, is used to deal with the ambiguous SSA case. sin x sin p x 2 1 Supplementary Angle Identity If 0 U 90, then sin U sin(180 U). D h E y r θ 180° − θ O x Figure 10.2-9 Proof Place the angle in standard position and choose a point D on its terminal side. Let r be the distance from D to the origin. The situation is shown in Figure 10.2-9. 180° u Because h is the second coordinate of D, then sin 180° u 1 h r . 2 Right tri- angle OED shows that sin u opposite hypotenuse h r sin 1 180° u . 2 Example 2 Solve a Triangle with SSA Information Given a possible triangle ABC with a 6, b 7, and A 65°, find angle B. Solution Use a proportion of the Law of Sines involving angle B and three known quantities. Section 10.2 The Law of Sines 629 a sin A 6 b sin B 7 sin B sin 65° sin B 7 sin 65° 6 sin B 1.06 There is no angle B whose sine is greater than 1. Therefore, there is no triangle satisfying the given data. ■ Example 3 Solve a Triangle with SSA Information An airplane A takes off from carrier B and flies in a straight line for 12 kilometers. At that instant, an observer on destroyer C, located 5 kilometers from the carrier, notes that the angle determined by the carrier, the destroyer (the vertex), and the plane is How far is the plane from the destroyer? 37°. Solution The given information is organized in Figure 10.2-10. B 5 37° C 12 b A Figure 10.2-10 Because there is no proportion of the Law of Sines using B that can be solved, use a proportion involving A to find a second angle of the triangle. This will provide you with enough information to find angle B and hence allowing you to use th
e Law of Sines to find b. c sin C 12 a sin A 5 sin 37° sin A sin A 5 sin 37° 12 sin A 0.2508 A 14.5° or A 180° 14.5° 165.5° 630 Chapter 10 Trigonometric Applications C 37°, Because that is impossible, then the sum of angles A, B, and C would be is the only possible 14.5° A 165.5° If and greater than 180°. measure of angle A. Therefore, B 180° 37° 14.5° 1 2 128.5°. All of the angles are known, and b can be found using either the Law of Cosines or the Law of Sines. b sin B b sin 128.5° c sin C 12 sin 37° b 12 sin 128.5° b 15.6 sin 37° The plane is approximately 15.6 kilometers from the destroyer. ■ Example 4 Solve a Triangle with SSA Information Solve triangle ABC when a 7.5, b 12, and A 35°. Solution Use a proportion of the Law of Sines involving the known quantities. a sin A 7.5 b sin B 12 sin B sin B 12 sin 35° 7.5 sin B 0.9177 sin 35° B 66.6° or B 180° 66.6° 113.4° Because the sum of angles A and B is less than are two possible cases, as shown in Figure 10.2-11. 180° in each case, there C 12 7.5 7.5 35° A c 113.4° B c Figure 10.2-11 66.6° B Section 10.2 The Law of Sines 631 B 66.6° Case 1 C 180° 35° 66.6° 78.4° 2 1 B 113.4° Case 2 C 180° 35° 113.4° 31.6° 2 1 Use the Law of Sines. Use the Law of Sines. c sin C c sin 78.4° a sin A 7.5 sin 35° c 7.5 sin 78.4° c 12.8 sin 35° c sin C c sin 31.6° a sin A 7.5 sin 35° c 7.5 sin 31.6° c 6.9 sin 35° Thus, in Case 1, B 113.4°, C 31.6°, B 66.6°, C 78.4°, c 6.9. and and c 12.8; and in Case 2, ■ Example 5 Solve a Triangle with ASA Information A plane flying in a straight line parallel to the ground passes directly over point A and later directly over point B, which is 3 miles from A. A few minutes after the plane passes over B, the angle of elevation from A to the plane is How high is the plane at that moment? and the angle of elevation from B to the plane is 67°. 43° Solution If C represents the plane, then the situation is represented in Figure 10.2-12. The height of the plane is h. C h a 67° 43° A 3 B D Figure 10.2-12 Angle ABC measures 180° 67° 113° mBCA 180° 1 . So 43° 113° 24°. 2 632 Chapter 10 Trigonometric Applications Use the Law of Sines to find side a of triangle ABC. a sin 43° 3 sin 24° a 3 sin 43° sin 24° a 5.03 Now use sin 67° h a to find h in right triangle CBD. sin 67° h 5.03 h 5.03 sin 67° h 4.63 The plane is about 4.63 miles high. ■ The Area of a Triangle The proof of the Law of Sines leads to the following formula for the area of a triangle. Area of a Triangle The area of a triangle containing an angle C with adjacent sides of lengths a and b is 1 2 ˛ ab sin C. y c B h a D C b Figure 10.2-13 x A Proof Place the vertex of angle C at the origin, with side b on the positive x-axis, as in Figure 10.2-13. Then b is the base and h is the height of the triangle. area of triangle ABC 1 2 base height 1 2 bh. The proof of the Law of Sines shows that area of triangle ABC 1 2 h a sin C. bh 1 2 ab sin C. Therefore, Figure 10.2-13 is the case when C is greater than C is less than is similar. 90° 90°; the argument when Example 6 Find Area with SAS Information Find the area of the triangle shown in Figure 10.2-14. Section 10.2 The Law of Sines 633 8 cm 130° 13 cm Figure 10.2-14 Solution Use the new formula for the area of a triangle. 1 2 ab sin C 1 2 1 8 13 2 21 sin 130° 39.83 Thus, the area is about 39.83 square centimeters. ■ An alternate formula for the area of a triangle, Heron’s formula, gives the area in terms of its sides. Heron’s Formula The area of a triangle with sides a, b, and c is 2s(s a)(s b)(s c), where s 1 2 (a b c). This formula is proved in Exercise 62. Example 7 Find Area with SSS Information Find the area of the triangle whose sides have lengths 7, 9, and 12. Solution Let a 7, b 9, and c 12. To use the Heron’s formula, first find s 12 14 2 Now, use Heron’s formula. s a s b 2s s c 1 21 21 2 14 7 214 21 1 2980 31.3 14 9 14 12 2 21 The area is about 31.3 square units. ■ 634 Chapter 10 Trigonometric Applications Exercises 10.2 Standard notation for triangle ABC is used throughout. Use a calculator and round off your answers to one decimal place at the end of each computation. In Exercises 1–8, solve triangle ABC under the given conditions. 1. A 48°, B 22°, a 5 2. B 33°, C 46°, b 4 3. A 116°, C 50°, a 8 4. A 105°, B 27°, b 10 5. B 44°, C 48°, b 12 6. A 67°, C 28°, a 9 7. A 102.3°, B 36.2°, a 16 8. B 97.5°, C 42.5°, b 7 In Exercises 9–16, find the area of triangle ABC under the given conditions. 9. a 4, b 8, C 27° 10. b 10, c 14, A 36° 11. c 7, a 10, B 68° 12. a 9, b 13, C 75° 13. a 11, b 15, c 18 14. a 4, b 12, c 14 15. a 7, b 9, c 11 16. a 17, b 27, c 40 In Exercises 17 – 36, solve the triangle. The Law of Cosines may be needed in Exercises 27–36. 17. b 15, c 25, B 47° 18. b 30, c 50, C 60° 19. a 12, b 5, B 20° 20. b 12.5, c 20.1, B 37.3° 21. a 5, c 12, A 102° 22. a 9, b 14, B 95° 23. b 11, c 10, C 56° 24. a 12.4, c 6.2, A 72° 25. A 41°, B 67°, a 10.5 26. a 30, b 40, A 30° 27. b 4, c 10, A 75° 28. a 50, c 80, C 45° 29. a 6, b 12, c 16 30. B 20.67°, C 34°, b 185 31. a 16.5, b 18.2, C 47° 32. a 21, c 15.8, B 71° 33. b 17.2, c 12.4, B 62.5° 34. b 24.1, c 10.5, C 26.3° 35. a 10.1, b 18.2, A 50.7° 36. b 14.6, c 7.8, B 40.4° In Exercises 37 and 38, find the area of the triangle with the given vertices. 37. 38. 1 1 , 0, 0 2 4, 2 2, 5 5, 1 2 3, 0 2 In Exercises 39 and 40, find the area of the polygonal region. Hint: Divide the region into triangles. 39. 120 55 89° 96° 103° 68.4 72° 40. 40 135 80° 20 120° 135° 23 1 3 75° 130° 30 30 41. A surveyor marks points A and B 200 meters apart on one bank of a river. She sights a point C on the opposite bank and determines the angles shown in the figure below. What is the distance from A to C? C 57° A 42° B 42. A forest fire is spotted from two fire towers. The triangle determined by the two towers and the fire has angles of and If the towers are 3000 meters apart, which one is closer to the fire? at the tower vertices. 28° 37° 43. A visitor to the Leaning Tower of Pisa observed that the tower’s shadow was 40 meters long and that the angle of elevation from the tip of the The shadow to the top of the tower was tower is now 54 meters tall, measured from the ground to the top along the center line of the a tower (see the figure). Approximate the angle that the center line of the tower makes with the vertical. 57°. α Section 10.2 The Law of Sines 635 angle of elevation from the end of the pole’s shadow to the top of the pole is the pole? 53°. How long is 45. A side view of a bus shelter is shown in the following figure. The brace d makes an angle of with the back and an angle of with 37.25° the top of the shelter. How long is the brace? 34.85° d 8 ft 5 ft 46. A straight path makes an angle of with the 6° horizontal. A statue at the higher end of the path casts a 6.5-meter shadow straight down the path. The angle of elevation from the end of the shadow to the top of the statue is the statue? How tall is 32°. 47. A vertical statue 6.3 meters high stands on top of a hill. At a point on the side of the hill 35 meters from the statue’s base, the angle between the hillside and a line from the top of the statue is 10°. What angle does the side of the hill make with the horizontal? 48. A fence post is located 36 feet from one corner of a building and 40 feet from the adjacent corner. Fences are put up between the post and the building corners to form a triangular garden area. The 40-foot fence makes a building. What is the area of the garden? angle with the 58° 57° 49. Two straight roads meet at an angle of 40° in Harville, one leading to Eastview and the other to Wellston (see the figure on the next page). Eastview is 18 kilometers from Harville and 20 kilometers from Wellston. What is the distance from Harville to Wellston? 44. A pole tilts at an angle 9° from the sun, and casts a shadow 24 feet long. The from the vertical, away 636 Chapter 10 Trigonometric Applications Harville 40° Eastview Wellston class. He has a long tape measure, but no way to measure angles. While pondering what to do, he paces along the side of the river using the five paths joining points A, B, C, and D (see the following figure). If he does not determine the width of the river, he will not pass the course. 50. Each of two observers 400 feet apart measures the angle of elevation to the top of a tree that sits on the straight line between them. These angles are 51° How far is the base of its trunk from each observer? respectively. How tall is the tree? and 65° 51. From the top of the 800-foot-tall Cartalk Tower, Tom sees a plane; the angle of elevation is 67°. the same instant, Ray, who is on the ground 1 mile from the building, notes that his angle of elevation to the plane is elevation to the top of Cartalk Tower is Assume that Tom, Ray, and the airplane are in a plane perpendicular to the ground. (See the following figure.) How high is the airplane? and that his angle of 8.6°. 81° At E A C B D a. Save Charlie from disaster by explaining how he can determine the width AE simply by measuring the lengths AB, AC, AD, BC, and BD and using trigonometry. b. Charlie determines that AD 90 AC 25 BD 22 BC 80 feet, feet. How wide is the river between A and E? feet, feet, and AB 75 feet, 54. A plane flies in a direction of 85° from Chicago. [Note: Aerial navigation directions are explained in Exercise 34 of Section 10.1.] It then turns and flies in the direction of 200° then 195 miles from its starting point. How far did the plane fly in the direction of for 150 miles. It is 85°? 81° 67° 800 ft 8.6° 1 mile 52. A plane flies in a direction of 105° from airport A. 55. A hinged crane makes an angle of 50° with the ground. A malfunction causes the lock on the hinge to fail and the top part of the crane swings down (see the figure). How far from the base of the crane does the top hit the ground? [Note: Aerial navigation directions are explained in Exercise 34 of Section 10.1.] After a time, it turns and proceeds in a direction of it lands at airport B, 120 miles directly
south of airport A. How far has the plane traveled? 267°. Finally, 53. Charlie is afraid of water; he can’t swim and refuses to get in a boat. However, he must measure the width of a river for his geography 14.6 m 19 m 50° Section 10.3 The Complex Plane and Polar Form for Complex Numbers 637 56. A triangular lot has sides of 120 feet and 160 feet. The angle between these sides is Adjacent to this lot is a rectangular lot whose longest side is 200 feet and whose shortest side is the same length as the shortest side of the triangular lot. What is the total area of both lots? 42°. 57. If a gallon of paint covers 400 square feet, how many gallons are needed to paint a triangular deck with sides of 65 feet, 72 feet, and 88 feet? 58. Critical Thinking Find the volume of the pyramid in the figure below. The volume is given by the formula V 1 3 Bh, where B is the area of the base and h is the height. h 34° 36° 46° 10 59. Critical Thinking A rigid plastic triangle ABC rests on three vertical rods, as shown in the figure. What is its area Horizontal plane 60. Critical Thinking Prove that the area of triangle ABC, in standard notation is given by a2 sin B sin C 2 sin A . 61. Critical Thinking What is the area of a triangle whose sides have lengths 12, 20, and 36? Hint: Drawing a diagram may be helpful. 62. Critical Thinking Use the area formula 1 ab sin C 2 sin2C 1 cos2C ab sin C and the Pythagorean identity to show that 1 2 Then use the Law of Cosines to show that 1 s c 2 1 cos C 1 cos C s 1 2 , where s ab ab ab cos and ab 1 2 1 cos C 2 s a 1 2 s b 2 21 . Combine the facts to prove Heron’s Formula. 10.3 The Complex Plane and Polar Form for Complex Numbers* Objectives • Graph a complex number in the complex plane • Find the absolute value of a complex number • Express a complex number in polar form • Perform polar multiplication The real number system is represented geometrically by the number line. The complex number system can be represented geometrically by the coordinate plane: The complex number a bi corresponds to the point (a, b) in the plane. For example, the point (2, 3) shown in Figure 10.3-1 is labeled by The other points shown are labeled similarly. 2 3i . and division *Section 4.5 is a prerequisite for this section. 638 Chapter 10 Trigonometric Applications Technology Tip Recall that complex numbers are entered by using the special i key on the TI keyboard or in the CPLX submenu of the Casio OPTN menu. i 2 + 3i −6 + 2.3i 2i = 0 + 2i −5 − 3i Figure 10.3-1 real 5.5 = 5.5 + 0i 4 − 3i When the coordinate plane is used to graph complex numbers in this way, it is called the complex plane. Each real number corresponds on the horizontal axis; so this axis is called the real axis. to the point The vertical axis is called the imaginary axis because every imaginary on the vertical axis. number corresponds to the point bi 0 bi a a 0i 0, b a, 0 2 1 1 2 is defined to be the distance from The absolute value of a real number c is the distance from c to 0 on the number line. So the absolute value (or modulus) of the complex number a bi to the origin in the complex plane. a bi 2 represents the distance from , which is given by a bi 2 2a2 b2. 0 a 0 b 0 0, 0 2 a, b to 2 1 2 1 0 1 2 1 2 Absolute Value of a Complex Number Technology Tip Use the ABS key to find the absolute value of a complex number. It is in the NUM submenu of TI and in the CPLX submenu of the Casio OPTN menu. The absolute value (or modulus) of the complex number a bi is a bi 00 00 2a2 b2. Example 1 Find Absolute Value of a Complex Number Find the absolute value of each complex number. a. 3 2i b. 4 5i Solution 3 2i a. 0 232 22 213 0 b. 0 4 5i 0 242 5 1 2 241 ■ 2 Absolute values and trigonometry lead to a useful way of representing be a nonzero complex number and denote complex numbers. Let a bi by r. Then r is the length of the line segment joining (a, b) and 0 (0, 0) in the plane. Let be the angle in standard position with this line segment as terminal side, as shown in Figure 10.3-2. a bi u 0 Section 10.3 The Complex Plane and Polar Form for Complex Numbers 639 i θ (a, b) r real Figure 10.3-2 NOTE It is customary to sin u place i in front of rather than after it. Some books abbreviate cos u i sin u r as r cis u. 1 2 Polar Form of a Complex Number Using the definitions of sine and cosine, the coordinates a and b can be expressed in terms of r and u. cos u a r a r cos u and sin u b r b r sin u Consequently, a bi r cos u 1 a bi r sin u i r 1 2 cos u i sin u . 2 is written in this way, it is said to be in When a complex number u polar form or trigonometric form. The angle is called the argument and a bi is called is usually expressed in radian measure. The number the modulus (plural, moduli). The number 0 can also be written in polar notation by letting and be any angle. r 0 r u 0 0 Every complex number can be written in polar form a bi r(cos U i sin U) 2a2 b2, a r cos U, and b r sin U. where r a bi 00 00 When a complex number is written in polar form, the argument uniquely determined because conditions in the box. is not and so on, all satisfy the u ± 4p, u ± 2p, u, u Example 2 Find Polar Form Express 23 i in polar form. Solution In this case, a 23 r 2a2 b2 2 and b 1. Therefore, 13 1 2 2 12 23 1 2. The angle must satisfy the following two conditions. u cos u a r 23 2 and sin u b r 1 2 13 i is represented by the Because point in the complex plane, it lies in the second quadrant, as shown in Figure 10.3-3. Therefore, must be a second- 13, 1 A u B quadrant angle. So, conditions. u 5p 6 satisfies these – 3 + i i 1 2 3 5π 6 real Thus, 23 i 2 cos a 5p 6 i sin 5p 6 b . Figure 10.3-3 ■ 640 Chapter 10 Trigonometric Applications Example 3 Find Polar Form Express 2 5i in polar form. Solution In this case, a 2 and b 5. r 2a2 b2 2 Therefore, 2 1 2 2 52 229. The angle must satisfy u cos u a r 2 229 and sin u b r 5 229 , so that i −2 + 5i 29 θ real Figure 10.3-4 Polar Multiplication and Division tan u sin u cos u 5 129 2 129 5 2 2.5. lies in the second quadrant (see Figure 10.3-4), u lies Because 2 5i p 2 solution to between p. and Using the tan u 2.5 TAN u 1.1903. is 1 key, the calculator indicates that a Because that angle is in the fourth quadrant, the only solution between p 2 and p is Thus, 2 5i 229 1 u 1.1903 p 1.9513. cos 1.9513 i sin 1.9513 . 2 ■ Multiplication and Division of Complex Numbers z1 i sin U1) r1(cos U1 If two complex numbers, then r1r2[cos(U1 z1z2 U2) i sin(U1 U2)] and z2 r2(cos U2 i sin U2) are any and z1 z2 r1 r2 [cos(U1 U2) i sin(U1 U2)], z2 0. That is, given two complex numbers written in polar form, • to multiply the two numbers multiply the moduli and add the arguments. • to divide the two numbers divide the moduli and subtract the arguments. Section 10.3 The Complex Plane and Polar Form for Complex Numbers 641 The proof of the multiplication statement, which is given at the end of this section, uses the addition identities for sine and cosine. Example 4 Multiplication of Numbers in Polar Form Find z1z2 when z1 2 cos a 5p 6 i sin 5p 6 b and z2 3 cos a 7p 4 i sin 7p 4 b . NOTE A complex number written in polar form can be written in rectangular form by evaluating each term and simplifying. For example, Solution In this case, r1 2, u1 5p 6 , r2 3, and u2 z1z2 r1r23 2 3 1 21 cos u1 1 cos 2 c 2 cos a 5p 6 i sin 2 23 2 a 23 i 5p cos cos 10p a 12 31p 12 u1 1 i sin . 7p 4 u22 4 5p 6 10p 12 a a i sin a i sin 7p 4 b u22 5p 6 21p 12 b i sin 31p 12 b Therefore, 7p 4 b d 21p 12 b d Example 5 Division of Numbers in Polar Form z1 Find when z2 10 z1 p 3 cos Q i sin p 3 R and z2 2 cos a p 4 i sin p 4 b . Solution In this case, r1 10, u1 p 3 , r2 2, and u2 p 4 . Therefore, z1 z2 10 2 c cos p 3 a p 4 b i sin p 3 a p 4 b d 5 cos a p 12 i sin p 12b ■ ■ z1z2 Proof of the Polar Multiplication Rule r21 Let cos u1 and z2 r11 z1 r11 3 r1r21 r1r21 r1r23 1 i sin u12 i sin u12 4 3 i sin u121 cos u1 cos u1 cos u1 cos u2 cos u1 cos u2 cos u2 r21 cos u2 i sin u1 cos u2 sin u1 sin u22 cos u2 i sin u22 4 i sin u22 i sin u22 . i sin u2 cos u1 i sin u1 cos u2 1 i2 sin u1 sin u22 sin u2 cos u12 4 Recall from Section 9.2 that u22 cos u1 1 cos u1 cos u2 sin u1 sin u2 642 Chapter 10 Trigonometric Applications and Therefore, sin u1 1 u22 sin u1 cos u2 cos u1 sin u2. z1z2 r1r23 1 r1r23 cos u1 cos u2 u22 cos sin u1 sin u22 i sin u22 4 u1 1 This completes the proof of the multiplication rule. The division rule is proved similarly. See Exercise 51. sin u2 cos u12 4 sin u1 cos u2 i u1 1 1 Exercises 10.3 In Exercises 1–8, plot the point in the complex plane that corresponds to each number. 23. 1. 3 2i 2. 7 6i 4. 22 7i 7. 2i 3 5 a 2 i b 5. 8. 1 i 1 i 2 21 1 4 3 i 1 6 3i 2 3. 6. 8 3 5 3 i 2 i 1 21 1 2i 2 In Exercises 9–14, find each absolute value. 9. 12. 5 12i 0 2 3i 0 0 0 10. 13. 2i 0 12i 0 0 0 11. 14. 0 0 1 i22 0 i 7 0 15. Give an example of complex numbers z and w z such that z w w . 0 0 0 0 16. If z 3 4i, find z 2 and where zz z is the 0 0 0 0 conjugate of z. See Section 4.5 for the definition of a complex conjugate. In Exercises 17–24, sketch the graph of the equation in the complex plane (z denotes a complex number of the form a bi ). 17. 18. 19. 20. 22. 4 z 0 lie 4 units from the origin. 0 Hint: The graph consists of all points that z 1 0 0 z 1 0 complex plane. What does the equation say about the distance from z to 1? Hint: 1 corresponds to (1, 0) in the 10 0 z 3 1 0 z 3i 2 z 2 3i 0 1 0 0 0 3 2 0 21. 0 z 2i 4 0 Hint: Rewrite it as 3. 1 2 z Re 2 z a bi denoted Re(z).] [The real part of the complex number is defined to be the number a and is 24. Im 5 2 z 2 1 [The imaginary part of z a bi is defined to be the number b (not bi) and is denoted Im(z).] In Exercises 25–32, express each number in polar form. 25. 3 4i 26. 4 3i 28. 27 3i 29. 1 2i 27. 5 12i 30. 3 5i 31. 5 2 7 2 i 32. 25 i211 In Exercises 33–38, perform the indicated multiplication or division. Express your answer in both recr(cos U i sin U). tangular form and polar form a bi 33. 3 cos a 34. 3 cos a p 12 p 8 i sin p 12b 2 cos a 7p 12
i sin 7p 12 b i sin p 8 b 12 cos a 3p 8 i sin 3p 8 b 7 2 a cos p 4 i sin p 4 b 35. 12 cos a 11p 12 i sin 11p 12 b 36. 37. 8 cos a 5p 18 cos 4 a p 9 6 cos a 7p 20 4 cos a p 10 i sin i sin i sin i sin 5p 18 b p 9 b 7p 20 b p 10b Section 10.3 The Complex Plane and Polar Form for Complex Numbers 643 254 38. cos a 26 cos a 9p 4 7p 12 i sin 9p 4 b i sin 7p 12 b In Exercises 39–46, convert to polar form and then multiply or divide. Express your answer in polar form. 1 i23 B A B 39. 41. 1 i A 1 i 1 i 43. 3i A 223 2i B 40. 42. 44. 3 3i 2 1 i 21 1 2 2i 1 i 4i 23 i 45. 46. i 1 i 1 1 1 i 23 i 21 223 2i 2 21 4 4i23 2 21 47. Explain what is meant by saying that multiplying z r a complex number 1 amounts to rotating z 90° around the origin. Hint: Express i and iz in polar form; what are their relative positions in the complex plane? cos u i sin u counterclockwise by i 2 48. Describe what happens geometrically when you multiply a complex number by 2. a bi 49. Critical Thinking The sum of two distinct complex c di, can be found a bi in the complex plane and form the and numbers, geometrically by means of the so-called parallelogram rule: Plot the points c di parallelogram, three of whose vertices are 0, and vertex of the parallelogram is the point whose coordinate is the sum a bi as in the figure below. Then the fourth c di and a bi 1 2 1 c di a c 1 2 1 2 b d i. 2 a + bi i 0 c + di real i 0 c + di a + bi real Complete the following proof of the parallelogram c 0. rule when a. Find the slope of the line K from 0 to a bi. Hint: K contains the points (0, 0) and (a, b). a 0 and c di. b. Find the slope of the line N from 0 to c. Find the equation of the line L, through a bi and parallel to line N of part b. Hint: The point (a, b) is on L; find the slope of L by using part b and facts about the slope of parallel lines. d. Find the equation of the line M, through c di and parallel to line K of part a. e. Label the lines K, L, M, and N in each figure. f. Show by using substitution that the point 2 2 satisfies both the equation of line a c, b d 1 L and the equation of line M. Therefore, a c, b d lies on both L and M. Because 1 the only point on both L and M is the fourth vertex of the parallelogram, this vertex must be coordinate a c Hence, this vertex has 2 b d a c, b d c di a bi i . 1 . 1 2 1 50. Critical Thinking Let number and denote its conjugate 2 z z. Prove that z 0 0 2 1 z a bi 2 1 2 be a complex z. a bi by and cos u1 Let z2 z1 r21 z1 z2 51. Critical Thinking Proof of the polar division rule. i sin u12 . i sin u12 i sin u22 i sin u12 i sin u22 a. Multiply the denominators and use the i sin u22 cos u1 cos u2 cos u1 cos u2 r11 cos u2 r11 r21 r11 r21 cos u2 cos u2 i sin u2 i sin u2 Pythagorean identity to show that it is the number r2. b. Multiply the numerators; use the subtraction identities for sine and cosine (Section 9.2) to show that it is cos u1 1 u22 i sin u1 1 u22 4 . r1 3 Therefore, z1 z2 r1 r2b 3 a cos u1 1 u22 i sin u1 1 u22 4 . r a. If cos u i sin u 52. Critical Thinking s explain 1 must be true. Hint: Think distance. why r explain , 2 1 Hint: See why Property 5 of the complex numbers in Section 4.5. cos b i sin b s r cos b i sin b cos b cos u cos u i sin u r 1 sin b sin u. and b. If , 1 2 2 2 644 Chapter 10 Trigonometric Applications c. If cos b cos u b u angles and same terminal side. Hint: cos u, sin u 1 2 sin b sin u, and show that in standard position have the 1 are points on the unit circle. 2 cos b, sin b and d. Use parts a–c to prove this equality rule for polar form: cos b i sin b s r cos u i sin u 2 1 for some exactly when integer k. Hint: Angles with the same terminal side must differ by an integer multiple of 2p. b u 2kp 2 and s r 1 10.4 DeMoivre’s Theorem and n th Roots of Complex Numbers Objectives • Calculate powers and roots of complex numbers • Find and graph roots of unity Polar form provides a convenient way to calculate both powers and roots cos u i sin u , of complex numbers. If then the multiplication for2 mula from Section 10.3 shows the following: z2 z z r r i sin z r u u 1 cos u u cos 2u i sin 2u 2 1 3 1 2 4 and r2 1 z3 z2 z r2 r cos 2u u i sin 2u u 1 2 4 3 2 1 cos 3u i sin 3u r3 1 2 2 and so on. Repeated application of the multiplication formula proves DeMoivre’s Theorem. DeMoivre’s Theorem For any complex number positive integer n, z r(cos U i sin U) and any zn r n(cos nU i sin nU). NOTE 1 tan 1 23 5p 6 Example 1 Find Powers of Complex Numbers Evaluate 23 i A B 5. Solution First express the complex number 2 of Section 10.3.) 23 i in polar form. (See Example 23 i 2 cos a 5p 6 i sin 5p 6 b Section 10.4 DeMoivre’s Theorem and nth Roots of Complex Numbers 645 Apply DeMoivre’s Theorem. 23 i 5 25 B cos c A 5 5p 6 b a 25p 6 i sin 5 5p a 25p 6 b d 32 cos a i sin 6 b Because 25p 6 p 6 24p 6 p 6 4p, p 6 can be substituted for 25p 6 . 23 i A B i sin 25p 6 b 5 32 32 cos a 25p 6 p 6 23 1 2 2 1623 16i cos a 32 a i sin p 6 b polar form i b rectangular form ■ NOTE tan 1 1 p 4 Example 2 Find Powers of Complex Numbers Evaluate 1 i 10. 2 1 Solution Express the complex number 1 i in polar form. 1 i 22 cos a p 4 i sin p 4 b Apply DeMoivre’s Theorem. 1 i 2 1 10 22 10p 4 i sin 10p 4 b 10 B cos a 5p cos a 2 0 i 2 A 32 32 1 32i i sin 5p 2 b polar form rectangular form ■ NOTE With complex a bi, numbers, it is not possible to choose the positive root as the nth root of as is done with real numbers, because “positive” and “negative” are not meaningful terms in the complex numbers. For 3 2i instance, should called positive or negative? be Nth Roots a bi is a complex number and n is a positive integer, the equation If zn a bi may have n different solutions in the complex numbers. Furthermore, there is no obvious way to designate one of these solutions as the nth root of (see note). Consequently, any solution of the equazn a bi tion a bi is called an nth root of a bi. Every real number is a complex number. When the definition of nth root of a complex number is applied to a real number, the terminology for real numbers no longer applies. For instance, in the complex numbers, 16 has 2, 2, 2i, z4 16, four fourth roots because each of whereas in the real numbers, 2 is the fourth root of 16. is a solution of 2i and 646 Chapter 10 Trigonometric Applications tan NOTE p 3 is in 823 1 8 8 8i˛23 Because Quadrant II of the complex plane, the angle is p p 2p 3 3 . Although nth roots are not unique in the complex numbers, the radical symbol will be used only for nonnegative real numbers and will have the same meaning as before. That is, if r is a nonnegative real number, then 2n denotes the unique nonnegative real number whose nth power is r. r Example 3 Find Roots of Complex Numbers Find the fourth roots of 8 8i23. Solution Express the complex number 8 8i23 in polar form. 8 8i23 16 cos a 2p 3 i sin 2p 3 b To solve z s z4 16 cos a cos b i sin b 2 1 2p 3 i sin 2p 3 b , find s and b such that is a solution. In other words, find s and b such that cos b i sin b s 1 3 4 16 2 4 cos a 2p 3 i sin 2p 3 b . Use DeMoivre’s Theorem to rewrite the left side. cos 4b i sin 4b s4 1 16 2 cos a 2p 3 i sin 2p 3 b The equality rules for complex numbers (proved in Exercise 52 of Section 10.3) show that the above equation is true if s4 16 and s 24 16 2 2kp 4b 2p 3 kp 2 b p 6 Therefore, the solutions of z4 16 cos a 2p 3 i sin 2p 3 b are z 2 cos c p a 6 i sin kp 2 b k 0, 1, 2, p a 6 kp 2 b d , and 3 produces four distinct where k is any integer. Letting solutions. k 0: z 2 cos a p 6 i sin p 6 b 23 i k 1: z 2 cos c 2 cos a k 2: z 2 cos c 2 cos a p a 6 2p 3 p a 6 7p 6 p 2 b i sin i sin p 6 a p 2 b d 2p 3 b 1 i23 p b i sin p 6 a p b d i sin 7p 6 b 23 i Section 10.4 DeMoivre’s Theorem and nth Roots of Complex Numbers 647 k 3: z 2 cos c 2 cos a 3p 2 b i sin p a 6 5p 3 i sin p 6 a 3p 2 b d 5p 3 b 1 i 23 Any other value of k produces an angle with the same terminal side as one of the four angles already used and is the same solution. For instance, b when k 4, then b p 6 4p 2 p 6 2p, so has the same terminal side b as p 6 . Therefore, all fourth roots of 8 8i23 have been found. ■ zn r cos u i sin u The general equation same method used in Example 3: substitute n for 4, r for 16, and 2p 3 can be solved exactly by the for as follows. A solution is a number cos b i sin b such that: u s , 2 1 2 cos b i sin b 2 4 1 cos nb i sin nb 1 n r r 2 cos u i sin u cos u i sin u 2 2 1 1 s 3 sn 1 Therefore, sn r s 2n r and nb u 2kp b u 2kp n where k is any integer. Letting angles b . This is stated in the following formula for nth roots. k 0, 1, 2, p , n 1 produces n distinct Formula for nth Roots For each positive integer n, the nonzero complex number r(cos U i sin U) has exactly n distinct nth roots. They are given by U 2kP n U 2kP n i sin cos 2n r a , b d c b where k 0, 1, 2, 3, p , n 1. a Example 4 Find Roots of Complex Numbers Find the fifth roots of 4 4i. Solution Express the complex number 4 4i in polar form. 4 4i 422 cos a p 4 i sin p 4 b 648 Chapter 10 Trigonometric Applications Apply the root formula with n 5, r 422, u p 4 , and k 0, 1, 2, 3, and 4. 2n r 25 412 422 1 5 1 2 222 1 5 2 The fifth roots have the following form. 1 A B 1 5 2 5 10 2 1 2 22 5 2 2 1 2 22 scos a p 4 2kp 5 b i sin p 4 a 2kp 5 bt , for k 0, 1, 2, 3, and 4 Therefore, the five distinct roots are as follows. p 20 b a i sin p 20 bt a 9p 20 b a i sin 9p 20 b T a 17p 20 b a i sin 17p 20 b T a 5p 4 b a i sin 5p 4 b T a k 0: 22 scos k 1: 22 scos a a p 4 5 b p 4 i sin p 4 5 bt a 2p 5 i sin b k 2: 22 scos a p 4 4p 5 b i sin k 3: 22 scos a p 4 6p 5 b i sin k 4: 22 scos a p 4 8p 5 b i sin Roots of Unity 22 scos 2p 5 p 4 bt a 22 cos a 22 bt cos p 4 p 4 S 4p 5 S 6p 5 a 22 bt cos p 4 S 8p 5 a 22 bt 33p 20 b a cos S i sin 33p 20 b T a ■ The n distinct nth roots of 1 (the solutions of roots of unity. Because number 1 is u 0 produces a formula for roots of unity. cos 0 i sin 0. sin 0 0, Applying the root formula with ) are called the nth the polar form of the and cos 0 1 r 1 and zn 1 Formula for Roots of Unity For eac
h positive integer n, there are n distinct nth roots of unity, which have the following form. cos 2kp n i sin 2kp n , for k 0, 1, 2, p , n 1. Section 10.4 DeMoivre’s Theorem and nth Roots of Complex Numbers 649 Example 5 Find Roots of Unity Find the cube roots of unity. Solution Apply the formula for roots of unity with k 0: cos 0 i sin 0 1 2p 2p 3 3 4p 4p 3 3 k 2: cos k 1: cos i sin i sin n 3 and k 0, 1, and 2. 1 2 1 2 23 2 23 2 i i ■ All roots of unity can be found from the first nonreal root. Let the first nonreal cube root of unity obtained in Example 5 be denoted by v. v cos 2p 3 i sin 2p 3 Using DeMoivre’s Theorem to find roots of unity. v2 and produces the other two cube v3 v2 cos a 2p 3 i sin 2 2p 3 b cos 4p 3 i sin 4p 3 v3 cos a 2p 3 i sin 3 2p 3 b cos 6p 3 i sin 6p 3 cos 2p i sin 2p 1 In other words, all the cube roots of unity are powers of true in the general case, as stated below. v. The same is All Roots of Unity Let n be a positive integer with n 77 1. Then the number z cos 2P n i sin 2P n is an nth root of unity and all the nth roots of unity are z, z2, z3, z4, p , zn1, zn 1. The nth roots of unity have an interesting geometric interpretation. Every nth root of unity has absolute value of 1: 2kp n cos ` i sin 2kp n ` cos B a 2 2kp n b 2 2kp n b sin a 2kp n sin2 2kp n cos2 B 21 1 650 Chapter 10 Trigonometric Applications Therefore, in the complex plane, every nth root of unity is exactly 1 unit from the origin. That is, the nth roots of unity all lie on the unit circle in the complex plane. Example 6 Find nth Roots of Unity Find the fifth roots of unity. Solution The fifth roots of unity have the following form. cos 2kp 5 i sin 2kp 5 , for k 0, 1, 2, 3, and 4 Therefore, the five roots of unity are i sin k 1: cos k 2: cos k 0: cos 0 i sin 0 1 2p 2p 5 5 4p 4p 5 5 6p 6p 5 5 8p 8p 5 5 k 3: cos k 4: cos i sin i sin i sin ■ These five roots can be plotted in the complex plane by starting at 1 1 0i, and moving counterclockwise around the unit circle, moving through an angle of 2p 5 at each step, as shown in Figure 10.4-1. If you connect these five roots, they form the vertices of a regular pentagon, as shown in Figure 10.4-2. cos 4π 5 + i sin 4π 5 i 1 y cos 2π 5 + i sin 2π 5 i 2π 5 2π 5 2π 5 2π 5 2π 5 cos 0 + i sin 0 real 1 cos 6π 5 + i sin 6π 5 cos 8π 5 + i sin 8π 5 Figure 10.4-1 Figure 10.4-2 real Section 10.4 DeMoivre’s Theorem and nth Roots of Complex Numbers 651 NOTE On wide screen calculators, you may 2 x 2 choose to use so that or the unit circle looks like a circle. 1.7 x 1.7 Graphing Exploration With your calculator in parametric graphing mode, use the following window settings. 0 t 2p t-step 0.067 2.2 x 2.2 1.5 y 1.5 Graph the unit circle, whose parametric equations are x cos t and y sin t Reset the t-step to be 2p 5 , and graph again. Your screen should now look exactly like the solid lines in Figure 10.4-2 because the calculator plotted only the five points corresponding to t 0, 2p 5 , 4p 5 , 6p 5 , and 8p 5 , and connected them with the shortest possible segments. Use the trace feature to move along the graph. The cursor will jump from vertex to vertex, that is, from one fifth root of unity to the next. Example 7 Graph Roots of Unity Graph the tenth roots of unity, and estimate the two tenth roots of unity in the first quadrant. Solution With a graphing calculator in parametric mode, set the range values as in the graphing exploration above. Because n 10, 2p 10 ular decagon whose vertices are the tenth roots of unity. By using the trace feature, you can approximate each tenth root of unity. and graph. The result is a reg- reset the t-step to p 5 1.5 1.5 2.2 2.2 2.2 2.2 1.5 Figure 10.4-3 1.5 Figure 10.4-4 Figures 10.4-3 and 10.4-4 show the two approximate tenth roots of unity in the first quadrant. 0.8090 0.5878i and 0.3090 0.9511i. ■ 5910ac10_616-687 9/21/05 2:31 PM Page 652 652 Chapter 10 Trigonometric Applications Exercises 10.4 In Exercises 1 – 10, calculate each given product and express your answer in the form a bi . 1. cos a p 12 i sin 6 p 12 b 2. cos a p 5 i sin 20 p 5 b 3. 3 cos a c 7p 30 i sin 5 7p 30 b d 4. 5. 23 4 c cos a 7p 36 i sin 12 7p 36 b d 12 1 i 1 theorem. 2 Hint: Use polar form and DeMoivre’s 6. 1 2 2i 8 2 8. 1 a 2 20 23 2 i b 10. 1 1 23 i 8 2 7. 9. 23 2 a 1 2 i b 10 1 22 a 14 i 22 b In Exercises 11 and 12, find all indicated roots of unity and express your answers in the form a bi. 11. fourth roots of unity 12. sixth roots of unity In Exercises 13–22, find the nth roots of each given number in polar form. 13. 64 cos a p 5 i sin p 5 b ; n 3 14. 8 cos a p 10 i sin p 10 b ; n 3 15. 81 cos a p 12 i sin p 12 b ; n 4 16. 16 cos a p 7 i sin p 7 b ; n 5 29. x4 1 i 23 30. x4 8 8i 23 In Exercises 31–35, represent the roots of unity graphto obtain ically. Then use feature approximations of the form for each root (round to four places). trace a bi the 31. seventh roots of unity 32. fifth roots of unity 33. eighth roots of unity 34. twelfth roots of unity 35. ninth roots of unity 36. Solve the equation x3 x2 x 1 0. find the quotient when then consider solutions of x4 1 0. x4 1 is divided by Hint: First x 1, 37. Solve x5 x4 x3 x2 x 1 0 Hint: Consider x6 1 and x 1 and see Exercise 36. 38. What are the solutions of xn1 xn2 p x3 x2 x 1 0? Exercises 36 and 37.) (See 39. Critical Thinking In the complex plane, the unit 0 1. circle consists of all numbers (points) z such that Suppose v and w are two points z 0 (numbers) that move around the unit circle in such a way that made one complete trip around the circle, how many trips has v made? Hint: Think polar and DeMoivre. at all times. When w has v w12 40. Critical Thinking Suppose u is an nth root of unity. Show that 1 u is also an nth root of unity. Hint: Use the definition, not polar form. 41. Critical Thinking Let u1, u2, p , un be the distinct nth roots of unity and suppose v is a nonzero cos u i sin u solution of the equation vu1, vu2, p ,vun are n distinct solutions Show that of the equation. Hint: Each ui is a solution of xn 1. zn r . 2 1 17. 1; n 5 18. 1; n 7 19. i; n 5 42. Critical Thinking Use the formula for nth roots and 20. i; n 6 21. 1 i; n 2 22. 1 i23; n 3 the identities In Exercises 23 – 30, solve the given equation in the complex number system. 23. x6 1 24. x6 64 0 25. x3 i 26. x4 i 27. x3 27i 0 28. x6 729 0 cos 1 x p 2 cos x sin x p 2 1 sin x cos u i sin u to show that the nonzero complex number r has two square roots and that 1 these square roots are negatives of each other. 2 Section 10.5 Vectors in the Plane 653 10.5 Vectors in the Plane Objectives • Find the components and magnitude of a vector • Perform scalar multiplication of vectors, vector addition, and vector subtraction Once a unit of measure has been agreed upon, quantities such as area, length, time, and temperature can be described by a single number. Other quantities, such as an east wind of 10 miles per hour, require two numbers to describe them because they involve both magnitude and direction. Quantities that have magnitude and direction are called vectors and are represented geometrically by a directed line segment or arrow, as shown in Figures 10.5-1 and 10.5-2. u Q v w P Figure 10.5-1 Figure 10.5-2 When a vector extends from a point P to a point Q, as in Figure 10.5-1, P is called the initial point of the vector, and Q is called the terminal point, and the vector is written Its length is denoted by ! . PQ ! PQ . When the endpoints are not specified, as in Figure 10.5-2, vectors are denoted by boldface letters such as u, v, and w. The length of a vector u is denoted by and is called the magnitude of u. u If u and v are vectors with the same magnitude and direction, the vectors u and v are said to be equivalent, written Some examples and nonexamples are shown in Figure 10.5-3. u v same magnitude, different directions u ≠ v same direction, different magnitudes u ≠ v different directions, different magnitudes Figure 10.5-3 654 Chapter 10 Trigonometric Applications Example 1 Confirm Equivalent Vectors P Let Show that 1 1, 2 , Q 5, 4 ! ! 1 2 OR PQ . , O 2 0, 0 , 2 1 and R 4, 2 , 2 1 as in Figure 10.5-4. y Solution 5 4 3 2 1 Q (5, 4) P (1, 2) R (4, 2) x O 1 2 3 4 5 Figure 10.5-4 Equivalent Vectors y P (x1, y1) O Q (x2, y2) R (x2 – x1, y2 – y1) x Figure 10.5-5 The distance formula shows that and ! OR have the same length. ! PQ 4 2 2 0 ! 2 PQ ! OR 2 5 1 1 4 0 1 ! PQ 2 2 and 2 2 ! OR 1 1 2 242 22 220 2 242 22 220 2 2 have the same slope: The lines containing › 4 2 5 1 ! OR and ! OR ‹ PQ slope ! PQ ! PQ Because slope, and 2 4 1 2 ‹ slope OR › 2 0 4 0 2 4 1 2 both point to the upper right on lines of the same have the same direction. Therefore, ! PQ ! OR . ■ According to the definition of equivalence, if two vectors are equivalent, then one of the vectors may be moved from one location to another, provided that its magnitude and direction are not changed, and the two vectors will remain equivalent. ! PQ Every vector point at the origin. If is equivalent to a vector ! PQ ! OR , P (x1, y1) and where R (x2 ! OR Q (x2, y2), x1, y2 y1). with initial then and ! OR Proof The proof is similar to the one used in Example 1. It follows ! PQ have the same length: from the fact that ! 2 y12 x12 OR x2 3 1 x12 2 x2 ! 1 PQ 0 4 y2 0 2 1 2 4 2 y2 3 1 y12 2 ! OR and either the lines containing the same slope: ! PQ and are both vertical or they have slope ‹ OR 0 0 y12 y2 › 1 x12 x2 1 y1 y2 x1 x2 ‹ slope › , PQ as shown in Figure 10.5-5. Every vector can be written as a vector with the origin as its initial point. The magnitude and direction of a vector with the origin as its initial point Section 10.5 Vectors in the Plane 655 are completely determined by the coordinates of its terminal point. Consequently, the vector with initial point (0, 0) and terminal point (a, b) is The numbers a and b are called the components of the denoted by a, b . vector I a, b . I H H a, b is the distance from (0, 0) to (a, b), Because the length of the vector I the distance formula gives its magnitude,
which is also called its norm. H Magnitude The magnitude (or norm) of the vector 2a2 b2. v 7 7 v a, b I H is CAUTION Example 2 Find Components and Magnitude of a Vector The order in which the coordinates of the initial point and terminal point are subtracted to a, b obtain I H cant. For the points x2, y22 Q P and : x1, y2 y1I x2, y1 y2I x1, y12 1 ! PQ x2 ! H QP x1 H is signifi- 1 Scalar Multiplication and ! PQ 3: y2 ! OR , Find the components and the magnitude of the vector with initial point P and terminal point 4, 3 2, 6 Q . 1 2 1 2 Solution According to the properties of equivalent vectors, where x2 4, x1 2, y1 6, where , 9 . 2 1 2 In other words, ! PQ ! OR 6, 9 H . I Therefore, the magnitude is ! PQ ! OR 262 9 1 2 2 236 81 2117 ■ Vector Arithmetic Vectors can be added, can be subtracted, and can be multiplied in three different ways. Addition, subtraction, and one type of multiplication are discussed in this section. Another type of multiplication is presented in the Excursion 10.6.A. Scalar Multiplication When dealing with vectors, it is customary to refer to ordinary real numbers as scalars. Scalar multiplication is an operation in which a scalar k is “multiplied” by a vector v to produce another vector denoted by kv. If k is a real number and is a vector, then v a, b HH II kv is the vector ka, kb . II HH The vector kv is called a scalar multiple of v. 656 Chapter 10 Trigonometric Applications Example 3 Perform Scalar Multiplication Find the components of 3v and 2v. Let v 3, 1 H I . Solution 3v 3 3, 1 I H 3 3, 3 1 H 9, 3 H I I 2v 2 3, 1 H I 2 3, 2 1 H 6, 2 H I I ■ The graphs of v, 3v, and illustrates that 3v has the same direction as v and direction. from Example 3, shown in Figure 10.5-6, has the opposite 2v 2v y 〈3, 1〉 v 〈9, 3〉 3v x −2v 〈−6, −2〉 Figure 10.5-6 Also note that 2v 7 7 Therefore, v 7 7 6, 2 H I 3, 1 I H 2 232 12 210 6 1 2 2 22 240 2210 3v Similarly, it can be verified that 7 an illustration of the following facts. 7 2v 7 7 2210 . Figure 10.5-6 is Geometric Interpretation of Scalar Multiplication The magnitude of the vector kv is is, 00 k 00 times the length of v, that kv k v . 00 The direction of kv is the same as that of v when k is positive and opposite that of v when k is negative. 7 7 7 7 00 Section 10.5 Vectors in the Plane 657 Vector Addition Vector addition is an operation in which two vectors u and v are added, u v resulting in a new vector denoted . Vector Addition If u a, b II HH and v then , c, d II HH u v a c, b d . II HH Example 4 Perform Vector Addition Let u 5, 2 H I and v 3, 1 . I H Find the components of u v. Solution u v H 5, 2 3, 1 I I H 5 3, 2 1 H 2, 3 H I I ■ Geometric Interpretation of u v The graph of u v from Example 4 is shown in Figure 10.5-7. y 〈−5, 2〉 〈−2, 35 –2 –1 〈3, 1〉 x v 1 3 Figure 10.5-7 Figure 10.5-7 illustrates the following geometric interpretation of vector addition. Geometric Interpretation of Vector Addition 1. If u and v are vectors with the same initial point P, then is the diagonal of the where ! u v , PQ parallelogram with adjacent sides u and v. is the vector ! PQ 2. If the vector v is moved without changing its magnitude or direction so that its initial point lies on the endpoint of the vector u, then is the vector with the same initial point P as u and the same terminal point Q as v. u v 658 Chapter 10 Trigonometric Applications Exercise 33 asks for the proof of the geometric interpretation of vector addition stated above. Vector Subtraction The negative (or opposite) of a vector c, d c, d tor H is defined using the negative of a vector as follows. v and is denoted c, d H v 1 1 , I 2H I I 2 1 1 is defined to be the vecv. Vector subtraction Vector Subtraction If u a, b II HH and v , c, d II HH u (v) then u v a, b II HH a c, b d HH II is the vector c, d II HH Example 5 Perform Vector Subtraction Let u 2, 5 I H and v 6, 1 H . I Find the components of u v. Solution NOTE The vectors v and v have the same magnitude, and lines that contain them have the same slope, but v and v have opposite directions. u v I 6, 1 2, 5 I H H 2 6, 5 1 H 4, 4 H I I ■ Geometric Interpretation of The graph of u v u v from Example 5 is shown in Figure 10.5-8. 〈−4, 4〉 u y 〈2, 5〉 u − v 1 u −v −1 1 Figure 10.5-8 〈6, 1〉 x v The Zero Vector The vector 0, 0 H I is called the zero vector and is denoted 0. Example 6 Perform Combined Vector Operations Let u 1 , and w 2, h 5 2 i . Find the components of each vector. Section 10.5 Vectors in the Plane 659 a. 2u 3v b. 4w 2u Solution a. 2u 3v 2 I 3 2 1, 6 h H 3 2, 12 2, 12 I H 2 2, 12 12 H 0. i 4w 2u 4 I I 2, 2 5 1, 6 h H 2 i 2, 12 8, 10 H 8 2, 10 12 H 10, 2 H I I H I I ■ Properties of Vector Addition and Scalar Multiplication Vector Properties Operations on vectors share many of the same properties as arithmetical operations on numbers. For any vectors u, v, and w and any scalars r and s, 1. 2. 3. 4. 5. 6. 7. 8. 9. u (v w) (u vv) 0 r(u v) ru rv (r s)v rv sv (rs)v r(sv) s(rv) 1v v 0v 0 and r0 0 associative for addition commutative additive identity additive inverse distributive distributive associative for multiplication multiplicative identity multiplication by 0 u v v u Proof of u a, b Let and I H therefore, v c, d H Addition of real numbers is commutative; . I H u v c, d a, b I I H a c, b d H I c a, d b H I c, d a, b H I H I v u The other properties are proved similarly. See Exercises 26–31. 660 Chapter 10 Trigonometric Applications Exercises 10.5 In Exercises 1–4, find the magnitude of the vector ! PQ . force and find an additional force v, which, if added to the system, produces equilibrium. 2 7, 11 2 4, 5 1 1 24. u1 25. u1 2, 5 , u2 I 3, 7 , u2 I H H 2 6, 1 I H 8, 2 H I , u3 , u3 I 4, 8 H 9, 0 H , u4 I 5, 4 H I 1. P 2. P 3. P 4. P 1 1 1 1 , Q 5, 9 2, 3 2 3, 5 1 , Q 7, 0 , Q 2 2 30, 12 2 , Q 25, 5 1 2 In Exercises 5–10, find a vector equivalent to the vector with its initial point at the origin. ! PQ 5. P 6. P 7. P 8. P 1 1 1 1 7, 11 2 2, 9 , Q 1, 5 2 1 , Q 2, 7 2 4, 8 1 , Q 2 10, 2 2 , Q 1 7, 9 2 1 5, 6 2 9 17 5 a 2 , 12 5 b 10. P 22, 4 A , Q B A 23,1 B In Exercises 11–15, find u v, u v, and 3u 2v. 11. u 12. u 13. u I 4, 0 2, 4 , v 6, 1 H I 1, 3 , v H I , v 3, 312 I H H H H I 412, 1 I 14, h 19 3 i 15. u 2 2, 12 I In Exercises 26 – 31, let w stated property holds. e, f HH , II and HH and let r and s be scalars. Prove that the c, d II , v a, b II u HH 26. v 0 v 0 v 28. 30. r 1 u v 2 v r rs 2 1 ru rv sv 2 1 s rv 2 1 27. 29. 31. v v 0 2 1 r s v rv sv 2 1 1v v and 0v 0 32. Let v be the vector with initial point x1, y12 and let k be any real 1 and 1 x2, y22 terminal point number. a. Find the component form of v and kv. b. Calculate and c. Use the fact that v 7 7 kv . 7 7 2k2 kv 7 7 k 0 0 to verify the v k 0 0 7 7 following equation: 33. Let u a, b I H and v c, d H I . Verify the accuracy of the two geometric interpretations of vector addition given on page 657 as follows: a. Show that the distance from to is the same as a c, b d 2 b. Show that the distance from (c, d) to a c, b d is the same as u . c. Show that the line through (a, b) and a, b d 2 1 they have the same slope. is parallel to v by showing that d. Show that the line through (c, d) and a c, b d is parallel to u. 2 1 1 1 In Exercises 16 – 23, let w . 3, 1 HH Find the magnitude of each vector. II II , 8, 4 HH , v u and 7 34. Let u u v 7 6, 2 HH u v II 16. 18. 3u v 20. 22 17. u v 19. v w 21. 23. 2 2 w 2u 1 v 2 3 7 6 v u1, u2, p , uk act on an object at the origin, the If forces resultant force is the sum The forces are said to be in equilibrium if their resultant force is 0. In Exercises 24 and 25, find the resultant p uk. u2 u1 a, b I H w and . 7 7 v c, d H . I Show that y (a – b, c – d) (a, b) u – v u –v w x v (c, d) Section 10.6 Applications of Vectors in the Plane 661 10.6 Applications of Vectors in the Plane Objectives • Perform operations with linear combinations of vectors • Determine the direction angle of a vector • Determine resultant forces in physical applications In the previous section, vectors were introduced and vector arithmetic was defined. In this section, vectors are applied to real-world situations. Unit Vectors A vector with length 1 is called a unit vector. For instance, 3 5 , 4 5 i h is a unit vector because 5b 2 4 5b a B a 9 25 16 25 B 25 25 B 1. Example 1 Unit Vectors Find a unit vector u with the same direction as the vector v 5, 12 H . I Solution Multiplying vector v by a scalar that is the reciprocal of its length produces a unit vector. The length of v is v 7 7 H 5, 12 252 122 2169 13 . I Let u 1 13 v 1 13 5, 12 H I 5 13 , 12 13 i . h The vector u 5 13 , 12 13 i h is a unit vector because u 1 13 v g g 1 13 ` ` v 1 13 13 1 ■ Multiplying a vector by a positive scalar produces a vector with the same direction. Thus, u the vector v 5, 12 H h . I 5 13 , 12 13 i is a unit vector with the same direction as Multiplying a vector by the reciprocal of its length to produce a unit vector, as in Example 1, works in the general case, as stated below. Unit Vectors If v is a nonzero vector, then same direction as v. 1 v 7 7 v is a unit vector with the 662 Chapter 10 Trigonometric Applications Alternate Vector Notation I u 7 H 0, 1 H I , I then 5i 7j. i It can be verified that the vectors are unit vectors. The vectors i and j play a special role because they lead to a useful alternate notation for vectors. For example, if 0, 1 H 5, 7 1, 0 H and j I Similarly, if 0, 7 H 5 1, 0 H I I is any vector, then I u v 5, 0 H a, b I H v a, b a, 0 H I H v ai bj I 0, b I H a 1, 0 H b 0, 1 H is said to be a linear combination of i and j. When The vector vectors are written as linear combinations of i and j, then the properties of vector addition and scalar multiplication, given in Section 10.5, can be used to write the rules for vector addition and scalar multiplication in terms of i and j. ai bj. I I ai bj 1 2 1 ci dj and cai cbj ai bj c 1 2 Example 2 Perform Operations with Linear Combinations If u 2i 6j and v 5i 2j, find 3u 2v. Solution 3u 2v 3 2i 6j 2 5i 2j 1 2 6i 18j 10i 4j
16i 22j 1 2 ■ Direction Angles ai bj v H a, b I If u determined by the standard position angle between terminal side is v, as shown in Figure 10.6-1. is a vector, then the direction of v is completely whose 360°, and 0° b v y θ x 〈a, b〉 a Figure 10.6-1 Section 10.6 Applications of Vectors in the Plane 663 The angle definitions of the trigonometric functions, u is called the direction angle of the vector v. According to the cos u a v and sin u b v 7 7 . Rewriting each of these equations gives the following fact. If v a, b II HH ai bj, then a v cos U and b 7 is the direction angle of v. 7 sin U v 7 7 where U Example 3 Find Velocity Vectors Find the component form of the vector that represents the velocity of an airplane at the instant its wheels leave the ground, if the plane is going angle with the 60 miles per hour and the body of the plane makes a horizontal. 7° Solution The velocity vector u 7°, v ai bj as shown in Figure 10.6-2. Therefore, has magnitude 60 and direction angle 7 i v cos u v 1 7 i 60 cos 7° 1 59.55i 7.31j 59.55, 7.31 H 2 2 I 7 sin u v 1 7 60 sin 7° 1 j 2 j 2 ■ Components of the Direction Angle y 10 60 7° v x 10 30 50 Figure 10.6-2 If v ai bj is a nonzero vector with direction angle u, then tan u sin u cos u b v a v b a . This fact provides a convenient way to find the direction angle of a vector. Example 4 Find Direction Angles Find the direction angle of each vector. v 10i 7j a. u 5i 13j b. x Solution a. The direction angle of u satisfies u tan u b a 13 5 2.6. y u θ 16 12 8 4 〈5, 13〉 13 4 5 8 a Figure 10.6-3 664 Chapter 10 Trigonometric Applications TAN1 Using the vector u is shown in Figure 10.6-3. b. The direction angle of v satisfies u key on a calculator indicates that u 68.96°. The y tan u b a 7 10 0.7. 〈–10, 7〉 8 4 v Because v lies in the second quadrant, must be between 180°. approximately equal to 90° has a tangent that is so u t 34.99° The period of tangent is t 180° A calculator shows that 0.7. tan t tan 180°, and 1 2 145.01° x for every t. Therefore, –12 –8 –4 –34.99° b Figure 10.6-4 u 34.99° 180° 145.01° is the angle between v is shown in Figure 10.6-4. 90° and 180° such that tan u 0.7. The vector ■ Vector Applications A common application of vectors is in modeling a system of forces acting on an object. Every force has direction and magnitude, therefore, each can be represented by a vector. The sum of all the forces acting on an object is called the resultant force. Example 5 Resultant Force An object at the origin is acted upon by two forces. A 150-pound force with the positive x-axis, and the other force of 100 makes an angle of with the positive x-axis, as shown in Figpounds makes an angle of ure 10.6-5. Find the direction and magnitude of the resultant force. 20° 70° R y 100 Q 50 70° P 20° x O 50 100 150 Figure 10.6-5 Section 10.6 Applications of Vectors in the Plane 665 Solution The forces acting upon the object are: ! OP ! OQ ! OR 1 1 150 cos 20° 100 cos 70° 2 2 is the sum of The resultant force ! OR ! OR 150 cos 20° 1 175.16i 145.27j i 2 1 150 sin 20° The magnitude of the resultant force i i ! OP j 2 150 sin 20° 100 sin 70° j j 2 2 1 1 and ! OQ . 100 cos 70° 1 i 2 1 100 sin 70° j 2 ! OR 2175.162 145.272 227.56. is ! OR 7 7 The direction angle of the resultant force satisfies u tan u 145.27 175.16 0.8294 A calculator in degree mode shows that u 39.67°. ■ Example 6 Resultant Force A 200-pound box lies on a ramp that makes an angle of with the horizontal. A rope is tied to the box from a post at the top of the ramp to keep it in position (see Figure 10.6-6). Ignoring friction, how much force is being exerted on the rope by the box? 24° P α T 24° C S θ Q R Figure 10.6-6 Solution ! TR Because of gravity, the box exerts a 200-pound weight straight down (vec! TR ). As Figure 10.6-6 shows, The force tor ! TP , the vector of the force pulling the box on the rope is represented by is the sum of ! TQ . ! TP and 666 Chapter 10 Trigonometric Applications down the ramp, and 7 ! TP TSC, a 24° 90°, 7 represents the magnitude of the force. In right TRP, a u 90°. triangle Therefore, The box weighs 200 pounds, so and in right triangle Use sin u to find ! TP 7 . 7 7 200. a u a 24° u 24° ! TR ! TP ! TR ! TP 200 7 7 7 7 sin u 7 7 sin 24° 7 ! TP 7 7 200 sin 24° 81.35 The force on the rope is about 81.35 pounds. ■ Example 7 Resultant Force 50° with an air speed of 300 miles An airplane is flying in the direction 50° per hour. If there was no wind, the course of the airplane would be . 120°, However, there is a 35-mile-per-hour wind from the direction as represented by the vectors p and w in Figures 10.6-7, which shows the angles using aerial navigation orientation. Find the course and ground speed of the plane (that is, its direction and speed relative to the ground taking the effect of the wind into consideration). y 240 120 50° w 120° –60 p x 240 Figure 10.6-7 y p + w p Solution 40° x 30° 240 Figure 10.6-8 shows p, w, and p w tion angle of the vector p w. p w. The course of the plane is the direc, and its ground speed is the magnitude of 240 120 60° w –60 Figure 10.6-8 The direction angle of p (the angle it makes with the positive x-axis) is 90° 50° 40°. The angle that w makes with the positive y-axis is Section 10.6 Applications of Vectors in the Plane 667 so the direction angle of w, as measured from the pos- 180° 120° 60°, itive x-axis, is p w p w 60° 90° 150°. Therefore, j 300 sin 40° 2 j 35 sin 150° 2 300 sin 40° 300 cos 40° 2 35 cos 150° 2 300 cos 40° 1 1 3 1 2 300 cos 40° 35 cos 150° 2 35 sin 150 199.50i 210.34j p w is The direction angle of j 4 3 1 35 cos 150° i 2 i 2 1 300 sin 40° 35 sin 150° j 2 tan u 210.34 199.50 tan u 1.0543 u 46.5° The course of the plane is the angle between 90° 46.5° 43.5° p w and true north. The ground speed of the plane is p w . p w 2199.502 210.342 289.9 Thus, the plane’s course is about miles per hour. 43.5° and its ground speed is about 289.9 ■ Exercises 10.6 In Exercises 1–5, find u v, u v, and 3u 2v. 1. u i j, v 2i j In Exercises 12–19, find the component form of the vector v whose magnitude and direction angle are given. U 2. 3. u 8i, v 2 3i 2j 1 i j 2 , v 3i u 4 2 1 2i . u 5. u 22j, v 23i 12. 14. 16. 18. 4, u 0° 10, u 225° v v 7 7 7 7 v 6, u 40° 7 7 v 1 2 , u 250° 13. 15. 17. 19, u 30° 20, u 120° 8, u 160° 3, u 310° In Exercises 6–11, find the components of the given w 4i j. vector, where u i 2j, v 3i j, and In Exercises 20–27, find the magnitude and direction angle of the vector v. 6. u 2w w 8. 10. 1 2 1 4 7. 1 2 1 3v w 2 9. 2u 3v 20. 22. 24. v 4, 4 H v I 8, 0 H v 6j 21. v 5, 523 H I I 23. 25. v 4, 5 H v 4i 8j I 8u 4v w 2 1 11. 3 1 u 2v 2 6w 26. v 2i 8j 27. v 15i 10j 668 Chapter 10 Trigonometric Applications In Exercises 28–31, find a unit vector that has the same direction as the given vector. 28. 30. 4, 5 H 5i 10j I 29. 7i 8j 31. 3i 9j In Exercises 32–35, an object at the origin is acted upon Uv , by two forces u and v, with direction angle respectively. Find the direction and magnitude of the resultant force. and Uu 32. u 30 pounds, uu 0°; v 90 pounds, uv 60° 33. u 6 pounds, uu 45°; v 6 pounds, uv 120° 34. 35. u 12 kilograms, uu uv 250° u 30 kilograms, uu uv 40° 130°; v 20 kilograms, 300°; v 80 kilograms, 36. Two ropes are tied to a wagon. A child pulls one with a force of 20 pounds, while another child pulls the other with a force of 30 pounds. See the figure. If the angle between the two ropes is 28°, how much force must be exerted by a third child, standing behind the wagon, to keep the wagon from moving? Hint: Assume the wagon is at the origin and one rope runs along the positive x-axis. Proceed as in Example 5 to find the resultant force on the wagon from the ropes. The third child must use the same amount in the opposite direction. 2 0 l b 28˚ 30 lb 37. Two circus elephants, Bessie and Maybelle, are dragging a large wagon, as shown in the figure. If Bessie pulls with a force of 2200 pounds and Maybelle with a force of 1500 pounds and the wagon moves along the dashed line, what is angle u? Bessie 24° θ Maybelle Exercises 38 – 41 deal with an object on an inclined plane. The situation is similar to that in Figure 10.6-6 ! is the component of the of Example 6, where TP 7 weight of the object parallel to the plane and is the component of the weight perpendicular to the plane. ! TQ 7 7 7 38. An object weighing 50 pounds lies on an inclined angle with the horizontal. plane that makes a Find the components of the weight parallel and perpendicular to the plane. Hint: Solve an appropriate triangle. 40° 39. Do Exercise 38 when the object weighs 200 20° pounds and the inclined plane makes a with the horizontal. angle 40. If an object on an inclined plane weighs 150 pounds and the component of the weight perpendicular to the plane is 60 pounds, what angle does the plane make with the horizontal? 41. A force of 500 pounds is needed to pull a cart up 15° a ramp that makes a Assuming that no friction is involved, find the weight of the cart. Hint: Draw a picture similar to Figure 10.6-6; the 500-pound force is parallel to the ramp. angle with the ground. In Exercises 42–47, find the course and ground speed of the plane under the given conditions. See Example 7. All angle measurements are given as aerial navigation directions. See Exercise 55 of Section 6.2. 42. air speed 250 miles per hour in the direction 60°; wind speed 40 miles per hour from the direction 330° 43. air speed 400 miles per hour in the direction 150°; wind speed 30 miles per hour from the direction 60° 44. air speed 300 miles per hour in the direction 300°; wind speed 50 miles per hour in (not from) the direction 30° 5910ac10_616-687 9/21/05 1:59 PM Page 669 45. air speed 500 miles per hour in the direction wind speed 70 miles per hour in the direction 180°; 40° Section 10.6 Applications of Vectors in the Plane 669 46. The course and ground speed of a plane are 70° and 400 miles per hour respectively. There is a 60-mile-per-hour wind blowing from the south. Find the approximate direction a
nd air speed of the plane. 47. A plane is flying in the direction 200° with an air u speed of 500 miles per hour. Its course and and 450 miles per hour, ground speed are respectively. What is the direction and speed of the wind? 210° 48. A river flows from east to west. A swimmer on the south bank wants to swim to a point on the opposite shore directly north of her starting point. She can swim at 2.8 miles per hour, and there is a 1-mile-per-hour current in the river. In what direction should she swim in order to travel directly north (that is, what angle should the swimmer make with the south bank of the river)? 49. A river flows from west to east. A swimmer on the north bank swims at 3.1 miles per hour along a line that makes a angle with the north bank of the river and reaches the south bank at a point directly south of his starting point. How fast is the current in the river? 75° 50. A 400-pound weight is suspended by two cables (see the following figure). What is the force (tension) on each cable? Hint: Imagine that the weight is at the origin and that the dashed line is the x-axis. Then cable v is represented by the vector c cos 65° i c sin 65° j 1 2 1 which has magnitude c. (Why?) Represent cable u similarly, denoting its magnitude as d. Use the fact that of two equations in the unknowns c and d. (why?) to set up a system u v 0i 400j 2 32° v 65° 400 51. A 175-pound high-wire artist stands balanced on a tightrope, which sags slightly at the point where he is standing. The rope in front of him makes a angle with the horizontal and the rope behind 6° him makes a force on each end of the rope. Hint: Use a picture and procedure similar to that in Exercise 50. angle with the horizontal. Find the 4° 52. Let v be the vector with initial point x1, y12 and let k be any real 1 and terminal point number. a. Show that x2, y22 tan u tan b , 1 where u is the direction b is the direction angle of kv. angle of v and t tan Use the fact that tan conclude that v and kv have either the same or opposite directions. t 180° to 1 2 b. Use the fact that (c, d) and c, d lie on the 1 2 same straight line on opposite sides of the origin to verify that v and kv have the same and opposite directions if direction if k 6 0. k 7 0 53. Let 7 and u v In Exercise 34 of was shown (see the u v a, b c, d . I H I H w Section 10.5, 7 figure with Exercise 34 of Section 10.5). Show that u v is equivalent to the vector w with initial point (c, d) and terminal point (a, b) by now showing that and w have the same direction. u v 7 7 670 Chapter 10 Trigonometric Applications 10.6.A Excursion: The Dot Product Objectives • Find the dot product of two vectors and the angle between two vectors • Determine projection and component vectors and use them in physical applications Dot Product NOTE The dot product of two vectors is found by multiplying corresponding components and finding the sum of the products. Properties of the Dot Product Unlike multiplication of real numbers where there is only one type of multiplication, vector operations include three types of multiplication: scalar multiplication, dot products, and cross products. This section discusses the vector operation called the dot product. Unlike scalar multiplication of vectors, the dot product is not a vector. The dot product of two vectors is a real number. The dot product of vectors ci dj is denoted ac bd. Thus, uv u ai bj c, d II HH and is defined to be the real number a, b II HH v and uv ac bd. Example 1 Find Dot Product of Two Vectors for the given vectors u and v. a. u v 2, 6 Find the dot product and v u I u 4i 2j and v 3i j u 5, 3 H b. c. H I and v I 2, 4 H 6, 3 H I a. Solution u v u v u v b. c. 2, 6 5, 3 H I H 4i 2j 2 2, 4 H 1 6, 3 5 I 3i 21 0 1 1 21 2 2 14 2 ■ The dot product has a number of useful properties. If u, v, and w are vectors, and k is a real number, then: 1. 2. 3v w) u v u w ku v k(u v) u kv 4. 5. 0 u 0 commutative distributive Section 10.6.A Excursion: The Dot Product 671 Proof Let a, b, c, and d be real numbers. 1. If 2. If 7 a u , u then a, b a, b I H I H v a, b and I H a, b I H c, d H I 2a2 b2. Therefore, a b b a2 b2 2 2 , I 1 then 1 c, d H ac bd ca db 2a2 b2 a 2 b 2. u 7 7 c, d H I a, b I H v u. v θ u Figure 10.6.A-1 The proofs of the last three statements are asked for in the exercises. Angles Between Vectors u v I u c, d H a, b I H are any nonzero vectors, then the angle between If and u and v is the smallest angle formed by these two vectors, as shown in Figure 10.6.A-1. The clockwise or counterclockwise rotation is ignored, and the angle between v and u is considered to be the same as the angle between u and v. Thus, the radian measure of . 4 Nonzero vectors u and v are said to be parallel if the angle between them radians. In other words, u and v are parallel if they lie on is either 0 or the same straight line through the origin and have either the same or opposite directions. The zero vector 0 is considered to be parallel to every vector. is in the interval 0, p p u 3 Any scalar multiple of u is parallel to u because it lies on the same straight line as u. Conversely, if v is parallel to u, it is easy to show that v must be a scalar multiple of u. This is shown in the exercises. Parallel Vectors Vectors u and v are parallel exactly when v ku, for some real number k. Example 2 Determine Parallel Vectors Determine whether the vectors u 2, 3 H I and v 8, 12 H I are parallel. Solution Vector v is a scalar multiple of u. I Thus, vectors u and v are parallel. v 8, 12 H 4 2, 3 H I 4u ■ The angle between nonzero vectors u and v is closely related to their dot product. 672 Chapter 10 Trigonometric Applications Angle Theorem If u is the angle between the nonzero vectors u and v, then u v u v 7 7 7 7 cos U, or equivalently, cos a, b) x u θ v (c, d) Figure 10.6.A-2 Proof Let a, b, c, and d be real numbers, and suppose that , v p, u a, b is not 0 or 7 I H then u and v form two sides of a triangle, as shown in Figure 10.6.A-2. and the angle 0 and 0. c, d H u If , I v u 7 7 7 The lengths of two sides of the triangle are 2c2 d2. v 7 side (opposite angle Cosines produces the following result. The distance formula shows that the length of the third b d Therefore, the Law of a c u ) is 2 2 2. u 1 1 2 2 7 7 7 2a2 b2 and c2 d2 a2 b2 a2 2ac c2 b2 2bd d2 a2 b2 c2 d2 2 2ac 2bd cos u cos u cos u cos ac bd ac bd cos u 7 7 The proof when u 7 is 0 or 7 7 7 is exercise 41. p v 7 2 u cos u u 7 v 7 7 7 7 7 cos u v 7 7 cos u u v ac bd y 〈–3, 1〉 2 1 u v 〈5, 2〉 –3 –1 1 5 Figure 10.6.A-3 Example 3 Find the Angle Between Vectors x Find the angle between the vectors in Figure 10.6.A-3. u 3, 1 H I and 5, 2 H I , which are shown Solution Apply the formula from the Angle Theorem with v u 3, 1 H I and . 5, 2 I H cos u u v u v COS1 Using the 3 5 21 2 2 12 1 3 1 2 1 2 252 22 2 1 1 key shows that 2 21 13 210 229 13 2290 2 u 2.4393 radians, or 139.76°. ■ The Angle Theorem has several useful consequences. For instance, taking u v and using the fact the absolute value of each side of cos u, u v 7 7 7 7 Section 10.6.A Excursion: The Dot Product 673 v 7 7 7 (because u v 7 7 7 7 is always positive), produces the 7 0 v u u that 7 7 7 0 7 following results. u v u 0 0 For any angle v 7 7 0 u, 7 7 cos u 0 0 u v u cos u 7 7 0 7 0 1; therefore, u v 7 This proves the Schwarz inequality. 7 7 7 0 0 0 cos cos u 0 ‘ u ‘ ‘ v ‘ cos u 0 0 Schwarz Inequality For any vectors u and v Vectors u and v are said to be orthogonal (or perpendicular) if the angle between them is p 2 fact about orthogonal vectors follows. radians 90° , 1 2 or if at least one of them is 0. The key Orthogonal Vectors Let u and v be vectors. Then u and v are orthogonal exactly when uv 0. u v 0. Proof If u or v is 0, then vectors, then by the Angle Theorem: If u and v are nonzero orthogonal u v u v cos u u v cos Conversely, if u and v are vectors such that asks for a proof that u and v are orthogonal. u v p 2 u v 0, 0 0 1 2 then Exercise 42 Example 4 Find Orthogonal Vectors Determine whether the given vectors are orthogonal. a. u 2,6 H and v I 9, 3 H I b. u 1 2 i 5j and v 10i j Solution a. u v 2,6 H I 9, 3 H I b. 2 9 6 1 2 2 18 18 5j b a 1 1 2 5 5 0 10 5 2 10i j 1 2 1 2 1 Vectors u and v are orthogonal. Vectors u and v are orthogonal. ■ 674 Chapter 10 Trigonometric Applications Projections and Components u If u and v are nonzero vectors, and is the angle between them, construct the perpendicular line segment from the terminal point P of u to the straight line on which v lies. This perpendicular segment intersects the line at a point Q. The three possibilities are shown in Figure 10.6.A-4. is called the projection of u onto v and is denoted projvu. ! OQ The vector A useful description of P u v Q O projvu projvu follows. P u Q O v projvu Figure 10.6.A-4 P u v O Q projvu Projection of u onto v If u and v are nonzero vectors, then the projection of u onto v is the vector projvu u v v2 b v. a projvu projvu kv Proof Because and v lie on the same straight line, they are parallel. for some real number k. Construct the orthogonal Therefore, vector from a point Q on v through point P, the terminal point of u. Let w be the vector with its initial point at the origin and the same length and direction as as in the two cases shown in Figure 10.6.A-5. ! QP , P u v Q projvu w O P u w v Q O projvu Figure 10.6.A-5 Because w is parallel to 10.6.A-5, the dot product: u projvu w kv w. it is orthogonal to v. As shown in Figure Consequently, by the properties of ! QP Section 10.6.A Excursion: The Dot Product 675 2 kv w u v 1 kv 1 2 v v k kv2 substitution distributive v v v2 But w v 0 because w and v are orthogonal. So, u v kv2 or equivalently, k u v v2 . Finally, multiplying both sides of the last statement by v, and substituting projvu for kv, the desired result is proved. u v v2 b projvu kv v a Example 5 Find Projection Vectors If u 8i 3j and v 4i 2j, find projvu and projuv. Solution v2 v v 42 1 2 u v 8 2 26 v u 2 20 and u2 u u 82 32 73 3 2 4 1 1 2 2 Therefore, and projvu u v
v2 b a v 26 20 a b 1 4i 2j 26 5 i 13 5 j 2 projuv v u u2 b a u 26 73 a b 1 8i 3j 208 73 i 78 73 j 2 ■ The projection vectors from Example 5 are shown in Figure 10.6.A-6. y projuv v projvu u x Figure 10.6.A-6 676 Chapter 10 Trigonometric Applications Projections and Components Recall from Section 10.6 that 1 v v is a unit vector in the direction of v. Then, projvu can be expressed as a scalar multiple of this unit vector. projvu u v v2 The scalar compvu. is called the component of u along v, and is denoted projvu compvu 1 v v b a Because 1 v v is a unit vector, it can be used to find the length of projvu . projvu compvu g 1 v v b g a 0 compvu 1 v v b g 0 0 g a compvu . 0 Also, because u v u v cos u, is the angle between u and v, u where compvu u v v u v cos u v u cos u This result is stated formally as follows. Projections and Components If u and v are nonzero vectors, and them, then u is the angle between and compvu u v v u cos U projvu compvu . 00 00 Example 6 Find Component Vectors If u 2i 3j and v 5i 2j, find compvu and compuv. Solution v 2 5 1 2 u v 2 2 22 229 and u 222 32 213 Section 10.6.A Excursion: The Dot Product 677 Therefore, compvu u v v 4 229 and compuv v u u 4 213 ■ Applications Vectors and the dot product can be used to solve a variety of problems. Example 7 Find Forces Due to Gravity A 4000-pound automobile is on an inclined ramp that makes a angle with the horizontal. Find the force required to keep it from rolling down the ramp, assuming that the only force that must be overcome is that due to gravity. 15° Solution The situation is shown in Figure 10.6.A-7, where the coordinate system is chosen so that the car is at the origin, the vector F representing the downward force of gravity is on the y-axis, and v is a unit vector from the origin down the ramp. 90° 15° 75°. The angle between v Because the car weighs 4000 pounds, and F is is the force pulling the car The vector down the ramp, so a force of the same magnitude in the opposite direction is needed to keep the car motionless. F 4000j. projvF projvF compvF F cos 75° 0 4000 cos 75° 1035.3 0 0 0 Therefore, a force of 1035.3 pounds is required to hold the car in place. ■ If a constant force F is applied to an object, pushing or pulling it a distance d in the direction of the force, as shown in Figure 10.6.A-8, the amount of work done by the force is defined to be the product of the magnitude of the force and the distance. W 1 magnitude of force distance 2 1 F d 1 2 2 If the magnitude of F is measured in pounds and d in feet, then the units for W are foot-pounds. For example, if you push a car for 35 feet along a level driveway by exerting a constant force of 110 pounds, the amount of foot-pounds. work done is 3850 110 35 1 2 When a force F moves an object in the direction of a vector d rather than in the direction of F, as shown in Figure 10.6.A-9, then the motion of the which is a force object can be considered as the result of the vector in the same direction as d. projdF, projvF 15° y v 75° F R a m p 15° x Figure 10.6.A-7 F d Figure 10.6.A-8 F θ d projdF Figure 10.6.A-9 678 Chapter 10 Trigonometric Applications NOTE This formula cos u cos 0 1, reduces to the previous one when F and d have the same direction because in that case, so that W F d of force times distance moved. magnitude Therefore, the amount of work done by F is the same as the amount of work done by projdF, as shown below. W projdF d 0 d d 2 compdF cos u 0 F 1 F d See note. Consequently, work can be described as follows. Work The work W done by a constant force F as its point of application moves along the vector d is W 00 compdF 00 d or equivalently, W F d. Example 8 Compute Work How much work is done by a child who pulls a sled 100 feet over level 45° ground by exerting a constant 20-pound force on a rope that makes a angle with the ground? F 45° d Figure 10.6.A-10 Solution The situation is shown in Figure 10.6.A-10, where force F on the rope has magnitude 20, and the sled moves along vector d of length 100. W F d F d cos u 22 2 20 100 100022 1414.2 Therefore, the work done is 1414.2 foot-pounds. ■ Exercises 10.6.A In Exercises 1–6, find u v, u u, and v v. 1. u , v 3, 4 II HH 5, 2 II HH 2. u , v 1, 6 II HH hh 4, 1 3 ii 3. u 2i j, v 3i 4. u i j, v 5j 5. u 3i 2j, v 2i 3j 6. u 4i j, v i 2j In Exercises 7–12, find the dot product when u w and , II 2, 5 HH u , v 4, 3 HH v w II 2 v w 1 u v 2 3u v 1 1 2 1 2w 7. 9. 11. 8. 10. 12. 2 2, 1 HH u . II v w 1 u v 2 u 4v 1 1 1 1 2 2 u v 2 2u w 2 In Exercises 13–18, find the angle between vectors u and v. 13. 14. 15. H u , v 4, 3 u 1, 2 H 0, 5 H I , v I u 2i 3j, v i 2, 4 H I I 16. u 2j, v 4i j 17. u 22 i 22 j, v i j 18. u 3i 5j, v 2i 3j In Exercises 19–24, determine whether the vectors u and v are parallel, orthogonal, or neither. 19. u 20. u 21. u I 2, 6 3, 3 I , v H 9, 6 2, 6 I H 6, 4 H I H I 22. u i 2j, v 2i 4j 23. u 2i 2j, v 5i 8j 24. u 6i 4j, v 2i 3j Section 10.6.A Excursion: The Dot Product 679 In Exercises 25–28, find a real number k such that vectors u and v are orthogonal. 25. u 2i 3j, v 3i kj 26. u 3i j, v 2ki 4j 27. u i j, v ki 22 j 28. u 4i 5j, v 2i 2kj In Exercises 29–32, find projuv and projvu. 29. u 3i 5j, v 6i 2j 30. u 2i 3j, v i 2j 31. u i j, v i j 32. u 5i j, v 2i 3j In Exercises 33–36, find compvu. 33. u 10i 4j, v 3i 2j 34. u i 2j, v 3i j 35. u 3i 2j, v i 3j 36. u i j, v 3i 2j In Exercises 37 – 39, let and w Verify that the given property of dot products is valid by calculating the quantities on each side of the equal sign. r, s II HH c, d II , v a, b II u HH HH . 37. 38. u v w u v u w 1 ku v k 2 u v 1 u kv 2 39. 0 u 0 40. Suppose u parallel vectors. c 0, a. If a, b I H and v c, d H I are nonzero show that u and v lie on the same nonvertical straight line through the origin. v c show that a 0, (that is, v is a scalar b. If a u multiple of u). Hint: The equation of the line on y mx which u and v lie is m (why?), which implies that d mc. c 0 Hint: If (otherwise u 0 , show that v is a scalar multiple of u. c 0 b 0 for some constant b ma (why?) and so then . a 0 and c. If 2 680 Chapter 10 Trigonometric Applications 41. Prove the Angle Theorem in the case when u is 0 or p. 42. If u and v are nonzero vectors such that show that u and v are orthogonal. Hint: If angle between u and v, what is does this say about cos u u and what u? u v 0, is the 43. Show that , 1, 2 , 1 2 right triangle by considering the sides of the triangle as vectors. are the vertices of a 3, 4 5, 2 2 1 2 1 44. Find a number x such that the angle between the vectors 1, 1 H I and x, 1 H I is p 4 radians. 45. Find nonzero vectors u, v, and u v u w , orthogonal to u. v w w , and neither v nor w is such that 46. If u and v are nonzero vectors, show that the uv vu , uv vu are orthogonal. vectors 30° 47. A 600-pound trailer is on an inclined ramp that makes a angle with the horizontal. Find the force required to keep it from rolling down the ramp, assuming that the only force that must be overcome is due to gravity. 48. In Example 7, find the vector that represents the force necessary to keep the car motionless. In Exercises 49–52, find the work done by a constant force F as the point of application of F moves along the vector ! PQ . 52. F 5i j, P 1, 2 2 1 , Q 1 4, 3 2 53. A lawn mower handle makes an angle of with the ground. A woman pushes on the handle with a force of 30 pounds. How much work is done as she moves the lawn mower a distance of 75 feet on level ground? 60° 54. A child pulls a wagon along a level sidewalk by exerting a force of 18 pounds on the wagon handle, which makes an angle of horizontal. How much work is done as she pulls the wagon 200 feet? with the 25° 20° 55. A 40-pound cart is pushed 100 feet up a ramp that makes a angle with the horizontal. How much work is done against gravity? Hint: The amount of work done against gravity is the negative of the amount of work done by gravity. Position the cart on a coordinate plane so that the cart is at the origin. Then the cart moves along vector i d downward force of gravity is and the F 0i 40j . 100 cos 20° 100 sin 20° j 2 1 1 2 20° 49. F 2i 5j, P 1 0, 0 50. 51. F i 2j, P 0, 0 1 1 , Q F 2i 3j, P ! component form of PQ . 2, 1 2 1 5, 2 2 5, 9 1 2 Hint: Find the 56. Suppose the child in Exercise 54 is pulling the wagon up a hill that makes an angle of the horizontal, and all other conditions remain the same. How much work is done in pulling the wagon 150 feet? with 20° C H A P T E R 10 R E V I E W Important Concepts Section 10.1 Section 10.2 Section 10.3 Section 10.4 Section 10.5 Standard notation for triangles . . . . . . . . . . . . . . 617 Law of Cosines . . . . . . . . . . . . . . . . . . . . . . . . . . 617 Law of Sines . . . . . . . . . . . . . . . . . . . . . . . . . . . . 625 Ambiguous case . . . . . . . . . . . . . . . . . . . . . . . . . 627 Area formulas for triangles . . . . . . . . . . . . . . . . . 632 Complex plane . . . . . . . . . . . . . . . . . . . . . . . . . . 638 Real axis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 638 Imaginary axis. . . . . . . . . . . . . . . . . . . . . . . . . . . 638 Absolute value of a complex number . . . . . . . . . 638 Argument . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 639 Modulus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 639 Polar form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 639 Polar multiplication and division . . . . . . . . . . . . 640 DeMoivre’s Theorem. . . . . . . . . . . . . . . . . . . . . . 644 Formula for nth roots . . . . . . . . . . . . . . . . . . . . . 647 Roots of unity . . . . . . . . . . . . . . . . . . . . . . . . . . . 648 Vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 653 Magnitude. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 655 Components . . . . . . . . . . . . . . . . . . . . . . . . . . . . 655 Scalar multiplication . . . . . . . . . . . . . . . . . . . . . . 655 Vector addition and subtraction . . . . . . . . .
. . . . 657 Section 10.6 Unit vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 661 Linear combination of i and j . . . . . . . . . . . . . . . 662 Direction angle of a vector . . . . . . . . . . . . . . . . . 663 Section 10.6.A Dot product. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 670 Angle between vectors . . . . . . . . . . . . . . . . . . . . 671 Parallel vectors . . . . . . . . . . . . . . . . . . . . . . . . . . 671 Angle theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 672 Schwarz inequality . . . . . . . . . . . . . . . . . . . . . . . 673 Orthogonal vectors . . . . . . . . . . . . . . . . . . . . . . . 673 Projection of u on v. . . . . . . . . . . . . . . . . . . . . . . 674 Component of u along v . . . . . . . . . . . . . . . . . . . 676 Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 678 681 682 Chapter Review Important Facts and Formulas Law of Cosines a2 b2 c2 2bc cos A Law of Cosines, Alternate Form: cos A b2 c2 a2 2bc Law of Sines a sin A b sin B c sin C Area of triangle ABC: ab sin C 2 Heron’s Formula Area of triangle a b c where . s 1 2 1 2 ABC 2s s a s b s c , 2 21 21 1 2a2 b2 a bi 0 0 a bi r 1 cos u i sin u , where 2 cos u1 r11 i sin u12 r 2a2 b2, a r cos u, b r sin u i sin u22 r1r23 u22 cos r21 cos u2 u1 1 i sin u1 1 u22 4 cos u1 cos u2 r11 r21 i sin u12 i sin u22 r1 r2 3 DeMoivre’s Theorem cos u1 1 u22 i sin u1 1 u22 4 cos u i sin u r 3 The distinct nth roots of 1 r 3 2 4 n rn cos u i sin u u 2kp n b d a 1 i sin 2n r cos a c u 2kp n b cos nu i sin nu 4 are: 2 k 0, 1, 2, p , n 1 1 2 The distinct nth roots of unity are: cos 2kp n i sin 2kp n 1 k 0, 1, 2, p , n 1 Q x2, y22 1 , then PQ › x1, y2 x2 H 2 y1I . and x1, y12 1 2a2 b2 If P a, b I H u If a, b I H and k is a scalar, then ku ka, kb . I H If u a, b and I H u v v c, d then , I H a c, b d H I and u v a c, b d H . I Chapter Review 683 Properties of Vector Addition and Scalar Multiplication For any vectors u, v, and w and any scalars r and s: u v w 2 1. 2. 3. 4. 5. 6. 7. 8 rs 2 1 1v v 0v 0 r0 0 2 ru rv 2 v rv sv rv 2 v r s sv 1 1 2 1 2 9. u If a, b I H ai bj, then a u cos u and b u sin u, where u is the direction angle of u. If u a, b I H ai bj and v c, d H u v ac bd. ci dj, then I If u is the angle between nonzero vectors u and v, then u v u v cos u. Schwarz Inequality u v 0 0 u v Vectors u and v are orthogonal exactly when u v 0. projvu u v v2 b a v compvu u v v u cos u, where u is the angle between u and v. Properties of Dot Products If u, v, and w are vectors, and k is a real number, then: 2. 1. u u u2 u v v u u v w 1 ku v k 4. 5. 0 u 0 3 kv 2 684 Chapter Review Review Exercises Note: Standard notation is used for triangles. Section 10.1 In Exercises 1–6, use the Law of Cosines to solve triangle ABC. 1. a 12, b 10, c 15 2. a 7.5, b 3.2, c 6.4 3. a 10, c 14, B 130° 4. a 7, b 8.6, C 72.4° 5. a 5, c 8, B 76° 6. a 90, b 70, c 40 7. Two trains depart simultaneously from the same station. The angle between their two tracks is 45 miles per hour and the other at 70 miles per hour. How far apart are the trains after 3 hours? One train travels at an average speed of 120°. 8. A 40-foot flagpole sits on the side of a hill. The hillside makes a angle with the horizontal. How long is a wire that runs from the top of the pole to a point 72 feet downhill from the base of the pole? 17° 9. Find angle ABC in the figure below. A B 18 12 C 10 10. A surveyor stakes out points A and B on opposite sides of a building. Point What is C is 300 feet from A and 440 feet from B. Angle ACB measures the distance from A to B? 38°. Section 10.2 In Exercises 11–18, use the Law of Sines to solve triangle ABC. 11. B 124°, C 31°, c 3.5 12. A 96°, B 44°, b 12 13. a 75, c 84, C 62° 14. a 5, c 2.5, C 30° 15. a 3.5, b 4, A 60° 16. a 3.8, c 2.8, C 41° 17. A 48°, B 57°, b 47 18. A 67°, c 125, a 100 19. Find the area of triangle ABC if b 24, c 15, and A 55°. 20. Find the area of triangle ABC if a 10, c 14, and B 75°. 21. A boat travels for 8 kilometers in a straight line from the dock. It is then sighted from a lighthouse which is 6.5 kilometers from the dock. The angle determined by the dock, the lighthouse (vertex), and the boat is far is the boat from the lighthouse? How 25°. 22. A pole tilts 12° from the vertical, away from the sun, and casts a 34-foot shadow on level ground. The angle of elevation from the end of the 64°. shadow to the top of the pole is How long is the pole? Chapter Review 685 23. Two surveyors, Joe and Alice, are 240 meters apart on a riverbank. Each sights a flagpole on the opposite bank. The angle from the pole to Joe (vertex) to Alice is 54°. The angle from the pole to Alice (vertex) to Joe is How far are Joe and Alice from the pole? 63°. 24. A straight road slopes at an angle of angle of elevation of the sun is 62.5°, road casts a 15-foot shadow downhill, parallel to the road. How high is the telephone pole? with the horizontal. When the 10° a telephone pole at the side of the 25. A woman on the top of a 448-foot building spots a small plane. As she 62°. views the plane, its angle of elevation is At the same instant a man at the ground-level entrance to the building sees the plane and notes that its angle of elevation is a. How far is the woman from the plane? b. How far is the man from the plane? c. How high is the plane? 65°. 26. Use the Law of Sines to prove Engelsohn’s equations given below: For any triangle ABC (standard notation), a b c sin A sin B sin C and a b c sin A sin B sin C . In Exercises 27–30, find the area of the triangle described. 27. angle of 30°, adjacent side lengths 5 and 8 28. angle of 40°, adjacent side lengths 3 and 12 29. side lengths 7, 11, and 14 30. side lengths 4, 8, and 10 Section 10.3 31. Simplify 0 4 2i i 1 3 2i 2 0 3 i 0 1 2i . 0 32. Simplify 0 33. Graph the equation 0 0 . in the complex plane. 0 2 z 0 z 3 0 0 34. Graph the equation 1 0 in the complex plane. 35. Express 1 i23 in polar form. 36. Express 4 5i in polar form. In Exercises 37–41, express the given number in the form a bi. 37. 2 cos a p 12 i sin p 12b 4 cos a p 6 i sin 38. 3 cos a p 8 i sin p 8 b 2 cos a 3p 8 i sin p 6 b 3p 8 b 686 Chapter Review 12 39. cos a i sin 3 cos a i sin 7p 12 5p 12 7p 12 b 5p 12 b Section 10.4 40. cos a p 12 i sin 18 p 12b 41. 13 3 c cos a 5p 36 i sin 12 5p 36b d In Exercises 42–46, solve the given equation in the complex number system, and express your answers in polar form. 42. x3 i 45. x4 i 43. x6 1 46. x3 1 i 44. x8 23 3i Section 10.5 In Exercises 47–50, let u 3, 2 HH II and v 8, 1 HH II . Find each of the following: 47. u v 48. 3v 49. 2v 4u 50. 3u 1 2 v In Exercises 51–54, let u 2i j and v 3i 4j. Find each of the following: 51. 4u v 52. u 2v 53. u v 54. u v 55. Find the components of the vector v such that v 5 and the direction angle of v is 45°. Section 10.6 56. Find the magnitude and direction angle of 3i 4j. 57. Find a unit vector whose direction is opposite the direction of 3i 6j. 58. An object at the origin is acted upon by a 10-pound force with direction 90° angle magnitude and direction of the resultant force. and a 20-pound force with direction angle 30°. Find the 59. A plane flies in the direction 120°, with an air speed of 300 miles per hour. The wind is blowing from north to south at 40 miles per hour. Find the course and ground speed of the plane. 60. An object weighing 40 pounds lies on an inclined plane that makes a 30° angle with the horizontal. Find the components of the weight parallel and perpendicular to the plane. Section 10.6.A In Exercises 61 –64, let following: u 3,4 , v II , and w 2, 5 II HH HH 0, 3 HH . II Find each of the 61. u v 62. u u v v 63. u v 2 1 w 64. u w 1 2 1 w 3v 2 65. What is the angle between the vectors 5i 2j and 3i j? 66. Is 3i 2j orthogonal to 4i 6j? Chapter Review 687 In Exercises 67 and 68, let ing: u 4i 3j and v 2i j. Find each of the follow- 67. projvu 68. compuv 69. If u and v have the same magnitude, show that u v and u v are orthogonal. 70. If u and v are nonzero vectors, show that the vector u kv is orthogonal to v, where k u v v2 . 71. A 3500-pound automobile is on an inclined ramp that makes a angle with the horizontal. Find the force required to keep it from rolling down the ramp, assuming the only force that must be overcome is due to gravity. 30° 72. A sled is pulled along level ground by a rope that makes a angle with the horizontal. If a force of 40 pounds is used to pull the sled, how much work is done in pulling it 100 feet? 50° C H A P T E R 10 Euler’s Formula One of the most interesting and surprising identities in all of mathematics is one that relates the exponential function to the trigonometric functions sin x and cos x and the imaginary number i. e x ix cos x i sin x e The identity, known as Euler’s formula, is named after the mathematician Leonhard Euler (1707–1783), who discovered it in 1748. The formula is used in many areas of calculus—most notably differential equations. Euler's formula is true for any real number x, and many real numbers produce surprising results. Example 1 Evaluating Euler’s Formula Evaluate eip. Solution Substitute p for x in the formula and simplify. eip cos p i sin p 1 i 1 Therefore, eip 1. 1 0 2 ■ Rewriting the last equation connects the five most common constants of mathematics: e, p, i, 0, and 1. eip 1 0 One of the most surprising aspects of this displayed equation is that raising an irrational number to an irrational power results in an integer. In fact, raising an imaginary number to an imaginary power can also give a real number, as shown in the next example. Example 2 Imaginary Numbers Raised to Imaginary Powers Show that i e p 2 i and find ii. Solution Substitute x p 2 into Euler’s formula and simplify. p 2 i cos e p 2 i sin p 2 0 i i 1 1 2 688 To find ii, raise both sides of the identity p 2 i2 e 2 1 p e 2 i e ii p 2 i e A B p A calculator computes the value of Figure 10.C-1. So ii 0.2079. p 2 i i e 1 2 e to the power i. p 2 to be 0.2078795764, as shown in Figu
re 10.C-1 Euler’s formula can be used to define complex powers of e, that is, xiy e x e cos y i sin y The equation terms of a real power of e and the cosine and sine of a real number. 2 1 defines a complex power of e in ez e xiy e cos y i sin y iy e e 1 2 x x ■ xiy. e Example 3 Complex Power of e Find the exact value and approximate value of ep2i. Solution Substitute x p and y 2 ep2i ep into the formula cos 2 i sin 2 z e e x 1 1 0.4161 0.9093i 23.14 9.629926 21.041772i 2 1 cos y i sin y . 2 exact value 2 approximate value ■ Exercises Find the exact value and the approximate value of the following powers of e. 1. pie e pi 3. ie e i 3i B 2. e3i A p 4 i 4. e 5. e 1ip 7. e1 p 3 i 6. e 2pi 8. epi 689 C H A P T E R 11 Analytic Geometry You are there! All planets travel in elliptical orbits around the sun, moons travel in elliptical orbits around planets, and satellites follow elliptical paths around the earth. Parabolic reflectors are used in spotlights, radar antennas, and satellite dishes. The long-range navigation system (LORAN) uses hyperbolas to enable a ship to determine its exact location. See Exercise 68 in Section 11.4. 690 Chapter Outline 11.1 Ellipses 11.2 Hyperbolas 11.3 Parabolas 11.4 11.5 11.6 11.7 Translations and Rotations of Conics 11.4.A Excursion: Rotation of Axes Polar Coordinates Polar Equations of Conics Plane Curves and Parametric Equations 11.7.A Excursion: Parameterizations of Conic Sections Chapter Review can do calculus Arc Length of a Polar Graph Interdependence of Sections ⎫ 11.1 ⎪ ⎬ 11.2 ⎪ ⎭ 11.3 11.5 11.7 > 11.4 >> 11.6 Sections 3.1, 3.2, and 3.4 are prerequisites for this chapter. Except for the discussion of standard equations for conics in Sections 11.1–11.4, Chapter 6 (Trigonometry) is also a prerequisite. When a right circular cone is cut by a plane, the intersection is a curve called a conic section, as shown in the figure below. (A point, a line, or two intersecting lines are sometimes called degenerate conic sections.) Conic sections were studied by the ancient Greeks and are still of interest. For example, the orbits of planets are ellipses, parabolic mir- rors are used in telescopes, and certain atomic particles follow hyperbolic paths. P Circle Ellipse Hyperbola Parabola Point Line Two intersecting lines Although the Greeks studied conic sections from a purely geometric point of view, the modern approach is to describe them in terms of the coordinate plane and distance, or as the graphs of certain types of equations. The study of the geometric properties of objects using a coordinate system is called analytic geometry. Circles are discussed in the appendix and used in prior sections. In this chapter, ellipses, hyperbolas, and parabolas are defined in terms of points and distances, and their equations are determined. The standard form of the equation of a conic includes the key information necessary for a rough sketch of its graph. Techniques for graphing conic sections with a calculator are discussed in the sections that define each conic section, and applications are given. 691 692 Chapter 11 Analytic Geometry 11.1 Ellipses Objectives • Define an ellipse • Write the equation of an ellipse • Identify important characteristics of ellipses • Graph ellipses An ellipse is a closed figure that can be thought of as a circle that has been elongated along a line of symmetry through its center. In this section, ellipses are defined in terms of points and distances, and then their equations are derived from the definition. Definition of an Ellipse Let P and Q be points in the plane and k a number greater than the distance from P to Q. The ellipse with foci (singular: focus) P and Q is the set of all points X such that the sum of the distance from X to P and the distance from X to Q is k. Written algebraically with X representing a point XP XQ k x, y 1 2 on the ellipse, To draw the ellipse, pin the ends of a string of length k at points P and Q, as shown in Figure 11.1-1. Place a pencil against the string, and keep the string taut while moving the pencil. center major axis P Q P Q Figure 11.1-1 Figure 11.1-2 vertex foci vertex minor axis PQ joining the foci is the center of the ellipse. The midpoint of the segment The points where the line through the foci intercept the ellipse are its vertices. The segment connecting the vertices is the major axis, and the segment through the center of the ellipse perpendicular to the major axis is the minor axis, as shown in Figure 11.1-2. If the points P and Q coincide, the ellipse generated is a circle with radius k 2 Thus, a circle is a special case of an ellipse. . Equation of an Ellipse The simplest case is an ellipse centered at the origin, with its foci on the c, 0 x- or y-axis. Suppose that the foci are on the x-axis at the points P 1 2 Section 11.1 Ellipses 693 Let a k 2 , so that k 2a. Then x, y 1 2 is on the and Q c, 0 , where 1 2 ellipse exactly when c 7 0. distance from 3 x, y 1 2 to P 4 3 distance from x, y 1 2 to Q 4 k 2a . Written algebraically 2a 2 2 1 1 which can be rewritten as x c 2 1 2 y˛ 2 2a 2 x c 1 2 2 y˛ 2 . 2 Square both sides and simplify the result. Square both sides again and simplify. a2 1 x c 2 2 y˛ 2 a˛ 2 cx 2 c˛ a˛ 2 1 2 2 a˛ x˛ a˛ 1 2 c˛ 2 a˛ 2 2 y˛ b 2a˛ 2 c˛ 2 2 a˛ 2 c˛ 2 2 so that 2, To simplify the last equation, let Equation [1] then becomes b˛ [1] 2˛x˛ b˛ 2 a˛ 2˛y˛ 2 a˛ 2 2˛b˛ Dividing both sides by the ellipse satisfy the equation 2 2b˛ a˛ shows that the coordinates of every point on 2 x˛ 2 a˛ 2 y˛ 2 b˛ 1 Standard form of an ellipse Conversely, it can be shown that every point whose coordinates satisfy 0, c the equation is on the ellipse. The equation for an ellipse with foci and on the y-axis is developed similarly. 0, c 2 1 1 2 Standard Equation of an Ellipse Centered at the Origin Let a and b be real numbers such that graph of each of the following equations is an ellipse centered at the origin. 0 66 b 66 a. Then the Foci on the x-axis: y˛ 2 b˛ 2 x˛ 2 a˛ 1 2 Foci on the y-axis: y˛ 2 a˛ 2 x˛ b˛ 1 2 2 y b −b x c a −a −c y a c −b x b −c −a 694 Chapter 11 Analytic Geometry When the equation is in standard form, the x- and y-intercepts are easily found. x-intercepts 2 2 y 0 2 x˛ 2 a˛ 1 1 0˛ 2 b˛ 2 a˛ 2 x˛ x –a y-intercepts 2 2 x 0 2 0˛ 2 a˛ 1 1 y˛ 2 b˛ 2 b˛ 2 y˛ y –b The characteristics of the graph of an ellipse centered at the origin are shown in the following box. Characteristics of Ellipses For 0 66 b 66 a, Foci on the x-axis: y˛ 2 1 2 2 2 x˛ a˛ b˛ Foci on the y-axis: y˛ 2 2 x˛ 2 b˛ 2 a˛ 1 foci −a −c minor axis y b −b center x c a major axis y a c foci x b major axis center −b minor axis −c −a ±±a ±±b x-intercepts: y-intercepts: major axis is on the x-axis vertices foci c 2a˛ (a, 0) and 2 and (c, 0), (c, 0) 2 b˛ (a, 0) where ±±b ±±a x-intercepts: y-intercepts: major axis is on the y-axis vertices foci c 2a˛ (0, a) and 2 and (0, c), (0, c) 2 b˛ (0, a) where Notice the following facts in both cases. • the foci are within the ellipse • the major axis always contains the foci and is determined by which denominator is larger • the distance between the foci is 2c • the center is the midpoint between the foci and the midpoint between the vertices • the distance between the vertices is 2a • 2 b˛ 2 a˛ c˛ 2 When the equation of an ellipse centered at the origin is in standard form, the denominator of the x term always gives the x-intercepts and the denominator of the y term always gives the y-intercepts. Section 11.1 Ellipses 695 Graphing an Ellipse Example 1 Graph an Ellipse Show that the graph of the equation the foci, the vertices, the major axis, and the minor axis. 25x˛ 2 16y˛ 2 400 is an ellipse. Label y 10 5 major axis vertices −10 −5 0 foci −5 x 5 10 minor axis −10 Figure 11.1-3 Solution Put the equation in standard form by dividing both sides by 400. 2 x˛ 16 2 y˛ 25 1 This is the equation of an ellipse with its center at the origin. The foci are on the y-axis because the denominator of and 2 16, b˛ To graph the ellipse, plot the intercepts and draw the ellipse, as shown in Figure 11.1-3. is larger. Since and the y-intercepts are 2 25 a˛ ± a ± 5. the x-intercepts are ± b ± 4 2 y˛ To locate the foci, note that on the y-axis. Therefore, the foci are c 125 16 19 3 0, 3 0, 3 and and that the foci are . 1 2 2 and the major axis lies on the y-axis 1 The vertices are 1 with endpoints at the vertices. and 2 1 0, 5 0, 5 , 2 The minor axis lies on the x-axis, with endpoints 4, 0 1 and 4, 0 . 2 1 2 ■ Example 2 Graph an Ellipse on a Calculator Graph the ellipse with equation 4x˛ 2 9y˛ 2 36 on a calculator. Solution Solve the equation for y. 3.1 9y˛ 2 36 4x˛ 2 2 36 4x˛ 2 y˛ −4.7 4.7 y ± B 9 36 4x2 9 ± 2 3 29 x˛ 2 The ellipse is defined by the two functions −3.1 Y1 2 3 29 x˛ 2 and Y2 2 3 29 x˛ 2, Figure 11.1-4 whose graphs are shown in Figure 11.1-4. ■ NOTE Graph all conic sections using a square window to see the correct shape. 696 Chapter 11 Analytic Geometry Ellipse Equations Example 3 Find the Equation of an Ellipse Find the equation of the ellipse that has vertices at 0, –216 Then sketch its graph by using the intercepts. . 1 0, –6 and foci at 2 A B Solution The foci of the ellipse lie on the y-axis, and its center is the origin. Thus, the equation has the form 2 x˛ 2 b˛ c 216 2 y˛ 2 a˛ 1 and by using the relationship among Find b by letting the values for an ellipse. a 6 and A x 4 8 Thus, the equation is 216 2 c˛ 2 2 b2 2 2 b˛ 2 a˛ 2 6 B 1 24 36 b˛ 2 12 b˛ b 212 3.5 2 x˛ 112 2 y˛ 2 6˛ 1 or 2 x˛ 12 2 y˛ 36 1 2 B and the intercepts are A 0, ± 6 1 2 and A ± 212, 0 B . See Figure 11.1-5. ■ Applications of Ellipses Elliptical surfaces have interesting reflective properties. A sound wave or light ray that passes through one focus and reflects off an ellipse will always pass through the other focus, as shown in Figure 11.1-6. Example 4 Finding the Foci The Whispering Gallery at the Museum of Science and Industry in Chicago is elliptical in shape, with a parabolic dish at each focus. (Parabolas are discussed in Section 11.3.
) The shape of the room and two parabolic dishes carry the quietest sound from one focus to the other. The width of the ellipse is 13 feet 6 inches and the length of the ellipse is 47 feet 4 inches. Assume that the ellipse is centered at the origin. Find its equation, sketch its graph, and locate the foci. Solution Because the length of the ellipse is 47 feet 4 inches, the value of a is half that amount, or 23 2 3 feet. Because the width is 13 feet 6 inches, the value y 8 4 −4 −4 −8 −8 Figure 11.1-5 foci Figure 11.1-6 where 2a NOTE The area of an A pab, ellipse is is the length of the major axis and 2b is the length of the minor axis. See Exercise 24 for a method to estimate the circumference of an ellipse. Section 11.1 Ellipses 697 of b is half that, or 6 3 4 feet. Therefore, the equation of the ellipse is 2 x˛ 23 a 2 2 3b 2 x˛ 5041 9 2 2 y˛ 3 6 4b a 2 y˛ 729 16 16y˛ 729 2 1 1 1 2 9x˛ 5041 The distance from the center to each focus is c, which is given by 2 a˛ 2 b˛ c˛ 2 5041 9 729 16 74,095 144 c 74,095 144 B 22.7 y 20 10 F1 F2 x −20 −10 0 10 20 −10 −20 Figure 11.1-7 Therefore, the foci are approximately in Figure 11.1-7. F11 22.7, 0 and F21 2 22.7, 0 2 , as shown ■ The planets and many comets have elliptical orbits, with the Sun at one focus. The Moon travels in an elliptical orbit with Earth at one focus, and man-made satellites usually travel in elliptical orbits around Earth. Example 5 Elliptical Orbits Earth’s orbit around the Sun is an ellipse that is almost a circle. The Sun is at one focus, the major axis is 299,190,000 km in length, and the minor axis is 299,148,000 km in length. What are the minimum and maximum distances from Earth to the Sun? Solution Choose a coordinate system with the center of the ellipse at the origin and the Sun at the point to get a diagram of the orbit. See Figure 11.1-8a. c, 0 2 1 Earth Sun Figure 11.1-8a Figure 11.1-8b 698 Chapter 11 Analytic Geometry The length of the major axis is 2a and the length of the minor axis is 2b, so 2a 299,190,000 a 149,595,000 2b 299,148,000 b 149,574,000 Therefore, the distance from each focus to the center is given by c 2a2 b2 2,507,000 It can be proved algebraically that the minimum and maximum distances from a focus to a point on the ellipse are at the endpoints of the major axis. That is, the maximum distance is and the minimum distance is See Figure 11.1-8b. a c. a c • minimum distance • maximum distance a c a c 147,088,000 km 91.2 million miles 152,102,000 km 94.3 million miles ■ Exercises 11.1 In Exercises 1–6, find the equation of the ellipse centered at the origin that satisfies the given conditions. 7. 1. foci on x-axis; x-intercepts ± 7, y-intercepts ± 2 2. foci on y-axis; x-intercepts ± 1, y-intercepts ± 8 3. foci on x-axis; major axis of length 12; minor axis of length 8 4. foci on y-axis; major axis of length 20; minor axis of length 18 y 4 2 0 −2 −4 −4 −2 x 2 4 5. endpoints of major and minor axes: 3, 0 1 3, 0 , 1 2 2 0, 7 1 0, 7 , 2 1 , 2 6. vertices 8, 0 2 1 and 1 8, 0 2 , minor axis of length 8 In Exercises 7–12, match one of the following equations to the given graph. 2 x˛ 9 2 x˛ 4 2x˛ 2 y˛ 16 2 y˛ 25 2 y˛ 1 1 2 12 2 1 y˛ 2 x˛ 9 16 2 y˛ 2 x˛ 25 4 2 6y˛ 2 18 x˛ 1 8. y 4 2 0 −4 −2 −2 −4 x 2 4 9. y 4 2 −4 −2 0 −2 −4 4 2 −4 0 −2 −2 −4 4 2 0 −4 −2 −2 −4 y y y 10. 11. 124 −2 −2 −4 2 4 Section 11.1 Ellipses 699 Calculus can be used to show that the area of the ellipse 2 with equation y˛ 2 b˛ the area of each ellipse in Exercises 17 – 22. Pab. 2 x˛ 2 a˛ 1 is Use this fact to find 2 x˛ 16 2 y˛ 4 1 3x˛ 2 4y˛ 2 12 6x˛ 2 2y˛ 2 14 17. 19. 21. 18. 20. 22. 2 x˛ 9 2 y˛ 5 1 7x˛ 2 5y˛ 2 35 5x˛ 2 y˛ 2 5 23. Washington, D.C. has a park located next to the White House called The Ellipse. Letting the center of the ellipse be at the origin of a coordinate system, the equation that defines the boundary of the park is 2 x˛ 562,500 2 y˛ 409,600 1 Find how many square feet of grass is needed to cover the entire park. 24. The Indian mathematician Ramanujan is credited with developing the following formula that approximates the circumference of an ellipse. If 2a and 2b are the lengths of the major and minor axes of the ellipse, the circumference can be approximated by p 3a 3b 2 a 3b b 3a 2 1 1 Estimate the amount of fencing needed to enclose The Ellipse park described in Exercise 23. 2 2 1 25. Consider the ellipse whose equation is Show that if circle. a b, then the graph is actually a 2 x˛ 2 a˛ 2 y˛ 2 b˛ 1. 26. Complete the derivation of the equation of the ellipse as follows. a. By squaring both sides, show that the equation 2 x c 2 y˛ 2 2a 2 1 2 may be simplified as x c 2 y˛ 2 2 ˛ 1 In Exercises 13–16, find a complete graph of the equation. b. Show that the last equation in part a may be further simplified as a2 1 x c 2 y˛ 2 ˛ 2 a˛ 2 cx. 2 x˛ 25 2 y˛ 4 1 14. 2 x˛ 6 2 y˛ 16 1 4x˛ 2 3y˛ 2 12 16. 9x˛ 2 4y˛ 2 72 13. 15. 2 c˛ 2 a˛ 1 x˛ 2 2 a˛ 2 y˛ 2 a˛ 2 c˛ 2 a˛ 2 1 . 2 27. Sketch the graph of b 8, b 12, and 2 2 y˛ 4 b 20. 1 x˛ 2 b˛ What happens to the b 2, b 4, for 700 Chapter 11 Analytic Geometry ellipse as b takes larger and larger values? Could the graph ever degenerate into a vertical line? 28. Halley’s Comet has an elliptical orbit with the sun at one focus and a major axis of 1,636,484,848 miles. The closest the comet comes to the sun is 54,004,000 miles. What is the maximum distance from the comet to the sun? 29. The orbit of the Moon around Earth is an ellipse with Earth at one focus. If the length of the major axis of the orbit is 477,736 miles and the length of the minor axis is 477,078 miles, find the minimum and maximum distances from Earth to the Moon. 30. Critical Thinking An arched footbridge over a 100-foot river is shaped like half an ellipse. The maximum height of the bridge over the river is 20 feet. Find the height of the bridge over a point in the river exactly 25 feet from the center of the river. 31. Critical Thinking Find the length of the sides (in terms of a, b, and c) of triangle FOC in the following ellipse. Its equation is 2 x˛ a2 F is one focus. Justify your answer. 2 y˛ 2 b˛ 1 and y C O x F 11.2 Hyperbolas Objectives • Define a hyperbola • Write the equation of a hyperbola • Identify important characteristics of hyperbolas • Graph hyperbolas Like an ellipse, a hyperbola has two foci, two vertices, and a center; but its shape is quite different. Definition of a Hyperbola Let P and Q be points in the plane and k be a positive number. The hyperbola with foci P and Q is the set of all points X such that the absolute value of the difference of the distance from X to P and the distance from X to Q is k. That is, 0 XP XQ 0 k, where X represents the point (x, y) and k is called the distance difference. Section 11.2 Hyperbolas 701 As shown in Figure 11.2-1, a hyperbola consists of two separate branches (shown in red). The distances XP and XQ are shown in blue. The dotted straight lines are the asymptotes of the hyperbola. The asymptotes are not part of the hyperbola but are useful in graphing. A hyperbola approaches its asymptotes, but it never touches them. is the center of the The midpoint of the segment joining the foci, hyperbola, and the line through P and Q is called the focal axis. The points where the focal axis intercepts the hyperbola are its vertices. PQ, center X P Q vertices foci Figure 11.2-1 Equation of a Hyperbola The simplest case is a hyperbola centered at the origin with its foci on the x- or y-axis. The equation of a hyperbola is derived by using the distance formula, and it is left as an exercise. Standard Equation of a Hyperbola Centered at the Origin Let a and b be positive real numbers. Then the graph of each of the following equations is a hyperbola centered at the origin. foci on the x-axis: y˛ b˛ 1 x˛ a˛ 2 2 2 2 foci on the y-axis: 2 x˛ 2 b˛ 2 y˛ 2 a˛ 1 b −c −a −b y y = xb a x c a y = − xb a y c a b −b −a −c y = xa b x y = − xa b The characteristics of the graph of a hyperbola centered at the origin are shown in the following list. Notice in both cases that • the hyperbola bends toward the foci • the positive term determines which way the hyperbola opens • the distance between the foci is 2c • the distance between the vertices is 2a • the center is the midpoint between the foci and the midpoint between the vertices 2 a˛ c˛ 2 b˛ 2 • When the equation is in standard form with the x term positive and y term negative, the hyperbola intersects the x-axis and opens left and right. When the x term is negative and the y term is positive, the hyperbola intersects the y-axis and opens up and down. 702 Chapter 11 Analytic Geometry Characteristics of Hyperbolas For positive numbers a and b, Foci on the x-axis: y˛ 2 b˛ 1 x˛ a vertices −c −a a c x foci a x-intercepts: y-intercepts: none focal axis is on the x-axis (a, 0) (a, 0) vertices (c, 0) foci and c 2a2 b2 and (c, 0), where Foci on the y-axis: x˛ b˛ 2 y˛ 2 a˛ 1 2 2 y c a −a −c vertices y = − x a b foci x a y = x b x-intercepts: none ±± a y-intercepts: focal axis is on the y-axis (0, a) (0, a) vertices (0, c) foci and c 2a2 b2 and (0, c), where y b a x and asymptotes: y b a x y a b x and asymptotes: y a b x Graphing a Hyperbola Example 1 Graph a Hyperbola 9x˛ Show that the graph of the equation is a hyperbola. Graph it and its asymptotes. Find the equations of the asymptotes, and label the foci and the vertices. 2 4y˛ 2 36 Solution Put the equation in standard form by dividing both sides by 36 and simplifying. 2 2 2 9x˛ 36 x˛ 4 2 x˛ 22 4y˛ 36 2 y˛ 9 2 y˛ 32 1 1 1 Section 11.2 Hyperbolas 703 b 3 Applying the fact in the box with is a hyperbola centered at the origin with vertices a 2 and shows that the graph and 2, 0 2, 0 and has asymptotes y 3 2 x and y 3 2 x. 2 First plot the vertices and sketch 1 2 1 the auxiliary rectangle determined by the vertical lines and y ± b ± 3. the horizontal lines The asymptotes go through the origin and the corners of this rectangle, as shown on the left in Figure 11.2-2. It is then easy to sketch the hyperbola by drawing curves that
are asymptotic to the dashed lines. x ± a ± 2 y (0, 3) y = 3 2 x y y = − x3 2 y = x3 2 − x2 4 y2 9 = 1 (–2, 0) (2, 0) (– 13, 0) (–2, 0) (2, 0) ( 13, 0) x x (0, –3) y = − x3 2 Figure 11.2-2 Locate the foci by using the formula c 222 33 14 9 113 3.6 c 2a2 b2 with a 2 and b 3. Therefore, the foci are the right in Figure 11.2-2. A 113, 0 and A B 113, 0 B , as shown on the graph on ■ Example 2 Graph a Hyperbola on a Calculator Identify the graph of 4x˛ 2 9y˛ 2 36, and then graph it on a calculator. Solution Dividing both sides of of a hyperbola. 4x˛ 2 9y˛ 2 36 by 36 shows that it is the equation 2 x˛ 9 2 y˛ 4 1 704 Chapter 11 Analytic Geometry To graph this hyperbola on a calculator, solve the original equation for y. 6.2 −9.4 9.4 9y˛ y˛ 2 4x˛ 2 4x˛ 2 36 2 36 9 y ± 4x˛ 2 36 9 B ± 2 3 2x2 9 The hyperbola is defined by the two functions −6.2 Y1 2 3 2x˛ 2 9 and Y2 2 3 2x˛ 2 9 Figure 11.2-3 whose graphs are shown in Figure 11.2-3. ■ NOTE The two branches of the hyperbola in Figure 11.2-3 do not correspond to the two functions shown in Example 2. One function gives the part above the x-axis and the other gives the part below the x-axis. y (0, 10) (3, 2) 4 2 −4 −2 0 2 x 4 −2 −4 (0, − 10) Figure 11.2-4 Writing the Equation of a Hyperbola Example 3 Find the Equation of a Hyperbola Find the equation of the hyperbola that has vertices at and passes through the point A asymptotes, and label the foci. 0, 1 1 2 Then sketch its graph by using the 3, 12 and 0, 1 B . 1 2 Solution The vertices are on the y-axis and the equation has the form 2 y˛ 2 a˛ 2 x˛ 2 b˛ 1, with a 1. Because 3, 12 A B A 2 B 1 is on the graph, 12 32 b2 12 2 9 b2 b2 9. 1 Therefore, b 3 and the equation of the hyperbola is 2 y˛ 12 2 x˛˛ 32 1 or y˛ 2 2 x˛ 9 1 The asymptotes of the hyperbola are the lines y ± 1 3 x. The foci are on the y-axis c units away from the center, 0, 0 1 2 , where Thus, the foci are at c 212 33 11 9 110 0, 110 and , 0, 110 A B B A as shown in Figure 11.2-4. ■ Section 11.2 Hyperbolas 705 Applications of Hyperbolas Applications modeled by hyperbolas occur in science, business, and economics. Unlike Halley’s comet, which as an elliptical orbit, some comets have hyperbolic orbits. These comets pass through the solar system once and never return. Additionally, the reflective properties of hyperbolas are used in the design of camera and telescope lenses. The Hubble Space Telescope incorporates a Cassegrain telescope (invented in 1672), which has both a hyperbolic mirror and a parabolic mirror. If a light ray passes through one focus of a hyperbola and reflects off the hyperbola at a point P, then the reflected ray moves along the straight line determined by P and the other focus, as shown in Figure 11.2-5. focus focus focus P focus P Figure 11.2-5 The next example illustrates the way hyperbolas are used in location systems. Example 4 Determine Locations An explosion was heard on a passenger ship and on a naval ship that are mile apart. Passengers on Ship A heard the sound 1 2 1 2 second before sailors on Ship B. The speed of sound in air is approximately 1100 feet per second. Describe the possible locations of the explosion. Solution In 1 2 second, the sound traveled 1 2 1 1100 2 , or 550 feet. Therefore, the explo- sion occurred at a point 550 feet closer to ship A than to ship B. That is, the difference between the distance from the explosion to ship B and from the explosion to ship A is 550 feet. Whenever a difference is constant, a hyperbola is usually a good model. Draw a coordinate system and place A and B on the x-axis equidistant from the origin. The locations of the ships are the foci of the hyperbola, and the hyperbola contains all possible locations of the explosion. 706 Chapter 11 Analytic Geometry y 1000 A tB −1000 0 1000 −1000 Figure 11.2-6a of A and B are 1 Because the distance from A to B is mile, or 2640 feet, the coordinates 2 1320, 0 as shown in Figure 11.2-6a. 1 The explosion occurred on one branch of the hyperbola with foci 0 XB XA 0 and 1 hyperbola. 1320, 0 2 for every point X on the 1320, 0 such that 550 1320, 0 and , 1 1 2 2 2 V1 V1 Let to focus A. Because the difference between the distances BV1 be the vertex of the hyperbola closer is on the hyperbola, and is 550 feet. As shown in Figure 11.2-6b, AV1 0 c a 0 1 2 0 2a 0 0 BV1 2 550 550 550 a 275 Because a 7 0, c a AV1 0 1 y c − a A V1 c + a B x a a c c NOTE For every point X on the branch of the hyperbola closer to A, XB XA 550. For every point X on the branch of the hyperbola closer to B, XB XA 550. Thus, a 275, c 1320, and b2 c2 a2 13202 2752 Figure 11.2-6b 1,742,400 75,625 b2 1,666,775 Therefore, the equation of the hyperbola is x2 a2 y2 b2 x2 75,625 y2 1,666,775 1. The explosion occurred somewhere on the branch of the hyperbola closer to the passenger ship at A, as illustrated in Figure 11.2-6c. y x2 75,625 − y2 1,666,775 = 1 A B x (−1320, 0) (−275, 0) (1320, 0) (275, 0) Figure 11.2-6c To find the exact location of the explosion, the sound must be detected at a third location that is the focus of another hyperbola that shares one of the two foci given in the original problem. The intersection of the two hyperbolas will identify the precise location of the explosion, but the second hyperbola will not have its center at the origin. See Example 10 in Section 11.4, which gives the procedure for finding the exact location. ■ Section 11.2 Hyperbolas 707 9. y 8 4 x 0 −8 −4 −4 4 8 Exercises 11.2 In Exercises 1–6, find the equation of the hyperbola centered at the origin that satisfies the given conditions. 1. x-intercepts ± 3, asymptote y 2x 2. y-intercepts ± 12, asymptote y 3x 2 , passing through A B B 10. A B 213, 6 3. vertex passing through 4, 13 , 2, 0 2 1 0, 112 A 3, 0 4. vertex 5. focus 6. focus 1 1 and vertex 2 2, 0 2 1 0, 4 2 and vertex 0, 112 A B In Exercises 7–12, match one of the following equations to the given graph. 2 y˛ 16 2 x˛ 9 1 8x2 y 2 8 16 y˛ 2 2 25x˛ 9 1 11. 12. 2 x˛ 9 2 y˛ 4 1 8x˛ 2 y˛ 2 8 2 x˛ 9 2 y˛ 4 1 8 4 0 −8 −4 −4 −8 8 4 y y 7. 8. 4 8 x t 4 8 −8 0 −4 −4 −8 −8 2 1 −2 0 −1 −1 −2 4 2 y y −4 −2 0 −2 −4 y 4 2 −4 −2 0 −2 − In Exercises 13–18, sketch a complete graph of the equation. Label the foci and the asymptote equations. 2 x˛ 6 2 y˛ 16 1 4x˛ 2 y˛ 2 16 14. 16. x2 4 y 2 1 3y˛ 2 5x˛ 2 15 18y˛ 2 8x˛ 2 2 0 18. x˛ 2 2y˛ 2 1 13. 15. 17. 708 Chapter 11 Analytic Geometry In Exercises 19–24, graph each equation using a graphing calculator. 30. y2 8 x2 36 1 at 6, 4 1 2 19. 4x2 y 2 16 21. 2 2y 2 1 x 20. 22. x2 4 x2 6 y2 1 y2 16 1 23. 3y 2 5x 2 15 24. 18y 2 8x 2 2 0 y2 25. Sketch the graph of 4 b 20. b 8, b 12, hyperbola as b takes larger and larger values? Could the graph ever degenerate into a pair of horizontal lines? x2 b2 What happens to the 1 and for b 2, b 4, 26. Sketch the graph of y2 a2 a 0.5. 1 x2 16 What happens to the a 8, a 4, for and a 2, a 1, hyperbola as a takes smaller and smaller values? Could the graph ever degenerate into a pair of horizontal lines? 27. April and Marty, 2 miles apart, are talking on the phone when lightning strikes nearby. They each hear the thunder, but April hears it 2.4 seconds after Marty. Sketch a graph of the locations where the lightning could have struck. [Sound travels at approximately 1100 feet per second.] 28. In Exercise 27, suppose that later in the conversation Marty hears the thunder 3 seconds after April. Sketch a graph of the locations where the lightning could have struck. For Exercises 29–30, write the equation of the tangent line to the given curve at the given point by using the following facts. The slope m of the tangent line to a hyperbola at the point (x, y) is m b2x a2 y m a2x b2 y for for x2 a2 y2 a2 y2 b2 x2 b2 1 1 29. x2 8 y2 4 1 at 4, 2 1 2 31. Show that the difference between the distance from each focus to any point on a hyperbola is equal to the distance between the vertices. 32. Derive the equation of a hyperbola centered at the be a point on the hyperbola with origin as follows. a. Let x, y P 2 1 c, 0 F11 foci F1 P 7 F2 P. the distance formula, and Exercise 31, and 2 By the definition of a hyperbola, Assume that F21 c, 0 . 2 F1 P F2 2a 2 Show that the last equation simplifies as shown. cx a2 a2 x c 2 1 2 y 2 b. Show that the last equation in part a may be further simplified as shown. c2 a2 2 x 1 a2 y2 a2 c2 a2 2 1 and show that the equation in 2 c. Let b2 c2 a2 b simplifies to the standard form of a hyperbola centered at the origin. 33. Critical Thinking The following hyperbola is centered at the origin with vertex V and the auxiliary rectangle as shown. Use the length of one of the sides of triangle POV to locate the foci. Justify your answer. y P x O V Section 11.3 Parabolas 709 11.3 Parabolas Objectives • Define a parabola • Write the equation of a parabola • Identify important characteristics of parabolas • Graph parabolas Parabolas appeared in Section 3.3 as the graphs of quadratic functions, which are a special case of the following more general definition. Definition of a Parabola Let L be a line in the plane and P be a point not on L. If X is any point not on L, the distance from X to L is defined to be the length of the perpendicular line segment from X to L. The parabola with focus P and directrix L is the set of all points X such that distance from X to P distance from X to L as shown in Figure 11.3-1. L X a xis P focus vertex directrix Figure 11.3-1 The line through P perpendicular to L is called the axis. The intersection of the axis with the parabola, which is the midpoint of the segment of the axis from P to L, is the vertex of the parabola, as shown in Figure 11.3-1. Equation of a Parabola Suppose that the focus is on the y-axis at the point nonzero constant, and that the directrix is the horizontal line x, y 2 1 horizontal line x, p , where p is a y p. If x, y to the x, y is the length of the vertical segment from to is any point on the parabola, then the distance from as shown in Figure 11.3-2. y p 0 By the definition of a parabola, y (x, y) (0, p) (0, −p) x (x, −p) Figur
e 11.3-2 2 1 distance from distance from x, y 1 x, y 1 x 0 to to 2 2 2 0, p 2 1 0, p 2 1 y p 2 1 2 distance from 1 distance from 1 2 2 2 x x x, y x, y to y p x, p to 710 Chapter 11 Analytic Geometry Square both sides of the equation and simplify. x 0 1 2 y˛ x 2py p2 02 y˛ x˛ 2 4py 1 y p 2 2 2 1 2 2 2py p2 standard form of a parabola Conversely, it can be shown that every point whose coordinates satisfy this equation is on the parabola. A similar argument works for the on the x-axis and directrix the vertical line parabola with focus x p, and leads to the following conclusion. p, 0 1 2 Standard Equation of a Parabola with Vertex at the Origin Let p be a nonzero real number. Then the graph of each of the following equations is a parabola with vertex at the origin. focus at directrix: (0, p) y p focus at directrix: (p, 0) x p x2 4py y2 4px The equations of a parabola in standard form can also be y 1 2 4 p x˛ NOTE written as and x 1 4 p y˛ 2 . One of the variables is quadratic and the other variable is linear. When the y term is linear, the parabola opens up or down; when the x term is linear, the parabola opens left or right. The characteristics of the graph of a parabola with vertex at the origin are shown in the following box. Notice the following facts in both cases. • the parabola bends toward the focus and away from the directrix • the linear term determines the orientation of the parabola and the axis of symmetry—left/right or up/down • the sign of p determines which way the parabola opens • the distance between the focus and the directrix is 2p • the distance from the vertex to the focus and the distance from the vertex to the directrix is p • the vertex is the midpoint of the line segment joining the focus and the directrix Characteristics of Parabolas Section 11.3 Parabolas 711 For a nonzero real number p, parabolas have the following characteristics. x2 4py y2 4px • focus on the y axis at (0, p) • focus on the x-axis at (p, 0) • directrix y p • directrix x p • axis of symmetry is the • axis of symmetry is the y-axis x-axis • If p 77 0 , then the parabola: opens up opens right y focus vertex x y directrix focus x vertex directrix • If p 66 0 , then the parabola: opens down opens left y directrix vertex x focus y directrix focus vertex x Example 1 Graphing a Parabola Show that the graph of the equation graph, and then find and label its focus and directrix. 2 8y 0 x˛ is a parabola. Draw its Solution Write the equation in standard form: form 2 8y. This equation is of the so the graph is a parabola. To find the value of p, note that 2 4py, x˛ x˛ 4p 8 p 2 712 Chapter 11 Analytic Geometry y 4 y = 2 x 8 4 F(0, −2) −8 −4 0 −4 −8 Therefore, the following statements are true, as shown in Figure 11.3-3. • the focus is 0, p 1 0, 2 2 1 2 y p • the directrix is • the parabola opens downward because p is negative 1 2 2 2 ■ Example 2 Graph a Parabola on a Calculator Figure 11.3-3 On a calculator, graph the parabola y2 12x. 10 1 Solution Solve the equation for y and enter both functions into a calculator. y ± 112x 10 The graphs of Y1 112x and Y2 112x are shown in Figure 11.3-4. ■ 10 Figure 11.3-4 Writing the Equation of a Parabola Example 3 Find the Equation of a Parabola y 8 4 x = 3 F(−3, 0) −8 −4 0 4 x 8 −4 −8 Figure 11.3-5 Find the focus, directrix, and equation of the parabola that passes through and has its focus on the x-axis. 0, 0 the point 1 Sketch the graph of the parabola, and label its focus and directrix. has vertex 1, 112 A B , , 2 Solution Because the focus is on the x-axis, the equation has the form Since is on the graph, 1, 112 A B 112 A 1 2 2 4p B 1 12 4p 3 p 2 4˛px. y˛ Therefore, the focus is The equation of the parabola is Figure 11.3-5. 2 1 3, 0 and the directrix is the vertical line x 3. whose graph is shown in 2 12x, y˛ ■ Applications of Parabolas Projectiles follow a parabolic curve, a fact that is used in the design of water slides in which the rider slides down a sharp incline, then up and over a hill, before plunging downward into a pool. At the peak of the hill, the rider shoots up along a parabolic arc several inches above the slide, experiencing a sensation of weightlessness. Section 11.3 Parabolas 713 Certain laws of physics show that sound waves or light rays from a source at the focus of a parabola will reflect off the parabola in rays parallel to the axis of the parabola, as shown in Figure 11.3-6. This is the reason that parabolas are used in automobile headlights and searchlights. focus axis axis focus Figure 11.3-6 Figure 11.3-7 Conversely, a sound wave or light ray coming toward a parabola will be reflected into the focus, as shown in Figure 11.3-7. This fact is used in the design of radar antennas, satellite dishes, and field microphones used at outdoor sporting events to pick up conversations on the field. Example 4 Parabola Application A radio telescope in the Very Large Array at Socorro, New Mexico, shown in Figure 11.3-8a, has the shape of a parabolic dish (a cross section through the center of the dish is a parabola). It is approximately 12 feet deep at the center and has a diameter of 82 feet. How far from the vertex of the parabolic dish should the receiver be placed in order to “catch” all the radio waves that hit the dish? Figure 11.3-8a 714 Chapter 11 Analytic Geometry y 40 20 −40 0 −20 Figure 11.3-8b 20 (41, 12) x 40 All radio waves hitting the dish are reflected into the focus, so the receiver should be located there. To find the focus, draw a cross section of the dish, with the vertex at the origin, as shown in Figure 11.3-8b. The equation of Because the point (41, 12) is on the this parabola is of the form parabola, x2 4py. 12 x2 4py 412 4p 1 412 48p p 412 48 p 1681 48 2 35 feet Substitute Simplify Divide both sides by 48 The focus is the point 0, 0 2 fore, the receiver should be placed about 35 feet from the vertex. which is p units from the vertex 0, p , 1 1 2 . There- ■ Exercises 11.3 In Exercises 1–6, find the equation of the parabola with vertex at the origin that satisfies the given condition. 7. 1. axis x 0, passing through 2. axis y 0, passing through 2, 12 2, 12 2 2 1 1 3. focus 4. focus 1 1 5, 0 2 0, 3.5 5. directrix 2 x 2 6. directrix y 3 In Exercises 7–10, match one of the following equations to the given graph. 8. 6x y˛ 2 2 4x y˛ y x2 4 2 8y x˛ y 8 4 −8 −4 0 4 −4 −8 8 4 y −8 −4 0 4 −4 −8 x 8 x 8 5910ac11_690-775 9/21/05 2:00 PM Page 715 9. 10. y 8 4 −8 −4 0 4 −4 −8 8 4 y −8 −4 0 4 −4 −8 x 8 x 8 In Exercises 11–14, sketch a complete graph of the equation. 11. x 6y˛ 2 13. 8x 2y˛ 2 12. 1 2 y2 2x 14. 4y x˛ 2 In Exercises 15–18, find the focus and directrix of the parabola. 15. y 3x2 17. y 1 4 x2 16. x 1 2 y2 18. x 6y˛ 2 Section 11.3 Parabolas 715 In Exercises 19–22, find the equation of the parabola with vertex at the origin passing through the given points. 2 2 8x y˛ 19. 21. 1 1 1, 2 and 1, 2 1 2 2 20. 3,3 2 b a and 3 2 b 3, a 1, 2 and 1 2 1, 2 2 22. 1 2, 5 and 2, 5 2 1 23. Find the point on the graph of that is closest to the focus of the parabola. 24. Find the point on the graph of 2 3y x˛ that is closest to the focus of the parabola. 25. The receiver in a parabolic television dish is 2 feet from the vertex and is located at the focus. Find an equation of a cross section of the receiver. (Assume that the dish is directed to the right and that the vertex is at the origin.) 26. The receiver in a parabolic television dish is 1.5 feet from the vertex and is located at the focus. Find an equation of a cross section of the receiver. (Assume that the dish is directed upward and that the vertex is at the origin.) 27. The filament of a flashlight bulb is located at the focus, which is 1 2 inch from the vertex of a parabolic reflector. Find an equation of a cross section of the reflector. (Assume that the flashlight is directed to the left and that the vertex is at the origin.) 28. The filament of a flashlight bulb is located at the focus, which is 1 4 inch from the vertex of a parabolic reflector. Find an equation of a cross section of the reflector. (Assume that the flashlight is directed downward and that the vertex is at the origin.) 716 Chapter 11 Analytic Geometry 11.4 Translations and Rotations of Conics Objectives • Write the equation of a translated conic • Graph a translated conic • Determine the shape of a translated conic without graphing • Apply translated conics to real-world problems Horizontal and Vertical Shifts Now that you are familiar with conic sections centered at the origin, the discussion will be expanded to include both conics centered at other points in the plane and ones with axes that may not be parallel to the coordinate axes. 2 1 x As you saw in Section 3.4, replacing a variable x with y f a function x 5, whereas replacing x with to the left. Similarly, if the rule of a function is given by replacing y with y 4 f is equivalent to tions, a similar result applies. in the rule of shifts the graph of the function 5 units to the right, 5 shifts the graph 5 units y f then shifts the graph 4 units upward, because For equations that are not func- that is, y 4 y f 4 and y with in an equation shifts the graph of the equation: • Let h and k be constant. Replacing x with y k 00 h 00 negative h 00 k 00 negative k • units upward for positive k and downward for units to the right for positive h and to the left for The process of writing the equation of a conic is the same as that discussed in Sections 11.1 through 11.3, except that x and y are replaced with x h y k, where (h, k) is the vertex of a parabola or the center of an ellipse or a hyperbola. and Example 1 Graph a Translated Conic Identify and sketch the graph of 36 2 1 and find its center, major axis, and minor axis. Solution The given equation can be obtained from the equation x2 9 y 2 36 1, whose graph is known to be an ellipse, as follows: y and replace y by replace x by x 5 1 This is the situation described in the previous box with 4 y 4. 2 h 5 and k 4. Therefore, the graph is the ellipse
x2 9 y 2 36 1 shifted 5 units to the right Section 11.4 Translations and Rotations of Conics 717 and 4 units downward, as shown in Figure 11.4-1. • The center of the ellipse is 5, 4 . 2 1 • The major axis lies on the vertical line • The minor axis is on the horizontal line y 4. x 5. y 8 4 −8 −4 0 4 8 x −4 −8 Figure 11.4-1 ■ Before identifying a conic section and determining its characteristics, the corresponding equation should be rewritten in standard form. Example 2 Identify a Conic Identify and sketch the graph of 4x 2 9y 2 32x 90y 253 0. Solution Rewrite the equation as 253 253 1 4x 4 1 2 32x x2 8x 1 9 2 90y 9y y2 10y 2 2 1 2 10y. y and 2 10y 25 y 2 2 2 8x 9 2 1 2 253 ? ? x Complete the square in 2 8x 16 x 4 1 4 16 64 Be careful here: 16 and 25 were not added to the left side of the equation. Actually were added, when the left side is multiplied out. Therefore, to leave the original equation unchanged, 64 and 225 must be added to the right side. 9 25 225 and x 4 1 2 8x 16 2 4 4 1 253 64 225 2 2 36 36 x 4 9 2 10y 25 36 NOTE Review the technique of completing the square in Section 2.2, if needed. 718 Chapter 11 Analytic Geometry The graph of this equation is the ellipse 2 x˛ 9 2 y˛ 4 1 shifted 4 units to the right and 5 units upward. Its center is at (4, 5), its major axis lies on the as horizontal line shown in Figure 11.4-2. and its minor axis lies on the vertical line x 4, y 5, y 8 4 −4 0 4 8 x −4 Figure 11.4-2 ■ Example 3 Writing the Equation of a Translated Conic Find the equation of an ellipse with center at points of its major and minor axes are Find the coordinates of the foci. 5, 2 , 1 2 1 1 5, 10 5, 4 2 , 1 2 such that the end2, 4 8, 4 and . , 1 2 2 Solution The major axis has length 12 and is parallel to the y-axis. The minor axis has length 6 and is parallel to the x-axis. Therefore, and the equation of the ellipse has the form a 6, b 3, 2 1 x h 32 2 1 2 y k 62 2 1 Because it has its center at form 1 5, 4 , 2 the equation of the ellipse has the y 8 4 1 −8 −4 0 4 8 x Since a 6 and 2 2 x 5 32 22 1 4 1 62 y 4 36 c 2a2 b2, 1. 2 2 and in an ellipse, b 3, c 262 32 136 9 127 313 −4 −8 The foci of 36 2 1 Figure 11.4-3 on the major axis. That is, the foci are as shown in Figure 11.4-3. A are 313 5, 4 313 units from the center 5, 4 5, 4 313 A 1 2 , and B B ■ Section 11.4 Translations and Rotations of Conics 719 Example 4 Identify a Translated Conic Identify and sketch the graph of istics of the conic. x 2y˛ 2 12y 14. Label all character- Solution Rewrite the equation and complete the square in y, being careful to add the appropriate amounts to both sides of the equation. 2 1 1 2 2 12y x 14 2y˛ x 14 2 6y y˛ 2 2 x 14 2 2 6y 9 y˛ 2 2 x 4 y 22 2 1 1 2 1 1 1 2 22 9 1 2 Thus, the graph is the graph of the parabola y˛ 2 1 2 x shifted 4 units to the left and 3 units downward, as shown in Figure 11.4-4. y 2 directrix focus directrix x 2 axis −4 −2 0 focus −2 −4 −6 Figure 11.4-4 2 1 2 x y˛ has its vertex at 0, 0 2 1 , the x-axis as its axis of sym- The parabola metry, 1 8 a , 0 b as its focus, and x 1 8 shifted, the parabola will have its vertex at line as its axis. y 3 as its directrix. After the graph is 4, 3 1 2 and the horizontal The translated parabola has its focus at 1 8 a 4, 3 , or b 31 8 a , 3 b and directrix x 1 8 4 ˛ 33 8 . ■ 720 Chapter 11 Analytic Geometry Standard Equations of Conic Sections The following is a summary of the standard equations of conic sections whose axes are parallel to the coordinate axes. Let (h, k) be any point in the plane. • If a and b are real numbers with then the graph of each of the following equations is an ellipse with center (h, k). a 77 b 77 0, (x h)2 a2 (y k)2 b2 1 (x h)2 b2 (y k)2 a2 1 ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ major axis on the horizontal line minor axis on the vertical line vertices: foci: (h c, k), (h a, k) (h c, k) and c 2a2 b2 where y k x h major axis on the vertical line minor axis on the horizontal line vertices: foci: (h, k c), (h, k a) (h, k c) and c 2a2 b2 where x h y k • If a and b are positive real numbers, then the graph of each of the following equations is a hyperbola with center (h, k). (x h)2 a2 (y k)2 b2 1 (y k)2 a2 (x h)2 b2 focal axis on the horizontal line (h a, k) vertices: foci: and (h c, k), (h a, k) (h c, k) and c 2a2 b2 asymptotes: y ±±± b a (x h) k where focal axis on the vertical line (h, k a) vertices: (h, k c) foci: and c 2a2 b2 asymptotes: y ±±± and (h, k c), (x h) k (h, k a) x h where a b • If p is a nonzero real number, then the graph of each of the following equations is a parabola with vertex (h, k). (x h)2 4p(y k) (y k)2 4p(x hh, k p) x h focus: directrix: the horizontal line axis: the vertical line p 77 0, opens upward if (h p, k) focus: directrix: the vertical line axis: the horizontal line p 77 0, opens to right if y k p downward if p 66 0 x h p y k to left if p 66 0 When the equation of a conic section is in standard form, the techniques in previous sections can be used to obtain its graph on a calculator. Technology Tip Casio has a conic section grapher, on the main menu, that produces the graphs of equations in standard form when the various coefficients are entered. Section 11.4 Translations and Rotations of Conics 721 Example 5 Calculator Graph of a Conic Graph the equation . Solution The graph is a hyperbola centered at solve the equation for y. 1 3, 1 . 2 To graph on a calculator 13 Y1 or Y2 10 Figure 11.4-5 Graph the last two functions on the same screen. The graph of the first is the top half and the graph of the second is the bottom half of the graph of the original equation, as shown in Figure 11.4-5. ■ When a second-degree equation is not in standard form, the fastest way to graph it is to use the method in Example 5, modified as in the next example. Example 6 Graph a Conic Not in Standard Form 2 8y˛ 2 6x 9y 4 0 x˛ without putting it in Graph the equation standard form. Solution Write the equation as This is a quadratic equation of the form ay˛ with 8y˛ 2 9y 1 2 6x 4 x˛ 2 0. 2 by c 0, 2 6x 4, a 8, b 9, and c x˛ which can be solved by using the quadratic formula. y b ± 2b2 4ac 2a 9˛ ± 292 4 8 2 8 9 ± 281 32 16 1 2 6x 4 x˛ 2 1 x˛ 2 6x 4 2 722 Chapter 11 Analytic Geometry 2.3 The graphs of both of the functions 8 2 4.3 Figure 11.4-6 Y1 Y2 9 281 32 16 9 281 32 16 1 1 2 6x 4 x˛ 2 6x 4 x˛ and 2 2 are shown on the same screen in Figure 11.4-6. The conic is an ellipse. ■ Rotations and Second-Degree Equations A second-degree equation in x and y is one that can be written in the form Ax˛ 2 Bxy Cy˛ 2 Dx Ey F 0 [1] for some constants A, B, C, D, E, and F, with at least one of A, B, or C nonzero. Every conic section is the graph of a second-degree equation. The terms Dx, Ey, and F determine the translation of the conic from the the term Bxy determines a rotation of the conic so origin. When that its axes are no longer parallel to the coordinate axes. For instance, 1 the ellipse equation can be written as B 0, 1 2 2 2 x˛ 4 y 3 6 12 2 2 1 2 a 12 12 x˛ 4 b y 3 6 T S 2 12 y 3 2 2 3x˛ 2 1 12 2 6y 9 2 2 y˛ 3x˛ 1 2 12y 18 12 2 2y˛ 3x˛ 2 12y 6 0 2 2y˛ 3x˛ 2 last equation above has The A 3, B 0, C 2, D 0, E 12, the and form of equation F 6. [1] with Conversely, it can be shown that the graph of every second-degree equation is a conic section (possibly degenerate—see page 691). When the equation has an xy term, the conic may be rotated from standard position such that its axis or axes are not parallel to the coordinate axes. Example 7 Identify a Conic Graph the equation conic. 3x˛ 2 6xy y˛ 2 x 2y 7 0 and identify the Solution Rewrite the equation as y˛ 2 2 6xy 2y 3x˛ 3x˛ 6x 2 y y by c 0, 1 2 a 1, b 6x 2, It can be solved for y by using the quadratic with 1 2 ay˛ The last equation has the form 2 x 7. and formula. c 3x˛ Section 11.4 Translations and Rotations of Conics 723 y y b ± 2b 2a 6x 2 1 2 4ac – 2 1 6x 2 2 4 1 2 2 1 3x 2 x 7 1 2 2 8 Half of the graph is obtained by graphing 8 8 6x 2 1 2 2 Y1 and the other half by graphing 2 6x 2 Y2 1 2 6x 2 2 4 1 2 2 6x 3x 2 x 7 3x 2 x 7 2 2 10 Figure 11.4-7 The graph is a hyperbola whose focal axis is tilted. ■ The Discriminant The following fact makes it easy to identify the graphs of second-degree equations without graphing. Graphs of Second-Degree Equations The graph of the equation Ax2 Bxy Cy2 Dx Ey F 0, • is a circle, an ellipse, or a point, if with A, B, C not all zero, B2 4AC 66 0 • is a parabola, a line, or two parallel lines, if B2 4AC 0 • is a hyperbola or two intersecting lines, if B2 4AC 77 0 The expression B2 4AC is called the discriminant. Example 8 Identify a Conic Identify the graph of your conclusions by graphing. 2x 2 4xy 3y 2 5x 6y 8 0 and confirm 5 Solution 20 5 15 Figure 11.4-8 Compute the discriminant with 2 4AC B 4 2 1 A 2, B 4, C 3. 2 4 2 3 16 24 8 and Hence, the graph is an ellipse, a circle, or a single point. Use the quadratic formula to solve for y. 2 3y 4x 6 y 2x 2 5x 8 0 2 Graph both solutions on the same screen, as shown in Figure 11.4–8. 1 2 1 1 y 4x 6 ± 2 1 2 4x 6 2 4 3 2 2 3 2x 2 5x 8 1 2 ■ 724 Chapter 11 Analytic Geometry Example 9 Use the Discriminant 10 Identify the graph of plete graph. Solution 3x˛ 2 5xy 2y˛ 2 8y 1 0 , and sketch a com- 10 10 10 Figure 11.4-9a 300 300 300 300 Figure 11.4-9b y d2 S d1 d3 B˛ So the The discriminant of the equation is graph is a hyperbola—or two intersecting lines in the degenerate case. 2 4 3 2 1. 2 4AC 5˛ To graph the equation, write the equation in quadratic form in y. 3x˛ 2y˛ 2 2 5xy 2y˛ 2 5xy 8y 3x˛ 3x˛ 2 8y 1 0 2 1 0 2 1 0 5x 8 y 2y˛ 1 2 1 2 Then use the quadratic formula to solve for y. 5x 8 1 2 ± 2 y 5x 8 4 1 2 2 4 2 3x˛ 2 1 1 2 Graphing these two functions in the standard window produces Figure 11.4-9a, which looks like a parabola. This cannot be correct: because the discriminant is positive, the graph must be a hyperbola. A different viewing window is needed for a complete graph of this hyperbola, which is shown in Figure 11.4-9b. ■ Applications The long-range navigation system (LORAN) uses hyperbolas to enable a ship to d
etermine its exact location by radio, as illustrated in the following example. Example 10 LORAN Application Three LORAN radio transmitters Q, P, and R are located 200 miles apart along a straight line and simultaneously transmit signals at regular intervals. These signals travel at a speed of 980 feet per microsecond, the speed of light. Ship S receives a signal from P and, 528 microseconds later, a signal from Q. It also receives a signal from R 305 microseconds after the one from P. Determine the ship’s location. Solution Let the x-axis be the line through the LORAN stations, with the origin located midway between Q and P, so that the situation looks like Figure 11.4-10. If the signal takes t microseconds to go from P to S, then d1 980t and d2 980 t 528 1 2 Q −100 P 100 R x 300 so that Figure 11.4-10 d1 0 d2 0 0 980t 980 t 528 1 2 0 980 528 517,440 feet. Section 11.4 Translations and Rotations of Conics 725 Since one mile is 5280 feet, ƒ d1 d2 ƒ 517,440 5280 98 miles. In other words, distance from P to S 0 1 1 2 distance from Q to S 98 miles. 2 0 This is the definition of a hyperbola given in Section 11.2; thus, S is on the hyperbola with foci and distance difr 98. ference This hyperbola has an equation of the form 2 100, 0 100, 0 Q P and 1 2 2 1 2 x˛ 2 a˛ y˛ 2 b˛ 1, where 2 a˛ c˛ ± a, 0 1 2 b˛ 2. 2 are the vertices, –c, 0 2 1 1 –100, 0 2 are the foci and Figure 11.4-11 and the fact that the vertex a, 0 is on the hyperbola show that distance from P to 0 3 a, 0 1 2 4 3 1 2 distance Q to 100 a 2 0 1 1 1 r 98 98 98 49 a, 0 2 4 0 100 a 2 0 2a 0 a 0 0 2 49˛ 2 100˛ 0 2 7599. Consequently, Thus, the ship lies on the hyperbola a˛ 2 2401 2 49˛ and 2 c˛ 2 a˛ b˛ 2 x˛ 2401 2 y˛ 7599 1. [2] A similar argument using P and R as foci shows that the ship also lies on and center (200, 0), the hyperbola with foci 1 whose distance difference r is 300, 0 100, 0 R P and 2 1 2 d3 0 As before, you can verify that 0 d1 980 305 298,900 feet 56.61 miles. a 56.61 2 28.305 and a˛ 2 28.305˛ 200 c, k 1 801.17. 100, 0 This hyperbola has center (200, 0) and its foci are and , 300, 0 2 100˛ The ship also lies on the hyperbola which implies that 2 801.17 9198.83 200 c, k 1 2 2 a˛ b2 c˛ 2 1 2 1 c 100. 2 x 200 801.17 1 2 2 y˛ 9198.83 1. 2 2 [3] 2, y˛ Since the ship lies on both hyperbolas, its coordinates are solutions of both equations [2] and [3]. They can be found algebraically by solving each of setting the results equal, and solving for x. They can the equations for be found geometrically by graphing both hyperbolas and finding the points of intersection. Since the signal from P was received first, the ship is closer to P. So it is located at the point S in Figure 11.4–12 or at the intersection point directly below it. A graphical intersection finder shows that point S is at approximately (130.48, 215.14), where the coordinates are in miles from the origin. ■ y Q −100 (a, 0) a P x 100 100 + a 100 − a Figure 11.4-11 900 S 500 Q P R 500 900 Figure 11.4-12 726 Chapter 11 Analytic Geometry Exercises 11.4 In Exercises 1–16, find the equation of the conic sections satisfying the given conditions. 1. ellipse with center (2, 3); endpoints of major and 0, 3 minor axes: 2, 1 2, 7 4, 3 , 1 2 2 1 minor axes: 2. ellipse with center , 1 3. ellipse with center 0, 2 1 2 2 2 endpoints of major and 10, 2 , 1 , 1 foci on the line 2 x 7; 5, 13 2 7, 4 major axis of length 12; minor axis of length , 2 1 5, 17 4. ellipse with center 2 major axis of length 15; minor axis of 1 ; foci on the line 3, 9 y 9; length 7 5. hyperbola with center passing through A 2, 3 2 3110, 11 2 1 ; vertex 6. hyperbola with center passing through A 5, 1 ; 2 1 1, 1 413 2, 1 3, 1 ; 2 ; 2 1 1 B vertex B 7. hyperbola with center (4, 2); vertex (7, 2); asymptote 3y 4x 10 8. hyperbola with center asymptote 1 6y 5x 15 3, 5 ; vertex 2 3, 0 ; 2 1 9. parabola with vertex (1, 0); axis x 1; passing through (2, 13) 10. parabola with vertex through 1 1, 1 2 3, 0 1 2 ; axis y 0; passing 11. parabola with vertex (2, 1); axis y 1; passing through (5, 0) 1, 3 1 2 ; axis y 3; passing 31. 1 3, 2 1 2 ; passing through 3, 6 2 1 12. parabola with vertex through 1 1, 4 2 13. ellipse with center 9, 2 and 1 2 2 3, 5 and 1 14. ellipse with center (2, 5); passing through (2, 4) 15. parabola with vertex 47 16 a , 2 b 16. parabola with vertex 5, 99 20b a 3, 2 5, 5 2 2 1 1 and focus and focus In Exercises 17–22, assume that the graph of the equation is a nondegenerate conic section. Without graphing, determine whether the graph is a circle, ellipse, hyperbola, or parabola. 17. 2 2xy 3y˛ x˛ 2 1 0 18. xy 1 0 19. 2 2xy y˛ x˛ 2 212x 212y 0 20. 2x˛ 2 4xy 5y˛ 2 6 0 21. 17x˛ 2 48xy 31y˛ 2 50 0 22. 2x˛ 2 4xy 2y˛ 2 3x 5y 10 0 In Exercises 23–34, sketch a complete graph of each conic section. 23. 24. 25. 26 16 2 1 2 y 3 12 2 1 2 1 x 1 16 12 2 1 27. y 4 x 1 1 29 28. 30 25 2 2 1 2 x 1 16 2 1 32. 33. 34 25 In Exercises 35–52, use the discriminant to identify the conic section whose equation is given, and find a viewing window that shows a complete graph. 35. 9x2 4y2 54x 8y 49 0 36. 4x˛ 2 5y˛ 2 8x 30y 29 0 Section 11.4 Translations and Rotations of Conics 727 37. 4y˛ 2 x˛ 2 6x 24y 11 0 38. 2 16y˛ x˛ 2 0 39. 3y˛ 2 x 2y 1 0 40. 2 6x y 5 0 x˛ 41. 41x˛ 2 24xy 34y˛ 2 25 0 42. 2 213xy 3y˛ x˛ 2 813x 8y 32 0 43. 17x˛ 2 48xy 31y˛ 2 49 0 44. 52x˛ 2 72xy 73y˛ 2 200 45. 9x˛ 2 24xy 16y˛ 2 90x 130y 0 46. 2 10xy y˛ x˛ 2 1 0 47. 23x˛ 2 2613xy 3y˛ 2 16x 1613y 128 0 48. 2 2xy y˛ x˛ 2 1212x 1212y 0 49. 17x˛ 2 12xy 8y˛ 2 80 0 50. 11x˛ 2 24xy 4y˛ 2 30x 40y 45 0 51. 52. 3x˛ 2 213xy y˛ 2 4x 413y 16 0 3x˛ 2 212xy 2y˛ 2 12 0 In Exercises 53 and 54, find the equations of two distinct ellipses satisfying the given conditions. 53. Center at 5, 3 axis of length 8. 1 54. Center at 2, 6 axis of length 6. 1 ; major axis of length 14; minor ; major axis of length 15; minor 2 2 55. Critical Thinking Show that the asymptotes of the 2 x˛ 2 a˛ 2 y˛ 2 a˛ 1 hyperbola other. are perpendicular to each 56. Find a number k such that of 3x˛ 2 ky˛ 2 4. 2, 1 Then graph the equation. 2 1 is on the graph 57. Find the number b such that the vertex of the 2 bx c lies on the y-axis. parabola y x˛ 58. Find the number d such that the parabola . passes through 2 dx 4 y 1 6, 3 1 2 1 2 59. Find the points of intersection of the parabola 4y˛ 2 4y 5x 12 and the line x 9. 60. Find the points of intersection of the parabola 4x˛ 2 8x 2y 5 and the line y 15. In Exercises 61–64, write the resulting equation in standard form. 61. Translate the hyperbola defined by the equation 3y2 20x 23 5x2 12y right 5 units. up 3 units and to the 62. Translate the hyperbola defined by the equation 16x2 9y2 64x 89 18y the right 3 units. down 2 units and to 63. Translate the hyperbola defined by the equation up 1 unit and to the 4x2 9y2 8x 54y 113 left 4 units. 64. Translate the hyperbola defined by the equation down 5 units and to 7x2 5y2 48 20y 14x the left 4 units. 65. Suppose a golf ball driven off the tee travels 210 yards down the fairway. During flight it reaches a maximum height of 55 yards. Find an equation that describes the ball’s parabolic path if the tee is at the origin and the positive x-axis is along the ground in the direction of the drive. 66. Suppose a golf ball driven off the tee travels 175 yards down the fairway. During flight it reaches a maximum height of 40 yards. Find an equation that describes the ball’s parabolic path if the tee is at the origin and the positive x-axis is along the ground in the direction of the drive. 67. Two listening stations 1 mile apart record an explosion. One microphone receives the sound 2 seconds after the other does. Use the line through the microphones as the x-axis, with the origin midway between the microphones, and the fact that sound travels at 1100 feet/second to find the equation of the hyperbola on which the explosion is located. Can you determine the exact location of the explosion? 68. Two transmission stations P and Q are located 200 miles apart on a straight shoreline. A ship 50 miles from shore is moving parallel to the shoreline. A signal from Q reaches the ship 400 microseconds after a signal from P. If the signals travel at 980 feet per microsecond, find the location of the ship (in terms of miles) in the coordinate system with x-axis through P and Q, and origin midway between them. 728 Chapter 11 Analytic Geometry 11.4.A Excursion: Rotation of Axes Objectives The graph of an equation of the form • Write the equation of a rotated conic section in terms of u and v • Determine the angle of rotation of a rotated conic section y u Ax˛ 2 Bxy Cy˛ 2 Dx Ey F 0, B 0, is a conic section that is rotated so that its axes are not parwith allel to the coordinate axes, as in Figure 11.4.A-1. Although the graph is readily obtained with a calculator, as in Examples 7–9 of Section 11.4, useful information about the center, vertices, etc., cannot be read directly from the equation, as it can be with an equation in standard form. However, if the xy coordinate system is replaced by a new coordinate system, as indicated by the blue uv axes in Figure 11.4.A-1, then the conic is not rotated in the new system and has a uv equation in standard form that will provide the desired information. v Rotation Equations x In order to use this approach, first determine the relationship between the xy coordinates of a point and its coordinates in the uv system. Suppose the uv coordinate system is obtained by rotating the xy axes about the oriIf a point P has coordinates gin, counterclockwise through an angle (x, y) in the xy system, its coordinates (u, v) can be found in the rotated coordinate system by using Figure 11.4.A-2. u. Figure 11.4.A-1 y y v P (x, y) (u, v Figure 11.4.A-2 Section 11.4.A Excursion: Rotation of Axes 729 Triangle OPQ shows that cos b OQ OP Therefore, u r and sin b PQ OP v r u r cos b and v r sin b Similarly, triangle OPR shows that u b cos 1 OR OP 2 x r and sin u b 1 PR OP 2 y r so that x r cos u b and y r sin 1 Applying the addition identity for cosin
e shows that 1 2 u b 2 2 1 u b cos u cos b sin u sin b r cos b x r cos r˛1 cos u 2 1 u cos u v sin u y r sin u b r sin b 2 1 2 sin u A similar argument with sine leads to the following result. 1 and the addition identity for 2 The Rotation Equations If the xy coordinate axes are rotated through an angle to produce the uv coordinate axes, then the coordinates (x, y) and (u, v) of a point are related by the following equations. U x u cos U v sin U y u sin U v cos U Example 1 A Rotated Conic in the uv System If the xy axes are rotated axes of the graph of p 6 radians, find the equation relative to the uv NOTE Recall that p 6 represents about 0.5236 radians, or 30°. 3x˛ 2 213xy y˛ 2 x 13y 0 Identify and graph the equation. Solution Because sin p 6 1 2 and cos p 6 13 2 , the rotation equations are x u cos y u sin p 6 p 6 v sin v cos p 6 p 6 13 2 u 1 2 v 2 u 13 2 v 1 730 Chapter 11 Analytic Geometry Substitute these expressions into the original equation. 3x˛ 2 213xy y˛ 13 2 u 1 2 v a 2 3 a 13 2 u 1 2 v b 213 2 u 13 2 v b Then multiply out the result. 1 a 2 a 1 2 x 13y 0 2 u 13 2 v b a 13 2 u 1 2 v b b 13 1 2 u 13 2 v b a 0 3 a 3 4 u2 13 2 uv 4 u2 13 1 1 4 v2 b 2 uv 3 4 v˛ a 213 2 b 13 4 u2 1 13 2 u 1 2 v b a a 2 uv 13 4 v2 b 13 a 1 2 u 13 2 v b 0 You can verify that the last equation simplifies to 4u2 2v 0 or equivalently, u2 1 2 v 4 1 8b a v In the uv system, u2 1 2 v is the equation of an upward-opening parabola y v 2 2 0 −2 − 2 −2 − 2 u π 6 x 2 2 Figure 11.4.A-3 Figure 11.4.A-3. with vertex at (0, 0), focus at 0, a 1 8b , and directrix v 1 8 , as shown in ■ Rotation Angle Rotation Angle to Eliminate xy Term Rotating the axes in the preceding example changed the original equation, which included an xy term, to an equation that had no uv term. This can be done for any second-degree equation by choosing an angle of rotation that will eliminate the xy term. 2 Dx Ey F 0 (B 0) Au2 Cv2 Du Ev F 0 by Ax˛ 2 Bxy Cy˛ The equation can be rewritten as rotating the xy axes through an angle cot 2U A C B U such that 0 66 U 66 P 2 b a The restriction 0 6 u 6 p 2 insures that 0 6 2u 6 p. Example 2 Find the Rotation Angle What angle of rotation will eliminate the xy term in the equation 153x˛ 2 192xy 97y˛ 2 1710x 1470y 5625 0, and what are the rotation equations? y 2θ 7 24 x Figure 11.4.A-4 Section 11.4.A Excursion: Rotation of Axes 731 Solution A 153, B 192, Letting through an angle of where and C 97, the figure should be rotated u, cot 2u 153 97 192 56 192 7 24 0 6 2u 6 p is positive, the terminal side of the angle lies in the first quadrant, as shown in Figure 11.4.A-4. The hypotenuse Because 2u of the triangle shown has length 272 242 1625 25. and cot 2u Hence, cos 2u 7 25 . The half-angle identities show that sin u 1 cos 2u 2 B Q 1 7 25 2 9 25 3 5 B cos u 1 cos 2u 2 B Q 1 7 25 2 16 25 4 5 B Using sin u 3 5 and the SIN1 key on a calculator, the angle of rotation u is approximately 0.6435 radians, or about 36.87°. The rotation equations are x u cos u v sin u 4 y u sin u v cos . ■ Identifying Rotated Conics The rotation equations can be used to find the equation of the rotated conic in the uv coordinate system. Substitute the x rotation equation for x and the y rotation equation for y, and simplify the result to eliminate the xy term. Example 3 Graph a Rotated Conic Graph the equation without using a calculator. 2 192xy 97y˛ 153x˛ 2 1710x 1470y 5625 0 Solution u The angle and the rotation equations for eliminating the xy term were found in the preceding example. Substitute the rotation equations into the given equation and simplify the result to eliminate the xy term. 732 Chapter 11 Analytic Geometry 153x˛ 153 a 97 153 a 2 2 b a a 3 4 192 1710 25 uv 9 2 24 25 v˛ 25 u2 24 16 25 u˛ 97 9 a a 2 4 2 192xy 97y 1710x 1470y 5625 0 5 u 4 5 v 3 b 1470 5625 0 3 a 5 u 4 5 v b 25 uv 12 25 v˛ 2 b 192 12 25 u˛ a b 2 7 25 uv 16 225u˛ 2250u 150v 5625 0 25 v2 b 2 25v˛ 2 2250u 150v 5625 0 9u2 v2 90u 6v 225 0 u2 10u v2 6v 9 1 2 1 225 2 Finally, complete the square in u and v by adding the appropriate amounts to the right side so as not to change the equation. 2 10u 25 u˛ 9 1 9 2 6v 9 v 225 9 2 2 9 2 2 1 9 25 2 1 Therefore, the graph is an ellipse centered at (5, 3) in the uv coordinate system, as shown in Figure 11.4.A-5. y v u x 5 3 1 36.87° 1 Figure 11.4.A-5 ■ Exercises 11.4.A In Exercises 1–4, rotate the axes through the given angle to form the uv coordinate system. Express the given equation in terms of the uv coordinate system. 1. 2. 3. 4 ; xy 1 ; 13x˛ 2 10xy 13y˛ 2 72 ; 7x˛ 2 613xy 13y˛ 2 16 0 sin u 1 15 ; x˛ 2 4xy 4y˛ 2 515y 1 0 In Exercises 5–8 find the angle of rotation that will eliminate the xy term of the equation and list the rotation equations in this case. 5. 41x2 24xy 34y˛ 2 25 0 6. 2 213xy 3y˛ 2 813x 8y 32 0 x˛ 7. 17x2 48xy 31y2 49 0 8. 52x˛ 2 72xy 73y˛ 2 200 9. Critical Thinking a. Given an equation Ey F 0 Ax2 Bxy Cy2 Dx B 0 u and an angle with equations to rewrite the equation in the form A¿u2 B¿uv C¿v2 D¿u E¿v F¿ 0 , , use the rotation are expressions involving and the constants A, . . . , F. where sin u, A¿, . . . , F¿ cos u b. Verify that B¿ 2 C A cos2 u sin2 u 2 2 1 c. Use the double-angle identities to show that sin 2u B cos 2u 1 C A sin u cos u B¿ B 1 2 is chosen so that d. If u cot 2u A C This proves the rotation angle B , show B¿ 0. that formula. Section 11.4.A Excursion: Rotation of Axes 733 10. Critical Thinking Assume that the graph of or C¿ (with at least nonzero) in the uv coordinate C¿v2 D¿u E¿v F¿ 0 A¿ A¿u2 one of system is a nondegenerate conic. Show that its A¿C¿ 7 0 have graph is an ellipse if and 1 A¿C¿ 6 0 A¿ the same sign), a hyperbola if and 1 A¿C¿ 0. C¿ have opposite signs), or a parabola if A¿ C¿ 11. Critical Thinking Assume the graph of Ax˛ 2 Bxy Cy˛ 2 Dx Ey F 0 2 1 B¿ is a nondegenerate conic section. Prove the statement in the box on page 723 as follows. a. In Exercise 9 a. show that 2 4A¿C¿ B2 4AC. b. Assume has been chosen so that B¿ 0. Exercise 10 to show that the graph of the original equation is an ellipse if a parabola if a hyperbola if B2 4AC 6 0, B2 4AC 0, B2 4AC 7 0. and u Use 12. Critical Thinking Suppose an xy- and uv- is u coordinate system have the same origin and the angle between the positive x-axis and the positive u-axis. Show that the point 1 rotated system is related to the point following equations. Hint: If the rotation from the xy-coordinate system to the uv-coordinate system is positive, then the rotation from the uv-coordinate system to the xy-coordinate system is negative. by the in the 2 x, y u, v 2 1 u x cos u y sin u v y cos u x sin u. In Exercises 13–16, find the new coordinates of the point when the coordinate axes are rotated through the given angle by using the equations in Exercise 12. 13. 15. 1 1 3, 2 ; u p 4 2 1, 0 ; u p 6 2 14. 16. 1 1 2, 4 ; u p 3 2 3, 3 ; sin u 5 13 2 734 Chapter 11 Analytic Geometry 11.5 Polar Coordinates Objectives • Locate points in a polar coordinate system • Convert between coordinates in rectangular and polar systems • Create graphs of equations in polar coordinates • Recognize equations and graphs of: cardioid rose circle lemniscate limaçon NOTE The coordinates of the origin can be written 0, u as 1 angle. where is any u , 2 The coordinate system most commonly used is the rectangular coordinate system, which is based on two perpendicular axes. However, other coordinate systems are possible. The Polar Coordinate System Choose a point O in the plane, called the origin, or pole. The horizontal ray extending to the right with endpoint O is called the polar axis. A point P in the plane has polar coordinates where r is the length of initial side and otherwise stated, will be measured in radians. is the angle with the polar axis as its as its terminal side, as shown in Figure 11.5-1. Unless OP u and OP r, u 2 1 u P θ (r, ) polar axis r θ O Origin Pole Figure 11.5-1 u -coordinate may be either positive or negative, depending on The whether it is measured as a counterclockwise or clockwise rotation. Figure 11.5-2 shows some points in a polar coordinate system, and the “circular grid” that a polar coordinate system imposes on the plane. π 2) ( 5, π 6) ( 5, (2, π) polar axis 3, − ( π 4) 4π 3 ) ( 4, Figure 11.5-2 Section 11.5 Polar Coordinates 735 The polar coordinates of a point P are not unique. For example, because 5p 3 7p 3 2, a and , 5p 3 b ˛ are coterminal angles, the coordinates p 3b 2, a , 2, a 7p 3 b , and all represent the same point, as shown in Figure 11.5-3. p 3 , π 3) ( 2, ( 2, 7π 3 ) 2, – 5π ( 3 ) π 3 O polar axis O 7π 3 polar axis Figure 11.5-3 O – 5π 3 polar axis the point The r-coordinate may also be negative, as shown in Figure 11.5-4. For r 7 0, u at a distance r from the origin—but on the opposite side of the origin from the point lies on the line containing the terminal side of r, u r2, π 4) −2.5, ( − π 2 ) π 4) ( 2, ( –3, 7π 6 ) 7π 6 polar axis O polar axis O − π 2 polar axis ( 3, 7π 6 ) Figure 11.5-4 ( 2.5, − π 2 ) Example 1 Polar Coordinates of a Point Determine if the given coordinates represent the same point as in a polar coordinate system. p 6 b 3, a a. 3, a 13p 6 b b. 3, a 5p 6 b 3, c. a 7p 6 b 736 Chapter 11 Analytic Geometry Solution 7π 6 Q π 6 P −5π 6 polar axis 13π 6 Figure 11.5-5 The point labeled Q in Figure 11.5-5 can be represented by the coordi- nates p 6 b 3, a , 3, a 13p 6 b , or 3, a . The point labeled P can be repre- sented by the coordinates 3, a . Thus, the coordinates in a and c represent the same point as but the coordinates in b do not. 7p 6 b 5p 6 b p 6 b , 3, a ■ Polar and Rectangular Coordinates Suppose that a polar and rectangular system of coordinates are drawn in the same plane, with the origins at the same point, so that the polar axis is the positive x-axis, and the polar line u p 2 nates of point P in the plane can be written as in Figure 11.5-6. 1 is the y-axis. The coordi- r, u 2 or as x, y 1 2 , as shown y-axis =θ π 2 (r, ) P θ (x, y) r θ polar axis Figure 11.5-6 x-axis = 0 θ Let r be as shown
in Figure 11.5-6, with r positive. Since r is the distance the distance formula shows that from (0, 0) to x, y , 1 2 r 2x2 y2 Also, by the definitions of the trigonometric functions in the coordinate plane, cos u x r sin u y r tan u y x These equations can be used to obtain the relationship between polar and rectangular coordinates. Coordinate Conversion Formulas Technology Tip Keys to convert from rectangular to polar coordinates, or vice versa, are in the TI ANGLE menu and in the ANGLE submenu of the Casio OPTN menu. y 2 A( 3, 1) x −2 0 2 −2 B(−1.97, −2.27) Figure 11.5-7 Section 11.5 Polar Coordinates 737 S Polar Rectangular Point P with polar coordinates (x, y), coordinates where x r cos U (r, U) has rectangular and y r sin U S Rectangular Polar Point P with rectangular coordinates coordinates where (r, U), (x, y) has polar 2 x˛ 2 y˛ 2 r˛ and tan U y x The conversion formulas from polar to rectangular coordinates give a unique solution for all values of r and u . Example 2 Polar S Rectangular Convert each point from polar coordinates to rectangular coordinates. a. p 6 b 2, a Solution a. For p 6 b 2, a b. 3, 4 1 2 , apply the conversion formulas using r 2 and u p 6 . x 2 cos y 2 sin p 6 p 6 2 13 2 2 1 2 1 13 So the rectangular coordinates are 13, 1 A . B b. For (3, 4), apply the conversion formulas using r 3 and u 4. x 3 cos 4 1.96 y 3 sin 4 2.27 So the approximate rectangular coordinates are 1.96, 2.27 . 2 1 Figure 11.5-7 displays the rectangular coordinates of the points in part a and part b, along with both a rectangular grid and a polar grid. ■ The conversion formulas from rectangular to polar coordinates do not tan u y x and infinitely many solutions when x 0. have unique solutions for r and . In particular, the equation has no solutions when x 0 u 738 Chapter 11 Analytic Geometry u tan 1 y x kp (k is any integer) Not every solution works for a specific point P. To find solutions that represent the point P, you need to know which quadrant contains P. u tan can be used for points in Quadrants I and IV because it would 1 y x be an angle between p 2 and p 2 . Similarly, u tan 1 y x p can be used for points in Quadrants II and III because it would be an angle between p 2 3p 2 and . Example 3 Rectangular S Polar Convert each point from rectangular coordinates to polar coordinates. 2, 4 3, 5 2, 2 0, 5 d. b. c. a. 1 2 1 2 1 2 1 2 Solution a. For , 2 2, 2 1 conversion r 222 the second set of equations of formulas with 2 18 212 2 x 2 and y 2 and tan u the coordinate that shows 1, or 2 2 u tan 1 The point is in Quadrant IV, so two possible y D(2 5, 2.03) answers are b. For 3 212 or 212, a 2 1 1 2 5 5 3 tan . 7p 4 b 2 134 1.0304 . A calculator shows that x A(2 2, − )π 4 B( 34, 4.17) C(5, )3π 2 Figure 11.5-8 . 2 1 tive y-axis, 5, p a 2 b 2, 4 2 1 tan u 4 2 , d. For 3, 5 Because the point Therefore, one possible answer is A 2 5. 0, 5 5 2 1 c. For is in Quadrant III, . 134, 4.17 B r 202 , u 3p 2 . 1 2 Because the point is on the nega- Therefore, two possible answers are 3p 2 b 5, a or u 1.0304 p 4.17. r 2 2 1 2 2 42 14 16 120 215 . 2, and 1 tan 2 2 1 1.11, which is in Quadrant IV. Because 2, 4 1 2 is in Quadrant II, u 1.11 p 2.03. Therefore, one possible answer is 215, 2.03 . 2 1 Figure 11.5-8 displays the polar coordinates of the points in parts a–d, along with both a rectangular grid and a polar grid. ■ Section 11.5 Polar Coordinates 739 Polar Graphs r 2 cos u , where r and are the variables, is a polar An equation like equation. Equations in x and y are called rectangular or Cartesian equations. Many useful curves have simple polar equations, although they may have complicated rectangular equations. u u Like other graphs, the graph of a polar equation in r and is the set of points (r, ) in the plane that make the equation true. It is possible to write a polar equation in rectangular form or a rectangular equation in polar form by using the coordinate conversion formulas, definitions, and basic facts about trigonometric functions. u Graphs of the Form r a and U b Several types of graphs have equations that are simpler in polar form than in rectangular form. For example, the graph of consists of points of u the form (1, ), which is all points that are 1 unit from the pole. That is, r 1 the graph of is the unit circle centered at the pole. The equation u p 4 consists of all points of the form . That is, all points that lie r 1 p 4 R r, Q on the line that makes a p 4 , or 45° , angle with the polar axis. Example 4 Polar Graphs Graph each polar equation below. a. r 3 Solution 2 a. The graph consists of all points 3, u , that is, all points whose 1 distance from the pole is 3. So, the graph is a circle of radius 3 with its center at the pole. b. u p 6 b. The graph consists of all p 6 R points r, Q These points . lie on the straight line that contains the terminal side of an angle of p 6 initial side is the polar axis. radians, whose r = 3 1 2 3 4 5 O polar axis =θ π 6 π 6 O polar axis Figure 11.5-9 Figure 11.5-10 ■ 740 Chapter 11 Analytic Geometry Graphs of Other Polar Equations Other types of graphs have simple equations in polar form. Several types of polar graphs have specific forms that can be classified by special names like cardioid, limaçon, and rose. Many polar equations are functions, with independent variable dependent variable r. Therefore, polar functions are written as u r f , and u . 1 2 Example 5 Polar Graphs Graph r 1 sin u. Solution Consider the behavior of sin u in each quadrant. θ As increases from 0 to , sin increases from 0 to 1. So r = 1 + sin increases from 1 to 2. θ π 2 θ θAs increases from to π, sin decreases from 1 to 0. So r = 1 + sin decreases from 2 to 1= π 1 O 1 θ As increases from π to , sin decreases from 0 to −1. So r = 1 + sin decreases from 1 to 0. θ 3π 2 θ θ As increases from to 2π, sin increases from −1 to 0. So r = 1 + sin increases from 0 to 1. θ 3π 2 θ 1 O θ = 3π 2 1 1 O θ= 2π 1 Figure 11.5-11 2.55 2.35 2.35 0.55 Figure 11.5-12 Section 11.5 Polar Coordinates 741 u 6 0, u 7 2p sin and For repeats the same pattern, so the full graph, as shown in the lower right drawing of Figure 11.5-11, is called a cardioid. The graph of displayed on a graphing calculator is shown in Figure 11.5-12. r 1 sin u u ■ The easiest way to graph a polar function of the form is to use a calculator in polar graphing mode. The following Tip should be helpful. 2 1 r f u Technology Tip The viewing window of a calculator in polar graphing mode has the usual settings for the x- and y-values, and also min, max, and (Casio has pitch instead of tion that contains a trigonometric function, the interval from min to max should be at least as large as the period of the function. step. step.) For a complete graph of a polar equa- u u u u u u u u The value of tor. A smaller value of but will also take longer to graph. u step determines the number of points plotted by the calculastep will generally result in a more accurate curve To view the polar coordinates using the Trace feature, from the Format menu choose PolarGC on TI models. Casio models automatically display coordinates for the type of graph shown. Graphing Exploration r 2 4 cos u on a calGraph culator in polar graphing mode. Experiment with the size of the window and the values of min, u max, and step to find (approximately) the graph shown in Figure 11.5-13. What window settings did you use to obtain the graph? u u ? ? ? Figure 11.5-13 ? Common Polar Graphs The following is a summary of commonly encountered polar graphs. In each case, a and b are constants, and is measured in radians. u Depending on the plus or minus sign and whether sine or cosine is used, the basic shape of each graph may differ from those shown by a rotation, reflection, or horizontal or vertical shift. 742 Chapter 11 Analytic Geometry Equation Name of Graph Shape of Graph π 2 π 2 r au r au u 0 u 0 1 1 2 2 Archimedian spiral π 0 π r a r a 1 – sin u 2 1 1 – cos u 1 2 cardioid r a sin nu r a cos nu n 2 1 2 rose For n odd, there are n petals. For n even, there are 2n petals. 3π 2 r = a θθ ≥( 0) 3π 2 r = a θθ ≤( 0) π π 2 3π 2 0 π π 2 3π 2 r = a(1 + cos ) θ r = a(1 − sin ) 3π 2 r = a cos n a θ r a sin u r a cos u circle π 2 π a π 0 n = 5 3π 2 r = a sin n θ π 2 a 3π 2 r = a cos θ 3π 2 r = a sin θ 0 0 a 0 0 Equation Name of Graph Shape of Graph Section 11.5 Polar Coordinates 743 2 – a2 sin 2u r˛ 2 – a2 cos 2u r˛ lemniscate r a b sin u r a b sin u a, b 7 0, a b 1 2 limaçon π π 2 π 3π 2 a π 0 π 2 3π 2 π 2 3π 2 a 0 r2 = a2 sin 2θ r2 = a2 cos 2θ cos θ 3π 2 b < a < 2b r = a + b sin θ 3π 2 a ≥ 2b r = a − b sin θ Exercises 11.5 1. What is one possible pair of polar coordinates of each of the points P, Q, R, S, T, U, V in the figure? π 2 2π 3 π 4 In Exercises 2–6, list four other pairs of polar coordinates for the given point, each with a different r 77 0, combination of signs (that is, U 66 0; r 77 0, r 66 0, U 66 0). r 66 0, U 77 0; U 77 0; and 2. 5. p 3 b 3, a 3. 1 5, p 2 4. 2, 2p a 3 b 1, p 6 b a 6. 13, a 3p 4 b π R Q P 1 7π 6 S T 3 5 V 7 polar axis In Exercises 7–10, convert the polar coordinates to rectangular coordinates. 7. 9. p 3 b 3, a 1, a 5p 6 b 8. 2, a p 4 b 10. 2, 0 1 2 U − π 3 744 Chapter 11 Analytic Geometry In Exercises 11–16, convert the rectangular coordinates to polar coordinates. 48. a. Find a complete graph of b. Predict what the graph of r 1 3 sin 2u. r 1 3 sin 3u will 11. 14. 313, 3 A B 3, 2 2 1 12. 15. 213, 2 A B 5, 2.5 1 2 13. 16. 2, 4 2 6.2, 3 1 1 2 In Exercises 17–22, sketch the graph of the equation without using a calculator. 17. r 4 18. r 1 19. u p 3 20. u 5p 6 21. u 1 22. u 4 In Exercises 23–46, sketch the graph of the equation. 23. 25. r u u 0 1 r 1 sin u 2 24. 26. r 3u u 0 1 2 r 3 3 cos u 27. r 2˛ cos u 28. r 6˛ sin u 29. r cos 2u 31. r sin 3u 30. r cos 3u 32. r sin 4˛u 33. r2 4 cos 2u 34. r2 sin 2u look like. Then check your prediction with a calculator. c. Predict what the graph of r 1 3 sin 4u will look like. Then check your prediction with a calculator. 49. If a and b
are constants such that r a sin u b cos u that the graph of Hint: Multiply both sides by r and convert to rectangular coordinates. ab 0 , show is a circle. x, y 50. Critical Thinking Prove that the coordinate r 6 0. conversion formulas are valid when P has coordinates and verify that the point Q with rectangular coordinates x, y 2 1 r 6 0, r proved in the text apply to Q. For instance, x r cos u, has polar coordinates is positive and the conversion formulas which implies that 2 r, u x r cos u. r 6 0, Hint: If Since with r, u , . 1 2 1 2 1 51. Critical Thinking Distance Formula for Polar is Coordinates: Prove that the distance from 1 s, b . Hint: If 2 1 s 7 0, r, u has an angle of , 1 sides have lengths r and s. Use the Law of Cosines. r, u to 2 r 7 0, 2 then the triangle with vertices u b, 2 s2 2rs cos u 7 b, 0, 0 2r˛ and s, b , u b whose 1 2 2 2 1 1 35. r 2 4 cos u 36. r 1 2 cos u 52. Critical Thinking Explain why the following 37. r sin u cos u 38. r 4 cos u 4 sin u 39. r sin u 2 40. r 4 tan u 41. r sin u tan u (cissoid) 42. r 4 2 sec u (conchoid) 43. r eu (logarithmic spiral) 44. r2 1 u 45. r 1 u u 7 0 1 2 46. 2 u r˛ 47. a. Find a complete graph of b. Predict what the graph of r 1 2 sin 3u. r 1 2 sin 4u will symmetry tests for the graphs of polar equations are valid. a. If replacing by produces an equivalent equation, then the graph is symmetric with u 0 respect to the line p u (the x-axis). produces an equivalent b. If replacing by u u u c. If replacing r by equation, then the graph is symmetric with u p2 respect to the line r (the y-axis). produces an equivalent equation, then the graph is symmetric with respect to the pole (origin). look like. Then check your prediction with a calculator. c. Predict what the graph of r 1 2 sin 5u will look like. Then check your prediction with a calculator. Section 11.6 Polar Equations of Conics 745 11.6 Polar Equations of Conics Objectives • Define eccentricity of an ellipse, a parabola, and a hyperbola • Develop and use the general polar equation of a conic section NOTE Do not confuse the eccentricity of a conic section, which is denoted as e and whose value varies, with the number e, which is the constant 2.718281828.... The meaning should be clear in context. In a rectangular coordinate system, each type of conic section has a different definition. By using polar coordinates, it is possible to give a unified treatment of conics and their equations. A key concept in this development is eccentricity. Eccentricity Recall that ellipses and hyperbolas are defined in terms of two foci, and both have two vertices that lie on the line through the foci. The eccentricity, e, of an ellipse or a hyperbola is the ratio e distance between the foci distance between the vertices If a conic is centered at the origin with foci on the x-axis, the situation is as follows. Ellipse y2 b2 1 x2 a2 Hyperbola y2 x2 b2 a2 1 For a 7 b, foci: 1 foci ± c, 0 2 vertices: c 2a2 b2 ± a, 0 2 1 foci: 1 ± c, 0 2 vertices: c 2a2 b2 ± a, 0 2 1 2c 2c foci −a −c c a −c −a a c vertices vertices e 2c 2a c a 2a 2a2 b2 a Figure 11.6-1 2a 2a2 b2 a c a e 2c 2a Figure 11.6-2 746 Chapter 11 Analytic Geometry A similar analysis shows that the formulas for e are also valid for conics whose foci lie on the y-axis. These formulas can be used to compute the eccentricity of any ellipse or hyperbola whose equation is in standard form. As Figure 11.6-1 shows, the distance between the foci of an ellipse is always less than the distance between its vertices, so that 0 66 e 66 1 for all ellipses. Similarly, the distance between the foci of a hyperbola is greater than the distance between its vertices, as shown in Figure 11.6-2, so that e 77 1 for all hyperbolas. Example 1 Eccentricity of a Conic Find the eccentricity of each given conic. a. 2 y˛ 4 x˛2 21 1 Solution a. 2 y˛ 4 x˛2 21 1 b. 4x˛2 9y˛2 32x 90y 253 0 represents a hyperbola with a2 4 and b2 21, so e 2a˛2 b2 a 14 21 5 2 2 4x2 9y2 32x 90y 253 0 125 2 2.5 can b. From Example 2 in Section 11.4, be written in standard form, as The graph of this equation has the same shape as the ellipse 2 x˛ 9 2 y˛ 4 1 with a horizontal and a vertical shift. Since the distances between the foci or vertices are the same in both ellipses, the eccentricity is the same for both. Using a2 9 and e 2a2 b2 a b2 4, 19 4 3 the eccentricity is 15 3 0.745 ■ The eccentricity of an ellipse measures its distortion from a circle, which is a special case of an ellipse. In a circle, the two foci coincide at the center, so the distance between the foci is 0 and its eccentricity is 0. As the foci move farther apart, the eccentricity increases, and the ellipse becomes more distorted from a circle. In Figure 11.6-3, the foci and eccentricities are shown in the same color as the corresponding ellipse. e = 0.5 y e = 0 1 0.5 e = 0.9 −1 −0.5 0 0.5 1 x −0.5 −1 Figure 11.6-3 Section 11.6 Polar Equations of Conics 747 .1 x 5 −5 0 In Figure 11.6-4, colors of the foci and the eccentricities correspond to each hyperbola. As the foci move farther from the vertices, the branches of the hyperbola become straighter and approach vertical lines. The preceding discussion does not apply to parabolas because they have only one focus. For reasons explained below, the eccentricity of any parabola is defined to be the number 1. −5 Figure 11.6-4 Alternate Definition of Conics The following description, whose proof is omitted, is sometimes used to define the conic sections because it provides a unified approach instead of the variety of descriptions given in Sections 11.1, 11.2, and 11.3. It is also used to determine the polar equations of conic sections. Alternate Definition of Conic Sections Let L be a fixed line called a directrix, P a fixed point not on L, and e a positive constant. The set of all points X in the plane such that distance between X and the fixed point distance between X and the fixed line XP XL e is a conic section with P as one focus. NOTE Recall that the distance from a point to a line is measured along the perpendicular segment from the point to the line. • For • For • For 0 66 e 66 1, e 1, e 77 1, the conic is an ellipse. the conic is a parabola. the conic is a hyperbola. Recall that the definition of a parabola was all points equidistant from a fixed point and a fixed line. Therefore, the alternate definition coincides with the original definition given in Section 11.3. Examples of this definition are shown in the following diagram XP XL 1 S XP XL XP XL 2 S XP 2XL The distance from X to P equals the distance from X to L. The conic is a parabola. The distance from X to P is twice the distance from X to L. The conic is a hyperbola. L X P e = 3 4 S XP 3 4 XL XP XL 3 4 The distance from X to P is 3 4 the distance from X to L. The conic is an ellipse. 748 L L Chapter 11 Analytic Geometry P d polar axis Polar Equations of Conics To generate polar equations of conics from the alternate definition, let P be the pole, and L be a vertical line d units to the left of the pole, as shown in Figure 11.6-5. A point X r, u 1 2 on a general conic satisfies the condition XP XL e By the definition of polar coordinates, the r-coordinate of X, shown in Figure 11.6–6, is the distance from the origin to X. Figure 11.6-5 Figure 11.6-6 also shows that XP r So the polar equation of the conic is given by XL d r cos u θ X(r, ) r θ P d r cos θ XP XL r d r cos u e r ed e r cos u r e r cos u ed ed 1 e cos u r˛1 2 r ed 1 e cos u If L is to the right of the pole, it can be shown that Figure 11.6-6 r ed 1 e cos u If L is a horizontal line, it can also be shown that r ed 1 e sin u or r ed 1 e sin u depending on whether L is below the pole or above it. If an equation has another value in place of 1, divide both numerator and denominator by that number to rewrite the equation in the desired form. When the constant term in the denominator is 1, the eccentricity is the coefficient of the trigonometric function. For example, suppose a conic is given by r 20 4 8 cos u Divide both numerator and denominator by 4. r 5 1 2 cos u The conic is a hyperbola because the eccentricity is 2. Section 11.6 Polar Equations of Conics 749 Polar Equations of Conic Sections Equations Graph Example 0 66 e 66 1 • Vertices and • One focus at (0, 0) Ellipse U 0 U P NOTE d is the distance from the focus at the pole to the directrix. r ed 1 e cos U e 1 • Vertex Parabola or U 0 U P ; r is not defined for the other value of • Focus at (0, 0) U r ed 1 e cos U e 77 1 Hyperbola U 0 • Vertices and • One focus at (0, 0) U P r ed 1 e sin U r ed 1 e sin U 0 66 e 66 1 • Vertices Ellipse U P 2 and U 3P 2 • One focus at (0, 0) e 1 • Vertices Parabola U P 2 or U 3P 2 ; r is not defined for the other value of U • Focus at (0, 0) e 77 1 • Vertices Hyperbola U P 2 and and U 3P 2 • One focus at (0, 0) 750 Chapter 11 Analytic Geometry Example 2 Polar Equations of Conic Sections Find a complete graph of r 3e 1 e cos u for the following eccentricities. a. e 0.7 Solution b. e 1 c. e 2 From the first equation in the preceding chart, with the graphs are an ellipse, a parabola, and a hyperbola, respectively, as shown in Figure 11.6-7. In each case, 0 u 2p . d 3, a. e 0.7 b. e 1 c. e 2 r 3 0.7 1 1 0.7 cos u 2 2.1 1 0.7 cos u r 1 3 1 1 1 cos u 2 3 1 cos u r 3 2 1 1 2 cos u 2 6 1 2 cos u ellipse 10 parabola 10 15 15 15 15 −15 10 10 Figure 11.6-7 hyperbola 10 −10 15 ■ Example 3 Polar Equations of Conic Sections Identify the conic section that is the graph of r 20 4 10 sin u and find its eccentricity and vertices. Solution First, rewrite the equation in one of the forms listed in the preceding box. r 20 4 10 sin u According to the equation, eccentricity 2.5. 4 1 e 2.5 20 1 2.5 sin u 5 1 2.5 sin u 2 . Thus, the graph is a hyperbola with Section 11.6 Polar Equations of Conics 751 The vertices occur when u p 2 and u 3p 2 substitute into the original equation. . To find the r-coordinates, u p 2 S r 20 4 10 sin u 3p 2 S r 20 4 10 sin p 2 3p 2 20 4 10 1 20 6 10 3 20 4 10 1 1 2 20 14 10 7 The vertices are 10
a 3 , p 2 b and 10 7 , 3p 2 b . a ■ Graphing Exploration Find a viewing window that shows a complete graph of the hyperbola in Example 3. Example 4 Polar Equations of Conic Sections (6, π) (3, 0) Find a polar equation of the ellipse with one focus at (0, 0) and vertices (3, 0) and 6, p . 1 2 Figure 11.6-8 Solution Because the vertices occur when of the form u 0 and u p, the polar equation is r ed 1 e cos u or r ed 1 e cos u. Select one of these equations, say r ed 1 e cos u and proceed as follows. If the selected equation leads to a contradiction, start again with the other form. Substitute the values of r and given by the vertices to obtain two equations. u (3, 0) 3 ed 1 e cos 0 6 6, p 1 2 ed 1 e cos p 3 ed 1 e ed 1 e 2 3 1 6 ed 1 e ed 1 e 2 and 6 1 752 Chapter 11 Analytic Geometry Therefore 3e 6 6e 9e 3 e 1 3 Substituting this value of e into the equation for d shows that and solving and the equation of the ellipse is d 12. ed 4 Hence, 3 1 2 1 e ed r 4 1 1 3 cos u or equivalently, r 12 3 cos u If you had started this process with the equation r ed 1 e cos u, you would have obtained is always positive. e 1 3 , which is impossible since the eccentricity ■ Exercises 11.6 a. d. b. e. c. f. In Exercises 1–6, which of the graphs a–f above could be the graph of the given equation? 3. r 6 2 4 sin u 4. r 15 1 4 cos u 1. r 3 1 cos u 2. r 6 2 cos u 5. r 6 3 2 sin u 6. r 6 3 2 3 2 sin u Section 11.6 Polar Equations of Conics 753 In Exercises 7–12, identify the conic section whose equation is given; if it is an ellipse or hyperbola, state its eccentricity. 7. r 12 3 4 sin u 8. r 10 2 3 cos u 9. r 8 3 3 sin u 10. r 20 5 10 sin u 11. r 2 6 4 cos u 12. r 6 5 2 cos u 25. r 10 4 3 sin u 26. r 12 3 4 sin u 27. r 15 3 2 cos u 28. r 32 3 5 sin u 29. r 3 1 sin u 30. r 10 3 2 cos u 31. r 10 2 3 sin u 32. r 15 4 4 cos u In Exercises 13–18, find the eccentricity of the conic whose equation is given. In Exercises 33–46, find the polar equation of the conic section that has focus (0, 0) and satisfies the given conditions. 13. 14. 15. 2 x˛ 100 y˛2 99 1 2 1 x 4 18 2 1 2 y 5 25 2 1 2 1 x 6 10 2 y˛2 40 1 16. 4x˛2 9y˛2 24x 36y 36 0 17. 16x˛2 9y˛2 32x 36y 124 0 18. 4x˛2 5y ˛2 16x 50y 71 0 19. a. Using a square viewing window, graph these ellipses on the same screen, if possible. y ˛2 x˛2 14 16 y˛2 6 b. Compute the eccentricity of each ellipse in part a. c. Based on parts a and b, how is the shape of an 1 x˛2 16 1 x˛2 16 y˛2 1 1 ellipse related to its eccentricity? 20. a. Graph these hyperbolas on the same screen, if possible. y˛2 x˛2 1 4 1 y˛2 4 x˛2 12 1 y˛2 4 x˛2 96 1 b. Compute the eccentricity of each hyperbola in part a. c. Based on parts a and b, how is the shape of a hyperbola related to its eccentricity? In Exercises 21–32, sketch the graph of the equation and label the vertices. 21. r 8 1 cos u 22. r 5 3 2 sin u 23. r 4 2 4 cos u 24. r 5 1 cos u 33. parabola; vertex 34. parabola; vertex 3, p 1 2 p 2 b 2, a 35. ellipse; vertices p 2 b 2, a and 8, a 3p 2 b 36. ellipse; vertices 2, 0 1 2 and 1 4, p 37. hyperbola; vertices 1, 0 1 2 and 1 2 3, p 2 38. hyperbola; vertices 2, a p 2 b and 4, a 3p 2 b 39. eccentricity 4; directrix r 2 sec u 40. eccentricity 2; directrix r 4 csc u 41. eccentricity 1; directrix r 3 csc u 42. eccentricity 1; directrix r 5 sec u 43. eccentricity 44. eccentricity 1 2 4 5 ; directrix r 2 sec u ; directrix r 3 csc u 45. hyperbola; vertical directrix to the left of the pole; eccentricity 2; 2p 3 b 1, a is on the graph. 46. hyperbola; horizontal directrix above the pole; eccentricity 2; 2p 3 b 1, a is on the graph. 47. A comet travels in a parabolic orbit with the sun as the focus. When the comet is 60 million miles from the sun, the line segment from the sun to the p 3 comet makes an angle of radians with the axis of the parabolic orbit. Using the sun as the pole 754 Chapter 11 Analytic Geometry and assuming the axis of the orbit lies along the polar axis, find a polar equation for the orbit. 48. Halley’s Comet has an elliptical orbit, with eccentricity 0.97 and the sun as a focus. The length of the major axis of the orbit is 3364.74 million miles. Using the sun as the pole and assuming the major axis of the orbit is perpendicular to the polar axis, find a polar equation for the orbit. 11.7 Plane Curves and Parametric Equations Objectives • Define plane curves and parameterizations • Find parametric equations for projectile motion and cycloids y f Many curves in the plane cannot be represented as the graph of a funcParametric graphing makes it possible to represent such tion curves in terms of functions and also provides a formal definition of a curve in the plane. x . 2 1 Plane Curves Consider an object moving in the plane during a particular time interval. In order to describe both the path of the object and its location at a particx, y ular time, three variables are needed: the time t, and the coordinates 1 of the object at time t. For example, the coordinates might be given by 2 x 4 cos t 5 cos 3t and y sin 3t t the object traces out the curve During the time interval shown in Figure 11.7-1. The points labeled on the graph show the location of the point at various times. Note that the object may be at the same location at different times, the points where the graph crosses itself. 0 t 12.5, y t = 12.9 –3 1 t = 0 x 9 Figure 11.7-1 Definition of a Plane Curve CAUTION Not every substitution of an expression for x gives a complete parameterization of the graph. For example, the parametric equations 2 x t y 2t 2 1 give a nonnegative xcoordinate and a negative y-coordinate for every value of t, so the parameterization produces only part of the graph. Section 11.7 Plane Curves and Parametric Equations 755 Let f and g be continuous functions of t on an interval I. The set of all points where (x, y), x f (t) and y g (t) is called a plane curve. The variable t is called a parameter and the equations that define x and y are called parametric equations. A pair of parametric equations that describe a given curve is called a parameterization of the curve. More than one parameterization is possible for a given curve. Example 1 Parameterizations of a Line Find three parameterizations of the line through 1, 3 1 2 with slope 2. Solution The equation of the line in rectangular coordinates is y 3 2 x 1 1 2 or equivalently y 2x 1 [1] Choose three expressions in terms of t to represent x, and substitute each into equation [1] to find corresponding expressions for y. a. x t y 2t 1 b. for any 2t 3 1 c. x tan t y 2 tan t 1 6 t 6 p 2 p 2 for any t Notice in a and b that when t runs through all real numbers, both x and , x tan t y take on all real numbers as well. In c when t runs from to p 2 p 2 takes all possible real number values, and hence so does y. Therefore, each parameterization represents the entire line. ■ Graphing Parametric Equations Parametric equations may be graphed by hand by plotting points, or by using a calculator in parametric mode. When choosing a viewing window, you must specify values not only for x and y, but also for t. You must also choose a t-step (or t-pitch), which determines how much t changes each time a point is plotted. A t-step between 0.05 and 0.15 usually produces a relatively smooth graph in a reasonable amount of time. 756 Chapter 11 Analytic Geometry Technology Tip For parametric graphing mode, choose PAR or PARM respectively in the TI MODE menu or the TYPE submenu of the Casio GRAPH menu. Graphing Exploration Graph the curve shown in Figure 11.7-1 at the beginning of this section using the window 0 t 12.5 10 x 10 0 y 15 and t-step 0.1. Does the graph look like Figure 11.7-1? Now change the t-step to 1.5 and graph the equations again. Now how does the graph look? Experiment with different t-steps to see how they affect the graph. Example 2 Parameterization of a Parabola By hand, graph the curve given by x 2t and y 4t˛ 2 4, 1 t 2. Confirm your sketch by using the parametric mode on a calculator. Solution 1 t 2, there is a value of x and a value of y that corresponds to For each specific value of t. Find several points by picking values for t, finding the corresponding values of x and y, plotting the points, and connecting the points in the order determined by the least to greatest values of t. t 1 0 1 2 x 2t y 4t˛ 2 4 0 4 0 12 2 0 2 4 y 12 8 4 (x, y) 2, 0 2 1 0, 4 1 2 2, 0 2 1 4, 12 2 1 14 −8 0 −4 −4 x 4 8 10 10 Figure 11.7-2a 6 Figure 11.7-2b Section 11.7 Plane Curves and Parametric Equations 757 The points are plotted and the direction of the curve is indicated in Figure 11.7-2a, and a calculator-generated graph is shown in Figure 11.7-2b. ■ The direction in which a parametric curve is traced is called its orientation. Eliminating the Parameter Some curves given by parametric equations can also be expressed as part of the graph of an equation in x and y. The process for doing this, called eliminating the parameter, is as follows. Solve one of the parametric equations for the parameter t and substitute this result in the other parametric equation. Example 3 Eliminate the Parameter Consider the curve given in Example 2. x 2t and y 4t˛ 2 4, 1 t 2 Find an equation in x and y whose graph includes the graph of the given curve. Solution Solve one of the parametric equations for t and substitute the result into the other equation. Solving x 2t for t shows that t x 2 . Substituting this into the equation for y and simplifying the result will eliminate t. x y 4t˛ 2 4 4 2 x 2b a 4 4 2 x˛ 4 b a 4 x˛ 2 4 4 The graph of y x˛ 2 4 is the parabola shown in Figure 11.7-3. y 12 8 4 −4 0 −4 Figure 11.7-3 y x˛ 2 4. Every point on the curve given by the parametric equations is also on the However, the curve given by the parametric equagraph of tion is not the entire parabola, but only the part shown in red, which joins 4, 12 the points These points correspond to the minimum 2, 0 and maximum values of t, 2 t 1 and t 2. and . 1 2 1 ■ Example 4 Parameterization of Transformations Given the parent relation represent the r
elation, and sketch the graph. x y˛ 2, write a set of parametric equations to Then write the parametric equations of the following successive transformations of the parent relation, and sketch each graph. 758 Chapter 11 Analytic Geometry y 4 2 0 −2 −4 a. a horizontal stretch by a factor of 5 b. then a horizontal shift 3 units to the right c. then a vertical shift down 2 units 2 4 6 8 10 x Finally, find the focus and a parameterization of the directrix of the parabola found in step c. Solution The parent relation can be parameterized as Figure 11.7-4a x t˛ 2 and y t t any real number, whose graph is shown in Figure 11.7-4a. a. A horizontal stretch by a factor of 5 2 x 5t˛ y t for any t b. Then a horizontal shift 3 units to the right x 5t˛ 2 3 y t for any t c. Then a vertical shift down 2 units x 5t˛ 2 3 y t 2 for any t y 4 2 0 −2 −4 x 2 4 6 8 10 y 4 2 0 −2 −4 x 2 4 6 8 10 y 4 2 0 −2 −4 x 2 4 6 8 10 Figure 11.7-4b Figure 11.7-4c Figure 11.7-4d Recall that the equation of a parabola has the form x, the equation can be written as 2 4px. y˛ Solving for x 1 4p y2 Therefore, the coefficient of the squared term can be used to find the value of p, which in turn can be used to find both the focus and the directrix of the parabola. Setting the coefficient of the squared term, 5, equal to 1 4p yields 5 1 4p or p 1 20 Section 11.7 Plane Curves and Parametric Equations 759 Note that the vertex has been translated to 3,2 1 2 and that the focus is of a unit to the right of the vertex. Thus, the focus is at 3 1 20 a , 2 b ˛, 1 20 or 61 20 a , 2 b . The directrix is the vertical line that is 1 20 of a unit to the left of the vertex, that is, x 3 1 20 59 20 . ■ Applications Example 5 Application of Parameterization of a Parabola A golfer hits a ball with an initial velocity of 140 feet per second so that with the horizontal. its path as it leaves the ground makes an angle of 31° a. When does the ball hit the ground? b. How far from its starting point does it land? c. What is the maximum height of the ball during its flight? NOTE In the applications in this section, air resistance is ignored and some facts about gravity that are proved in physics are assumed. y Solution (x, y) y x 140 t 31° x Figure 11.7-5a 100 0 0 600 Figure 11.7-5b Imagine that the golf ball starts at the origin and travels in the direction of the positive x-axis. If there were no gravity, the distance traveled by the ball in t seconds would be 140t feet. As shown in Figure 11.7-5a, the coordinates of the ball would satisfy x, y 1 2 x 140t cos 31° y 140t sin 31° x 1 140 cos 31° t 2 y 1 140 sin 31° t. 2 However, gravity at time t exerts a force of 16t2 feet per second per second downward, that is, in the negative direction on the y-axis. Consequently, the coordinates of the golf ball at time t are x 1 140 cos 31° 2 t and y 140 sin 31° 1 t 16t˛ 2. 2 The path given by these parametric equations is shown in Figure 11.7-5b. a. The ball is on the ground when y 0, that is, at the x-intercepts of the graph, which can be found geometrically by using trace and zoom-in. To find the intercepts algebraically, set t 16t˛ 2 140 sin 31° 16t y 0 2 0 0 140 sin 31° t and solve for t. 1 1 2 t 0 or 140 sin 31° 16t 0 Thus, the ball hits the ground after approximately 4.5066 seconds. t 140 sin 31° 16 4.5066 760 Chapter 11 Analytic Geometry Technology Tip The graphical root finder and maximum finder do not operate in parametric mode. b. The horizontal distance traveled by the ball is given by the x-coordinate of the second intercept. The x-coordinate when t 4.5066 is x 1 140 cos 31° 4.5066 2 21 540.81 feet. c. The graph in Figure 11.7-5 looks like a parabola—and it is, as you can verify by eliminating the parameter t (see Exercise 40). The y-coordinate of the vertex is the maximum height of the ball. It can be found graphically by using trace and zoom-in, or algebraically as follows. The vertex occurs halfway between the two x-intercepts at x 0 540.81 x 540.81, that is, when 270.405. x 0 and 2 t x 270.405 140 cos 31° 1 2 t 270.405 140 cos 31° 2.2533 Therefore, the y-coordinate of the vertex, which is the maximum height of the ball, is y 2 81.237 feet. 140 sin 31° 2.2533 2.2533 16 1 21 2 1 2 ■ Example 6 Projectile Motion y (x, y) 138 t 26° x 3 y − 3 x Figure 11.7-6a A batter hits a ball that is 3 feet above the ground. The ball leaves the bat with an initial velocity of 138 feet per second, making an angle of with the horizontal and heading toward a 25-foot fence that is 400 feet away. Will the ball go over the fence? 26° Solution Suppose that home plate is at the origin and that the ball travels in the direction of the positive x-axis. The vertical and horizontal distances traveled by the ball, disregarding gravity, are 75 0 0 (400, 25) Figure 11.7-6b x 138t cos 26° x 1 138 cos 26° 2 as shown in Figure 11.7-6a. sin 26° y 3 138t t y 138 sin 26° 1 t 3, 2 Allowing for the effect of gravity on the y-coordinate, the ball’s path is given by the parametric equations 600 x 1 138 cos 26° 2 t and y 138 sin 26° 2 1 t 3 16t˛ 2. The graph of the ball’s path, shown in Figure 11.7-6b, was made with the grid-on feature and vertical tick marks 25 units apart. It shows that the y-coordinate of the ball is greater than 25 when its x-coordinate is 400. So, the ball goes over the fence. ■ Section 11.7 Plane Curves and Parametric Equations 761 The procedure used in Example 6 applies to the general case. Replacing 3 by k, and 138 by v leads to the following conclusion. 26° by u, Projectile Motion When a projectile • is fired from the position (0, k) on the positive y-axis at an angle with the horizontal, U • in the direction of the positive x-axis, • with initial velocity v feet per second, • with negligible air resistance, then its position at time t seconds is given by the parametric equations x (v cos U)t and y (v sin U)t k 16t˛ 2. Graphing Exploration Will the ball in Example 6 go over the fence if its initial velocity is 135 feet per second? Use degree mode and the viewing window of Figure 11.7-6b with to graph the ball’s and path. You may need to use trace if the graph is hard to read. If the answer still is not clear, try changing the t step to 0.02. t step 0.1 0 t 4 Cycloids Imagine a bug that is sitting at a point P at the edge of a wheel. The path traced out by the bug as the wheel rolls is a curve called a cycloid, as shown in Figure 11.7-7. P P P Figure 11.7-7 x Technology Tip In parametric graphing zoom-in can be very time-consuming. It is often more effective to limit the t range to the values near the points you are interested in and set the t step very small. The picture may be hard to read, but trace can be used to determine coordinates. P cycloid Q Figure 11.7-8 Cycloids have a number of interesting applications. For example, of all the possible paths joining P and Q in Figure 11.7-8, an arch of an inverted cycloid (in red) is the curve along which a particle subject only to gravity will slide from P to Q in the shortest possible time. The Dutch physicist Christiaan Huygens, who invented the pendulum clock, proved that a particle takes the same amount of time to slide to the bottom point, Q, of an inverted cycloid (see Figure 11.7-9) starting from any point P on the curve. 762 Chapter 11 Analytic Geometry P P P P P Q Figure 11.7-9 Example 7 Parameterization of a Cycloid Find a parameterization of a cycloid generated by point P on a circle of radius 3 that rolls along the x-axis. Solution Begin with P at the origin and the center C of the circle at (0, 3). As the circle rolls along the x-axis, the segment rotates through an angle of t radians, as shown in Figure 11.7-10. CP 12 3 3t Figure 11.7-10 y O 3 The distance from T to the origin is the length of the arc of the circle from T to P. From the formula for arc length in Section 6.3, ru 3t. Therefore, the center C has coordinates (3t, 3). For PQC in Figure 11.7-11 shows that 0 6 t 6 p 2 , triangle 3 P t C Q (x, y) x T or, equivalently, sin t PQ 3 and cos t CQ 3 PQ 3 sin t and CQ 3 cos t Figure 11.7-11 Thus, the point (x, y) in Figure 11.7-11 has the following coordinates. x OT PQ 3t 3 sin t 3 y CT CQ 3 3 cos t 3 t sin t 1 1 cos t 1 2 2 Section 11.7 Plane Curves and Parametric Equations 763 20 0 −10 14π Figure 11.7-12 A similar analysis for other values of t shows that these equations are valid for all values of t. (See Exercises 44–46.) Therefore, the parametric equations of the cycloid are x 3 and y 3 1 cos t t sin t , for any t. 1 2 1 2 ■ In general, a cycloid generated by a point on a circle of radius r has the following parameterization. x r (t sin t) y r (1 cos t) , for any t and Exercises 11.7 In Exercises 1–14, find a viewing window that shows a complete graph of the curve. 1. x t 2 4, y t 2 , 2 t 3 2. x 3t 2, y 2 5t, 0 t 2 3. x 2t, y t 2 1, 1 t 2 4. x t 1. x 4 sin 2t 9, y 6 cos t 8, 0 t 2p 6. x t 3 3t 8, y 3t 2 15, 4 t 4 7. x 6 cos t 12 cos 0 t 2p 2t, y 8 sin t 8 sin t cos t, 8. x 12 cos t, y 12 sin 2t, 0 t 2p 9. x 6 cos t 5 cos 3t, y 6 sin t 5 sin 3t , 0 t 2p 10. x 3t 2 10, y 4t 3, t any real number 11. x 12 cos 3t cos t 6, y 12 cos 3t sin t 7, 0 t 2p 12. x 2 cos 3t 6, y 2 cos 3t sin t 7, 0 t 2p 13. x t sin t, y t cos t, 0 t 8p 14. x 9 sin t, y 9t cos t, 0 t 20 15. x t 3, y 2t 1, t 0 16. x t 5, y 1t, t 0 17. x 2 t 2, y 1 2t 2 , for any t 18. x t 2 1, y t 2 1 , for any t 19. x e t, y t , for any t 20. x 2e t, y 1 e t, t 0 21. x 3 cos t, y 3 sin t, 0 t 2p 22. x 4 sin 2t, y 2 cos 2t, 0 t 2p 23. x 3 cos t, y 4 sin t, 0 t 2p 24. x 2 sin t 3, y 2 cos t 1, 0 t p In Exercises 25 and 26, sketch the graphs of the given curves and compare them. Do they differ? If so, how? 25. a. b. x 4 6t, x 2 6t, y 7 12t, y 5 12t, 0 t 1 0 t 1 26. a. b. c. x t, x 1t , x e t 2t , for any t for any t for any t 27. By eliminating the parameter, show that the curve with parametric equations x a c a t, y b 1 2 for any t d b t 2 1 is a straight line. In Exercises 15–24, the given curve is part of the graph of an equation in x and y. Eliminate the parameter by solv
ing one equation for t and substituting the result into the other equation. In Exercises 28–30, find a parameterization of the given curve. Confirm your answer by graphing. 28. line segment from Exercise 27. 14, 5 1 to 1 2 5, 14 2 Hint: See 2 (within one degree) needed so that the ball travels at least 150 feet. 764 Chapter 11 Analytic Geometry 29. line segment from 6, 12 2 1 30. line segment from (18, 4) to 1 to 12, 10 1 16, 14 2 In Exercises 31–34, locate all local maxima and minima (other than endpoints) of the curve. 31. x 4t 6, y 3t˛ 2 2, 10 t 10 32. x t˛ 3 sin t 4, y cos t, 1.5 t 2 33. 34. x 4t˛ 3 t 4, y 3t˛ 2 5, 2 t 2 x 4t˛ 3 cos t 5, y 3t˛ 2 8, 2 t 2 35. Show that the ball’s path in Example 5 is a parabola by eliminating the parameter in the parametric equations below. x 140 cos 31° y 140 sin 31° t 2 t 16t˛ 2 2 1 1 In Exercises 36–41, use a calculator in degree mode, and assume that air resistance is negligible. 36. A ball is thrown from a height of 5 feet above the ground with an initial velocity of 60 feet/second at an angle of a. Graph the ball’s path. b. When and where does the ball hit ground? with the horizontal. 50° 37. A medieval bowman shoots an arrow which leaves the bow 4 feet above the ground with an initial velocity of 88 feet/second at an angle of with the horizontal. a. Graph the arrow’s path. b. Will the arrow go over the 40-foot-high castle 48° wall that is 200 feet from the archer? 38. A golfer at a driving range stands on a platform 2 feet above the ground and hits the ball with an initial velocity of 120 feet/second at an angle of 39° with the horizontal. There is a 32-foot-high fence 400 feet away. Will the ball fall short, hit the fence, or go over it? 39. A golf ball is hit off the tee at an angle of and lands 300 feet away. What was its initial velocity? y 0. Hint: The ball lands when Use this fact and the parametric equations for the ball’s path to find two equations in the variables t and v. Solve for v. x 300 and 30° 40. A football kicked from the ground has an initial velocity of 75 feet/second. a. Set up the parametric equations that describe the ball’s path. Experiment graphically with different angles to find the smallest angle and x 150 b. Use algebra and trigonometry to find the angle needed for the ball to travel exactly 150 feet. y 0. Hint: The ball lands when Use this fact and the parametric equations for the ball’s path to find two equations in the variables t and and substitute this result into the other one; then solve for The double-angle identity may be helpful for putting this equation into a form that is easy to solve. Solve the “x equation’’ for t u. u. 41. A skeet is fired from the ground with an initial velocity of 110 feet/second at an angle of a. Graph the skeet’s path. b. How long is the skeet in the air? c. How high does it go? 28°. 42. A golf ball is hit off the ground at an angle of u degrees with an initial velocity of 100 feet/second. a. Graph the path of the ball when u 30° and In which case does the ball land u 60°. when farthest away? b. Do part a when c. Experiment further and make a conjecture as to the results when the sum of the two angles is 90°. u 65°. u 25° and d. Prove your conjecture algebraically. Hint: Find the value of t at which a ball hit at angle hits the ground (which occurs when value of t will be an expression involving Find the corresponding value of x (which is the distance of the ball from the starting point). Then do the same for an angle of use the cofunction identities (in degrees) to show that you get the same value of x. u this u. 90° u y 0 , and ; 2 43. A golf ball is hit off the ground at an angle of u degrees with an initial velocity of 100 feet/second. a. Graph the path of the ball when u 20°, u 40°, u 60°, and b. For what angle in part a does the ball land u 80°. farthest from where it started? c. Experiment with different angles, as in parts a and b, and make a conjecture as to which angle results in the ball landing farthest from its starting point. In Exercises 44–46, complete the derivation of the parametric equations of the cycloid in Example 7. 44. a. If p 2 6 t 6 p has measure , verify that angle t p 2 and that u in the figure Section 11.7 Plane Curves and Parametric Equations 765 t p x OT CQ 3t 3 cos 2 R Q t p y CT PQ 3 3 sin 2 R Q . b. Use the addition and subtraction identities for sine and cosine to show that in this case x 3 . and y 3 1 cos t t sin t 1 2 1 2 y P (x, y 3t 45. a. If 3p 2 6 t 6 2p, verify that angle u in the figure has measure t 3p 2 and that x OT CQ 3t 3 cos y CT PQ 3 3 sin t 3p a 2 b t 3p a 2 b . b. Use the addition and subtraction identities for sine and cosine to show that in this case t sin t 2 and y 3 1 1 cos x, y) x O T 3t 46. a. If , verify that angle u in the figure p 6 t 6 3p 2 t measures 3p 2 and that x OT CQ 3t 3 cos y CT PQ 3 3 sin 3p a 2 3p 2 a t b t b . b. Use the addition and subtraction identities for sine and cosine to show that in this case x 3 1 t sin t 2 and y 3 1 cos t . 2 1 y 3 P (x, y) 3 θ t C Q x O T 3t x 8 cos t and y 5 sin t x 3t and y 5t x 3t and y 4t 47. Critical Thinking Set your calculator for radian mode and for simultaneous graphing mode. Check your instruction manual for how to do this. Particles A, B, and C are moving in the plane, with their positions at time t seconds given by: A: B: C: a. Graph the paths of A and B in the window 0 t 2. with paths intersect, but do the particles actually collide? That is, are they at the same point at the same time? For slow motion, choose a very small t-step, such as 0.01. 0 x 12, 0 y 6, and The b. Set t-step 0.05 and trace to estimate the time at which A and B are closest to each other. c. Graph the paths of A and C and determine geometrically, as in part b, whether they collide. Approximately when are they closest? d. Confirm your answers in part c as follows. 1 Explain why the distance between particles A and C at time t is given by 8 cos t 3t d 2 2. 2 2 A and C will collide if at some time. Using function graphing mode, graph this distance function when . Zoom-in if necessary, and show that d is always positive. Find the value of t for which d is smallest. 5 sin t 4t 0 t 2 1 d 0 2 48. Critical Thinking A particle moves on the y 3. horizontal line seconds is given by exercise explores the motion of the particle. a. Graph the path of the particle in the viewing Its x-coordinate at time t 2 23t 8. 3 13t˛ x 2t˛˛ This 10 x 10, 2 y 4, window with 0 t 4.3, t-step 0.05. calculator seems to pause before completing the graph. Note that the and 766 Chapter 11 Analytic Geometry b. Use trace (starting with t 0 and watch the c. At what times t does the particle change direction? What are its x-coordinates at these times? 2 path of the particle as you press the right arrow key at regular intervals. How many times does it change direction? When does it appear to be moving the fastest? 11.7.A Excursion: Parameterizations of Conic Sections Objectives • Define parametric equations for a circle, an ellipse, a hyperbola, and a parabola Conic sections can often be graphed more conveniently in parametric mode. Parameterizations for conic sections can be found by using Pythagorean identities, as shown in the following examples. Circles Example 1 Parameterization of a Circle The equation of the circle with center (4, 1) and radius 3 is x 4 2 9. y 1 2 1 Show that the following equations provide a parameterization of this circle. 1 2 2 x 3 cos t 4 and y 3 sin t 1, 0 t 2p [1] Solution To show that the parametric equations satisfy the circle equation, substitute into the equation of the circle and use the Pythagorean identity. 10 sin sin t 1 1 1 3 cos t 4 4 2 3 cos t 2 2 9 cos2 t 9 sin2 t cos2 t sin2 t 9 1 9 1 1 9 2 2 With this parameterization the circle is traced out in a counterclockwise direction from the point (7, 1), as shown in Figure 11.7.A-1. Another parameterization is given by x 3 cos 2t 4 and y 3 sin 2t 1, 0 t p Verify that this last parameterization traces out the circle in a clockwise direction twice as fast as the parameterization given in [1], because t runs from 0 to rather than to 2p. p, ■ 5 3 2 Figure 11.7.A-1 NOTE When the values of t are given in radian measure, such as sure your calculator is in radian mode when graphing. 2p, make Section 11.7.A Excursion: Parameterizations of Conic Sections 767 Parametric Equations of a Circle The circle with center (c, d) and radius r is given by the parametric equations x r cos t c and y r sin t d (0 t 2P) . The procedure used in Example 1 works in the general case. Example 1 is the special case where r 3 and 4, 1 c, d . 1 2 1 2 Ellipses Because an ellipse is a generalization of a circle, a similar parameterization can be used. Example 2 Parameterization of an Ellipse Find a parameterization of the following ellipse. 2 x˛ 25 2 y˛ 4 1 6.4 Solution Let 9.4 Use the Pythagorean identity to show that these parametric equations satisfy the equation of an ellipse. 9.4 x 5 cos t and y 2 sin t, 0 t 2p . 6.4 Figure 11.7.A-2 2 x˛ 25 2 y˛ 4 2 2 2 2 1 1 2 sin t 4 4 sin2 t 4 5 cos t 25 25 cos2 t 25 cos2 t sin2 t 1 The graph is shown in Figure 11.7.A-2. Its major axis has length and its minor axis has length 2 2 4. 2 5 10, ■ The parameterization in Example 2, where the center of the ellipse is at (0, 0), can be extended to the general case. Parametric Equations of an Ellipse The ellipse with center and a horizontal axis of length 2a and a vertical axis of length 2b is given by the parametric equations (c, d) x a cos t c and y b sin t d (0 t 2P). 768 Chapter 11 Analytic Geometry Hyperbolas The hyperbola centered at (c, d) with equation 2 1 x c 2 a˛ 2 1 2 y d 2 b˛ 2 1 can be obtained from the following parameterization. By a Pythagorean identity, x a sec t c y b tan t d, 0 t 2p 1 tan2 t sec2 t. a sec t c c a2 2 y d b2 b tan t d d b2 Therefore a2 1 a sec t a2 2 1 b tan t b2 2 Parametric Equations of a Hyperbola a2 sec2 t a2 sec2 t tan2 t 1 b2 tan2 t b2 A similar argument works for other hyperbolas and l
eads to the following conclusion. Hyperbolas with center at (c, d) have the following parameterizations. Equation (x c)2 a2 (y d)2 b2 1 (y d)2 a2 (x c)2 b2 1 Parameterization x a sec t c y b tan t d (0 t 2P) x b tan t c y a sec t d (0 t 2P) Example 3 Parameterization of a Conic Identify the conic section whose equation is given below, and find a parameterization for it 16 2 1 Solution The equation is a hyperbola with center at as the second equation in the preceding box, with c, d Therefore, its parametric equations are x 4 tan t 2 y 3 sec t 5, 0 t 2p. 2, 5 It has the same form a 3, b 4, and ■ Section 11.7.A Excursion: Parameterizations of Conic Sections 769 Parametric Equations of a Parabola When a parabola has an equation such as y 4 x 5 1 2 2 7, in which y is a function of x, then it can be graphed on a calculator either in function mode or in parametric mode with x t and y 4 t 5 1 2 2 7. A parabola with an equation such as x 2 y 3 1 2 2 4, in which x is a function of y, cannot be graphed (by using a single equation) in function mode on a calculator, but it can be graphed in parametric mode by letting x 2 t 3 2 1 2 4 and y t. Similar techniques work for other parabolas. Exercises 11.7.A In Exercises 1–4, find a parameterization of the given curve. Confirm your answer by graphing. 1. circle with center (9, 12) and radius 5 2. 2 14x 8y 29 0 2 y˛ x˛ in Section 11.4. Hint: see Example 2 3. 2 y˛ x˛ 2 4x 6y 9 0 4. circle with center 7, 4 1 2 and radius 6 In Exercises 5–26, find parametric equations for the curve whose equation is given, and use these parametric equations to find a complete graph of the curve. 5. 7. 9. x2 10 1 y2 36 4x˛ 2 4y˛ 2 1 x2 10 y2 36 1 11. 2 4y˛ 2 1 x˛ 6. y2 49 x2 81 1 8. 2 4y˛ x˛ 2 1 y2 9 x2 16 1 2x˛ 2 y˛ 2 4 10. 12. 13. 8x 2y˛ 2 14. 4y x˛ 2 15 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 2 1 x 2 16 2 1 2 y 3 12 2 1 2 1 x 1 16 12 ˛1 y 3˛ 25 2 1 2 2 2 x 1 16 25 11 R E V I E W Important Concepts Section 11.1 Section 11.2 Section 11.3 Section 11.4 Ellipse: foci, center, vertices, major and minor axes. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 692 Equation of an ellipse centered at the origin . . . 693 Characteristics of ellipses . . . . . . . . . . . . . . . . . . 694 Applications of ellipses . . . . . . . . . . . . . . . . . . . . 696 Hyperbola: foci, center, vertices, asymptotes, focal axis . . . . . . . . . . . . . . . . . . . . . . . . . . . 700-701 Equation of a hyperbola centered at the origin. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 701 Characteristics of hyperbolas . . . . . . . . . . . . . . . 702 Applications of hyperbolas . . . . . . . . . . . . . . . . . 705 Parabola: focus, directrix, vertex, axis. . . . . . . . . 709 Equation of a parabola with vertex at the origin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 710 Characteristics of parabolas . . . . . . . . . . . . . . . . 711 Applications of parabolas . . . . . . . . . . . . . . . . . . 712 Horizontal and vertical shifts . . . . . . . . . . . . . . . 716 Standard equations of conic sections . . . . . . . . . 720 Rotations of conics . . . . . . . . . . . . . . . . . . . . . . . 722 Discriminant . . . . . . . . . . . . . . . . . . . . . . . . . . . . 723 Applications of rotated conics. . . . . . . . . . . . . . . 724 Section 11.4.A Rotation equations . . . . . . . . . . . . . . . . . . . . . . . 729 Rotation angle to eliminate the xy term . . . . . . . 730 Section 11.5 Polar coordinates . . . . . . . . . . . . . . . . . . . . . . . . 734 Pole . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 734 Polar axis. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 734 Polar/rectangular coordinate conversion . . . . . . 737 Polar graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . 739 Section 11.6 Eccentricity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 745 Alternate definition of conic sections . . . . . . . . . 747 Polar equations of conic sections . . . . . . . . . . . . 749 770 Chapter Review 771 Section 11.7 Plane curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . 755 Parameter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 755 Parametric equations. . . . . . . . . . . . . . . . . . . . . . 755 Eliminating the parameter . . . . . . . . . . . . . . . . . 757 Cycloid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 761 Section 11.7.A Parametric equations of a circle . . . . . . . . . . . . . 767 Parametric equations of an ellipse . . . . . . . . . . . 767 Parametric equations of a hyperbola . . . . . . . . . 768 Parametric equations of a parabola. . . . . . . . . . . 769 Important Facts and Formulas Equation of an ellipse with center (h, k) and axes on the lines x h, y k: 2 1 x h a2 2 1 2 y k b2 2 1 Equation of a hyperbola with center (h, k) and vertices on the line y k: 2 1 x h a2 2 1 2 y k b2 2 1 Equation of hyperbola with center (h, k) and vertices on the line x h: 2 1 y k a2 2 1 2 x h b2 2 1 Equation of a parabola with vertex (h, k) and axis x h: Equation of a parabola with vertex (h, k) and axis x h 2 4p y k 2 1 2 1 y k: Rotation equations: y k 2 1 2 4p 1 x h 2 x u cos u v sin u y u sin u v cos u Rotation Angle: To eliminate the xy term in rotate the axes through an angle u such that Ax2 Bxy Cy2 Dx Ey F 0, cot 2u A C . B The rectangular and polar coordinates of a point are related by x r cos u r2 x˛ 2 y2 and and y r sin u tan u y x 772 Chapter Review If e and d are constants with tion of the form e 7 0, then the graph of a polar equa- r ed 1 – e cos u or r ed 1 – e sin u is an ellipse if e 7 1. 0 6 e 6 1, a parabola if e 1, and a hyperbola if Review Exercises In Exercises 1–10, find the foci and vertices of the conic, and find a viewing window that shows a complete graph of the equation. Section 11.1 1. x2 16 y2 20 1 3. 25x2 4y2 100 2. x2 4 y2 25 1 4. 4x2 9y2 36 5. Find the equation of the ellipse with center at the origin, one vertex at . (0, 4), passing through B 6. Find the equation of the ellipse with center at the origin, one vertex 212 13, 213 A at (3, 0), passing through 1, a . 3 b Section 11.2 7. x 2 9 y2 16 1 8. x 2 16 y2 4 1 9. Find the equation of the hyperbola with center at the origin, one vertex at . passing through 1, 212 0,2 , 1 2 A B 10. Find the equation of the hyperbola with center at the origin, one vertex at (3, 0), passing through 5, 8 3b a . Section 11.3 In Exercises 11–16, find the equation of the parabola with vertex at the origin that satisfies the given condition 11. axis x 0, passing through 12. axis y 0, passing through 1, 5 1 1, 5 2 2 1 13. focus 14. focus 1 1 4, 0 2 0, 3 15. directrix 2 x 4 16. directrix y 2 17. Find the focus and directrix of the parabola 10y 7x2. 18. Find the focus and directrix of the parabola 3y2 x 4y 4 0. Section 11.4 In Exercises 19–28, sketch the graph of the equation and identify the conic. If there are asymptotes, give their equations and label all characteristic points. Chapter Review 773 19. 21 16 2 2 y 5 4 2 1 1 23. 4x2 9y2 144 20. 3x2 1 2y2 22. 2 1 y 4 25 2 1 2 x 1 4 2 1 24. x2 4y2 10x 9 0 25. 27. 2 6 2y 4 x 3 1 x y˛2 2y 2 2 26. 28. 2 9 x 1 3y 6˛1 y x˛2 2x 3 2 29. What is the center of the ellipse 4x˛2 3y˛2 32x 36y 124 0? 30. Find the equation of the hyperbola with center at the origin, one vertex at (0, 5), passing through 1, 3˛15 A . B 31. Find the equation of the hyperbola with center at (3, 0), one vertex at (3, 2), passing through 1, 15 A . B 32. Find the equation of the parabola with vertex (2, 5), axis x 2, passing through (3, 12). 33. Find the equation of the parabola with vertex 3, 1 passing through . 1 2 3 2 , 1 2b a , axis y 1 2 , 34. Find the equation of the parabola with vertex (5, 2) that passes through the points (7, 3) and (9, 6). 35. Find the equation of the ellipse with center at (3, 1), one vertex at (1, 1), passing through 2, 1 a 3 2 A . b In Exercises 36–39, assume that the graph of the equation is a nondegenerate conic. Use the discriminant to identify the graph. 36. x˛2 y˛2 xy 4y 0 37. 4xy 3x˛2 20 0 38. 4x˛2 4xy y˛2 15x 215y 0 39. 3x˛2 212xy 2y2 12 0 In Exercises 40–45, find a viewing window that shows a complete graph of the equation. 40. x˛2 xy y˛2 3y 6 0 41. x˛2 xy 2 0 42. x˛2 4xy y˛2 5 0 43. x˛2 3xy y˛2 212x 212y 0 44. x˛2 2xy y˛2 412y 0 45. x˛2 xy y˛2 6 0 774 Chapter Review Section 11.4.A In Exercises 46–47, find the rotation equations when the x- and y-axes are rotated through the given angle. 46. 45° 47. 60° In Exercises 48–49, find the angle through which the x- and y-axes should be rotated to eliminate the xy term in the equation. 48. x ˛2 4xy y ˛2 5 0 49. x ˛2 xy y ˛2 3y 6 0 Section 11.5 50. List four other pairs of polar coordinates for the point 2, a p 4 b . 51. Plot the points 3p 4 b 2, a and 3, 2p a 3 b on a polar coordinate graph. In Exercises 52–61, sketch the graph of the polar equation. 52. r 2 54. u 5p 6 56. r 4 cos u 58. r cos 3u 60. r 1 2 sin u 53. u 2p 3 55. r 2u u 0 1 2 57. r 2 2 sin u 59. 2 cos 2u r˛ 61. r 5 62. Convert A 63. Convert a 3, 13 B 3, 2p 3 b from rectangular to polar coordinates. from polar to rectangular coordinates. Section 11.6 64. What is the eccentricity of the ellipse 24x˛2 30y˛2 120? 65. What is the eccentricity of the ellipse 3x˛2 y˛2 84? In Exercises 66–69, sketch the graph of the equation, labeling the vertices and identifying the conic. 66. r 14 7 7 cos u 68. r 10 3 4 sin u 67. r 24 3 9 cos u 69. r 12 2 sin u In Exercises 70–73, find a polar equation of the conic that has focus (0, 0) and satisfies the given conditions. 70. hyperbola; vertices p 2 b 5, a and 3, a 3p 2 b 71. eccentricity 1; directrix r 2 sec u 72. eccentricity 0.75; directrix r 3 csc u 73. ellipse; vertices (4, 0) and 6, p 1 2 Chapter Review 775 In Exercises 74–77, find a viewing window that shows a complete graph of the curve with the given parametric equations. 74. x 64 cos 3 p6 1 2 4 t and y 16t2 64 sin S p 6 T t, 0 t p 75. x t˛ 3 t 1 and y t2 2t, 3 t 3 76. x t2 t 3 and y t3 5t, 3 t 3 77. x 8
cos t cos 8t and y 8 sin t sin 8t, 0 t 2p Section 11.7 In Exercises 78–81, sketch the graph of the curve whose parametric equations are given, and by eliminating the parameter, find an equation in x and y whose graph contains the given curve. 78. x 3 cos t, y 5 sin t, 0 t 2p 79. x cos t, y 2 sin2 t, 0 t 2p 80. x et, y 1t 1, t 1 81. x 2t 1, y 2 t, 3 t 3 82. Which of the following is not a parameterization of the curve x y˛ 2 1? a. b. c. d. x t2 1, x sin2 t 1, x t˛4 1, x t˛6 1, y t, y sin t, y t˛2, y t 3, any real number t any real number t any real number t any real number t 83. Which of the curves in Questions 74–77 appear to be the graphs of ? functions of the form y f x 1 2 Section 11.7.A In Exercises 84–87, find a parameterization of the given curve. Confirm your answer by graphing. 84. circle with center 85. circle with center 3, 2 3, 5 2 2 1 1 and radius 4 and radius 5 86. 9x2 4y 2 54x 16y 61 0 87. 4x2 y 2 16x 6y 21 0 In Exercises 88–97, find parametric equations for the curve whose equation is given, and use these parametric equations to find a complete graph of the curve. 88. 9x2 9y2 1 90. 16x2 y2 1 92. 94 25 2 1 2 1 x 2 81 2 1 2 y 5 100 2 1 96. y 3 x 2 2 5 2 1 89. 4x2 9y2 1 91. x2 36y2 1 93. 95. 2 1 x 2 49 2 1 2 y 5 64 12 2 1 97. x 32 11 Figure 11.C-1 Arc Length of a Polar Graph Many applications of calculus involve finding the distance along a curve, or arc length. Although calculus is usually needed to find the exact value of the arc length, approximations are often sufficient. A curve may be approximated by straight segments with endpoints on the curve, as shown in Figure 11.C-1. In polar coordinates, the curve, r c for some conwhich is a circle, can be represented by the function 2pc 6.28c. stant c. The circumference of the circle is The length of each blue segment is also c, since the dashed lines form equilateral triangles. Thus, the approximate length of the curve is the total length of the blue segments, or 6c. In general, the Law of Cosines may be used to find the length of a segment with two endpoints on the curve, as shown in the following example. Example 1 Estimating the Length of a Curve Estimate the length of the spiral with the equation r 2u from 0 to 2p. Solution The dashed lines in Figure 11.C-2 divide the spiral into triangles with an Figure 11.C-2 angle of value of p 3 u at the origin. The table below shows the value of r for each that is an endpoint of a segment. u r 0 0 p 3 2p 3 2p 3 4p 3 p 2p 4p 3 8p 3 5p 3 10p 3 2p 4p a π 3 r2 The first segment has one endpoint at the origin, so its length is the value 2p 3 of r at the other endpoint, which is . r1 Every remaining segment is the side of a triangle that is opposite an angle Figure 11.C-3 at the origin of p 3 , as shown in Figure 11.C-3. By the Law of Cosines, 2 r 2 a2 r 1 a 2r 1 2 2 r1r2 cos 2 r1r2 2 r 2 p 3 776 The lengths of the segments can be calculated using this formula. Segment from u 0 to u p 3 : 2p 3 2.09 Segment from Segment from u p 3 to u 2p 3 : u 2p 3 to u p : BQ BQ 2p 3 R 2 4p 3 R 2 2 4p 3 R Q 2 2p 2 1 2p 3 R Q 4p 1 3 R Q 4p 3 R Q 3.63 5.54 2p 2 Segment from u p to Segment from Segment from u 4p 3 u 5p 3 : u 4p 3 u 5p 3 2 2p 2 B 1 2 8p 3 b a 2p 1 2 a 8p 3 b 7.55 : B a 2 8p 3 b 10p 2 a 3 b 8p 3 b a a 10p 3 b 9.60 to to u 2p : 10p 2 B a 3 b 2 4p 2 1 10p 3 b 1 a 4p 2 11.66 The approximate length of the spiral is the sum of the segments. arc length 2.09 3.63 5.54 7.55 9.60 11.66 40.07 ■ Exercises In Exercises 1 – 4, approximate each curve for 0 U 2P by six segments and estimate each arc length. 1. r 1.5u 2. r 3 2 cos u 3. r 1 cos u 4. r 1 2 sin u 5. Use the figure to estimate the length of the cardioid with the equation r 1 sin u π4 3 π 3 2 π5 3 6. Use Heron’s formula for the area of a triangle (see page 633) and your results from Exercise 5 to estimate the area of the cardioid with the equation r 1 sin u 777 C H A P T E R 12 Systems and Matrices Is this a diamond in the rough? The structure of certain crystals can be defined by a large system of linear equations with more than a hundred equations and variables. A variety of resource allocation problems involving many variables can be handled by solving an appropriate system of equations. The fastest solution methods involve matrices and are easily implemented on a computer or calculator. See Exercise 36 in Section 12.2. 778 Chapter Outline 12.1 Solving Systems of Equations 12.1.A Excursion: Graphs in Three Dimensions 12.2 Matrices 12.3 Matrix Operations 12.4 Matrix Methods for Square Systems 12.5 Nonlinear Systems 12.5.A Excursion: Systems of Inequalities Chapter Review can do calculus Partial Fractions Interdependence of Sections Readers who are familiar with solving systems of linear equations may omit section 12.1. 12.1 > 12.2 > 12.3 > 12.4 > 12.5 Real-world situations often require a common solution to several equations with multiple variables. Such a collection of equations is known as a system of equations. Solutions to a system of equations in two or three variables may be represented geometrically by intersections of lines or planes. In this chapter, systems will be solved graphically, alge- braically by substitution or elimination, and by two matrix methods: row reduction and inverse matrices. 12.1 Solving Systems of Equations Objectives • Solve systems of equations by graphing, substitution, and elimination • Recognize consistent and inconsistent systems • Solve applications using systems • Recognize consistent, inconsistent, and dependent systems. • Solve application s using systems. A system of equations is a set of two or more equations in two or more 3 2 variables. When a system has 3 equations in 2 variables, it is called a system. In the examples of systems of equations shown below, the first is 3 3 3 4 2 2 a system. system, and the third is a system, the second is a 2x 5y 3z 1 x 2y z 2 3x y 2z 11 2x 5y z w 0 2y 4z 41w 5 3x 7y 5z 8w 6 x2 y2 25 x2 y 7 Three equations in three variables Three equations in four variables Two equations in two variables The first two systems above are called linear systems because the variables in each equation are all to the power of one, thus they are all linear. The third is a nonlinear system because at least one equation is nonlinear— in this case, quadratic. 779 780 Chapter 12 Systems and Matrices Technology Tip Table setup is accessed from the TBLSET key of TI and RANG in the Casio TABLE menu. ¢ The increment is TBL on TI and PITCH on Casio. The table type is labeled INDPNT on TI. Figure 12.1-1 Solutions of a System of Equations A solution of a system of equations is a set of values that satisfy all the equations in the system. In the first system of equations on the previous page, substituting x 1, y 2, and 2x 5y 3z 2 x 2y z 3x y 2z 3 2 1 3 z 3 gives the following: 3 2 10 11 , makes all equations true, the Because the set of values y 7, set is a solution of the system. The set of values is a solution of the first two equations, but not the third, so it is not a solution of the system. z 12 x 0, Solutions of systems of equations in two variables can be found numerically by comparing tables of values for the equations. Example 1 Solving a System Numerically Find a solution of the system of equations below by using tables of values for the equations. 2x y 1 3x 2y 12 Solution First, solve each equation for y. Then create a table of values for each equation. The table in Figure 12.1-1 shows solutions to each equation. To solve the system, find a common output. y 2x 1 y 3 2 x 6 Notice that at x 2, y 3 x 2, is a solution of the system of equations. the y-values are the same for the two equations. Thus, ■ Solving systems numerically has several disadvantages. First, there is no way of knowing whether all possible solutions have been found. Second, many values may have to be checked before a solution is found. And third, if a solution lies between the values in the table, it may be missed. Solving Systems with Graphs One method of solving systems of equations in two variables is graphing the equations and finding the point(s) of intersection. Since the graph of each equation represents all possible solutions of that equation, a point of intersection of two graphs represents a solution of both equations. The advantage of solving a system graphically is that the solution is shown visually, but solving systems graphically is limited to two-variable systems. Section 12.1 Solving Systems of Equations 781 Example 2 Graphical Solutions of a Linear System Find a solution of the system of equations below by graphing the equations. 2x y 1 3x 2y 4 3.1 Solution Solve each equation for y, graph each equation, and find the coordinates of all points of intersection. 4.7 4.7 y 2x 1 y 3 2 x 2 3.1 Figure 12.1-2 The system of equations has exactly one solution, as shown in Figure 12.1-2. x 0.86 and y 0.71, ■ Type of System and Number of Solutions Systems of equations may be classified according to the number of solutions. A system with no solutions is called inconsistent, and a system with at least one solution is called consistent. Linear Systems Because the graphs in a linear system with two variables are lines, there are exactly three geometric possibilities. • the lines can be parallel and have no point of intersection • the lines can intersect at a single point • the lines can coincide Each of these possibilities leads to a different number of solutions for the system. The three types of linear systems are shown below. 2 2 Lines are parallel y Lines intersect at a single point y Lines coincide y x x x no solutions inconsistent system one solution consistent system Figure 12.1-3 infinitely many solutions consistent system 782 Chapter 12 Systems and Matrices Number of Solutions of a Linear System There are exactly three possibilities for the number of solutions of a system of linear equations. 2 2 • no solutions (inconsistent system) • one solution (consistent system) • infinitely many solutions (consistent system) Solving Systems Algebraic
ally Solving systems of equations by graphing often gives approximate solutions, while algebraic methods produce exact solutions. Furthermore, algebraic methods are often as easy to use as graphical methods. Two common algebraic methods are substitution and elimination. Substitution Method Solving Systems with Substitution To solve a system using the substitution method: 1. Solve one equation for x (or y). 2. Substitute the expression for x (or y) into the other equation. 3. Solve for the remaining variable. 4. Substitute the value found in Step 3 into one of the original equations, and solve for the other variable. 5. Verify the solution in each equation. Example 3 Solving a System by Substitution Solve the system of equations below by substitution. Solution Solve the first equation for y. 3x y 12 2x 3y 2 y 3x 12 Substitute the expression for x. 3x 12 for y in the second equation and solve 2 1 2 3x 12 2x 3 2x 9x 36 2 11x 38 x 38 11 3.45 CAUTION When solving a system of equations, remember to find values for all of the variables. Section 12.1 Solving Systems of Equations 783 To find the value of y, substitute 38 11 , the value of x, into y 3x 12 and simplify. 3.1 y 3 38 11b 12 18 11 a 1.64 4.7 4.7 3.1 Figure 12.1-4 Solving Systems by Elimination The exact solution to the system is y 1.64. mate solution is x 3.45 , x 38 11 , y 18 11 , and the approxi- ■ The solution may be confirmed by graphing, as shown in Figure 12.1-4, where y 1.64. x 3.45 and Elimination Method Elimination is another algebraic method used to solve systems. To solve a system using the elimination method: 1. Multiply one or both of the equations by a nonzero constant so that the coefficients of x (or y) are opposites of each other. 2. Eliminate x (or y) by adding the equations, and solve for the remaining variable. 3. Substitute the value found in Step 2 into one of the original equations, and solve for the other variable. 4. Verify the solution in each equation. Example 4 Solving a System by Elimination Solve the system of equations below by elimination. x 3y 4 2x y 1 Solution Multiply the first equation by 2. 2x 6y 8 2x y 1 Add the equations to eliminate x, and solve the resulting equation. 2x 6y 8 2x y 1 7y 7 y 1 784 Chapter 12 Systems and Matrices 3.1 Substitute 1 4.7 4.7 3.1 Figure 12.1-5 for y in one of the original equations and solve for x. 4 2 x 1 y 1. x 1, x 3 1 1 The solution is confirmed The solution of the system is graphically in Figure 12.1-5. ■ Solutions of Consistent and Inconsistent Systems The following examples show how the elimination method may be used to solve consistent systems with infinitely many solutions or inconsistent systems. Example 5 Recognizing an Inconsistent System Solve the system of equations below by elimination. 2x 3y 5 4x 6y 1 Solution Multiply the first equation by then add the two equations. 2, 4x 6y 10 4x 6y 1 0 9 The last statement, nal system has no solutions. Thus, the system is inconsistent. , is always false. This indicates that the origi- 0 9 Graphing Exploration Confirm the result of Example 5 geometrically by graphing the two equations in the system. Do the lines intersect, or are they parallel? ■ Example 6 Recognizing a System with Infinitely Many Solutions Solve the system of equations below by elimination. 2x 4y 6 3x 6y 9 Solution Multiply the first equation by 3 and the second equation by 2, then add the two equations. 3.1 4.7 4.7 3.1 Figure 12.1-6 Section 12.1 Solving Systems of Equations 785 6x 12y 18 6x 12y 18 0 0 0 0, The last equation, is always true. This indicates that the two equations represent the same line, and every ordered pair that satisfies the first equation must also satisfy the second equation. Thus, the system has infinitely many solutions. ■ Using a Parameter to Write Solutions It is common to represent solutions of consistent systems that have infinitely many solutions in terms of a variable called a parameter, which represents any real number. In Example 6, let and substitute this value into one of the equations. y t 2x 4t 6 t is any real number. x 3 2t Solve for x. The solutions can be written as solutions can be found by substituting real values for t, as follows. Individual numerical x 3 2t, y t. t 1 t 2 t 0 x 5, y 1 x 1, y 2 x 3, y 0 Solving Larger Systems by Elimination It is possible to use elimination to solve larger systems. Equations are combined in pairs to create a system of equations with one fewer variable that can be solved using the techniques discussed in this section. The solutions of the reduced system are then substituted back into the original equations to find the remaining variables. Example 7 Solving a 3 3 System by Elimination Solve the system of equations below by elimination. 2x y z 1 x 3y z 5 x 4y 2z 10 Solution Eliminate z by adding equations [1] and [2]. 2x y z 1 x 3y z 5 x 2y 4 [1] [2] [3] [1] [2] [4] 786 Chapter 12 Systems and Matrices Eliminate the same variable, in this case z, by combining two other equations. One possible way is to multiply equation [2] by 2 and add it to equation [3]. 2x 6y 2z 10 x 4y 2z 10 x 2y 0 [2] 2 [3] [5] The two resulting equations, [4] and [5], form a system of two equations in two variables, which can be solved by elimination, substitution, or graphing. x 2y 4 x 2y 0 4y 4 y 1 [4] [5] y 1 Find the value of x by substituting 1 4 x 2 1 x 2 2 into equation [4]. To find the value of z, substitute the values tion [1] from the original system, and solve. x 2 and y 1 into equa The solution is x 2, y 1, z 4. The solution should be checked in all equations of the original system. ■ Applications of Systems Systems of equations occur in many real-world applications. The simplest situations involve two quantities and two linear relationships between these quantities, as shown in the following example. Example 8 2 2 Linear System Application A ball game is attended by 575 people, and total ticket sales are $2575. If tickets cost $5 for adults and $3 for children, how many adults and how many children attended the game? Solution Let x be the number of adults and y be the number of children. The first equation is based on the total number of people at the game. number of adults x number of children y total attendance 575 The second equation is based on the ticket sales. Notice that the term for each type of ticket sales is found by multiplying the price per ticket by Section 12.1 Solving Systems of Equations 787 the number of tickets sold, and total ticket sales is the sum of the sales of the different types of tickets. adult ticket sales child ticket sales total ticket sales number of adults r price per ticket q r q 5x price per child q r q 3y number of children r 2575 1000 Solve the system of equations. 0 0 Figure 12.1-7 600 x y 575 5x 3y 2575 number of tickets total ticket sales Multiply the first equation by equation. 3 and add the result to the second 3x 3y 1725 5x 3y 2575 2x 850 x 425 So 425 adults and firm by graphing, as shown in Figure 12.1-7. y 575 425 150 children attended the game. Con- ■ Example 9 Mixture Application A cafe sells two kinds of coffee in bulk. The Costa Rican sells for $4.50 per pound, and the Kenyan sells for $7.00 per pound. The owner wishes to mix a blend that would sell for $5.00 per pound. How much of each type of coffee should be used in the blend? Solution Let x be the amount of Costa Rican coffee and y be the amount of Kenyan coffee in each pound of the blend. The first equation is based on the weight of the coffee. weight of Costa Rican x weight of Kenyan y one pound of blend 1 The second equation is based on the price of the coffee. price of Costa Rican price of Kenyan price of blend price per pound q r q weight of coffee r price per pound r weight of coffee q q r 4.50x 7.00y $5.00 788 1 Chapter 12 Systems and Matrices Solve the system of equations. x y 1 4.5x 7y 5 0 1 Figure 12.1-8 Multiply the first equation by and add it to the second equation. 7 7x 7y 7 4.5x 7y 5 2.5x 2 x 0.8 1 0.8 0.2 The owner should use 0.8 pounds of Costa Rican coffee and pounds of Kenyan coffee in each pound of blend, or 80% Costa Rican and 20% Kenyan. See Figure 12.1-8 for graphical confirmation. ■ Exercises 12.1 In Exercises 1–6, determine whether the given values of x, y, and z are a solution of the system of equations. 13. x y c d x y 2c d (where c, d are constants) 1. x 1, y 3 2x y 1 3x 2y 9 2. x 3, y 4 2x 6y 30 x 2y 11 3. x 2, y 1 4. x 0.4, y 0.7 14. x 3y c d 2x y c d (where c, d are constants) In Exercises 15–34, use the elimination method to solve the system. 3.1x 2y 0.16 5x 3.5y 0.48 15. 2x 2y 12 2x 3y 10 16. 3x 2y 4 4x 2y 10 . x 1 2 , y 3, z 1 6. 2x y 4z 6 3y 3z 6 2z 2 , z 1 2 x 2, y 3 2 3x 4y 2z 13 x 3y 5z 5 1 x 8z 3 2 In Exercises 7 – 14, use substitution to solve the system. 7. 9. x 2y 5 2x y 3 3x 2y 4 2x y 1 11. r s 0 r s 5 8. 3x y 1 x 2y 4 10. 5x 3y 2 x 2y 3 12. t 3u 5 t u 5 17. x 3y 1 2x y 5 19. 2x 3y 15 8x 12y 40 21. 3x 2y 4 6x 4y 8 23. 12x 16y 8 42x 56y 28 25. 9x 3y 1 6x 2y 5 27. x 3 2x 5 y 2 y 5 3 2 18. 4x 3y 1 x 2y 19 20. 2x 5y 8 6x 15y 18 22. 2x 8y 2 3x 12y 3 24. 1 y 1 x 2 6 5 3 20x 24y 10 26. 8x 4y 3 10x 5y 1 28. x 3 x 6 3y 5 y 2 4 3 29. 30 2y 3 x y 3 3x y 2 1 2 31. 3.5x 2.18y 2.00782 1.92x 6.77y 3.86928 32. 463x 80y 13781.6 0.0375x 0.912y 50.79624 33. 34. 2x 4y z 14 2x y 6z 31 x 3y 2z 14 x 3y 2z 8 4x 3y z 3 5x y 6z 20 In Exercises 35 and 36, find the values of c and d for which both given points lie on the given straight line. Hint: Substitute the x- and y-values of each of the 2 2 given points into the equation to create a system. 35. 36. cx dy 2; 1 cx dy 6; 0, 4 and 2 1 2, 16 1, 3 1 2 and 1 2 2, 12 2 37. Bill and Ann plan to install a heating system for their swimming pool. They have gathered the following cost information. System Installation cost Monthly operational cost Electric $2000 Solar $14,000 $80.00 $ 9.50 a. Write a linear equation for each heating system that expresses its total cost y in terms of x, the number of years of operation. b. What is the five-year total cost of
electric heat? of solar heat? c. In what year will the total cost of the two heating systems be the same? Which is the cheapest system before that time? Section 12.1 Solving Systems of Equations 789 38. One parcel of land is worth $100,000 now and is increasing in value at the rate of $3000 per year. A second parcel is now worth $60,000 and is increasing in value at the rate of $7500 per year. a. For each parcel of land, write an equation that expresses the value y of the land in year x. b. Graph the equations in part a. c. Where do the lines intersect? What is the significance of this point? d. Which parcel will be worth more in five years? in 15 years? 39. A toy company makes dolls, as well as collector cases for each doll. To make x cases costs the company $5000 in fixed overhead, plus $7.50 per case. An outside supplier has offered to produce any desired volume of cases for $8.20 per case. a. Write an equation that expresses the company’s cost to make x cases itself. b. Write an equation that expresses the cost of buying x cases from the outside supplier. c. Graph both equations on the same axes and determine when the two costs are the same. d. When should the company make the cases themselves, and when should they buy them from the outside supplier? 40. The sum of two numbers is 40. The difference of twice the first number and the second is 11. What are the numbers? 41. A 200-seat theater charges $3 for adults and $1.50 for children. If all seats were filled and the total ticket income was $510, how many adults and how many children were in the audience? 42. A theater charges $4 for main floor seats and $2.50 for balcony seats. If all seats are sold, the ticket income is $2100. At one show, 25% of the main floor seats and 40% of the balcony seats were sold, and ticket income was $600. How many seats are on the main floor and how many in the balcony? 43. An investor has part of her money in an account that pays 2% annual interest, and the rest in an account that pays 4% annual interest. If she has $4000 less in the higher paying account than in the lower paying one and her total annual interest income is $1010, how much does she have invested in each account? 790 Chapter 12 Systems and Matrices 44. The death rate per 100,000 population y in year x for heart disease and cancer is approximated by these equations: Heart Disease: Cancer: 6.9x 2y 728.4 1.3x y 167.5, recent month the store actually sold half of its deluxe models and two-thirds of the regular models and took in a total of $26,700. How many of each kind of tape recorder did they have at the beginning of the month? x 0 corresponds to 1970. If the equations where remain accurate, when will the death rates for heart disease and cancer be the same? (Source: U.S. Department of Health and Human Services) 45. At a certain store, cashews cost $4.40 per pound and peanuts cost $1.20 per pound. If you want to buy exactly 3 pounds of nuts for $6.00, how many pounds of each kind of nuts should you buy? Hint: If you buy x pounds of cashews and y pounds of peanuts, then equation by considering cost and solve the resulting system. Find a second x y 3. 46. A store sells deluxe tape recorders for $150. The regular model costs $120. The total tape recorder inventory would sell for $43,800. But during a 47. How many cubic centimeters of a solution cm3 1 2 that is 20% acid and of another solution that is cm3 45% acid should be mixed to produce 100 solution that is 30% acid? of a 48. How many grams of a 50%-silver alloy should be mixed with a 75%-silver alloy to obtain 40 grams of a 60%-silver alloy? 49. A machine in a pottery factory takes 3 minutes to form a bowl and 2 minutes to form a plate. The material for a bowl costs $0.25 and the material for a plate costs $0.20. If the machine runs for 8 hours straight and exactly $44 is spent for material, how many bowls and plates can be produced? 12.1.A Excursion: Graphs in Three Dimensions Objectives • Plot points in three dimensions • Graph planes in three dimensions In section 12.1, two-dimensional graphs were used to interpret and solve systems of equations in two variables. Systems of equations in three variables can be represented by three-dimensional graphs, as shown in this excursion. However, finding the solutions of such systems requires algebraic techniques that are presented in the following sections. • Use graphs of planes to visualize the number of solutions to a of equations• Solve application s using systems. 3 3 system z y x Figure 12.1.A-1 Three-Dimensional Coordinates Just as ordered pairs of real numbers (x, y) are identified with points in a plane, ordered triples (x, y, z) of real numbers can be identified with points in three-dimensional space. To do this, draw three coordinate axes as shown in Figure 12.1.A-1. The axes in three-dimensional space are usually called the x-axis, the y-axis, and the z-axis. In three-dimensional coordinates, the arrowhead on each axis indicates the positive direction. Each pair of axes determines a coordinate plane, which is named by the axes that determine it. There are three coordinate planes, the xy-plane, the yz-plane, and the xz-plane, which divide the three-dimensional space into eight regions, called octants, shown in Figure 12.1.A-2. The octant in which all coordinates are positive is called the first octant. Section 12.1.A Excursion: Graphs in Three Dimensions 791 NOTE The octants can be considered as the regions above and below each quadrant of the xy-plane. z xy-plane y yz-plane Figure 12.1.A-2 x xz-plane Plotting Points in Three Dimensions To plot the point (a, b, c) in a three-dimensional coordinate system, move a units from the origin along the x-axis, move b units parallel to the yaxis, then move c units parallel to the z-axis. Dashed lines are used to indicate the distances parallel to the y- and z-axes. Example 1 Points in Space Plot the given points in a three-dimensional coordinate system. Solution 1, 3, 4 , 2 1 1 0, 2, 0 , 2 1 2, 2, 3 2 z (1, 3, 4) (0, −2, 0) y x (2, −2, –3) Figure 12.1.A-3 ■ Graphs in Three Dimensions A function in three dimensions can be written in terms of x, y, and z, or as a function of two variables, x and y. The graph of a linear equation in three dimensions is a plane. A comparison of two-dimensional and threedimensional forms follows. 792 Chapter 12 Systems and Matrices Two dimensions Three dimensions slope-intercept form of a line slope y-intercept general form of a line point-slope form of a line y-axis x-axis vertical line horizontal line z mx ny b F mx ny b x, y 1 2 m z2 x2 n z2 y2 z1 x1 z1 y1 b Ax By Cz D z z0 m x x02 1 n y y02 slope-intercept form of a plane slope in x-direction slope in y-direction z-intercept general form of a plane point-slope form of a plane yz-plane xz-plane xy-plane plane: parallel to yz-plane parallel to xz-plane parallel to xy-plane Graphing Planes One method of graphing a plane in three dimensions is to find the x-, y-, and z-intercepts, plot the intercepts on the axes, and then sketch the plane. To find the x-intercept, set y and z equal to 0, and solve for x. The y-intercept and z-intercept are found in a similar manner. Example 2 Graphing a Plane in General Form Graph the plane 2x 3y 4z 12. y mx b f mx b x 2 1 m y2 x2 y1 x1 b Ax By C y y0 m x x02 0, 0, 3) Solution First, find the intercepts. x-intercept: 2x 3 x 0 1 2 y (0, 4, 0) y-intercept: 2 z-intercept: 2 0 0 1 1 2 2 3y 12 6 12 4 4z 12 z 3 1 2 x (6, 0, 0) Figure 12.1.A-4 Plot the intercepts, and sketch the plane that contains them, as shown in Figure 12.1.A-4. ■ Section 12.1.A Excursion: Graphs in Three Dimensions 793 Example 3 Graphing a Plane in Slope-Intercept Form Graph the plane F x, y 1 2 x 2y 2. Solution First, find the intercepts. Use the fact that z F x, y 2 1 x-intercept: y-intercept: z-intercept: 2 1 0 2 to find the z-intercept. 0 x 2 x 2 0 0 2y 2, 0, 0) x (0, 1, 0) (0, 0, −2) Figure 12.1.A-5 Plot the intercepts and sketch the plane that contains them, as shown in Figure 12.1.A-5. Example 4 Graphing a Plane Parallel to an Axis Graph the plane z 4 2 x 1 . 2 1 Solution First, find the intercepts. There is no y term; thus, the plane has no y-intercept. It is parallel to the y-axis. z (0, 0, 2 ) x-intercept (−1, 0, 0) z-intercept Plot the intercepts and sketch the plane that contains them, as shown in Figure 12.1.A-6. Figure 12.1.A-6 y ■ y ■ Graphical Representations of 3 3 Systems A linear system of equations in three variables is represented graphically 3 3 by three planes. A solution of a linear system is a point of intersection of all three planes. As in two variables, a linear system in three variables can have no solutions, one solution, or infinitely many solutions. 794 Chapter 12 Systems and Matrices No solutions One solution Infinitely many solutions Figure 12.1.A-7 All remaining possibilities not shown are listed below: • three planes coincide (infinitely many solutions) • two planes coincide and the third plane intersects them (infinitely many solutions) • two planes coincide and the third plane is parallel (no solutions) Exercises 12.1.A In Exercises 1–8, plot the given point in a threedimensional coordinate system. 1. (1, 4, 5) 3. 1 0, 2, 3 2 5. (3, 0, 0) 7. 1 2, 3, 1 2 2. 1 3, 2, 4 2 4. (4, 0, 6) 6. 8. 0, 0, 4 2 3, 1, 5 1 1 2 In Exercises 9–20, graph the plane described by the given equation. 9. x 3y z 6 10. 5x 2y 4z 10 11. 3x 4y 6z 9 12. 2x 4y 5z 8 13. z x y 3 14. z 2x y 6 2x 4y 8 3x 5y 9 15. 17. F x 19. 3x 2z 6 16. 18. 20. F x, y 1 2 z 8 4 1 4x z 2 y 3 2 21. Critical Thinking Describe the possibilities for the number of solutions of a linear system of equations with 2 equations in 3 variables. Explain your answers in terms of intersections of two planes. You may include a sketch in your answer. Section 12.2 Matrices 795 12.2 Matrices Objectives Augmented Matrices • Represent systems of equations by augmented matrices • Solve systems of equations by row reduction • Solve systems by using a calculator to obtain redu
ced row echelon form matrices • Solve applications by using matrices • Solve application s using systems. It is often convenient to use an array of numbers, called a matrix, as a method to represent a system of equations. For example, the system is written in matrix form as x 2y 2 2x 6y 2 1 2 a 2 6 2 2b In this shorthand, only the coefficients of the variables are written. This representation is called an augmented matrix where each row of the matrix represents an equation of the system. The numbers in the first column are coefficients of x, the numbers in the second column are coefficients of y, and the third column’s numbers are the constant terms. A vertical dashed line is often used to represent the equal signs. Example 1 Writing a System as an Augmented Matrix Write an augmented matrix for the system of equations. x 2y 3z 2 2y 5z 6 3x 3y 10z 10 2 6 2 . ¢ Solution The augmented matrix is Notice that 0 is the x-coefficient in the second equation. ■ Solving Systems Using Augmented Matrices Recall that in the elimination method, an equation may be multiplied by a nonzero constant, or two equations may be added together. Also, the order of the equations is irrelevant, so equations may be interchanged. Performing any of these operations produces an equivalent system, that is, a system with the same solutions. Augmented matrices can be used to solve linear systems. When dealing with matrices, operations similar to those used in the elimination method are called elementary row operations. 796 Chapter 12 Systems and Matrices Elementary Row Operations Performing any of the following operations on an augmented matrix produces an augmented matrix of an equivalent system: • Interchange any two rows. • Replace any row by a nonzero constant multiple of itself. • Replace any row by the sum of itself and a nonzero constant multiple of another row. Example 2 shows the use of the elementary row operations in solving a system. To solve a system of two equations in two variables using elementary row operations, produce an equivalent matrix that has one row with an x-coefficient of 1 and a y-coefficient of 0, and the other row with an x-coefficient of 0 and a y-coefficient of 1. The desired equivalent matrix and its corresponding system are shown below. 1 0 a 0 1 a bb > > xa yb Example 2 Using an Augmented Matrix Solve the system of equations x 2y 2 2x 6y 2 Solution The system is solved below by using elementary row operations in the elimination method on the left and the augmented matrix method on the right. Compare the steps performed in each method. x 2y 2 2 6y 2 2b 2x 1 2 2 6 a NOTE In previous methods, the step to replace a row with the sum of itself and a multiple of another row was done in two or more steps. Replace the second row by the sum of itself and times the first row. 2 x 2y 2 2y 6 Multiply the second row by . 1 2 x 2y 2 y 3 Replace the first row by the sum of itself and times the second row. 2 x 8 y 3 2r1 r2 S r2 1 0 2 2 2 6b r2 S r2 a 1 2 1 0 a 2 1 2 3b 2r2 r1 S r1 1 0 a 0 1 8 3b Section 12.2 Matrices 797 This last augmented matrix 1 0 a 0 1 8 3b represents the same solution of the system as the solution obtained by using the elimination method. The y 3. solution of the system is x 8 , ■ Reduced RowEchelon Form NOTE A matrix can represent a system of equations that has variables other than x, y, and z. When given a matrix without the corresponding system, the choice of letters used to represent the variables is arbitrary. Another common choice is x1, and x3. x2, Reduced Row-Echelon Form The last matrix of Example 2 is in reduced row-echelon form, which is summarized as follows. A matrix is in reduced row-echelon form if it satisfies the following conditions. • All rows consisting entirely of zeros (if any) are at the bottom. • The first nonzero entry in each nonzero row is a 1 (called a leading 1). • Any column containing a leading 1 has zeros as all other entries. • Each leading 1 appears to the right of leading 1’s in any preceding row. Gauss-Jordan Elimination The method of using elementary row operations to produce an equivalent matrix in reduced row-echelon form is called Gauss-Jordan elimination. When an augmented matrix is in reduced row-echelon form, the solutions of the system it represents can be read immediately, as in the last step of Example 2. Example 3 Using Gauss-Jordan Elimination The matrices below are in reduced row-echelon form. Write the system represented by each matrix, find the solutions, if any, and classify each system as consistent, consistent with infinitely many solutions, or inconsistent. a. 1 0 a 3 0 4 0b Solution b. 1 0 a 2 0 1 3b a. The system represented by the augmented matrix is x 3y 4 0x 0y 0 798 Chapter 12 Systems and Matrices The second equation, , is always true. The system is consistent with infinitely many solutions. All solutions lie on the line represented by x 3y 4. 0 0 Represent the solutions of the system using the parameter t. Letting x 4 3t. yields by substituting real values for t. y t Individual solutions may then be found x 3t 4, so a real number 3t x 7 x 2. The system represented by the matrix is x 3 y 7 z 4 y 7, The solution of the system is sistent. x 3, c. The system represented by the matrix is x 2y 1 0x 0y 3 z 4, so the system is con- The second equation, tem is inconsistent because it has no solutions. 0 3, is always false. This indicates that the sys- ■ Technology Tip Check your calculator manual to learn how to enter and store matrices in the matrix memory. To put a matrix in reduced row echelon form, use rref in the MATH submenu of the TI MATRIX menu. Calculators and Reduced Row-Echelon Form Most graphing calculators have a command that uses elementary row operations to put a given matrix into reduced row-echelon form. Example 4 Using Reduced Row-Echelon Form Solve the following systems of equations using a calculator’s reduced rowechelon form feature: a. 2x y 0 4x y 18 Solution b. x 2y 3z 5 3x y 5z 3 y 2z 6 Write the augmented matrix for each system, enter each system into a calculator as a matrix, then reduce to reduced row-echelon form (see Technology Tip). a. 2 4 a 1 1 0 18b . ° Section 12.2 Matrices 799 Figure 12.2-1a Figure 12.2-1b The solution is x 3, y 6 . The solution is x 4, y 0, z 3. ■ Example 5 Calculator Solution to an Inconsistent System Solve the system of equations ¢ Solution x y 2z 1 2x 4y 5z 2 3x 5y 7z 4 Write the augmented matrix for the system, enter the matrix into a calculator, then reduce to reduced row-echelon form, as shown in Figure 12.2-2. The last row of the reduced matrix represents the equation 0x 0y 0z 1. Figure 12.2-2 Because the equation has no solution, the original system has no solution and is therefore inconsistent. ■ Example 6 Calculator Solution of a System Solve the system of equations below. 2 0 2 ° 5 2 17 1 4 23 3 6 40 0 0 0 ¢ Solution 2x 5y z 3w 0 2y 4z 6w 0 2x 17y 23z 40w 0 Notice that all of the constant terms in the system are zero. A system like this has at least one solution, namely, which is called the trivial solution. However, there may be nonzero solutions as well. w 0, y 0, x 0, z 0, Write the augmented matrix for the system, then enter and reduce it to reduced row-echelon form using a calculator, as shown in Figure 12.2-3. Figure 12.2-3 800 Chapter 12 Systems and Matrices The system corresponding to the reduced matrix is x 5.5z 0 y 2z 0 w 0 The system is consistent with infinitely many solutions. The value of w is always 0, and the first two equations can be solved for x and y in terms of z. x 5.5z y 2z w 0 Letting z t, the solutions of the system all have the form x 5.5t, y 2t, z t, w 0. Individual solutions may be found by substituting real values for t. t 0 t 1 t 3 x 0, x 5.5, x 16.5, y 0, y 2, y 6, z 0, z 1, z 3, w 0 w 0 w 0 ■ Example 7 Application Using Calculator Reduced RowEchelon Form Charlie is starting a small business and borrows $10,000 on three different credit cards, with annual interest rates of 18%, 15%, and 9%, respectively. He borrows three times as much on the 15% card as he does on the 18% card, and his total annual interest on all three cards is $1244.25. How much did he borrow on each credit card? Solution Let x be the amount borrowed on the 18% card, y the amount borrowed on the 15% card, and z the amount borrowed on the 9% card. The total amount borrowed is $10,000. x y z 10,000 Total interest is the sum of the amounts of interest on the three cards. Interest on 18% card 0.18x Interest on 15% card 0.15y Interest on 9% card 0.09z Total interest 1244.25 The amounts on the cards are related by a third equation. Amount on 15% card y 3 times amount on 18% card 3x The equation equations is Section 12.2 Matrices 801 y 3x is equivalent to 3x y 0. Therefore, the system of x y z 10,000 0.18x 0.15y 0.09z 1244.25 3x y 0 The corresponding matrix and its reduced row-echelon form are shown in Figure 12.2-4. 1 0.18 3 ° 1 0.15 1 1 0.09 0 10,000 1244.25 0 ¢ Figure 12.2-4 The solution is on the 18% card, $3825 on the 15% card, and $4900 on the 9% card. Charlie borrowed $1275 and x 1275, y 3825, z 4900. ■ Exercises 12.2 In Exercises 1–4, write the augmented matrix of the system. 1. 2. 3. 4. 2x 3y 4z 1 x 2y 6z 0 3x 7y 4z 3 x 2y 3w 7z 5 2x y 3w 2z 4 3x 2y 7w 6z 0 z 0 y 7 4 y 5z 0 x 1 2 2x 3 2 2y 1 3 z 0 2x 1 2 w 6z 1 y 7 2 x 6y 2w z 2 1 4 4y 1 2 w 6z 3 2x 3y 2w 1 2 z 4 In Exercises 5–8, the augmented matrix of a system of equations is given. Express the system in equation notation. 5. 2 4 a 3 7 1 2b 6. 2 1 a 3 6 5 9 2 0b 7 ¢ In Exercises 9–12, the reduced row-echelon form of the augmented matrix of a system of equations is given. Find the solutions of the system. 1 0 0 0 § 11 ≤ 10. 12 ∂ 802 Chapter 12 Systems and Matrices In Exercises 13–20, use Gauss-Jordan elimination to solve the system. 13. 15. 17. 19. x 3y 2z 0 2x 3y 2z 3 x 2y 3z 0 x 2y 2z 1 x 2y 2z 4 2x 2y 3z 5 x 2y 4z 6 x y 13z 6 2x 6y z 10 x y z 200 x 2y 2z 0 2x 3y 5z 600 2x y z 200 14. 16. 18. 3x 7y 9z 0 x 2y 3z 2 x 4y z 2 2x y 2z 1 3
x y 2z 0 7x y 3z 2 x y 5z 6 3x 3y z 10 x 3y 2z 5 20. 3x y z 6 x 2y z 0 In Exercises 21–35, solve the system by any method. 21. 22. 23. 25. 27. 29. 31. 32. 11x 10y 9z 5 x 2y 3z 1 3x 2y z 1 x 2y 3z 4w 8 2x 4y z 2w 3 5x 4y z 2w 3 x y 3 5x y 3 9x 4y 1 x 4y 13z 4 x 2y 3z 2 3x 5y 4z 2 4x y 3z 7 x y 2z 3 3x 2y z 4 x y z 0 3x y z 0 5x y z 0 24. 26. 28. 30. 2x y 2z 3 x 2y z 0 x y z 1 2x 4y z 3 x 3y 7z 1 2x 4y z 10 x 4y z 3 x 2y 2z 0 2x 2y 2x y 3z 2w 6 4x 3y z w 2 x y z w 5 2x 2y 2z 2w 10 x y z w 1 x 4y z w 0 x 2y z 2w 11 x 2y z 2w 3 33. 34. x 2y z 3w 18 x y 3z 3w 7 4y 3z 2w 8 2x 2y 3z w 7 3x y 2z 5w 0 x 3y 2z 5w 0 x 2y 5z 4w 0 2x y 5z 3w 0 35 10 z 2 1 4 3 x 1 Hint: Let . 36. A matrix can be used to represent a set of points in space, with the x-coordinates in the first column, the y-coordinates in the second column, and the z-coordinates in the third column. Each row represents a point. A crystal lattice is used to represent the atomic structure of a crystal. The two matrices below represent simple cubic and A10 crystal lattices, in which the atoms of the crystal are at the points represented by the rows of the matrix. Simple Cubic x 0 3.35 0 3.35 0 3.35 0 3.35 ® y 0 0 3.35 3.35 0 0 3.35 3.35 z 0 0 0 0 3.35 3.35 3.35 3.35 A10 y 0 0.48 2.93 0.48 3.40 0.95 3.40 3.88 x 0 2.93 0.48 0.48 3.40 3.40 0.95 3.88 z 0 0.48 0.48 2.93 0.95 3.40 3.40 3.88 ∏ ∏ ® In two different three-dimensional coordinate systems, plot the points in each matrix and connect them to form two prisms. How are the two lattices alike? How are they different? 37. A collection of nickels, dimes, and quarters totals $6.00. If there are 52 coins altogether and twice as many dimes as nickels, how many of each kind of coin are there? 38. A collection of nickels, dimes, and quarters totals $8.20. The number of nickels and dimes together is twice the number of quarters. The value of the nickels is one-third of the value of the dimes. How many of each kind of coin are there? 39. Lillian borrows $10,000. She borrows some from her friend at 8% annual interest, twice as much as that from her bank at 9%, and the remainder from her insurance company at 5%. She pays a total of $830 in interest for the first year. How much did she borrow from each source? 40. An investor puts a total of $25,000 into three stocks. She invests some of it in stock A and $2000 more than one-half that amount in stock B. The remainder is invested in stock C. Stock A rises 16% in value, stock B 20%, and stock C 18%. Her investment in the three stocks is now worth $29,440. How much was originally invested in each stock? 41. An investor has $70,000 invested in a mutual fund, bonds, and a fast food franchise. She has twice as much invested in bonds as in the mutual fund. Last year the mutual fund paid a 2% dividend, the bonds 10%, and the fast food franchise 6%; her dividend income was $4800. How much is invested in each of the three investments? 42. Tickets to a concert cost $2 for children, $3 for teenagers, and $5 for adults. When 570 people attended the concert, the total ticket receipts were $1950. Three-fourths as many teenagers as children attended. How many children, adults, and teenagers attended? 43. A company sells three models of humidifiers: the bedroom model weighs 10 pounds and comes in an 8-cubic-foot box; the living room model weighs 20 pounds and comes in an 8-cubic-foot box; the whole-house model weighs 60 pounds and comes in a 28-cubic-foot box. Each of their delivery vans has 248 cubic feet of space and can hold a maximum of 440 pounds. In order for a van to be as fully loaded as possible, how many of each model should it carry? 44. Peanuts cost $3 per pound, almonds $4 per pound, and cashews $8 per pound. How many pounds of each should be used to produce 140 pounds of a mixture costing $6 per pound, in which there are twice as many peanuts as almonds? Section 12.2 Matrices 803 45. If Tom, George, and Mario work together, they can paint a large room in 4 hours. When only George and Mario work together, it takes 8 hours to paint the room. Tom and George, working together, take 6 hours to paint the room. How long would it take each of them to paint the room alone? Hint: If x is the amount of the room painted in 1 hour by Tom, y is the amount painted by George, and z the amount painted by Mario, then x y z 1 4 . 46. Pipes R, S, and T are connected to the same tank. When all three pipes are running, they can fill the tank in 2 hours. When only pipes S and T are running, they can fill the tank in 4 hours. When only R and T are running, they can fill the tank in 2.4 hours. How long would it take each pipe running alone to fill the tank? 47. A furniture manufacturer has 1950 hours available each week in the cutting department, 1490 hours in the assembly department, and 2160 in the finishing department. Manufacturing a chair requires 0.2 hours of cutting, 0.3 hours of assembly, and 0.1 hours of finishing. A chest requires 0.5 hours of cutting, 0.4 hours of assembly, and 0.6 hours of finishing. A table requires 0.3 hours of cutting, 0.1 hours of assembly, and 0.4 hours of finishing. How many chairs, chests, and tables should be produced in order to use all the available production capacity? 48. A stereo equipment manufacturer produces three models of speakers, R, S, and T, and has three kinds of delivery vehicles: trucks, vans, and station wagons. A truck holds 2 boxes of model R, 1 of model S, and 3 of model T. A van holds 1 box of model R, 3 of model S, and 2 of model T. A station wagon holds 1 box of model R, 3 of model S, and 1 of model T. If 15 boxes of model R, 20 of model S, and 22 of model T are to be delivered, how many vehicles of each type should be used so that all operate at full capacity? 49. A company produces three camera models: A, B, and C. Each model A requires 3 hours of lens polishing, 2 hours of assembly time, and 2 hours of finishing time. Each model B requires 2, 2, and 1 hours of lens polishing, assembly, and finishing time, and each model C requires 1, 3, and 1 hours, respectively. There are 100 hours available for lens polishing, 100 hours for assembly, and 65 hours for finishing each week. How many of each model should be produced if all available time is used? 804 Chapter 12 Systems and Matrices 12.3 Matrix Operations Objectives • Add and subtract matrices • Multiply a matrix by a scalar factor • Multiply two matrices • Use matrix multiplication to solve problems • Use matrices to represent directed networks Matrices were used in Section 12.2 to solve systems of linear equations. However, matrices are also useful for organizing data. The arithmetic of matrices has practical applications in the natural sciences, engineering, the social sciences, and management. Matrices are now considered in a more general setting. A matrix has been defined as an array of numbers. The dimensions of a m n matrix indicate the number of rows and columns in the matrix. An matrix has m rows and n columns. For example matrix 3 rows, 3 columns B 3 2 5 0 12 4 1 § matrix ¥ 4 rows, 1 column Each entry of a matrix can be located by stating the row and column in is the entry in row i and column j of its which it appears. An entry 5 corresponding matrix. In the matrices above, Two matrices are said to be equal if they have the same dimensions and the corresponding entries are equal. 12. and b41 a13 aij The general form of a matrix can be written as shown below. a11 a21 a31 o am1 a12 a22 a32 o am2 a13 a23 a33 o am3 p p p p A • a1n a2n a3n o amn µ Matrix Addition and Subtraction Matrices may be added or subtracted, but unlike real numbers, not all sums and differences are defined. It is only possible to add or subtract matrices that have the same dimensions. Matrix Addition and Subtraction Matrices that have the same dimensions may be added or subtracted by adding or subtracting the corresponding entries. For matrices that have different dimensions, addition and subtraction are not defined. Section 12.3 Matrix Operations 805 Example 1 Adding and Subtracting Matrices For the given matrices, find ¢ and A B ¢ Figure 12.3-1a Solution Both are 3 3 matrices, so add or subtract the corresponding entries. Figure 12.3-1b Scalar Multiplication 10 11 5 ¢ ¢ The results are confirmed in Figures 12.3-1a and 12.3-1b. ■ Multiplication and Matrices Scalar Multiplication There are two different types of multiplication associated with matrices: scalar multiplication and matrix multiplication. Scalar multiplication is the product of a real number and a matrix, while matrix multiplication is the product of two matrices. Scalar multiplication is the product of a scalar, or real number, and a matrix. If A is an number, then kA is the each entry of A by k. p matrix and k is a real matrix formed by multiplying m n m n k ° a11 o am1 a1n o amn ∞ p ¢ ° p ∞ p ka11 o kam1 ka1n o kamn ¢ Example 2 Scalar Multiplication For the matrix ¢ , find 3A. 806 Chapter 12 Systems and Matrices Solution Multiply each entry of A by 3. 3A 18 6 6 3 0 15 21 15 ¢ Figure 12.3-2 The results are confirmed in Figure 12.3-2. ■ Matrix Multiplication To multiply two matrices, multiply the rows of the first matrix by columns of the second matrix. The number of entries in each row of the first matrix must be the same as the number of entries in each column of the second matrix. To multiply a row by a column, multiply the corresponding entries, then add the results. In the following illustration, row 2 of the first matrix is multiplied by column 1 of the second matrix to produce the entry in row 2 column 1 of the product matrix. Note that the product of a row and a column is a single number * * * * * * ¢ ° ° ¢ 32 c a21b11 * 10 * c a22b21 21 c a23b31 * * * * * * ¢ ° * 8 * * * * * * * ¢ Matrix Multiplication The product AB is defined only when the number of columns of A is the same as the number of rows of B. If A is an m p m n matrix and B is an n p matrix, then AB is an matrix C where the entry in the ith row, jth column is cij ai1b1j ai2b2j ainbn
j The following diagram shows how the dimensions of the product matrix are related to the dimensions of the factor matrices: > m n matrix > 1 2 > > n p matrix 1 2 > m p matrix > 1 2 CAUTION Before finding the entries of a product matrix, check the dimensions of the factor matrices to make sure that the product is defined. Section 12.3 Matrix Operations 807 Example 3 Matrix Multiplication For the given matrices, find AB, BA, AC, and CA when defined. A 3 1 a 1 0 2 4b 1b Solution First, verify that each product is defined. A is > 2 3 and B is > 3 2, so AB is defined. > > B is 3 2 and A is 2 3, so BA is defined. A is > 2 3 and C is > 2 2, > C is 2 2 and A is > 2 3, so AC is not defined. so CA is defined. AB is a 2 2 matrix. AB 3 1 a 1 0 2 4b ° 2 0 1 3 5 8 ¢ row 1 of A column 1 of B ⎫⎪⎪⎪⎪⎬⎪⎪⎪⎪⎭ row 1 of A column 2 of B ⎫⎪⎪⎪⎪⎬⎪⎪⎪⎪⎭ row 2 of A column 1 of 21 row 2 of A column 2 of B 8 8 1 1 2 2 b ⎫⎪⎪⎪⎪⎬⎪⎪⎪⎪⎭ ⎫⎪⎪⎪⎪⎬⎪⎪⎪⎪⎭ 8 2 a 12 35b BA is a 3 3 matrix. BA 4b ° 21 21 21 20 34 ¢ Note that AB BA. CAUTION in general. AB BA Matrix multiplication is not commutative, that is, AB and BA may have different dimensions, as in Example 3, or BA may not be defined when AB is. Even when AB and BA are both defined and have the same dimensions, they may not be equal. 808 Chapter 12 Systems and Matrices AC is not defined. CA is a 2 3 matrix. CA 1 0 a 3 1b a 3 1 1 0 2 4b Technology Tip To multiply matrices on a calculator, enter the matrices in the memory and recall them as needed. If the product of two matrices is not defined, the calculator will give an error message 21 21 21 14 4b ■ Matrices can also be multiplied on a calculator, as shown below. Figure 12.3-4 Applications Figure 12.3-3 Matrices are a convenient way to handle data that is grouped into categories. If the categories of the rows of one matrix are the same as those of the columns of another, the matrices can often be multiplied to form a meaningful product. Example 4 Using Matrix Multiplication A furniture restorer refinishes chairs, tables, and dressers. The amount of time required to complete each step of refinishing and the cost per hour of each step are given by the following matrices. Find the product of the two matrices and interpret the result. removing finish 0.5 hr 1 hr 3 hr ° • chair table dresser Hours sanding 2.5 hr 4 hr 7 hr finishing 1 hr 1.5 hr 4.5 hr ¢ removing finish sanding finishing cost per hour $7 $18 $10 ° ¢ Figure 12.3-5 Section 12.3 Matrix Operations 809 Solution 3 3 so multiplication of the The first matrix is and the second is 3 1 first matrix by the second matrix is defined, and the product is a matrix. In the product of a row and a column, the amount of time for each step is multiplied by the cost per hour for that step, giving the cost for that step. The costs are then added, giving the total cost for the item. The product matrix will give the total cost for refinishing each item. 3 1, 0.5 hr 1 hr 3 hr ° 2.5 hr 4 hr 7 hr 1 hr 1.5 hr 4.5 hr $7 $18 $10 ¢ chair table dresser ¢ ° total cost $58.50 $94 $192 £ ≥ ■ Directed Networks The following figure is a directed network. The points that are labeled by capital letters are called vertices, and the arrows indicate the direction in which the paths between the vertices can be traveled. For example, from vertex L there is 1 path to M and 2 paths to K, but from K, there are 0 paths to L or to M. K L M Figure 12.3-6 From To Number of paths An adjacency matrix can be used to represent the connections between the vertices, as shown at right. The entries in the matrix are the number of direct paths, called one-stage paths, from one vertex to another. From: K L M To £ ≥ 810 Chapter 12 Systems and Matrices Multiplying Adjacency Matrices A A A2 If A is the adjacency matrix of a network, then the product is a matrix for the number of two-stage paths, that is, paths from one vertex to another through one intermediate vertex. Example 5 Using an Adjacency Matrix Find the matrix for the number of two-stage paths for the directed network on page 809, and interpret the result. Solution The matrix for the number of two-stage paths is ¢ The matrix gives the number of two-stage paths, which are given in the table below. From To Number of two-stage paths Paths none none none ; L S M S L (see Note; M S L S K; M S L S M; M S M S M ■ Example 6 Food Webs A food web shows the relationships between certain predators and prey in an ecosystem. A directed network can be used to represent a food web, with the arrows pointing in the direction of prey to predator. Write an adjacency matrix for the following food web. NOTE Two different two-stage paths may pass through the same vertices in the same order, as shown below. K L K L M M Section 12.3 Matrix Operations 811 fish shellfish sea lions sea otters killer whales Figure 12.3-7 Solution The matrix is given below. From so sl kw • To: so 1 0 0 0 0 sl 0 1 0 0 0 kw 0 1 1 1 0 µ ■ In Exercises 7–12, determine if the product AB or BA is defined. If a product is defined, state its dimensions. Do not calculate the products. 7. A 3 8 a 6 0 7 1b 8. A 9. A 10. A 11. A ° ° ° ° 1 9 10 2 2 34 15 7 10 ¢ 17 12 5 B 5 7 a 6 8 11 15b B ° 2 13 5 4 2 25 9 1 0 ¢ B 1 3 a 2 4b Exercises 12.3 In Exercises 1–6, refer to matrices A, B, and C below ¢ Find each of the following: 1. A B 3. A C 5. 2C 2. AB 4. 3A 6. 2B 3C 812 Chapter 12 Systems and Matrices 12. A ± 10 6 1 4 12 0 23 3 ≤ B 1 3 a 2 2 3 1b In Exercises 13–18, find AB. 13. A 3 2 a 2 4b B 1 0 a 2 3 3 1b 14. A 15. A 16. A 17 ≤ 18. A 10 1 a 0 1 1 0 0 1b 2b ¢ 23. Write an adjacency matrix for the directed network below. J K L 24. Write an adjacency matrix for the directed network below. N P M 25. Find a matrix for the directed network in Exercise 23 that represents the number of two-stage paths. 26. Find a matrix for the directed network in Exercise 24 that represents the number of two-stage paths. 27. A bakery sells giant cookies, sheet cakes, and 3tiered cakes. The time required for each step is given in the matrix below. baking decorating giant cookie sheet cake 3 -tiered cake 0.5 hr 0.75 hr 1.5 hr ° 0.25 hr 0.5 hr 1.25 hr ¢ In Exercises 19–22, show that AB is not equal to BA by computing both products. The cost per hour for baking and decorating is given by the matrix below. Find the product of the two matrices and interpret the result. 19. A 3 5 a 2 1b 20 6b B 3 2 1 1 5 2 21 ¢ 22 ≤ cost per hour baking decorating a $4 $7b 28. A boutique sells shirts, pants, and dresses. The time required for each step is given in the matrix below. cutting sewing shirt pants dress 1 hr 1.5 hr 2 hr ° 1 hr 1.25 hr 1.75 hr ¢ Section 12.3 Matrix Operations 813 The cost per hour for cutting and sewing is given by the matrix below. Find the product of the two matrices and interpret the result. 32. Write an adjacency matrix for the food web represented by the directed network below. cost per hour cutting sewing a $5 $9b 29. A small college offers lecture and lab courses. The class sizes are given in the matrix below. freshman sophomore level lecture 150 30 lab a level 100 25 junior senior level level 75 25 50 20 b The tuition for each course is given by the matrix below. Find the product of the two matrices. Interpret all meaningful entries in the product. birds spiders insects frogs 33. An airline offers nonstop flights between certain cities, as shown on the directed network below. Minneapolis lecture lab St. Louis Chicago Madison freshman level sophomore level junior level senior level $200 $240 $280 $320 ± $40 $48 $56 $64 ≤ 30. A store sells trail mixes made of nuts and dried fruit. The nutritional information per serving is given in the matrix below. nuts fruit fat protein carbohydrates 1 g 3 g 65 g ° 52 g 20 g 21 g ¢ Milwaukee a. Write an adjacency matrix A for the directed network above. b. Find the matrix A2, which represents the number of flights between cities with exactly 1 layover. Find the matrix the number of flights between cities with exactly 2 layovers. which represents A3, A A2 A3 , and interpret the c. Find the matrix result. The percent of fruit and nuts per serving in each mix is given below. Find the product of the two matrices, and interpret the result. 34. A delivery company ships packages between certain locations, as shown by the directed network below. mix A mix B nuts fruit 30% a70% 45% 55%b Dallas/Fort Worth Amarillo 31. Write an adjacency matrix for the food web represented by the directed network below. El Paso Austin Houston cheetahs gazelles lions zebras hyenas San Antonio a. Write an adjacency matrix A for the directed network above. b. Find the matrix result. A A2 A3 , and interpret the 814 Chapter 12 Systems and Matrices 12.4 Matrix Methods for Square Systems Objectives • Define the matrix n n identity • Find the inverse of an invertible matrix • Solve square systems of equations using inverse matrices A system of equations that has the same number of equations as variables is called a square system. There is a method of solving this type of system that does not require row reduction. This method only works if the system has a unique solution. Examine the system of equations below. x y z 2 2x 3y 5 x 2y z 1 ⎫ ⎪ ⎬ ⎪ ⎭ [1] Instead of using a single matrix to represent the system, we can consider the system as having three parts: the coefficients, the variables, and the constants. A matrix can be used to represent each part ¢ coefficients ° ¢ variables constants The relationship between the three matrices and the system is shown in the following example. Example 1 A Matrix Equation For the system of equations and the three matrices above, verify that AX B. Solution By the definition of matrix multiplication: AX 2x 3y x 2y z ¢ According to the system of equations [1], the entries of the matrix AX are equal to the corresponding entries of B, so the two matrices are equal, that is, AX B. ■ AX B, Example 1 shows that a square system can be represented by the matrix equation where A is the matrix of the system’s coefficients and B is the matrix
of the system’s constants. Thus, the system can be solved AX B by solving the corresponding matrix equation. Solving the equation means finding the entries of the matrix X, which are the variables of the system. Section 12.4 Matrix Methods for Square Systems 815 Identity Matrices and Inverse Matrices To solve a matrix equation it is necessary to “undo” the matrix multiplication. One method of solving similar equations with real numbers is by multiplying both sides of the equation by the inverse of a. AX B, ax b 1ax a a x a 1b 1b [2] The solution of equation [2] depends on the fact that which is the identity for multiplication of real numbers. Thus, in order to define the inverse of a matrix, we must first define the identity for matrix multiplication. aa 1 1, n n identity matrix is the matrix with n rows and n columns that The has 1’s on the diagonal from the top left to the bottom right and 0’s as all its other entries. In I2 1 0 a 0 1b I3 ° 1 0 0 0 1 0 0 0 1 ¢ I4 ± ≤ The identity matrix, matrices. In, is the identity for multiplication of n n Identity Matrix For any n n matrix A, AIn InA A. Example 2 Verifying the Identity Matrix Property For matrix C, verify that CI2 I2C C. C 3 5 a 2 7b Solution By the definition of matrix multiplication, CI2 I2C 3 5 1 0 a a 2 7b a 1 0 0 1b a 3 5 0 1b 2 7b 7b 2 7b C C ■ Inverse Matrices An n n n n matrix A is called invertible, or nonsingular, if there exists an (In this case it is also true that BA In. AB In. matrix B such that 2 816 Chapter 12 Systems and Matrices The matrix B is called the inverse of A, and is written as AA 1, Not all matrices have a multiplicative inverse. 1 A 1A In. A where Example 3 Verifying an Inverse Matrix For the given matrices, verify that B is the inverse of A. A 2 3 a 1 1b B 1 3 a 1 2b Solution By the definition of matrix multiplication, 1 3 a 2 3 AB 1 1b a Notice also that 1 3 BA a 1 2b 1b I2 1 2b a 2 3 1 1b a 1 1 2 3 1 2 21 2 1 2 1 21 3 3 2 2 1 I2 2 0 1b 21 2 1 2 1 21 1 1 2 2 1 b ■ There are several methods of finding the inverse of an invertible matrix. Example 4 Finding an Inverse Matrix Find the inverse of matrix A 2 1 a 6 . 4b Solution Suppose 1 A x y a u . vb Then AA 1 I2. 1 AA 2 1 a 6 4b a x y u vb a 2u 6v u 4vb 1 0 a 0 1b I2 2x 6y x 4y 1 Setting the corresponding entries of in two systems of equations, one for each column. AA I2 and equal to each other results 2x 6y 1 x 4y 0 2u 6v 0 u 4v 1 The solutions of the two systems are x 2, y 1 2 and u 3, v 1. Thus, 1 A 2 1 2 a 3 . 1b Check this by verifying that AA 1 I2. ■ Section 12.4 Matrix Methods for Square Systems 817 Most graphing calculators can also be used to find inverse matrices directly by using the key, as shown in Figure 12.4-1. 1 x Figure 12.4-1 Solving Square Systems Using Inverse Matrices Recall that the need for inverse matrices arose out of the matrix equation AX B , which represents a square system of equations. The equation can now be easily solved by multiplying both sides by the inverse of the coefficient matrix A. A AX B 1AX A InX A X A 1B 1B 1B Multiply both sides by A 1 . 1A In A InX X by definition of inverse. by definition of identity. Thus, the solution of a square system of equations with an invertible coefficient matrix A and constant matrix B is X A 1B. CAUTION Because matrix multiplication is not commutative, it is important to always multiply in the same order on both sides of the equation. Suppose that a square system of equations can be represented AX B, by the matrix equation coefficients, X is the matrix of the variables, and B is the matrix of the constants. If A is invertible, then the unique solution of the system is where A is the matrix of the X A1B If A is not invertible, then the system is either consistent with infinitely many solutions or inconsistent. Its solutions (if any) may be found by using Gauss-Jordan elimination. Example 5 Solving a 2 2 System Using a Matrix Equation Use an inverse matrix to solve x y 2 2x 9y 15 Solution The coefficient matrix A 1 2 a 1 9b and the constant matrix Then X A 1B 3 , 1b a so the solution is x 3, y 1. B 2 15b . a ■ Matrix Solution of a Square System Figure 12.4-2 818 Chapter 12 Systems and Matrices Example 6 Solving a 3 3 System Using a Matrix Equation Use an inverse matrix to solve Solution The coefficient matrix is A B ° . Then X A 1B ¢ 2 5 1 x y z 2 2x 3y 5 x 2y z 1 1 0 1 ¢ and the constant matrix is , so the solution is x 7, y 3, as shown in Figure 12.4-3. ■ Curve Fitting Just as two points determine a unique line, three noncollinear points determine a unique parabola. In general, a polynomial of degree n can be determined by noncollinear points, a circle by three noncollinear points, and a general conic by five noncollinear points. Matrix methods can be used to find curves that pass through a given set of points. n 1 Example 7 Finding the Parabola Through Three Points Find the equation of the parabola that passes through the points ( 1, 10 ( 4, 25 ), and ( ). 1, 4 ), Solution The equation of a parabola can be written as y ax2 bx c . Substitute the values of x and y from each point into the equation to form a system of equations with the variables a, b, and c. x, y 1 1, 4 1, 10 4, 25 1 1 y ax2 bx c 4 a b c 10 a b c 25 16a 4b c 2 2 2 2 1 Find the values of a, b, and c by solving the matrix equation AX B, The solution is the equation of the parabola is 1 1 16 a 2, ° ¢ X , a b c ° ¢ , and B ° 4 10 25 ¢ c 5 y 2x2 3x 5. , as shown in Figure 12.4-4. Thus, ■ Figure 12.4-3 Figure 12.4-4 Section 12.4 Matrix Methods for Square Systems 819 Exercises 12.4 In Exercises 1–4, write the identity matrix matrix, and verify that InC C. CIn In for each 1. C 3 4b 1 3 1 0 2 0 ¢ 2. C 6 3 a 4 2b 4 ¢ In Exercises 5–8, verify that B is the inverse of A. 5. A 6. A 3 5 4 2 a a 1 2b B 8 6b B a 2 5 3 4 1 4 1 3b 1 7 ¢ In Exercises 9–12, write a set of systems of equations that represent the solution of the matrix equation AA1 In . (See Example 4.) Do not solve the systems. 9. A 2 4 a 0 1b 10. A 1 2 a 3 5b 11 ¢ 12 ¢ In Exercises 13–20, find the inverse of the matrix, if it exists. 13. 1 3 a 2 4b 15. 3 6 a 1 2b 17 ¢ 14. 3 1 a 5 4b 16. 18 ¢ 19 ¢ 20 ¢ In Exercises 21–26, solve the system of equations by using inverse matrices. (See Examples 5 and 6.) 21. 3x 5y 4 2x 6y 12 22. 3x 5y 23 8x 6y 10 23. 25. 1 x y x z 2 6x 2y 3z 3 2x y 0 4x y 3z 1 3x y 2z 2 24. x 2y 3z 1 2x 5y 3z 0 x 8z 1 26. 3x 3y 4z 2 z 1 y 4x 3y 4z 3 In Exercises 27–34, solve the system by any method. 27. 28. 29. 30. 31. 32. 33. x y 2w 3 2x y z w 5 3x 3y 2z 2w 0 x 2y z 2 x 2y 3z 1 y z w 2 2x 2y 2z 4w 5 2y 3z w 8 x 2y 4z 6 y z 1 x 3y 5z 10 x 4y 5z 2w 0 2x y 4z 2w 0 x 7y 10z 5w 0 4x 2y z 5w 0 6 x 2y 3z 9 3x 2y 4z 2x 6y 8z w 17 2x 2z 2w 2 3w 2 x x 4y z 3w 7 4y z 5 x 12y 3z 3w 17 x 2y 2z 2w 23 4x 4y z 5w 7 2x 5y 6z 4w 0 5x 13y 7z 12w 7 820 Chapter 12 Systems and Matrices 34. x 2y 5z 2v 4w 0 2x 4y 6z v 4w 0 5x 2y 3z 2v 3w 0 6x 5y 2z 5v 3w 0 x 2y z 2v 4w 0 35. Critical Thinking Consider the two systems of equations below. x 2y 3 3x 6y 9 x 2y 4 3x 6y 7 a. Write a matrix equation that represents each system. Which matrices in the two equations are the same? Does either matrix equation have a solution? b. Solve each system by any method. Make a conjecture about the relationship between the existence of an inverse coefficient matrix and the nature of the solutions to any system with those coefficients. 36. Critical Thinking Consider the system of equations below. x 2y 3z 4 2x 4y z 1 x 5y 2z 15 a. Write an augmented matrix that represents the system, and reduce it to reduced row echelon form. b. Write a matrix equation that represents the system, and solve it using an inverse matrix. c. Describe your results in part a in terms of the matrix equation from part b. In Exercises 37–40, find constants a, b, c such that the y ax2 bx c. three given points lie on the parabola (See Example 7.) 37. (1, 0), (2, 3), (3, 8) 38. (1, 1), (2, 1), (3, 2) 39. (3, 2), (1, 1), (2, 1) 40. (1, 6), (2, 3), (4, 25) In Exercises 41–43, write a system of equations that determines the polynomial of the given type that passes through the given points. Do not solve the system. 41. cubic; (0, 5), (2, 1), (4, 7 ), (8, 3) 42. cubic; ( 3 , 1), ( 1 2 , ), (0, 6), (3, 0) 43. quartic; ( 5 1 , ), ( 2 , 0), (1, 3), (2, 5), (10, )4 44. Concentrations of the greenhouse gas carbon CO2 , have increased quadratically over CO2 dioxide, the past half century. The concentration y of in parts per million, in year x is given by an equation of the form , y ax 2 bx c. a. Let x 0 correspond to 1958 and use the following data to find a, b, and c. Year 1958 1979 2001 CO2 Concentration 315 337 371 b. Use this equation to estimate the CO2 concentration in the years 1983, 1993, and 2003. For comparison purposes, the actual concentrations in 1983 and 1993 were 343 ppm and 357 ppm respectively. 45. Find constants a, b, and c such that the points (0, 2) (ln 2, 1), and (ln 4, 4) lie on the graph of f . (See Example 7.) aex be x c x 46. Find constants a, b, and c such that the points (0, 1) (ln 2, 4), and (ln 3, 7) lie on the graph of f aex be x c. x 1 2 1 2 47. A conic section has the equation Ax2 Bxy Cy2 Dx Ey F 0. values of A, B, C, D, E, and F for the conic section that passes through the six given points: (3, 4), (6, 2), (2, 6), (12, 1), (4, 3), (1, 12). Write the equation and identify the type of conic section. Find the 48. A candy company produces three types of gift boxes: A, B, and C. A box of variety A contains 0.6 lb of chocolates and 0.4 lb of mints. A box of variety B contains 0.3 lb of chocolates, 0.4 lb of mints, and 0.3 lb of caramels. A box of variety C contains 0.5 lb of chocolates, 0.3 lb of mints, and 0.2 lb of caramels. The company has 41,400 lb of chocolates, 29,400 lb of mints, and 16,200 lb of caramels in stock. How many boxes of each kind should be made in order to use up all their stock? 49. Certain circus animals are fed the same three food mixes: R, S, and T. Lions receive 1.1 units of mix R, 2.4 units of mix S, and 3.7 u
nits of mix T each day. Horses receive 8.1 units of mix R, 2.9 units of mix S, and 5.1 units of mix T each day. Bears receive 1.3 units of mix R, 1.3 units of mix S, and 2.3 units of mix T each day. If 16,000 units of mix R, 28,000 units of mix S, and 44,000 units of mix T are available each day, how many of each type of animal can be supported? Section 12.5 Nonlinear Systems 821 12.5 Nonlinear Systems Objectives • Solve nonlinear systems algebraically • Solve nonlinear systems graphically The matrix methods discussed in Section 12.2 and Section 12.4 can only be used for linear systems. Sometimes, however, it is necessary to solve nonlinear systems. Some nonlinear systems may be solved algebraically, by substitution or elimination. Example 1 Solving a Nonlinear System by Elimination Solve the system of equations. x2 y2 5 x2 y2 13 Solution This system can easily be solved by elimination. Add the two equations to eliminate , and solve the resulting equation. y2 x2 y2 5 x2 y2 13 2x2 18 x2 9 x ± 3 NOTE Nonlinear systems can have any number of solutions, including solutions with the same x-value but different y-values—and vice versa. Thus, it is convenient to write the solutions as ordered pairs, (x, y). Substitute each value of x into one of the original equations and solve for y. The x term is squared in both equations, so it is not necessary to substitute both 3 and because 3, 2 2. 32 3 1 32 y2 5 y2 4 y ± 2 The system has four solutions: and 3, 2 x 3, y 2; Written as ordered pairs, the solutions are (3, 2), 3, 2 x 3, 3, 2 y 2. , and y 2; x 3, x 3, y 2; , . 1 2 1 2 1 2 ■ Example 2 Solving a Nonlinear System by Substitution Solve the system of equations. 2x2 y2 1 x y 12 822 Chapter 12 Systems and Matrices Solution This system can be solved by substitution. Solve the second equation for y, and substitute into the first equation. 2x2 y 12 x 2 1 1 2x2 144 24x x2 1 x2 24x 145 0 12 x 2 x 5 or x 29 If these values of x are substituted back into the second equation, the y 41, respectively. This gives two soluresulting solutions are tions, (5, 7) and y 7 . 29, 41 or 1 2 CAUTION When solving nonlinear systems algebraically, extraneous solutions can result. Check all solutions in all of the original equations. Notice that if the values for x were substituted into the first equation y ;7 This would instead, the resulting solutions would be or 29, 41 29, 41 , and give 4 solutions, (5, 7), However, the 5, 7 do not satisfy the second equation; they solutions are extraneous. 5, 7 , 2 29, 41 y ;41. and Thus, the solutions of the system are (5, 7) and 29, 41 . 2 1 ■ Solving Nonlinear Systems Graphically Consider the system below. y x4 4x3 9x 1 y 3x2 3x 7 Substitution may seem like an appropriate method for solving the system. However, if the expression for y in the first equation is substituted for y in the second equation, the result is x4 4x3 9x 1 3x2 3x 7 x4 4x3 3x2 12x 6 0 This fourth-degree equation cannot be readily solved algebraically, so a graphical approach is appropriate. Example 3 Solving a Nonlinear System Graphically Solve the following system of equations graphically. y x4 4x3 9x 1 y 3x2 3x 7 Solution Graph both equations on the same screen. Trace or use an intersection finder to determine the coordinates of the intersections. Section 12.5 Nonlinear Systems 823 36 4.7 4.7 12 Figure 12.5-1 The graphs intersect at four points. The approximate solutions are 1.5, 4.4 ( ), (2.1, 0.1), and (3.9, 26.3). 0.5, 4.8 ), ( ■ Example 4 Solving a Nonlinear System Graphically Solve the following system of equations graphically. 2x2 3y2 30 2x2 xy y2 8 Solution In order to graph the equations, they must both be solved for y. 2x2 3y2 30 x2 y2 10 2 3 10 2 3 y ± B x2 To solve the second equation for y, use the quadratic formula with y as the variable. 2x2 xy y2 8 0 2x2 8 y2 xy 1 a 1, b x, and c 2x2 8 2 5 4.7 4.7 5 Figure 12.5-2 y x ± 2x2 4 2 1 2x2 8 2 x ± 29x2 32 2 Graph the four equations and then trace or use an intersection finder. 3, 2 There are four solutions: (3, 2), ( ). The first two are exact solutions, a fact that can be confirmed by substituting the values into the original equations. 2.2, 2.6 2.2, 2.6 ), and ( ), ( ■ 824 Chapter 12 Systems and Matrices Example 5 Application of a Nonlinear System The revenue and cost (in dollars) for manufacturing x bicycles are given by the following equations: cost: revenue: y 85x 120,000 y 0.04x2 320x A solution of this system is called a break-even point, which occurs when the cost and revenue are equal. Find all break-even points of this system. Solution Graph both equations on the same screen. To choose a window, notice that the graph of the revenue equation is a parabola with vertex (4000, 640,000). 650,000 0 0 8000 Figure 12.5-3 There are two break-even points: approximately (565, 168,022) and (5310, 571,353). This means that if the manufacturer makes and sells 565 or 5310 bicycles, the cost and revenue will be the same. That is, the manufacturer will break even. From the graph, you can see that between 565 and 5310, the revenue is greater than the cost, so the manufacturer will make a profit when producing between 565 and 5310 bicycles. ■ Exercises 12.5 In Exercises 1–12, solve the system algebraically. 10. 1. 3. 5. 7. 9. x2 y 0 2x y 3 x2 y 0 x2 3y 6 x y 10 xy 21 xy 2y2 8 x 2y 4 x2 y2 4x 4y 4 x y 2 2. 4. 6. 8. x2 y 0 3x y 2 x2 y 0 x2 4y 4 2x y 4 xy 2 xy 4x2 3 3x y 2 x2 y2 4x 2y 1 x 2y 2 11. x2 y2 25 x2 y 19 12. x2 y2 1 x2 y 5 In Exercises 13–28, solve the system by any means. 13. y x3 3x2 4 y 0.5x2 3x 2 14. y x3 3x2 x 3 y 2x2 5 15. y x3 3x 2 y 3 x2 3 16. y 0.25x4 2x2 4 y x3 x2 2x 1 17. y x3 x 1 y sin x 18. y x2 4 y cos x 19. 25x2 16y2 400 9x2 4y2 36 20. 9x2 16y2 140 x2 4y2 4 21. 5x2 3y2 20x 6y 8 x y 2 22. 4x2 9y2 36 2x y 1 23. x2 4xy 4y2 30x 90y 450 0 x2 x y 1 0 24. 3x2 4xy 3y2 12x 2y 7 0 x2 10x y 21 0 25. 4x2 6xy 2y2 3x 10y 6 4x2 y2 64 26. 5x2 xy 6y2 79x 73y 196 0 x2 2xy y2 8x 8y 48 0 27. x2 3xy y2 2 3x2 5xy 3y2 7 28. 2x2 8xy 8y2 2x 5 0 16x2 24xy 9y2 100x 200y 100 0 In Exercises 29–32, find the center r of the circle through the three given points. (h, k) (x h)2 (y k)2 r 2 and radius that passes 29. (0, 5), (3, 4), (4, 3) 30. (3, 4), (2, 5), (3,6) 31. (5, 25), (17, 21), (2, 24) 32. (8, 12), (14, 4), (6.4, 12.8) 33. Find the break-even points for the following revenue and cost functions. (See Example 5.) cost: revenue: y 30x 25,000 y 0.03x2 100x 34. A 52-foot-long piece of wire is to be cut into three pieces, two of which are the same length. The two equal pieces are to be bent into circles and the third piece into a square. What should the length of each piece be if the total area enclosed by the two circles and the square is 100 square feet? Section 12.5 Nonlinear Systems 825 35. A rectangular box (including top) with square ends and a volume of 16 cubic meters is to be constructed from 40 square meters of cardboard. What should its dimensions be? 36. A rectangular sheet of metal is to be rolled into a circular tube. If the tube is to have a surface area (excluding ends) of 210 square inches and a volume of 252 cubic inches, what size of metal sheet should be used? (Recall that the circumference of a circle with radius r is that the volume of a cylinder with radius r and height h is pr2h. 2pr and 2 37. Find two real numbers whose sum is 16 and whose product is 48. 38. Find two real numbers whose sum is 34.5 and whose product is 297. 39. Find two positive real numbers whose difference is 1 and whose product is 4.16. 40. Find two real numbers whose difference is 25.75 and whose product is 127.5. 41. Find two real numbers whose sum is 3 such that the sum of their squares is 369. 42. Find two real numbers whose sum is 2 such that the difference of their squares is 60. 43. Find the dimensions of a rectangular room whose perimeter is 58 feet and whose area is 204 square feet. 44. Find the dimensions of a rectangular room whose perimeter is 53 feet and whose area is 165 square feet. 45. A rectangle has an area of 120 square inches and a diagonal 17 inches in length. What are its dimensions? 46. A right triangle has an area of 225 square centimeters and a hypotenuse 35 centimeters in length. To the nearest hundredth of a centimeter, how long are the legs of the triangle? 47. Find the equation of the straight line that y x2 intersects the parabola only at the point (3, 9). Hint: What condition on the discriminant guarantees that a quadratic equation has exactly one real solution? 826 Chapter 12 Systems and Matrices 12.5.A Excursion: Systems of Inequalities Objectives • Solve inequalities in two variables • Solve systems of inequalities by graphing • Solve linear programming problems Inequalities in two variables are solved by graphing. The solution to an inequality in two variables is the region in the coordinate plane consisting of all points whose coordinates satisfy the inequality. Example 1 Solving an Inequality in Two Variables Solve the inequality y 2x 2 . Solution y 8 4 −8 −4 0 −4 −8 x 4 8 y 2x 2. First, graph the line The solution to the inequality is the set of all points on the line, plus all points in either the region above or the region below the line. To determine which region is the solution, choose a point that is not on the line, such as (0, 0), and test it in the inequality. ? 2 0 2 0 1 2 False The inequality is false for the test point, so shade the region that does not contain that point—in this case, the region above the line. ■ Figure 12.5.A-1 The method used in Example 1 can be summarized as follows. Test-Point Method for Solving Inequalities in Two Variables Replace the inequality symbol by an equal sign, and graph the resulting line. inequalities, use a solid line to indicate that • For and the line is part of the solution. • For 77 and 66 inequalities, use a dashed line to indicate that the line is not part of the solution. Choose a test point that is not on the line, and substitute its coordinates in the inequality. • If the coordinates of the test point make the
inequality true, then the solution includes the region on the side of the line containing the test point. • If the coordinates of the test point make the inequality false, then the solution includes the region on the side of the line that does not contain the test point. The test-point method can be used to solve any inequality, but the following technique for linear inequalities with two variables is often easier, especially when using a calculator. Section 12.5.A Excursion: Systems of Inequalities 827 Example 2 Solving a Linear Inequality Solve the inequality 6x 3y 6 6 . Solution First, solve the inequality for y. 3y 6 6x 6 y 6 2x 2 If the point y 2x 2 that lie below the line y 2x 2 2 x, y satisfies this inequality, then it lies below the line 1 . Thus, the solution of the inequality is the set of all points , as shown in Figure 12.5.A-2. The line y 2x 2 is not part of the solution. 6.2 –9.4 9.4 Figure 12.5.A-2 –6.2 ■ CAUTION Graphing calculators do not display a dashed line with the shade feature. The line in Figure 12.5.A-2 is not part of the solution. The solution method used in Example 2 can be summarized as follows. Solve the inequality for y so that it has one of the following forms: y 77 mx b y mx b y 66 mx b y mx b • The solution of y 77 mx b is the half-plane above the line. • The solution of y 66 mx b is the half-plane below the line. y mx b For and part of the solution. y mx b , the line y mx b is also Solving Linear Inequalities in Two Variables 828 Chapter 12 Systems and Matrices Technology Tip To display shading on TI models, select the graph style for a function by moving the cursor to the left of the equal sign in the function editor to the graph’s style icon, shown in the first column. Press ENTER repeatedly to rotate through the graph styles until the desired style is shown. Then graph the function as usual. There is also a SHADE option in the DRAW menu. In the function memory of CASIO models, select Type(F3) the inequality sign desired. (F6) and then Systems of Inequalities in Two Variables The solution of a system of inequalities in two variables is found by graphing all the inequalities on the same coordinate plane. The solution is the region that is common to the solutions of all the graphs. Example 3 Solving a System of Inequalities Solve the system of inequalities. y x 3 y 7 x 1 Solution Graph the two inequalities together, as shown in Figure 12.5.A-3. Figure 12.5.A-4 shows a calculator screen with regions shaded. y 10 5 6.2 x −10 −5 0 5 10 9.4 9.4 −5 −10 Figure 12.5.A-3 6.2 Figure 12.5.A-4 Section 12.5.A Excursion: Systems of Inequalities 829 The area where the shaded regions intersect is the solution of the system. Any point in this shaded region of the plane is a solution of both inequalities. For example, the point (5, 2) in the solution satisfies both inequalities, as shown below ■ Linear Programming Linear programming is a process that involves finding the maximum or minimum output of a linear function, called the objective function, subject to certain restrictions, called constraints. For linear programming problems in two variables, the objective function and the constraints are linear inequalities. has the form The solution of the system of linear inequalities formed by the constraints is called the feasible region. ax by, x, y F 2 1 The feasible region is a region in the plane that is bounded by straight lines, such as those shown in Figure 12.5.A-5. The first figure is bounded on all four sides and the two other figures have one side that is not bounded. P Q S R Figure 12.5.A-5 A corner point of such a region is any point where two of the sides intersect, such as points P, Q, R, and S in the first region in Figure 12.5.A-5. The key to solving linear programming problems is the following theorem, which will not be proved here. Fundamental Theorem of Linear Programming The maximum value or minimum value of the objective function (if it exists) always occurs at one or more of the corner points of the feasible region. Thus, the solution may be found by graphing the feasible region, and testing the coordinates of the corner points in the objective function to find the maximum or minimum value of the function. 830 Chapter 12 Systems and Matrices To see why the Fundamental Theorem above is true, notice that the graph of the objective function is a plane in three dimensions. The portion of the plane that lies above the feasible region must have its high point and low point at a corner, as shown in Figure 12.5.A-6. z objective function y feasible region x Figure 12.5.A-6 Example 4 A Linear Programming Problem Find the maximum value of the function constraints below. F 1 x, y 2 3x 2y subject to the x 4y 18 2x y 9 x 0 y 0 µ Solution First, graph the feasible region. This is the solution of the system of inequalities. NOTE In linear programming problems with a feasible region that is unbounded on one or more sides, there may be a minimum or maximum value of the objective function, but (usually) not both. y (0, 4.5) 6 4 2 2x − y = 9 (6, 3) x + 4y = 18 x (0, 0) 2 6 (4.5, 0) Figure 12.5.A-7 Section 12.5.A Excursion: Systems of Inequalities 831 Second, find the corner points of the feasible region from the graph. They are (0, 0), (0, 4.5), (4.5, 0), and (6, 3). Third, evaluate the objective function at each of the corner points. Corner point F(x, y) 3x 2y (0, 0) (0, 4.5) (4.5, 0) (6, 3.. 13.5 24 2 2 2 2 The solution is the corner point (6, 3), which yields the largest value of the objective function, 24. This is the maximum value of the objective function. ■ Example 5 Application Carla is making earrings and necklaces to sell at a craft fair. The profit from each pair of earrings is $3, and the profit from each necklace is $5. She has 12 hours to make all of the jewelry she plans to sell. Each pair of earrings takes 15 minutes to make, and each necklace takes 40 minutes to make. She also has $80 for supplies. The supplies for a pair of earrings cost $2, and the supplies for a necklace cost $4. How many pairs of earrings and how many necklaces should she make to maximize her profit? Solution Let x be the number of pairs of earrings and y be the number of necklaces. 3x 5y. The objective function, which represents the profit, is The constraints are inequalities involving time and cost. In addition, there is an implied constraint that neither quantity can be negative. This gives the following linear programming problem: x, y P 2 1 Maximize Subject to: µ 3x 5y 1 x, y P 2 15x 40y 720 2x 4y 80 x 0 y 0 time cost y 21 18 15 12 9 6 3 (16, 12) x 0 −8 8 16 24 32 40 48 56 Figure 12.5.A-8 The feasible region is shown in Figure 12.5.A-8. 832 Chapter 12 Systems and Matrices The corner points are shown below, with the objective function evaluated at each point. Corner point P(x, y) 3x 5y (0, 0) (0, 18) (40, 0) (16, 12 40 16 3 1 2 5 2 5 0 1 18 1 5 2 5 0 1 12 1 0 90 120 108 2 2 2 2 The maximum value occurs at (40, 0). To make the maximum profit, Carla should make 40 pairs of earrings and 0 necklaces. Her maximum profit will be $120. ■ Summary: Solving Linear Programming Problems To solve a linear programming problem: • Graph the feasible region. This is the solution of the system of inequalities formed by the constraints. • Find the corner points of the feasible region from the graph. • Evaluate the objective function at each corner point. • Choose the corner point which yields the greatest (or least) value of the objective function. This is the maximum (or minimum) value of the function on the feasible region. Exercises 12.5.A In Exercises 1–12, solve the system of inequalities. 1. 3. y 2x 4 y 7 x 2 y 7 3x 1 y 6 1 x 3 2 2. y x 3 y 4x 2 4. y 1 x 1 4 y 7 4x 1 5. 7. 2x 3y 7 6 x 2y 5 7 2x 1 y x 6. 8. 2x 4y 5 3x 2y 8 y x y 6 2x 3 Section 12.5.A Excursion: Systems of Inequalities 833 9. y 7 2x 1 x y 6 16 11. y 1 x 1 10. y x 3 x y 6 1 12. 4x 9y 36 4y x 6 16 20. Minimize: Subject to: 1 F ⎧ ⎪ ⎨ ⎪ ⎩ 2 4x 2y x, y 3x 12y 60 8x 5y 56 x 0 y 0 In Exercises 13–16, graph the feasible region described by the system of inequalities, and find all of its corner points. 13. 15. 3x 4y 24 9x 2y 30 x 0 y 0 x 2y 8 3x y 6 x 0 y 0 14. 16. x y 7 7x 2y 24 x 0 y 0 2x 3y 12 5x 2y 10 x 0 y 0 Solve the following linear programming problems. 17. Maximize: Subject to: 18. Maxmimize: Subject to: 19. Minimize: Subject to ⎧ ⎪ ⎨ ⎪ ⎩ 2x 5y x 3x 7y x, y x 2y 10 7x 11y 105 , y 4 3x 4y 32 4x 2y 36 x 0 y 0 21. A lunch counter sells two types of sandwiches, roast beef and chicken salad. The profit on the sandwiches is $2 for chicken salad and $3 for roast beef. The amount of bread available is enough for 30 sandwiches. There are 4 hours available to prepare sandwiches. If chicken salad sandwiches take 7 minutes to prepare and roast beef sandwiches take 10 minutes, how many of each type of sandwich should be prepared to maximize the profit? 22. A dealer has a lot that can hold 30 vehicles. In this lot, there are two available models, A and B. The dealer normally sells at least twice as many model A cars as model B cars. If the dealer makes a profit of $1300 on model A cars and $1700 on model B cars, how many of each car should the dealer have in the lot? 23. A 30-acre orchard is to contain two types of trees, peach and almond. The profit per year is $16.80 per peach tree and $21.60 per almond tree. An acre can sustain 1080 peach trees or 900 almond trees. If the grower has available labor to plant a maximum of 30,000 trees, how many of each type of tree should be planted? 24. An investor has $12,000 to invest into two different funds. Fund A, which is a high-risk fund, yields an average return of 14%. Fund B, which is a low-risk fund, yields an average return of 6%. To reduce the risk, the investor wants the amount in fund B to be at least twice the amount in fund A. How much should be invested in each fund to maximize the return? What is the maximum return? C H A P T E R 12 R E V I E W Important Concepts Section 12.1 System of equations . . . . . . .
. . . . . . . . . . . . . . . 779 Solution of a system. . . . . . . . . . . . . . . . . . . . . . 780 Number of solutions of a system . . . . . . . . . . . . 782 Substitution method. . . . . . . . . . . . . . . . . . . . . . 782 Elimination method . . . . . . . . . . . . . . . . . . . . . . 783 Inconsistent system . . . . . . . . . . . . . . . . . . . . . . 784 Section 12.1.A z-axis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 790 Coordinate plane . . . . . . . . . . . . . . . . . . . . . . . . 790 Octants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 790 Section 12.2 Section 12.3 Section 12.4 Section 12.5 Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 795 Augmented matrix . . . . . . . . . . . . . . . . . . . . . . . 795 Equivalent systems. . . . . . . . . . . . . . . . . . . . . . . 795 Elementary row operations . . . . . . . . . . . . . . . . 796 Reduced row-echelon form . . . . . . . . . . . . . . . . 797 Gauss-Jordan elimination. . . . . . . . . . . . . . . . . . 797 m n matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . 804 Equality of matrices . . . . . . . . . . . . . . . . . . . . . . 804 Addition and subtraction of matrices . . . . . . . . 804 Scalar multiplication of matrices . . . . . . . . . . . . 805 Matrix multiplication . . . . . . . . . . . . . . . . . . . . . 806 Directed networks . . . . . . . . . . . . . . . . . . . . . . . 809 Adjacency matrix . . . . . . . . . . . . . . . . . . . . . . . . 809 Matrix equation . . . . . . . . . . . . . . . . . . . . . . . . 814 Identity matrices . . . . . . . . . . . . . . . . . . . . . . . . 815 Invertible matrix. . . . . . . . . . . . . . . . . . . . . . . . . 815 Inverse of a matrix . . . . . . . . . . . . . . . . . . . . . . . 816 Matrix solutions of a square system . . . . . . . . . 817 Curve fitting. . . . . . . . . . . . . . . . . . . . . . . . . . . . 818 Algebraic solutions of systems of nonlinear equations . . . . . . . . . . . . . . . . . . . . . . 821 Graphical solution of systems of nonlinear equations . . . . . . . . . . . . . . . . . . . . . . 822 Section 12.5.A Solutions of systems of linear inequalities . . . . 826 Linear programming . . . . . . . . . . . . . . . . . . . . . 829 834 Review Exercises In Exercises 1–10, solve the system of linear equations by any method. Chapter Review 835 Section 12.1 1. 5x 3y 4 2x y 3 3. 3x 5y 10 4x 3y 6 5. 7. 9. 3x y z 13 x y 2z 9 3x y 2z 9 4x 3y 3z 2 5x 3y 2z 10 2x 2y 3z 14 x 2y 3z 1 5y 10z 0 8x 6y 4z 8 2. 3x y 6 2x 3y 7 4. 6. 8. 1 4 1 10 2y 3z 1 4x 4y 4z 2 10x 8y 5z 4 x y 4z 0 2x y 3z 2 3x y 2z 4 10. 4x y 2z 4 x y 1 z 1 2 2x y z 8 11. The sum of one number and three times a second number is 20. The sum of the second number and two times the first number is 55. Find the two numbers. 12. You are given $144 in $1, $5, and $10 bills. There are 35 bills. There are two more $10 bills than $5 bills. How many bills of each type do you have? 13. Let L be the line with equation 4x 2y 6 and M the line with equation 10x 5y 15. a. L and M do not intersect. Which of the following statements is true? b. L and M intersect at a single point. d. All of the above are true. c. L and M are the same line. e. None of the above are true. 14. Which of the following statements about the given system of equations are false? x 4y z 2 6x 4y 14z 24 2x y 4z 7 z 0 z 1 x 2, x 1, x 1, y 3, y 1, y 3, is a solution. is a solution. a. b. c. d. The system has an infinite number of solutions. e. is not a solution. is a solution. z 1 x 2, y 5, z 3 15. Tickets to a lecture cost $1 for students, $1.50 for faculty, and $2 for others. Total attendance at the lecture was 460, and the total income from tickets was $570. Three times as many students as faculty attended. How many faculty members attended the lecture? 16. An alloy containing 40% gold and an alloy containing 70% gold are to be mixed to produce 50 pounds of an alloy containing 60% gold. How much of each alloy is needed? 836 Chapter 12 Systems and Matrices Section 12.2 For Exercises 17–20, write the system of linear equations represented by the augmented matrix. 17. 2 2 a 6 3 16 7b 19 10 2 1 3 ¢ 18. 2 3 a 1 2 4 1b 20 ¢ For Exercises 21–26, write the system represented by each matrix, find the solutions, if any, and classify each system as consistent or inconsistent. 21. 4 1 a 2 5 14 9b 23 ¢ 25. 2 1 4 3 1 1 a 4 7b 22. 9 12 a 6 8 3 4b 24 ¢ 26 ¢ Section 12.3 In Exercises 27–30, perform the indicated matrix multiplication or state that the product is not defined 1b 1 0 2 4b B 2 4 a E ° 3 1b 4b 3 3 1 ¢ 27. AB 28. CD 29. AE 30. DF In Exercises 31–34, find the inverse of the matrix, if it exists. 31. 33. 3 4 a 3 1 2 ° 7 9b 2 1 2 6 2 5 ¢ 32. 2 1 a 6 3b 34 ¢ Section 12.4 In Exercises 35–38, use matrix inverses to solve the system. 35. x 2y 3z 4 2x y 4z 3 3x 4y z 2 36. 2x y 2z 2u 0 x 3y 2z u 0 x 4y 2z 3u 0 x 4y 2z 3u 0 37. x 4y 2z 6w 2 3x 4y 2z w 0 5x 4y 2z 5w 4 4x 4y 2z 3w 1 39. Find the equation of the parabola passing through the points , (2, 17), (8, 305). 3, 52 1 2 Chapter Review 837 38. 2x y 2z u 2 x 3y 4z 2u 2v 2 2x 3y 5z 4u v 1 x 3y 2z 4u 4v 4 2x 3y 6z 4u 5v 0 40. The table shows the number of hours spent per person per year on home video games. Find a quadratic equation that contains x 6 this data, with to 1996. corresponding Year 1996 2000 2004 Hours 25 76 161 [Source: Statistical Abstract of the U.S.: 2001] Section 12.5 In Exercises 41–46, solve the system. 41. x2 y 0 y 2x 3 43. x2 y2 16 x y 2 45. x3 y3 26 x2 y 6 Section 12.5.A 47. Minimize and maximize F 2x 2y 4 x y 36 2x y 10 x 0 y 0 subject to ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ 42. x2 y2 25 x2 y 19 44. 6x2 4xy 3y2 36 x2 xy y2 9 46. x2 3xy 2y2 y x 0 5x2 10xy 5y2 8 30x 10y x, y 1 2 48. Minimize and maximize 8x 7y x, y 2 subject to F 1 4x 3y 24 3x 4y 8 x 0 y 0 ⎧ ⎪ ⎨ ⎪ ⎩ 49. Animal feed is to be made from corn and soybeans. One pound of corn has 30 units of fat and 20 units of protein, and one pound of soybeans has 20 units of fat and 40 units of protein. What is the minimum total weight of feed to supply a daily requirement of 2800 units of fat and 2200 units of protein? 50. A home supply store sells two models of dehumidifiers, standard and 3 deluxe. The standard model comes in a 10-ft box and weighs 10 lb, and 3 the deluxe model comes in a 9-ft box and weighs 12 lb. The store’s delivery van has 248 ft of space and can hold a maximum of 440 lb. If the store makes a profit of $20 on the standard model and $30 on the deluxe model, how many boxes of each model can the van carry to maximize the profit for each load? 3 C H A P T E R 12 Partial Fractions In calculus it is sometimes necessary to write a complicated rational expression as the sum of simpler ones. Two forms of rational expressions whose sum is a rational expression will be introduced in this section: denominators with nonrepeated factors and those with repeated factors. Example 1 Denominators with Nonrepeated Factors Find the constants A and B such that 7x 6 2 x 6 x˛ A x 2 B x 3 . Solution The denominators on the right side of the equation are linear factors of the denominator on the left side. Multiply both sides of the equation by the common denominator, x 3 x 2 7x and collect like terms. x 2 2 2 Ax 3A Bx 2B Ax Bx 3A 2B x 3A 2B A B 1 2 Because the polynomials on the left and right sides of the last equation are equal, their coefficients must be equal term by term. 2 1 1 A B 7 3A 2B 6 Coefficient of x Constant term The two equations above form a system of equations with unknowns A A 4 and B. Solving the system yields Therefore, B 3. and 7x ■ In Example 1, 4 x 2 3 x 3 is called the partial fraction decomposi- tion, or simply the partial fractions, of 7x 6 2 x 6 . x If the denominator of a rational expression can be expressed as a product of nonrepeated linear factors, each term of the decom- position has the form A x a. Figure 12.C-1 Nonrepeated Linear Factor Denominators 838 5 Solution 5 5 5 Figure 12.C-2 Figure 12.C-3 Nonrepeated Quadratic Factor Denominators When a factor of the denominator is repeated, every power less than or equal to the multiplicity of the factor must be considered. Example 2 Repeated Linear Factors Find the partial fraction decomposition of 2x x 2 15x 10 2 4 3 3x . The denominator can be factored into ure 12.C-2. Because 2 must be considered as possible denominators of the decomposition. The process of finding the numerators is the same as that shown in Example 1. x 2 x 1 2, 2 x 2 is a repeated factor, both as shown in Fig2 x 2 x 2 and 2 2 1 1 1 1 2 1 2 Multiply both sides of the equation by the common denominator, x 1 2 , and collect like terms on the right side. C x 2 1 2 2 15x 10 A x 1 x 2 2 B x 2 2x 4x 4 A x x 1 1 2 Ax 2 Bx 2B Cx C 2 4Ax 4A Bx 4A 2B C 1 The polynomials on the left and right sides are equal, so coefficients must be equal term by term. 4A 4A B C 15 4A 2B C 10 Coefficient of x 2 Coefficient of x Constant term This is a system of equations with unknowns A, B, and C. The augmented matrix of the system and an equivalent reduced row echelon form matrix are shown in Figure 12.C-3. A 3, Therefore, C 4, and B 1, 2 15x 10 2 x 2 2x ■ In theory, any polynomial with real coefficients can be written as a product of real linear factors and real quadratic factors. (See Section 4.2.) If the denominator contains a nonrepeated quadratic factor, then the decomposition will contain a term of the form where ax 2 bx c is irreducible over the set of Ax B ax2 bx c real numbers. , Like repeated linear factors, if an irreducible quadratic factor is repeated, every power less than or equal to the multiplicity of the factor must be considered. The numerator of all quadratic factors has the form Ax B. 839 Example 3 Nonrepeated Quadratic Factor Find the partial fraction decomposition of x 2 2 10x 8 3 6x . x Solution 10 The denominator has the nonrepeated linear factor as shown in x Figure 12.C-4, and the nonrepeated quadratic factor found by using synthetic division. Thus, the partial fraction decomposition has
the x 4, 2 2x 2 –10 10 form x 3 6x x 2 2 10x 8 A x 4 Bx C 2 2x 2 x . –10 Figure 12.C-4 1 2 1 x Multiplying both sides of the equation by the common denominator, x 4 , and collecting like terms yields x 4 2 Cx 4Bx 4C 2A 4C 2 2x 2 2 2 2x 2 x 2 A x 1 Ax 2 2Ax 2A Bx 2A 4B C Bx C A B 1 x Because there is no sponding coefficient must be 0. Therefore, 2 term in the original rational expression, the corre 4B C 1 2 A 4C 2 Coefficient of x 2 Coefficient of x Constant term Figure 12.C-5 Figure 12.C-5. The solution of the system is A 3 5 , B 3 5 , and C 1 5 , as shown in x 2 x3 6x2 10x x2 2x 2 1 5 3 x 4 a 3x 1 x2 2x 2b ■ All the rational expressions in the previous examples have been proper fractions, which means the degree of the numerator was less than the degree of the denominator. If the rational expression is improper, then divide the numerator by the denominator and decompose the remainder that is a proper fraction. Example 4 Decomposing an Improper Rational Expression Find the partial fraction decomposition of 3 x 3 2 x 2 x x . Solution The rational expression is an improper fraction because the degree of the numerator is greater than the degree of the denominator. Therefore, divide the numerator by the denominator (shown in the margin) and write the remainder as a fraction of the divisor. x 1 2 x 2x 2 x 3 3 0x x 2 2x 2x 1 840 Figure 12.C-6 Decomposition into Partial Fractions 3 x 3 x 2 x 2 x 2x 1 2 x 2 Now decompose x discussed in Example 1. x 1 2x 1 x 2 x 2 into A x 2 B x 1 by using the procedure 2x 1 A 1 B x 1 x 2 2 Ax A Bx 2B A B x A 2B 2B 1 A 1 and B 1. As shown in Figure 12.C-6 ■ When the denominator contains a power of a linear or a quadratic factor, every integral power of that factor must be taken into consideration when finding partial fractions, as outlined below. 1. Divide numerator by denominator if the fraction is improper, and find partial fractions of the remainder. 2. Factor the denominator into factors of the form (px q)m (ax2 bx c)n, and over the set of real numbers. where ax2 bx c is irreducible 3. For each linear factor of the form (px q)m , the partial fraction must include the following sum: A1 px q A2 (px q)2 p Am (px q)m (ax2 bx c)n, the 4. For each quadratic factor of the form partial fraction must include the following sum: B2x C2 (ax2 bx c)2 B1x C1 ax2 bx c Bnx Cn (ax2 bx c)n p Exercises In Exercises 1–7, find the partial fraction decomposition of each expression. 5. 5x x 2 1 3 1 x 2 2 10x 8 3 6x 6. x 1. 3. x 2 3x 2 x 2x 1 2 3x 18 3 4x x 2. 4. 1 2 1 x 2x 7. 2 x 3 3 4x x 2 2x 3 2 x 21 x 3 x 2x 2 8x 4 841 C H A P T E R 13 Statistics and Probability What are the odds? Suppose a dart player can hit the bulls-eye about 25% of the time. How likely is the event shown above? The number of bulls-eyes in a given number of tries can be described as a binomial experiment, and the probabilities can be easily calculated. See Exercises 5–8 in Section 13.4.A. 842 Chapter Outline Basic Statistics 13.1 13.2 Measures of Center and Spread 13.3 Basic Probability 13.4 Determining Probabilities 13.4.A Excursion: Binomial Experiments 13.5 Normal Distributions Chapter Review can do calculus Area Under a Curve Interdependence of Sections 13.1 > 13.2 13.3 > 13.4 13.5 > > S tatistics and probability are essential tools for understanding the modern world. Both involve studying a group of individuals or objects, known as a population, and a subset of the population, known as a sample. In statistics, information from a sample is used to draw conclusions about the population. In probability, information from the population is used to draw conclusions about a sample. Population statistics probability Sample 13.1 Basic Statistics Objectives • Identify data types • Create displays of qualitative and quantitative data • Describe the shape of a distribution data qualitative quantitative discrete continuous Statistics is used to make sense of information, or data, by using techniques to organize, summarize, and draw conclusions from the data. Most statistical data is gathered by taking a random sample of the population. In a random sample, all members of the population and all groups of members of a given size have an equal chance of being in the sample. Data can be divided into two types: qualitative and quantitative. Quantitative data is numerical, such as “the number of hours spent studying each night” or “the distance from home to school.” Qualitative data can be divided into categories, such as “liberal,” “moderate,” “conservative,” or “blue eyes,” “brown eyes.” Quantitative data can be further classified as either discrete or continuous. If the difference between two values can be arbitrarily small, the data is continuous. If there is a minimum increment between two different values, the data is discrete. 843 844 Chapter 13 Statistics and Probability Example 1 Types of Data In each example, identify the data as either qualitative or quantitative. If quantitative, then identify it as discrete or continuous. a. the height of each player on a basketball team b. the style of shoes worn by each student in a classroom c. the number of people in each household in the United States Solution a. The data is quantitative, because each value can be written as a number, such as 68.32 inches. It is continuous, because there is no minimum difference between two values; two players could have heights 68.32 inches and 68.33 inches, or 68.323 inches and 68.324 inches. b. The data is qualitative, since it can be grouped into categories, such as tennis shoes, sandals, and high heels. c. The data is quantitative, because each value is a number. It is discrete, because there is a minimum difference of 1 between two different values; two households could have 4 and 5 members, but not 4 and 4.1 members. ■ NOTE Continuous data is sometimes treated as discrete, and vice versa. For example, heights are usually rounded to the nearest inch, so there is a minimum difference of 1 inch between measurements. In the discrete case, for amounts of money, the minimum increment of $0.01 is so small that the data can often be treated as continuous. Data Displays One of the most important uses of statistics is to organize data and display it visually. Most displays show the data values or categories and some measure of how often each value or category occurs. The number of times a value occurs is known as the frequency of that value. If the frequency is divided by the total number of responses, the result is the relative frequency of that value, which can be expressed as a fraction, a decimal, or a percent. A frequency table displays the categories with frequencies, relative frequencies, or both. Example 2 Frequency Table A group of 30 people were asked their favorite flavor of ice cream. Of these, 6 chose vanilla, 12 chose chocolate, 4 chose butter pecan, and 8 chose mint chocolate chip. Create a table with frequencies and relative frequencies for each flavor. Section 13.1 Basic Statistics 845 Solution Flavor Frequency Relative frequency Vanilla Chocolate Butter pecan Mint chocolate chip 6 12 4 8 6 30 1 5 12 30 2 5 0.2 20% 0.4 40% 4 30 2 15 8 30 4 15 0.13 13% 0.27 27% ■ Displaying Qualitative Data Two common ways of displaying qualitative data are bar graphs and pie charts. A bar graph displays the categories on a horizontal axis and the frequencies or relative frequencies on a vertical axis, or vice versa. The height or length of each bar shows the frequency of the value. All bars should have the same width. Example 3 Bar Graph Use the data in the frequency table from Example 2 to make a bar graph. Solution y c n e u q e r F 14 12 10 8 6 4 2 0 Vanilla Chocolate Butter pecan Mint chocolate chip Flavor ■ A pie chart displays the categories and their relative frequencies. The “pie” is divided into sectors whose central angle measure equals the fraction of represented by the relative frequency of each category. 360° The central angle measure of the sector that represents a category with relative frequency r is r 360°. 846 Chapter 13 Statistics and Probability Example 4 Pie Chart Create a pie chart using the data in the frequency table from Example 2. Label each sector with the category and its relative frequency. Solution The central angle measures of the categories of the sectors are: Vanilla: 0.2 360° 72° Chocolate: 0.4 360° 144° Butter pecan: 0.13 360° 47° Mint chocolate chip: 0.27 360° 97° Mint chocolate chip 0.27 Vanilla 0.20 72° 97° 47° 144° Butter pecan 0.13 Chocolate 0.40 ■ Displaying Quantitative Data Numerical data can also be displayed in a variety of ways to indicate each value and its frequency. The shape of a smooth curve over the display indicates characteristics of the data values. The most common distribution shapes are shown below. Uniform: All the data values have approximately the same frequency. Symmetric: The right and left sides of the distribution have frequencies that are mirror images of each other. Skewed right: The right side of the distribution has much lower frequencies than the left. Skewed left: The left side of the distribution has much lower frequencies than the right. NOTE Before statistical analysis, numerical data is usually arranged in order from the lowest value to the highest value. An arrangement of numerical data is called a distribution. Common Distribution Shapes NOTE If an outlier is caused by an error in measurement or other type of error, it is usually removed from the data set before further analysis. In general, an outlier should not be removed without justification. Section 13.1 Basic Statistics 847 An outlier is a data value that is far removed from the rest of the data, which usually indicates that the value needs investigation. Outliers may be caused by errors or by unusual members of the population. Example 5 The Shape of Data From the four given shapes, choose the best distribution for the data. a. the last digit of each number in the phone book b. the salaries of the employees of a corporation c. the ag
e of retirement for all people in the U.S. d. the heights of all adult women in the U.S. Solution a. The last four digits of a phone number are assigned randomly, so all digits have about the same frequency. The distribution is uniform. b. In a typical corporation, most employees earn relatively low salaries while a few executives make high salaries. The distribution is skewed right. c. Few people make enough money to retire young, and most people retire in their 60’s or later. The distribution is skewed left. d. The average height of an adult woman is at the middle of the distribution, which is symmetric with respect to this value. ■ Two common displays of quantitative data are the stem plot and the histogram. A stem plot is commonly used to display small data sets. The data below shows 31 test scores for a class exam: 32, 67, 89, 90, 87, 72, 75, 88, 95, 83, 97, 72, 85, 93, 79, 63 70, 87, 74, 86, 98, 100, 97, 85, 77, 88, 92, 94, 81, 76, 64 The stem plot for the class test data is shown below: 3 4 5 6 7 8 9 10 Figure 13.1-1 6 0 3 The entry sents the scores 63, 64, and 67. There are 31 scores represented. represents the score 32. Similarly, the row 3 0 2 4 7 repre- 848 Chapter 13 Statistics and Probability Note that the score of 32 is far below the remaining data and so it could be an outlier in this distribution. It may be the score of a student who didn’t study for the test. However, it is most likely not an error in measurement, so the value cannot be removed from the data set. The distribution is skewed left (Figure 13.1-1). Creating a Stem Plot To create a stem plot: 1. Choose the leading digit or digits to be the stems. Arrange the stems vertically from lowest to highest value from top to bottom. 2. The last digit is the leaf. Record a leaf for each data value on the same horizontal line as its corresponding stem. Arrange the leaves from lowest to highest value from left to right. 3. Provide a key that indicates the total number of data elements and an interpretation of one stem and leaf indicating appropriate units. Example 6 Stem Plot A company uses a 3-minute recorded phone message to advertise its product. A random sample of 40 calls is used to determine how much of the message was heard before the listener hung up. Create a stem plot of the data below and discuss its shape. 2.4, 0.2, 3.0, 2.8, 1.5, 1.9, 0.7, 1.0, 2.5, 1.3, 0.8, 2.1, 3.0, 0.4, 1.2, 3.0, 1.1, 0.3, 0.7, 1.8, 0.3, 1.0, 2.1, 3.0, 2.9, 0.5, 1.4, 3.0, 2.8, 1.2, 0.5, 0.5, 1.5, 0.9, 1.8, 0.6, 0.6, 0.7, 0.8, 0.8 Solution Key: 0 8 represents a time of 2 2.8 minutes. 40 times are represented. The distribution is skewed right, as shown in Figure 13.1-2. The values on the right side have lower frequencies than the values on the left. Notice that the distribution is cut off at 3, because no phone calls last longer than 3 minutes, the length of the entire message. ■ A histogram, which can be thought of as a bar graph with no gap between adjacent bars, is often used with large sets of quantitative data. First, the Figure 13.1-2 Section 13.1 Basic Statistics 849 data is divided into a convenient number of intervals of equal width. The frequencies (or relative frequencies) of the data in the intervals are the heights of the rectangles. For example, the test scores given on page 847 can be represented by the histogram in Figure 13.1-3 12 10 8 6 4 2 0 0 10 20 30 40 50 60 70 80 90 100 Score Figure 13.1-3 Here, the test scores have been divided into 10-point intervals, 0 through 9, 10 through 19, and so on. The histogram indicates, for example, that there are no scores between 50 and 59, eight scores between 70 and 79, and one score of 100. Each bar on the graph has width 10, so the class interval is said to be 10. Creating a Histogram To create a histogram: Technology Tip Most graphing calculators will produce histograms. Check the Xmin, Xmax, and Xscl to make sure the class intervals are relevant to the data. If you do not know how to create a histogram with a graphing calculator, refer to the Technology Appendix. 1. Divide the range of the data into classes of equal width, so that each data value is in exactly one class. The width of these intervals is called the class interval. 2. Draw a horizontal axis and indicate the first value in each class interval. 3. Draw a vertical scale and label it with either frequencies or relative frequencies. 4. Draw rectangles with a width equal to the class interval and height equal to the frequency of the data within each interval. Example 7 Histogram Create a histogram of the following scores. 580, 490, 590, 390, 410, 370, 470, 540, 490, 660, 500, 670, 430, 670, 490, 720, 580, 680, 590, 480, 560, 480, 400, 440, 560, 540, 330, 490, 540, 540, 520, 650, 540, 600, 630, 580, 540, 500, 270, 600, 390, 540, 300, 350, 600, 540, 510, 410, 370, 390, 160, 500, 740, 510, 540, 560, 510, 430, 440, 590, 560, 510, 600, 460, 450, 510, 420, 430, 560, 680, 610, 600, 600, 520, 480, 490, 320, 450, 500, 490 850 Chapter 13 Statistics and Probability Solution The smallest value is 160 and the largest value is 740, so the range of the data is 580 points. A convenient choice for the classes is 150 through 199, 200 through 249, 250 through 299, and so on—a class interval of 50 points. The frequency table below shows how many data values are in each class. Class 150– 199 200– 249 250– 299 300– 349 350– 399 400– 449 450– 499 500– 549 550– 599 600– 649 650– 699 700– 749 Frequency 1 0 1 3 6 9 13 20 11 8 6 2 The histogram is shown in Figure 13.1-4. The shape is approximately symmetric 25 20 15 10 5 0 150 200 250 300 350 400 450 500 550 600 650 700 Score Figure 13.1-4 ■ Figure 13.1-5 shows three histograms of the data in Example 7, created on a calculator. Notice that the choice of the scale on the x-axis determines the width of the class intervals. Which do you think is the best representation of the data? 12 25 150 0 Xscl 25 800 150 0 800 150 Xscl 50 Figure 13.1-5 50 0 800 Xscl 100 Remember that it is more important to be able to interpret a distribution than it is to simply produce the display. Technology can easily produce a histogram, but the purpose of the display is to help interpret the data. Section 13.1 Basic Statistics 851 Exercises 13.1 In Exercises 1–4, identify the population and the sample. Exercise Frequency Relative frequency 1. There are three schedule options for classes at a high school: 90-minute classes every other day for a year, 90-minute classes every day for a semester, or 45-minute classes every day for a year. Out of 1200 students, 50 students from each grade level are chosen at random and asked their preference. Aerobics Kickboxing Tai chi Stationary bike 20 8 8 14 ? ? ? ? 2. The manager of a convenience store wishes to determine how many cartons of eggs are damaged in shipment and delivery. For ten shipments of 1000 cartons of eggs, she examines every cracked eggs. carton to see how many cartons contain 50th 3. A survey is taken to determine the number of pets in the typical American family. A computer is used to randomly select 5 states, then 10 counties in each state, then 50 families in each county. Each of these families is asked how many pets they own. 4. A biologist tranquilizes 400 wild elephants and measures the lengths of their tusks to determine their ages. 11. Complete the table to show the relative frequency for each category. 12. Create a bar graph for the data with the vertical axis showing the frequencies. 13. Create a bar graph with the vertical axis showing the relative frequencies. 14. Create a pie chart for the data. For Exercises 15–18, suppose 25 people are asked their favorite color. The results are: 6 red, 8 blue, 5 purple, 4 green, 1 yellow, and 1 orange. 15. Create a frequency table for the given data. Include relative frequencies. Use the descriptions in Exercises 1–4 to answer Exercises 5–10. 16. Create a bar graph for the data with the vertical axis showing the frequencies. 5. Determine whether the data in each description is qualitative or quantitative. If the data is quantitative, determine whether it is continuous or discrete. 17. Create a bar graph with the vertical axis showing the relative frequencies. 18. Create a pie chart for the data. 6. Describe two ways in which the data from Exercise 1 could be displayed. 7. How large is the sample in Exercise 2? 8. Describe two ways in which the data from Exercise 2 could be displayed. In Exercises 19–24, state whether the shape of the distribution is best described as uniform, symmetric, skewed right, or skewed left. 19. The scores of a national standardized test 20. The age at which students get a driver’s license 9. How large is the sample in Exercise 3? 21. 100 10. Which would be more appropriate to display the data from Exercise 4: a stem plot or a histogram? Explain your reasoning. The following frequency table gives the preferred type of exercise for 50 women at a local gym. 80 60 40 20 0 852 Chapter 13 Statistics and Probability 22 23. The position of the second hand of a clock at 100 randomly chosen times in a 12-hour period 24. 8 9 10 11 12 In Exercises 25–28, create a stem plot for the given data. 25. 23, 45, 38, 41, 24, 67, 42, 46, 51, 33, 43, 47, 54, 49, 47, 36, 27, 33, 41, 29 26. 1.8, 2.0, 1.4, 5.6, 1.1, 2.6, 0.8, 1.5, 1.4, 2.6, 0.7, 1.6, 0.4, 1.1, 0.5, 1.3 27. 98, 87, 100, 86, 92, 78, 56, 100, 90, 88, 93, 99, 76, 83, 86, 91, 72, 85, 79, 81, 82, 91, 86, 70, 84 During summer semester, a community college surveyed its students to determine travel time to campus. A random sample of 30 students gave the following times in minutes: 12 15 32 42 12 25 55 40 65 28 48 75 18 17 42 45 60 35 25 37 6 27 22 35 45 90 30 8 40 55 During fall semester, the survey was repeated. The new times in minutes are: 63 46 9 35 104 52 40 25 31 7 43 29 40 69 52 48 20 20 86 55 32 75 46 63 29 14 48 37 17 14 28. Create a stem plot of the summer semester data. 29. Does the data set of the summer semester times contain any outliers? Explain. 30. Create a stem plot o
f the fall semester data. 31. Compare the shape of the two data sets. Which type of distribution do you think best describes these data sets? What might explain the differences in these data sets? 32. Create a histogram with a class interval of 10 for the data below. 68, 84, 59, 72, 62, 76, 61, 63, 68, 56, 70, 79, 54, 65, 66, 71, 70, 58, 63, 68, 84, 63, 53, 68, 63, 76, 66, 66, 70, 72, 88, 68, 75, 63, 76, 58, 86, 65, 66, 73, 53, 76, 59, 81, 59, 65, 67, 73, 62, 75, 89, 58 33. Create a histogram for the following ACT scores. Be sure to choose an appropriate class interval. 14, 25, 15, 18, 17, 11, 15, 10, 25, 6, 11, 4, 12, 24, 19, 14, 20, 13, 23, 19, 13, 20, 14, 24, 10, 18, 30, 22, 16, 26, 10, 23, 22, 19, 23, 21, 16, 18, 18, 20, 25, 14, 19, 7, 16, 18, 31, 14, 7, 10, 16, 13, 18, 10 34. Create a histogram with a class interval of 5 for the data in the stem plot below 35. Critical Thinking How does the shape of the histogram you created in Exercise 34 compare to the shape of the data in the stem plot? Which do you think is a better representation of the data, and why? Section 13.2 Measures of Center and Spread 853 13.2 Measures of Center and Spread Objectives • Calculate measures of center • Calculate measures of spread • Choose the most appropriate measure of center or spread • Create and interpret a box plot While the shape of a stem plot or a histogram gives a picture of a data set, numerical measures are more precise and can be calculated easily (using technology for large data sets). These measures help to further summarize and interpret data. Two quantities are commonly used to describe a data set: a measure of the “center” of the data and a measure of how spread out the data is. Measures of Center A sample may contain hundreds, or even thousands of data values. This information is often summarized by one value that represents the center, or central tendency, of the data. The three most common measures of center are mean, median, and mode. Mean The mean is more commonly known as the average. The mean is calculated by adding all values and dividing by the total number of values. © Recall from Chapter 1, the symbol is used to indicate the sum of a set of values. mean: x x1 x2 p xn n ©xi n x, The mean is represented by x1, the data variable, while each data value is represented as so on. The sum is divided by n, the number of data elements. read as “x bar.” The x is used to represent and x3, x2, Example 1 Mean Number of Accidents A six-month study of a busy intersection reports the number of accidents per month as 3, 8, 5, 6, 6, 10. Find the mean number of accidents per month at the site. Solution x1 x 3, x2 ©xi n 8, x3 5, x4 3 8 5 6 6 10 6 6, x5 6, x6 38 6 10 6.3 The data shows an average of 6.3 accidents per month at the given intersection. ■ One problem with the mean as a measure of center is that it may be distorted by extreme values, as shown in the following example. 854 Chapter 13 Statistics and Probability Example 2 Mean Home Prices In the real-estate section of the Sunday paper, the following houses were listed: 2-bedroom fixer-upper: 2-bedroom ranch: 3-bedroom colonial: 3-bedroom contemporary: 4-bedroom contemporary: 8-bedroom mansion: $98,000 $136,700 $210,000 $289,900 $315,500 $2,456,500 Find the mean price, and discuss how well it represents the center of the data. Solution x ©xi n 98,000 136,700 210,000 289,900 315,500 2,456,500 6 3,506,600 6 584,433.33 In the data set, 5 out of the 6 values are below $350,000, but the mean is over $550,000, so the mean does not seem to be a very good representation of the center of the data set. Notice that the value $2,456,500 is more than twice the rest of the data combined. Thus, it has a very large effect on the mean, “pulling” it away from the other values. ■ Median As shown in Example 2, the mean is not always the best way to represent the center of a distribution. If the distribution is skewed or contains extreme values, the median, or middle value of the data set is often used. To determine the median, the data must be in order from smallest to largest (or largest to smallest). If the number of values is odd, then one number will be the middle number, as shown below. 3, 4, 7, 8, 9, 11, 15 There are 7 values, and the median, which is in the position, is 8. Notice that three values are less than the median and three values are greater than the median. 4th If the number of values is even, there are two middle numbers, as shown below. 17, 22, 24, 30, 35, 40 There are 6 values, and the median is the average of the middle numbers, 27. positions. So the median is which are in the and 3rd 4th 24 30 2 Notice that three values are less than the median and three values are greater than the median. Median Section 13.2 Measures of Center and Spread 855 p xn x3 x2, x1, If are ordered from smallest to largest, then the median is the middle entry when n is odd and the average of the two middle entries when n is even. for n odd, the value in the n 1 2 position median for n even, the average of the values in the n 2 and n 2 1 positions d The median is said to be a more resistant measure of center than the mean, since it is less affected by a skewed distribution or extreme values in a data set. Example 3 Median Home Prices Find the median of the data in Example 2, and discuss how well it represents the center of the data. Solution The data is already in order from smallest to largest, with n 6. 98,000 1st position 136,700 2nd position 210,000 3rd position 289,900 4th position 315,500 5th position 2,456,500 6th position The median is the average of the values in positions which are 210,000 and 289,900, so n 2 3 and n 2 1 4, median 210,000 289,900 2 499,900 2 249,950 A price of $249,950 is much more representative of the houses in this listing. The most expensive house does not have the same strong effect that it had in the calculation of the mean. ■ Mode The mode is the data value with the highest frequency. It is most often used for qualitative data, for which the mean and median are undefined. The mode can be thought of as the “most typical” value in the data set. NOTE If every value in a data set occurs the same number of times, there is no mode. If two or more scores have equal frequencies that are higher than those of all other values, the data set is called bimodal (two modes), trimodal (three modes), or multimodal. 856 Chapter 13 Statistics and Probability Example 4 Mode of a Data Set Find the mode of the data represented by the bar graph below 14 12 10 8 6 4 2 0 Purple Orange Red Favorite color Green Blue Solution The height of each bar represents the frequency, so the mode is the category with the tallest bar, which is red. ■ Mean, Median, and Mode of a Distribution Recall the shapes of symmetric, skewed left, and skewed right distributions. y median mode y y mode mode median median x x x mean symmetric mean skewed left Figure 13.2-1 mean skewed right The mean is the balance point of a distribution. Notice that on a skewed distribution, the mean moves toward the tail to balance out the “weight” of the outlying data. The median divides the area under the distribution into 2 equal areas. The mode is the highest point on the distribution. • If a distribution is symmetric, then the mean and median are equal. • If a distribution is skewed left, then the mean is to the left of the median. • If a distribution is skewed right, then the mean is to the right of the median. Technology Tip If necessary, see the Technology Appendix to learn how to find the mean and median of a data set. Section 13.2 Measures of Center and Spread 857 Calculator Exploration Use the statistics functions of your calculator to find the mean and median of the data represented by the following stem plot. 6\|7 represents a score of 67 points. 43 scores are represented. 3 4 5 6 7 8 9 10 Key Measures of Spread Finding the shape and center of a data set still gives an incomplete picture of the data. The following stem plots show three data sets with a symmetric distribution and center 105. NOTE If the mean and median of a distribution are the same value, as in the data sets represented by the stem plots at right, their value is often referred to as the center. 6 7 8 9 10 11 12 13 14 10 11 12 13 14 10 11 12 13 14 1 5 1 3 5 7 9 5 9 The data has a different spread in each stem plot. The spread of the data, or variability, is an important characteristic of a data set. The second plot has the most variability because the data is very spread out, while the third has the least because the date is clustered very near the center. The three most common measures of spread are the standard deviation, the range, and the interquartile range. Standard Deviation The standard deviation of a data set is the most common measure of variability. It is best used if the data is symmetric about a mean. Standard deviation measures the average distance of a data element from the mean. x. is the difference, from the mean x xi The deviation of a data value xi Consider the following data set: The mean of the data is 2, 5, 7, 8, 10 2 5 7 8 10 5 6.4. The points are shown with their deviations on a number line in Figure 13.2-2. 858 Chapter 13 Statistics and Probability NOTE The distance from xi x data value to the mean is the absolute value of the deviation. deviation = 0.6 deviation = −1.4 deviation = 1.6 deviation = −4.4 deviation = 3.6 0 1 2 3 4 5 6 6.4 7 8 9 10 Figure 13.2-2 The average of the deviations is 4.4 1.4 1 2 0.6 1.6 3.6 5 0, because the positive and negative values cancel each other out. To avoid this, each deviation is squared, then the average is found. This quantity is called the variance. The square root of the variance is the standard deviation, denoted by the Greek letter s (sigma). For the data set {2, 5, 7, 8, 10}, first square each deviation. 4.4 2, 2 1 1.4 2 51 2, 0.62, 1.62, 3.62 5 6 19.36, 1.96, 0.36, 2.56, 12.96 6 Average the squared deviations to find the variance, s2 19.36 1.96 0.36 2.56 12.96 5 s2 : 7.44 Take
the square root of the variance to find the standard deviation: s 27.44 2.73 Population versus Sample If data is taken from a sample instead of the instead of n when averentire population, it is common to divide by aging the squared deviations. The result is called the sample standard deviation and is denoted by s. For large data sets, the sample standard deviation is very close to the population standard deviation. n 1 Standard Deviation To find the standard deviation of a data set with n values, 1. Subtract each value from the mean to find the deviation. 2. Square each deviation, and find the mean of the squared n 1 deviations. If the data is from a sample, divide by instead of n. The result is called the variance. 3. Take the square root of the variance. These steps are summarized in the following formulas: Population standard deviation s B π(xi x)2 n Sample standard deviation π(xi x )2 s B n 1 Section 13.2 Measures of Center and Spread 859 Example 5 Standard Deviation Find the population standard deviation of the data in the first stem plot on page 857 using the formula. Then use a calculator to find the population standard deviations of the data in the other two plots. Solution For the first stem plot, squared deviations. x 105 and n 9. The following table shows the xi xi 85 91 x 20 14 xi 1 x 2 2 400 196 99 6 36 101 4 16 105 109 111 119 125 0 0 4 16 6 36 14 20 196 400 s 400 196 36 16 0 16 36 196 400 9 B 1296 9 B 12 An informal interpretation is that the average distance from the data values to the mean is 12 units. The population standard deviations of the data in the second and third stem plots are shown in Figure 13.2-3. Figure 13.2-3 Notice that the second stem plot, which is the most spread out, has the largest standard deviation, and the third stem plot, which is the most clustered together, has the smallest standard deviation. ■ Range The range is the difference between the maximum and minimum data values. The main advantage of the range is that it is easy to compute. Example 6 The Range Find the range of the data in each stem plot on page 857. 860 Chapter 13 Statistics and Probability Solution The range of the data in the first stem plot is The range of the data in the second stem plot is The range of the data in the third stem plot is 125 85 40. 145 65 80. 119 91 28. ■ Interquartile Range Like the mean, the standard deviation and range are strongly affected by extreme values in the data. The interquartile range is a measure of variability that is resistant to extreme values, yet gives a good indication of the spread of the data. Recall that the median is the middle value of the data set. Thus, the median divides the data into two halves, the lower half and the upper half. The quartiles further divide the data into fourths. The quartile, is the is the median of the upper median of the lower half. The half. (The median may be considered to be the quartile.) quartile, Q1, Q3, 2nd 3rd 1st Figure 13.2-4 shows the quartiles for n even or n odd. lower half upper half n even: 3, 5, 7, 9, 11, 12, 13, 16, 17, 19, 23, 27, 29, 31 Q1 lower half Q3 upper half n odd: 2, 3, 6, 8, 9, 14, 15, 16, 20, 21, 23, 26, 28, 30, 33 Q1 Q3 Figure 13.2-4 The interquartile range is the difference between the quartiles, IQR Q3 Q1 which represents the spread of the middle 50% of the data. A value that is less than considered an outlier, as shown in Figure 13.2-5. IQR Q1 1 2 or greater than 1.5 Q3 1.5 IQR is 2 1 1.5 IQR IQR median 1.5 IQR outlier Q1 Q3 Figure 13.2-5 outlier Section 13.2 Measures of Center and Spread 861 Example 7 Interquartile Range Find the interquartile range of the data in the first stem plot on page 857. Solution The quartiles of the data in the first stem plot are shown below. 85 91 99 101 105 109 111 119 125 = Q1 91 + 99 2 = 95 = Q3 111 + 119 2 = 115 The interquartile range is 115 95 20. ■ Calculator Exploration Use a graphing calculator to find the interquartile range of the data in each of the other two stem plots on page 857. Five-Number Summary and Box Plots The five-number summary of a data set is the following list: minimum, Q1, median, Q3, maximum These values are used to construct a display called a box plot, as follows: 1. Construct a number line and locate each value of the five-number sum- mary. minimum Q1 median Q3 maximum 2. Construct a rectangle whose length equals the interquartile range, with a vertical line to indicate the median. minimum Q1 median Q3 maximum 862 Chapter 13 Statistics and Probability Technology Tip See the technology appendix, if necessary, for instruction on constructing a box plot using a graphing calculator. 3. Construct horizontal whiskers to the minimum and maximum values. minimum Q1 median Q3 maximum If a data element is an outlier, it may be marked with a or other mark. The whiskers then extend to the farthest value on each side that is not an outlier. Example 8 Constructing a Box Plot Construct a box plot for the data in the first stem plot on page 857. Solution The five-number summary of the data is 85, 95, 105, 115, 125. The box plot is shown below. 80 85 90 95 100 105 110 115 120 125 ■ Calculator Exploration Use a graphing calculator to construct a box plot for the data in each of the other two stem plots on page 857. Exercises 13.2 In Exercises 1–4, find the mean of each data set. 6. Find the median of the data set in Exercise 2. 1. 23, 25, 38, 42, 54, 57, 65 7. Find the median of the data set in Exercise 3. 2. 3, 5, 6, 2, 10, 9, 7, 5, 11, 6, 4, 2, 5, 4 8. Find the median of the data set in Exercise 4. 3. 3.6, 7.2, 5.9, 2.8, 21.6, 4.4 4. 78, 93, 87, 82, 90 5. Find the median of the data set in Exercise 1. 9. Find the mean, median, and mode of the following data set: 13, 13, 12, 6, 14, 9, 11, 19, 13, 9, 7, 16, 11, 12, 15, 12, 11, 12, 14, 9, 11, 13, 17, 13, 13 5910ac13_842-903 9/21/05 2:32 PM Page 863 Section 13.2 Measures of Center and Spread 863 In Exercises 10–13, find the mode of the data set represented by each display. 10. 30 25 20 15 10 5 0 A 11 12. C y D 8 6 4 2 24. 6, 8, 4, 11, 8, 8, 9, 6, 6, 8, 8, 12, 10, 10, 7 25. 50, 72, 86, 92, 86, 77, 57, 80, 93, 74, 53, 69, 65, 57, 73, 60, 66, 94, 81, 81 26. Find the range of the data set in Exercise 22. 27. Find the range of the data set in Exercise 23. 28. Find the range of the data set in Exercise 24. 29. Find the range of the data set in Exercise 25. 30. Find the interquartile range of the data set in Exercise 22. 31. Find the interquartile range of the data set in Exercise 23. 32. Find the interquartile range of the data set in Exercise 24. 0 1 2 3 4 5 6 x 33. Find the interquartile range of the data set in Exercise 25. 13. Vegetables Fruit Grains Dairy Meat Fat For each distribution shape, indicate whether the mean is larger, the median is larger, or the mean and median are equal. 14. symmetric 15. skewed left 16. skewed right 17. uniform Find the population standard deviation of the following data sets without using a calculator. 18. 8, 9, 10, 11, 12 19. 6, 8, 10, 12, 14 34. Find the five-number summary of the data set in Exercise 22, and create a box plot for the data. 35. Find the five-number summary of the data set in Exercise 23, and create a box plot for the data. 36. Find the five-number summary of the data set in Exercise 24, and create a box plot for the data. 37. Find the five-number summary of the data set in Exercise 25, and create a box plot for the data. For Exercises 38–43, the wait times of 30 people in a doctor’s office are given below, rounded to the nearest five minutes: 40, 35, 65, 40, 40, 5, 50, 85, 30, 50, 60, 60, 10, 65, 15, 45, 20, 40, 45, 70, 70, 25, 40, 45, 70, 65, 45, 25, 15, 25 38. Construct a histogram of the data. Describe the shape of the data set. Based on the shape, discuss the relative positions of the mean and median. 39. Find the mean and median of the data set. 40. Which measure of central tendency is preferred 20. 10, 10, 10, 10, 10 21. 0, 5, 10, 15, 20 for this data set? Why? Use a calculator to find the population and sample standard deviations of the following data sets. 22. 3, 6, 3, 5, 7, 8, 2, 6, 3, 6, 8, 4, 8, 2, 6, 9 23. 24, 17, 18, 18, 19, 26, 19, 8, 25, 15, 17, 11, 27, 20 41. Find the sample standard deviation, range, and interquartile range of the data set. 42. Construct a box-plot of the data. 43. Explain why the sample standard deviation is or is not a good measure of dispersion for this data set. 864 Chapter 13 Statistics and Probability 44. During a baseball game, 9 players had 1 hit each, 3 players had 2 hits each, and 6 players had no hits. Find the mean number of hits per player. Find the sample standard deviation and the population standard deviation of the data, and interpret your results. 45. A teacher has two sections of the same course. The average on an exam was 94 for one class with 20 students, while the average was 88 for the other class with 30 students. Find the combined average exam score. 46. The mean score of a class exam was 78, and the median score was 82. Sketch a possible distribution of the scores. 47. Over the last year, 350 lawsuits for punitive damages were settled with a mean settlement of $750,000 and a median settlement of $60,000. Sketch a possible distribution of the settlements. 48. A restaurant employs six chefs with the salaries in dollars shown below: 25,000, 27,000, 35,000, 105,000, 40,000, 45,000 Determine the mean and median salaries. 49. Which measure of center more accurately describes the “typical” salary at the restaurant in Exercise 48? 50. The speed of a computer is primarily determined by a chip in the CPU. A manufacturer tested 12 chips and reported the following speeds in megahertz units: 11.6, 11.9, 12.0, 12.0, 14.0, 15.2, 13.0, 14.3, 13.6, 13.8, 12.8, 12.9 51. Create two data sets of five numbers each that have the same mean but different standard deviations. 52. Create two data sets of five numbers each that have the same standard deviations but different means. 53. Critical Thinking How is the mean of a data s
et affected if a constant k is added to each value? 54. Critical Thinking How is the standard deviation of a data set affected if a constant k is added to each value? 55. Critical Thinking How is the mean of a data set affected if each value is multiplied by a constant k? 56. Critical Thinking How is the standard deviation of a data set affected if each value is multiplied by a constant k? 57. Critical Thinking What must be true about a data set in order for the standard deviation to equal 0? 13.3 Basic Probability Objectives Definitions • Define probability and use properties of probability • Find the expected value of a random variable • Use probability density functions to estimate probabilities In the study of probability, an experiment is any process that generates one or more observable outcomes. The set of all possible outcomes is called the sample space of the experiment. Some examples of experiments and their sample spaces are shown in the following table. Section 13.3 Basic Probability 865 3 Figure 13.3-1 Experiment tossing a coin Sample space heads and tails, written as {H, T} rolling a number cube (Figure 13.3-1) {1, 2, 3, 4, 5, 6} choosing a name from the phone book all the names in the phone book counting the number of fish in a lake the set of non-negative integers An event is any outcome or set of outcomes in the sample space. For example, in the experiment of rolling a number cube, the set {1, 3, 5} is an event, which can be described as “rolling a 1, 3, or 5,” or simply “rolling an odd number.” The probability of an event is a number from 0 to 1 (or 0% to 100%) inclusive that indicates how likely the event is to occur. • A probability of 0 (or 0%) indicates that the event cannot occur. • A probability of 1 (or 100%) indicates that the event must occur. never happens 0 as likely as not 1 8 0.125 1 4 0.25 3 8 0.375 1 2 0.5 5 8 0.625 3 4 0.75 7 8 0.875 always happens 1 • The sum of the probabilities of all outcomes in the sample space is 1. • The probability of an event is the sum of the probabilities of the outcomes in the event. Probability Distributions The probability of an event E can be described by a function P, where the domain of the function is the sample space and the range of the function is the closed interval [0, 1]. denotes the probability of the outcome X, and X 2 denotes the probability of the event E. P P E 1 1 2 The rule of the function P can be described by a table, called a probability distribution. Example 1 Probability Distribution Suppose that 100 marbles are placed in a bag; 50 red, 30 blue, 10 yellow, and 10 green. An experiment consists of drawing one marble out of the bag and observing its color. a. What is the sample space of the experiment? 866 Chapter 13 Statistics and Probability NOTE Probabilities expressed as percents are often called chances. According to this probability distribution, there is a 50% chance of drawing a red marble, a 30% chance of drawing a blue marble, a 10% chance of drawing a yellow marble, and a 10% chance of drawing a green marble. b. Write out a reasonable probability distribution for this experiment, and verify that the sum of the probabilities of the outcomes is 1. c. What is the probability that a blue or green marble will be drawn? Solution a. The sample space is all possible outcomes: red, blue, yellow, green 6 5 b. A reasonable probability distribution is shown below, which is based on the relative frequency of marbles of each color. Color of marble Red Blue Yellow Green Probability 50 100 0.5 30 100 0.3 10 100 0.1 10 100 0.1 The sum of the probabilities of the outcomes is 0.5 0.3 0.1 0.1 1 c. The event “a blue or green marble will be drawn” can be written as the set of outcomes {blue, green}. The probability of the event is the sum of the probabilities for blue and green. blue, green P 15 P 1 62 blue P 1 2 green 2 0.3 0.1 0.4 ■ Mutually Exclusive Events Two events are mutually exclusive if they have no outcomes in common. Two mutually exclusive events cannot both occur in the same trial of an experiment. If two events E and F are mutually exclusive, then the probability of the event (E or F) is the sum of the individual probabilities The complement of an event is the set of all outcomes that are not contained in the event. The complement of event E can be thought of as “the event that E does not occur.” An event and its complement are always mutually exclusive, and together they contain all the outcomes in the sample space. Thus, the probability of an event and the probability of its complement must add to 1, which leads to the following fact. Probability of a Complement If an event E has probability p, then the complement of the event has probability 1 p. Section 13.3 Basic Probability 867 Example 2 Mutually Exclusive Events An experiment consists of spinning the spinner in Figure 13.3-2. The following table shows the probability distribution for the experiment. Outcome Probability A 0.4 S 0.3 C 0.2 E 0.1 a. Which of the following pairs of events E and F are mutually exclusive? E E A, C, E 6 5 a vowel 1 2 F F E 1 a vowel 2 F 5 6 6 C, S 5 in the first five letters 1 of the alphabet C 2 b. What is the complement of the event {A, S}? c. What is the probability of the event “the spinner does not land on A?” Solution A E C S Figure 13.3-2 a. The events E {A, C, E} and F {C, S} are not mutually exclusive because they have a common outcome, C. F E (a vowel) and The events alphabet) are not mutually exclusive because they have two common outcomes, A and E. E {C} are mutually exclusive (in the first five letters of the (a vowel) and The events because they have no common outcome. F b. The complement of the event {A, S} is the set of outcomes that are not in the event, {C, E}. c. The complement of the event “the spinner does not land on A” is {A}, which has a probability of 0.4. Thus, the probability of the event is 1 0.4 0.6. ■ Independent Events Two events are independent if the occurrence or non-occurrence of one event has no effect on the probability of the other event. For example, if an experiment is repeated several times under exactly the same conditions, the outcomes of the individual trials are independent. If two events E and F are independent, then the probability of the event is the product of the individual probabilities, P and 868 Chapter 13 Statistics and Probability NOTE The terms mutually exclusive and independent are often confused. Some important differences are detailed below. Mutually exclusive Independent The term often refers to two possible results for a single trial of a given experiment. The term often refers to the results from two or more trials of an experiment or from different experiments. The word “or” is often used to describe a pair of mutually exclusive events. The word “and” is often used to describe a pair of independent events. For mutually exclusive events E and F, E or For independent events E and F, E and If two events are mutually exclusive, they cannot be independent, because the occurrence of one would cause the other to have a probability of 0. Example 3 Independent Events The probability of winning a certain game is 0.1. Suppose the game is played on two different occasions. What is the probability of a. winning both times? b. losing both times? c. winning once and losing once? Solution a. The results of the two different trials are independent, so the probability of winning both times can be found by multiplying the probability of winning each time. winning both games P 1 2 0.1 0.1 0.01 b. Since losing is the complement of winning, the probability of losing 1 0.1 0.9. is by multiplying the probability of losing each time. The probability of losing both times can be found losing both games P 1 2 0.9 0.9 0.81 c. The complement of the event (winning once and losing once) is the set of the two events in parts a and b. The events in parts a and b are mutually exclusive, because it is impossible to win both times and lose both times, so their probabilities may be added. winning once and losing once P 1 1 1 2 0.01 0.81 0.18 2 ■ Section 13.3 Basic Probability 869 Random Variables In many cases, the characteristics of an experiment that are being studied are numerical, such as the total on a roll of two number cubes. In other cases, the outcomes of an experiment may be assigned numbers, such as heads 1, tails 0. A random variable is a function that assigns a number to each outcome in the sample space of an experiment. Example 4 Random Variable An experiment consists of rolling two number cubes. A random variable assigns to each outcome the total of the faces shown. a. Write out the sample space for the experiment. b. Find the range of the random variable. c. List the outcomes to which the value 7 is assigned. Solution a. The sample space may be written as a set of ordered pairs. (1, 1) (2, 1) (3, 1) (4, 1) (5, 1) (6, 1) (1, 2) (2, 2) (3, 2) (4, 2) (5, 2) (6, 2) (1, 3) (2, 3) (3, 3) (4, 3) (5, 3) (6, 3) (1, 4) (2, 4) (3, 4) (4, 4) (5, 4) (6, 4) (1, 5) (2, 5) (3, 5) (4, 5) (5, 5) (6, 5) (1, 6) (2, 6) (3, 6) (4, 6) (5, 6) (6, 6) b. The smallest possible value is 2, which is assigned to the outcome (1, 1). The largest possible value is 12, which is assigned to the outcome (6, 6). The range is the set of integers from 2 to 12. c. The value 7 is assigned to the outcomes (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), and (6, 1). ■ Expected Value of a Random Variable The expected value, or mean, of a random variable is the average value of the outcomes. In the experiment of rolling two number cubes, suppose the experiment was repeated 10 times, resulting in the following values for the random variable: 8, 5, 8, 6, 11, 11, 3, 9, 9, 7 The average value is 8 5 8 6 11 11 3 9 9 7 10 7.7. If the experiment is repeated a large number of times, the average approaches the expected value. A simulation was used to run a large number of trials, and the results are shown in Figure 13.3-3. The averages seem to be approaching 7, w
hich is a reasonable estimate of the expected value. Average sum of two number cubes 7.3 6.98 7.26 7.005 6.978 Number of trials 100 200 300 400 500 Figure 13.3-3 870 Chapter 13 Statistics and Probability To calculate the expected value of a random variable from a probability distribution, multiply each value by its probability, and add the results. Example 5 Expected Value A probability distribution for the random variable in the experiment in Example 4 is given below. Find the expected value of the random variable. Sum of faces Probability 2 1 36 3 1 18 4 1 12 5 1 9 6 5 36 7 1 6 8 5 36 9 1 9 10 1 12 11 1 18 12 1 36 Solution Multiply each value by its probability, and add. 1 36b 2 a 3 1 18b a 4 1 12b a 5 1 9b a 6 5 36b a 7 1 6b a 8 5 36b a 9 1 9b a 10 1 12b a 11 1 18b a 12 1 36b a 7 ■ The expected value is not always in the range of the random variable, as shown in the following example. Example 6 Expected Value of a Lottery Ticket The probability distribution for a $1 instant-win lottery ticket is given below. Find the expected value and interpret the result. Solution Win $0 $3 $5 Probability 0.882746 0.06 0.04 $10 0.01 $20 $40 $100 $400 $2500 0.005 0.002 0.0002 0.00005 0.000004 0.882746 0 1 100 0.0002 2 3 0.06 1 400 1 2 0.04 5 2 0.00005 1 1 2 10 1 2 2500 1 20 0.01 2 0.000004 0.005 2 1 0.71 2 40 0.002 2 1 The average amount won is $0.71, though it is not possible to win exactly 71 cents on one ticket. However, since the ticket costs $1, there is an average net loss of $1 $0.71 $0.29 per play. ■ Section 13.3 Basic Probability 871 Probability Density Functions In Example 1, colored marbles were drawn from a bag and the probability of each color being drawn was determined by its relative frequency. red: 0.5 blue: 0.3 yellow: 0.1 green: 0.1 This probability distribution is displayed in a bar graph in Figure 13.3-4, in which each bar is 1 unit wide. Thus the area of each rectangular bar represents the probability of the corresponding color. The sum of the areas of the bars is 1. If rectangles in the bar graph have width 1 unit, then the area of each rectangle represents the probability of the corresponding category ll o Y n e r e G A function with the property that the area under the graph corresponds to a probability distribution is called a probability density function. Figure 13.3-4 Example 7 Discrete Probability Density Functions Draw a probability density function for the distribution in Example 5. Solution The probability density function is a piecewise-defined function, shown in Figure 13.3-5, where the height of each piece is the probability of the value on the left endpoint of the interval. The area of the shaded rectangle represents the probability that the sum is 9. y 6 36 4 36 2 36 x 0 1 2 43 5 6 7 8 9 10 11 12 13 Figure 13.3-5 ■ y When a random variable has infinitely many values within a certain interval, the probability distribution can be represented by a continuous density function, as in the following example. Example 8 Continuous Probability Density Function The probability density function in Figure 13.3-6 can be used to estimate the probability that a customer calling a company’s customer service line will have to wait for a given amount of time. The area of each square on the grid is and the total area under the curve is 1. Estimate the probability that a customer will have to wait between 2 and 3 minutes. 0.5 0.05 0.025, x 0 1 2 43 5 6 7 (minutes) Figure 13.3-6 0.6 0.5 0.4 0.3 0.2 0..9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 872 Chapter 13 Statistics and Probability Solution is the area under The probability the curve between 2 and 3, which is shaded 1 2 in Figure 13.3-7. The area is approximately 3 squares on the grid, or 3 1 2 1 0.025 2 0.0875 Thus, the probability that a customer will have to wait between 2 and 3 minutes is about 0.0875. y 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 1 2 43 5 6 7 Figure 13.3-7 x ■ Exercises 13.3 Use the following probability distribution for Exercises 1–4. Suppose the experiment is repeated three times. Assume the trials are independent. Outcome Probability A 0.5 B 0.3 C ? D ? 1. List the sample space for the probability distribution. 2. Suppose that outcomes C and D have the same probability. Complete the probability distribution. 3. What is the probability of the outcome (A or B)? 4. What is the probability of the outcome (not A)? Exercises 5–8 refer to the spinner at the right. The probability of landing on black is 1 2 , and the proba- bility of landing on red is 1 3 . 5. Create a probability distribution for the experiment of spinning the spinner. 6. What is the probability it will land on black all three times? 7. What is the probability it will land on white all three times? 8. What is the probability of the outcome (black, white, red)? of the outcome (black, red, white)? What is the probability that it will land once on each color? A doctor has assigned the following chances to a medical procedure: full recovery condition improves no change condition worsens 55% 24% 17% 4% Suppose the procedure is performed on 5 patients. Assume that the procedure is independent for each patient. 9. What is the probability that all five patients will recover completely? 10. What is the probability that none of the patients will get worse? A bag contains red and blue marbles, such that the 21. What is the probability that at least one person is Section 13.3 Basic Probability 873 probability of drawing a blue marble is An exper- . 3 8 iment consists of drawing a marble, replacing it, and drawing another marble. The two draws are independent. A random variable assigns the number of blue marbles to each outcome. 11. What is the range of the random variable? absent? 22. Find the expected value of the random variable, and interpret the result. An experiment consists of planting four seeds. A random variable assigns the number of seeds that sprout to each outcome. 12. What is the probability that the random variable has an output of 2? 23. Complete the following probability distribution for the experiment. 13. What is the probability that the random variable has an outcome of 0? 14. What is the probability that the random variable has an outcome of 3? Sprouted Probability 0 ? 1 2 3 4 0.154 0.345 0.345 0.130 15. Create a probability distribution for the random 24. Find the expected value of the random variable, variable. and interpret the result. 16. Calculate the expected value of the random 25. Draw a probability density function for the variable. In Exercises 17–20, find the expected value of the random variable with the given probability distribution. 17. 18. 19. 20. Outcome 0 1 5 10 1000 Probability 0.43 0.32 0.24 0.10 0.01 Outcome 0 1 2 3 Probability 0.25 0.25 0.25 0.25 Outcome 15 16 17 18 19 20 Probability 0.1 0.3 0.2 0.2 0.1 0.1 Outcome Probability 2 1 4 3 1 2 4 1 4 An office employs 5 people. A random variable is assigned to the number of people absent on a given day. The probability distribution is given below. Absent 0 1 2 3 Probability 0.59 0.33 0.07 0.01 4 0 5 0 random variable. Shade an area of the graph that corresponds to the probability that 3 or more seeds will sprout. 26. Use the probability distribution to determine the probability that each seed will sprout, assuming that they are independent. Hint: if the probability that one seed will sprout is p, what is the probability that all four seeds will sprout? A random variable with a uniform distribution has a probability density function that is constant over the range of the variable, and 0 everywhere else. Use the graph of the probability density function shown below for Exercises 27–30 27. What is the range of the random variable? 28. What is the height h of the probability density function? 29. What is the probability that the random variable is between 2 and 4? 30. What is the probability that the random variable is greater than 4? 874 Chapter 13 Statistics and Probability y 0.5 0.4 0.3 0.2 0. 10 11 12 13 14 15 16 17 18 19 20 (inchesminutes) 9 10 The probability density function at left above models the number of inches of rainfall per year for a certain location. The area of each square on the grid is 0.01. 31. Estimate the probability that the rainfall for a certain year is between 14 and 16 inches. 32. Estimate the probability that the rainfall for a certain year is greater than 17 inches. 33. The median of a probability density function is the point that divides the area under the curve into two equal areas. Estimate the median rainfall, based on the given probability density function. In commuting to work, Joshua takes one bus, then transfers to another bus. The probability density function at right above models the total length of time that he has to wait for both buses. 34. What is the height of the probability density function at t 5 minutes? 35. What is the probability that Joshua has to wait for less than 5 minutes? 36. What is the probability that Joshua has to wait for between 3 and 6 minutes? 13.4 Determining Probabilities Objectives • Estimate probability using experimental methods • Estimate probability using theoretical methods The exact probability of a real event can never be known. Probabilities are estimated in two ways: experimentally and theoretically. Experimental Estimates of Probability Suppose an outcome of an experiment has a probability of 0.3. If the experiment were repeated many times, that outcome would occur in approximately 30% of the trials. In 100 trials, for example, it might occur 30 times, or maybe 28 times or 34 times. Statistical analysis shows, however, that it is unlikely that it would occur fewer than 20 times or more than 40 times. Figure 13.4-1 Section 13.4 Determining Probabilities 875 The basis for experimental estimates of probability may be summarized as follows: As the number of trials of an experiment increases, the relative frequency of an outcome approaches the probability of the outcome. Thus, if an experiment is repeated n times, the experimental estimate o
f the probability of an event is P E 2 1 number of trials with an outcome in E n Example 1 Experimental Estimate of Probability An experiment consists of throwing a dart at the target in Figure 13.4-1. Suppose the experiment is repeated 200 times, with the following results: red yellow blue 43 86 71 Write a probability distribution for the experiment. Solution The probabilities may be estimated using the experimental formula. Outcome Probability red 43 200 0.2 yellow 86 200 0.4 blue 71 200 0.4 ■ Probability Simulations In order to estimate probability using the experimental approach, a large number of trials is needed. Because this approach is often time-consuming, computer simulations that duplicate the conditions of a single trial are often used. Most graphing calculators have random number generators that can be used to simulate simple probability experiments. To simulate an experiment with a large number of trials, it is easiest to use a program that can keep track of the frequency of each outcome. Suppose an experiment consists of tossing three coins and counting the number of heads, and that the probability of heads for each coin is 0.5. The possible outcomes of the experiment are 0 heads, 1 head, 2 heads, or 3 heads. One trial can be simulated by a command that randomly generates three values, which can be either 0 or 1, and adds them (see the Technology Tip on page 876). The three random integers represent the three coins, where 1 represents heads and 0 represents tails. 876 Chapter 13 Statistics and Probability Technology Tip To randomly generate three values that can be 0 or 1 and add them: TI Casio sum (randInt (0, 1, 3)) Sum {Int 2Ran#, Int 2Ran#, Int 2Ran#} The following is a sample program for running a simulation with n trials of the experiment. The program displays a list of the probabilities for 0 heads, 1 head, 2 heads, and 3 heads. 1 22 S T 0, 1, 3 randInt Prompt N 0 S A: 0 S B: 0 S C: 0 S D For (K, 1, N, 1) sum 1 If T 0 A 1 S A If T 1 B 1 S B If T 2 C 1 S C If T 3 D 1 S D End N is the number of trials. T is the outcome of a single trial. A is the number of trials with 0 heads. B is the number of trials with 1 head. C is the number of trials with 2 heads. D is the number of trials with 3 heads. {A/N, B/N, C/N, D/N} Example 2 Probability Simulation Use the program to create a probability distribution for the experiment of tossing 3 coins and counting the number of heads. Assume that P(heads) P(tails) 0.5 for each coin. Solution The results will vary each time the program is run. Using the results from Figure 13.4-2, one approximate distribution is shown below. Outcome 0 heads 1 head 2 heads 3 heads Probability 0.14 0.38 0.36 0.12 Figure 13.4-2 ■ NOTE The assumption that all outcomes are equally likely was used in the probability simulation in Example 2. If P(heads) 1 2 , then heads P(tails) and tails are equally likely. Also, for the given calculator commands, the outcomes 0 and 1 are equally likely each time. Probability for Equally Likely Outcomes Section 13.4 Determining Probabilities 877 Calculator Exploration Run the probability simulation in Example 2, using 100 trials. Are your results similar to those in the example? Compare your results to those of your classmates. Theoretical Estimates of Probability In the theoretical approach, certain assumptions are made about the outcomes of the experiment. Then, the properties of probability are used to determine the probability of each outcome. The most common assumption is that all outcomes are equally likely, that is, they have the same probability of occurring. For example, in tossing a coin it is usually assumed that heads and tails are equally likely. Since the probabilities must add to 1, the probability of each outcome must equal 1 2 . This idea is used to develop the following formula. Suppose an experiment has a sample space of n outcomes, all of which are equally likely. Then the probability of each outcome is 1 n, and the probability of an event E is given by P(E) number of outcomes in E n Example 3 Rolling a Number Cube An experiment consists of rolling a number cube. Suppose that all outcomes are equally likely. a. Write the probability distribution for the experiment. b. Find the probability of the event that an even number is rolled. Solution a. The sample space consists of the 6 outcomes {1, 2, 3, 4, 5, 6}. If the outcomes are equally likely, then the probability of each is probability distribution for the experiment is 1 6 . The Outcome Probability 878 Chapter 13 Statistics and Probability b. The event that an even number is rolled consists of the 3 outcomes {2, 4, 6}. Thus, the probability that an even number is rolled is 3 6 1 2 . ■ Of course, the outcomes of an experiment are not always equally likely. Based on the probability simulation of tossing 3 coins in Example 2, the outcomes 0, 1, 2, and 3 heads do not seem to be equally likely. However, it is possible to determine the probability theoretically by considering each coin separately. Example 4 Theoretical Probability Use properties of probability to write a theoretical probability distribution for the experiment in Example 2. Solution Each coin has no effect on the other coins, so the outcomes of the coins are independent. Thus, the probabilities can be multiplied. The only outcome with 0 heads is TTT. The probability of tails for each coin is 0.5, so the probability of the outcome TTT is T T 0.5 0.5 1 T 0.5 2 21 0.125 There are three outcomes with 1 head: HTT, THT, and TTH. Each outcome has a probability of 0.125. For example, H T 0.5 0.5 1 T 0.5 2 21 0.125 The probabilities of the three outcomes can be added, so the probability of 1 head is 0.125 0.125 0.125 0.375. There are three outcomes with 2 heads: HHT, HTH, and THH. The probability of 2 heads is also 0.125 0.125 0.125 0.375. There is one outcome with 3 heads, HHH, which also has a probability of 0.125. The probability distribution is shown below. Outcome 0 heads 1 head 2 heads 3 heads Probability 0.125 0.375 0.375 0.125 ■ How do the theoretical probabilities obtained in Example 4 compare to the experimental ones you found in Example 2? Section 13.4 Determining Probabilities 879 Counting Techniques The probability formula for equally likely outcomes uses the size of the sample space. In simple experiments, this may be easily determined, but some experiments require more sophisticated counting techniques. The basis of most counting techniques is the Fundamental Counting Principle, which is also known as the Multiplication Principle. n1 Consider a set of k experiments. Suppose the first experiment outcomes, and so on. Then outcomes, the second has has n2 the total number of outcomes is experiments. p nk for all k n2 n1 Example 5 Using the Fundamental Counting Principle A catalog offers chairs in a choice of 2 heights, regular and tall. There are 10 colors available for the finish, and 12 choices of fabric for the seats. The chair back has 4 different possible designs. How many different chairs can be ordered? Solution Each option can be considered as an experiment. The number of choices for each option is the number of outcomes. According to the Fundamental Counting Principle, the number of different chairs is 2 10 12 4 960 ■ Consider the following experiment: Each letter of the alphabet is written on a piece of paper, and three letters are chosen at random. There are two important questions in determining the nature of the experiment: 1. Is each letter replaced before the next letter is chosen? 2. Does the order of the letters matter in the result? If the answer to Question 1 is yes, the letters are said to be chosen with replacement. In this case, letters may be repeated in the result. Also, the number of letters to choose from is always the same. If the answer is no, then the letters are said to be chosen without replacement. In this case, there will be no repeated letters, and the number of letters to choose from decreases by 1 for each letter chosen. If the answer to Question 2 is yes, the result is said to be order important. In this case, the outcome CAT is considered to be different from the outcome ACT. If the answer is no, the result is said to be in any order. In this case, the six outcomes CAT, CTA, TAC, TCA, ACT, ATC are considered to be the same. Fundamental Counting Principle NOTE Many times, it is necessary to use the context of the experiment to determine whether it is with replacement, or if order is important. 880 Chapter 13 Statistics and Probability Figure 13.4-3 NOTE equal 1. 0! is defined to Three of the four possible cases are shown in the table below. The fourth case, with replacement and in any order, will not be discussed. Use the Fundamental Counting Principle to explain the number of outcomes in each case. With replacement Order important Without replacement Order important Without replacement Any order 26 26 26 17,576 26 25 24 15,600 26 25 24 3 2 1 2600 Example 6 3 Coin Toss Use the Fundamental Counting Principle to verify the probability distribution in Example 4. Solution There are two possible outcomes for each coin: heads and tails. Thus, the To find the 8 number of outcomes for tossing three coins is different outcomes, it is helpful to use a tree diagram, as shown in Figure 13.4-3. The outcomes are given below. 2 2 2 8. HHH HHT 2 heads 3 heads HTH 2 heads HTT 1 head THH 2 heads THT 1 head TTH 1 head TTT 0 heads Each of these outcomes is equally likely, so the probability distribution is Outcome 0 heads 1 head 2 heads 3 heads Probability 1 8 0.125 3 8 0.375 3 8 0.375 1 8 0.125 ■ Permutations and Combinations The two cases without replacement are called permutations (order important) and combinations (any order). In order to write a formula for permutations and combinations, n!—read “n factorial”—is used to describe the product of all the integers from 1 to n. n! n n 1 n 2 p 2 21 1 2 3 1 21 21 1 2 In the example of drawing 3 letters without replacement where order is important, t
he number of permutations can be written using factorials as 26 25 24 26 25 24 23 22 p 3 2 1 23 22 p 3 2 1 26! 23! In the case where order is not important, the number of combinations can be written as 26 25 24 3 2 1 26! 3 2 1 1 23! 2 26! 3! 23! Section 13.4 Determining Probabilities 881 Permutation and combination formulas are also written using factorials. Permutations and Combinations Technology Tip Permutations and combinations are located on the PRB submenu of the MATH menu on TI and on the PROB submenu of the OPTN menu on Casio. Permutations If r items are chosen in order without replacement from n possible items, the number of permutations is nPr n! (n r)! Combinations If r items are chosen in any order without replacement from n possible items, the number of combinations is nCr n! r!(n r)! If each item is equally likely to be chosen, the permutations and combinations are all equally likely for a given value of r. Note: may also be written as nPr Pn,r or P(n, r), and may be nCr written as Cn,r , C(n, r), or n r b a . Example 7 Matching Problem Suppose you have four personalized letters and four addressed envelopes. If the letters are randomly placed in the envelopes, what is the probability that all four letters will go to the correct addresses? Solution For all four letters to go to the correct addresses, they must be chosen in the exact same order as the envelopes. The size of the sample space is the number of permutations, 4P4. Thus, the probability is 1 24 4P4 4! 4 4 1 0.04. 4! 0! 24 ! 2 ■ Example 8 Pick-6 Lottery In a “pick-6” lottery, 54 numbered balls are used. Out of these, 6 are randomly chosen. To win, at least 3 balls must be matched in any order. What is the probability of winning the jackpot (all 6 balls)? What is the probability of matching any 5 balls? any 4 balls? any 3 balls? 882 Chapter 13 Statistics and Probability Solution The size of the sample space is the number of combinations. The probability of winning the jackpot is 54C6 25,827,165 1 25,827,165 0.00000004 The event of matching 5, 4, or 3 numbers can be determined as follows: matching k numbers P 1 number of combinations that match k numbers 54C6 2 6Ck ways to match k numbers out of 6. The remaining 6 k There are numbers do not match any of the 6 winning numbers, so there are 48 numways to choose the remaining numbers. bers to choose from, giving 6Ck combinaBy the Fundamental Counting Principle, there are tions that match k numbers. 48C6k 48C6k Figure 13.4-4 matching k numbers P 1 6Ck 48C6k 2 48C1 54C6 6 48 6C5 6C4 48C2 6C3 48C3 54C6 54C6 54C6 25,827,165 15 1128 25,827,165 20 17296 25,827,165 0.00001 0.0007 0.01 ■ matching 5 numbers matching 4 numbers matching 3 numbers P P P 1 1 1 2 2 2 Exercises 13.4 For Exercises 1–4, an experiment consists of drawing a marble out of a bag, observing the color, and then placing it back in the bag. Suppose the experiment is repeated 75 times, with the following results: 4. Suppose it is known that there is a total of 300 marbles in the bag. Estimate the number of each color of marble. red blue green yellow 38 23 11 3 1. Write a probability distribution of the experiment using the experimental formula. 2. Based on your distribution from Exercise 1, what is the probability of drawing either a blue or green marble? 3. Based on your distribution from Exercise 1, what is the probability of drawing two yellow marbles in a row? A dreidel is a top with four sides, used in a Hanukah game. The sides are labeled with the Hebrew letters nun, gimel, hay, and shin. A dreidel is spun 100 times, with the following results: nun gimel hay shin 10 45 24 21 dreidel 5. Write a probability distribution for the dreidel. 6. A player wins tokens if the dreidel lands on either gimel or hay. What is the probability of winning tokens? 7. A player loses tokens if the dreidel lands on nun. What is the probability of losing four times in a row? Exercises 8–11 refer to the following experiment: two number cubes are rolled, and a random variable assigns the sum of the faces to each outcome. 8. The following commands generate two random integers from 1 to 6 and add them. TI-83/86: sum(randInt(1, 6, 2)) TI-89/92: sum({rand(6), rand(6)}) Sharp 9600: Casio 9850: HP-38: sum Sum 1 5 INT 6 RANDOM 6 RANDOM 1 ©LIST INT 15 2 1 1 1 2 1 2 62 1, 6random int 22 Int 6Ran# 1, Int 6Ran# 1 2 1 Run 5 trials of the experiment and list your results. 9. Run a simulation of the experiment with at least 50 trials, and create a probability distribution. 10. Use your probability distribution from Exercise 9 to find the probabilities for the following values of the random variable. a. greater than 9 b. less than 6 c. at least 4 11. Use your probability distribution from Exercise 9 to find the expected value of the random variable. 12. A bag contains 3 red marbles and 4 blue marbles. Suppose each marble is equally likely to be chosen. What is the probability of the event of drawing a red marble? 13. Suppose that a person’s birthday is equally likely to be any day of the year. What is the probability that a randomly chosen person has the same birthday as you? A teacher writes the name of each of her 25 students on a slip of paper and places the papers in a box. To call on a student, she draws a slip of paper from the box. Each paper is equally likely to be drawn, and the papers are replaced in the box after each draw. 14. What is the probability of calling on a particular student? 15. What is the probability of calling on the same student twice in a row? Section 13.4 Determining Probabilities 883 16. If there are 9 students in the last row, what is the probability of calling on a student in the last row? 17. If the class contains 11 boys and 14 girls, what is the probability of calling on a girl? What is the probability of calling on 3 girls in a row? 18. A clothing store offers a shirt in 5 colors, in long or short sleeves, with a choice of three different collars. How many ways can the shirt be designed? 19. A quiz has 5 true-false questions and 3 multiple choice questions with 4 options each. How many possible ways are there to answer the 8 questions? 20. A license plate has 3 digits from 0 to 9, followed by 3 letters. How many different license plates are possible? 6 21. A gallery has 25 paintings in its permanent collection, with display space for 10 at one time. How many different collections can be shown? 22. A committee of 8 people randomly chooses 3 people in order to be president, vice president, and treasurer. In how many ways can the officers be chosen? 23. A baseball team has 9 players. How many different batting orders are there? 24. A small library contains 700 novels. In how many ways can you check out 3 novels? 25. A manufacturer is testing 4 brands of soda in a blind taste test. The participants know the brands being tested but do not know which is which. What is the probability that a participant will identify all 4 brands correctly by guessing? 26. A researcher is studying the abilities of people who claim to be able to read minds. He chooses 6 numbers between 1 and 50 (inclusive), and asks each participant to guess the numbers in order. What is the probability of guessing all 6 correctly? A botanist is testing two kinds of seeds. She divides a plot of land into 16 equal areas numbered from 1 to 16. She then randomly chooses 8 of these areas to plant seed A, and she plants seed B in the remaining areas 10 11 12 13 14 15 16 884 Chapter 13 Statistics and Probability 27. How many ways are there to choose 8 plots out of 16? 28. What is the probability of the event shown at right 29. What is the probability that all of the areas with 4 2 rectangle on one side of seed A will form a the plot, and all of the areas with seed B will form a rectangle on the other side? 4 2 Critical Thinking Exercises 30–35 refer to a famous probability problem: suppose a certain number of people are in a room. What is the probability that two or more people in the room will have the same birthday? “everybody has a different birthday.” How many ways are there to name a different date for each of 3 people? of 20 people? of n people? 32. The probability that everybody has a different birthday can be written as Number of ways to name n different dates Number of ways to name n dates Use your results from Exercises 30 and 31 to write a formula in terms of n for the probability that everybody has a different birthday. 33. Use your results from Exercise 32 to write a formula in terms of n for the probability that two or more people have the same birthday. Find the n 35. for probability for n 20, and for n 3, 34. How many people must be in the room for the 30. Suppose each day of the year is equally likely to probability to be approximately 1 2 that two or be a person’s birthday. How many ways are there to name one date per person (not necessarily all different) for each of 3 people? of 20 people? of n people? 31. The complement of the event “two or more people will have the same birthday” is more have the same birthday? 35. How many people must be in the room for the probability to be 1 that two or more have the same birthday? Hint: do not use the formula. 13.4.A Excursion: Binomial Experiments Objectives • Calculate the probability of a binomial experiment Many experiments can be described in terms of just two outcomes, such as winning or losing, heads or tails, boy or girl. These experiments determine a group of problems called binomial or Bernoulli experiments, named after Jacob Bernoulli, a Swiss mathematician who studied these distributions extensively in the late 1600’s. Binomial Experiments Here is a typical binomial experiment: in a basketball contest, each contestant is allowed 3 free-throws. If a certain individual has a 70% chance of making each free-throw, what is the probability of making exactly 2 out of 3? Binomial Experiment NOTE The terms “success” and “failure” are often used in experiments to designate outcomes such as heads or tails, even if neithe
r outcome is preferred. Section 13.4.A Excursion: Binomial Experiments 885 The essential elements in a binomial experiment are given below. A set of n trials is called a binomial experiment if the following are true. 1. The trials are independent. 2. Each trial has only 2 possible outcomes, which may be designated as success (S) and failure (F). 3. The probability of success p is the same for each trial. The probability of failure is q 1 p. In the example of the basketball contest, the outcome SFS indicates that the first free-throw is a success, the second a failure, and the third a success. The trials are independent, so the probability of the outcome SFS is the product of the probabilities for each trial. SFS .7 1 21 0.3 21 0.7 2 0.147 Example 1 Basketball Contest Refer to the basketball contest described on page 884. Suppose that the probability of making each free-throw is 0.7. What is the probability of making exactly 2 free-throws in 3 tries? Solution The outcomes in the event “2 free-throws in 3 tries” are SSF, SFS, FSS. The probability of the event is the sum of the probabilities of the three outcomes SSF P P SFS P P FSS 2 free-throws .7 0.7 0. SSF 2 S 21 2 1 F 21 2 1 S 21 2 1 P SFS 0.147 0.147 0.147 0.441 0.147 2 0.147 2 0.147 2 FSS 0.3 21 0.7 21 0.7 21 P 1 1 1 P 0.7 0.3 0.7 1 1 1 2 2 2 ■ In Example 1, note that the probability of SSF is the same as the probability of SFS or FSS. In general, the probability of any outcome with r successes and n–r failures in n trials is qq p q 2 n r times 21 1 ⎧⎨⎩ ⎧⎨⎩ r times prqnr pp p p 1 2 1 2 To develop a general formula for the probability of r successes in n trials, it is necessary to determine how many different outcomes have r suc- 886 Chapter 13 Statistics and Probability cesses. Consider the number of outcomes with 3 successes in 5 trials, as shown below. For clarity, the F’s are left as blanks. SSS SS S S SS SSS SS SS S S S SS S S S S SS SSS The number of outcomes is the same as the number of ways to choose 3 positions for the S out of 5 possible positions. The order of the S’s does not matter, because they are all the same. This is the number of combinations, 5C3 10. Probability of a Binomial Experiment In a binomial experiment, P(r successes in n trials) nCr prqnr where p is the probability of success, and probability of failure. q 1 p is the Example 2 Lottery Tickets A lottery consists of choosing a number from 000 to 999. All digits of the number must be matched in order, so the probability of winning is 0.001. 1 1000 the lottery 1000 times in a row. A ticket costs $1, and the prize is $500. Suppose you play a. Write a probability distribution for the number of wins. b. What is the probability that you will break even or better? Solution a. The number of wins could be anything from 0 to 1000. However, the probability of winning more than a few times is so small that it is essentially 0. Thus, the sample space will be considered as 0, 1, 2, 3, 4, and 5 or more wins. The probabilities are calculated using the binomial probability formula, with 1000C01 0 wins 2 1000C11 1 win 2 1000C21 2 wins 2 1000C31 3 wins 2 1000C41 4 wins 2 5 or more wins 2 0 and p 0.001. n 1000 1000 0.3677 2 999 0.3681 998 0.1840 997 0.0613 996 0.0153 0.999 1 0.999 2 1 2 0.999 0.999 0.999 0.001 2 1 0.001 2 0.001 0.001 0.001 1 0.3677 0.3681 0.1840 0.0613 0.0153 0.0036 Outcome 0 wins 1 win 2 wins 3 wins 4 wins 5 or more wins Probability 0.3677 0.3681 0.1840 0.0613 0.0153 0.0036 Section 13.4.A Excursion: Binomial Experiments 887 b. In order to break even or better, you must win 2 or more times. The probability is the sum of the probabilities of winning 2, 3, 4, or 5 or more times. break even or better P 1 0.1840 0.0613 0.0153 0.0036 0.2642 ■ 2 Example 3 Multiple Choice Exam Morgan is taking a 10-question multiple choice test but has not studied. Each question has 4 possible responses, only one of which is correct. Find the probability of getting the results below if he answers all questions randomly. a. exactly 6 questions correct b. 4 or fewer questions correct c. 8 or more questions correct Solution The probability of getting each question correct is p 1 4 0.25. Technology Tip The command binompdf( in the DISTR menu of TI finds the probability for r successes in n trials of a binomial experiment, given n, p and r. The command binomcdf( finds the cumulative probability for r or fewer successes in n trials. a. b. c. Figure 13.4.A-1 6 P 1 6 correct 10C61 2 4 or fewer correct 8 1 8 or more correct 10 2 0.25 3 0.75 0.25 1 2 10C01 2 10C31 0.25 2 1 10C81 0.25 2 0 0.0004 0.25 0.75 0.25 0.75 2 2 2 2 P 1 10C21 P 1 10C101 2 1 2 0 4 0.016 0.75 7 1 2 0.75 2 8 0.75 1 10 2 10C41 2 0.25 4 10C11 0.25 2 1 10C91 0.25 2 0.75 9 2 1 2 1 9 0.75 2 1 6 0.922 2 1 0.75 2 ■ Binomial Distributions Consider the following binomial experiment: a coin is tossed 4 times, and the probability of heads on each toss is A probability distribution for . 1 2 the number of heads is shown below. Number of heads Probability 0 1 16 16 A probability density function that represents this distribution is shown in Figure 13.4.A-2. Notice that the shape of the graph is symmetric. The expected value of this distribution is x 1 16b 0 a 1 1 4b a 2 3 8b a 3 1 4b a 4 1 16b a 2 1 2 3 4 5 which is the (approximate) center of the probability distribution. y 0.5 0.25 0 Figure 13.4.A-2 888 Chapter 13 Statistics and Probability For large values of n or when p is near 1 2 , the shape of a binomial dis- tribution is approximately symmetric, with its center at the expected value. For small values of n if p is different from 1 2 , the distribution will be skewed. However, as n increases, the distribution becomes more symmetric. The graphs in Figure 13.4.A-3 show the distributions of a binomial experiment with p 1 3 for n 3, 10, and 30. The shape of the distribution approaches a special curve, called the normal curve, which is developed in the next section. Characteristics of a Binomial Distribution n = 3 n = 10 n = 30 Figure 13.4.A-3 The distribution of a binomial experiment with n trials and probability of success p is approximately symmetric for large values of n. The center of the distribution is the expected value. The expected value of the binomial distribution is np. The standard deviation of the binomial distribution is 1npq. Example 4 Multiple Choice Exam Find the expected value and standard deviation of the number of questions correct on the multiple choice exam in Example 3. Solution In this case, n 10, p 0.25, and q 0.75. The expected value is and the standard deviation is np 10 0.25 1 2 2.5 1npq 110 0.25 1 21 0.75 2 1.4. 11 This means that if a large number of students guessed on the exam, the average number correct would be 2.5, with a standard deviation of 1.4. A graph of the distribution is shown in Figure 13.4.A-4. ■ Figure 13.4.A-4 0.3 0 0 Section 13.4.A Excursion: Binomial Experiments 889 Exercises 13.4.A For Exercises 1–4, a binomial experiment consists of planting 4 seeds. The probability of success (that a given seed will sprout) is The sample outcome SFSS means that the first seed sprouted, the second seed did not sprout, and the third and fourth seeds sprouted. p 0.65. 1. What is the probability of failure? 2. Write out the two successes. 4C2 6 outcomes that have exactly 3. Complete the probability distribution below. 8. What is the expected value of the number of bulls-eyes in 4 tries? A true-false exam has 100 questions, and for each question What is the probability of answering P(true) P(false) 0.5. 9. 50 questions correct? 10. 70 questions correct? 11. 30 questions correct? 12. 90 questions correct? Sprouted Probability 0 ? 1 ? 2 ? 3 ? 4 ? 13. Find the expected value and standard deviation of the number of correct answers. 4. Find the expected value of the number of seeds that will sprout, and interpret the result. For Exercises 5–8, suppose the probability of a certain dart player hitting a bulls-eye is 0.25. 5. Write a probability distribution for the number of bulls-eyes in 4 tries. 6. What is the probability of hitting at least 3 bulls- eyes in 4 tries? 7. What is the probability of hitting less than 2 bulls- eyes in 4 tries? An experiment consists of rolling a number cube 30 times. The outcome 6 is considered a success, and all other outcomes are considered failures. Assume all p P(success) 1 6 faces are equally likely, so that . 14. Find the expected value of the probability distribution. What is the probability of e successes? m 15. What is the probability of m 1 is the probability of m 1 successes? successes? What 16. Find the standard deviation of the probability distribution. What is the probability that the outcome is between m s m s and s ? 13.5 Normal Distributions Objectives • Draw a normal distribution given its mean and standard deviation • Use normal distributions to find probabilities To statisticians, the most important probability density function is the normal curve (sometimes called the bell curve). Normal distributions are used to predict the outcomes of many events, such as the probability that a student scores a 750 on the SAT, or the probability that you will grow to be 67 inches tall. For statisticians, it is also a valuable tool for predicting if an outcome is statistically significant or just caused by chance. 890 Chapter 13 Statistics and Probability Properties of the Normal Curve A normal distribution is bell-shaped and symmetric about its mean. The x-axis is a horizontal asymptote, and the area under the curve and above the x-axis is 1. The maximum value occurs at the mean, and the curve has two points of inflection, at 1 standard deviation to the right and left of the mean. Because a normal distribution is symmetric, the mean, median, and mode have the same value. This value is also called the center. The Greek letter m (mu) is used to represent the mean of a normally distributed population. Also, the population standard deviation is used for the standard deviation. s Figure 13.5-1 shows t
hree normal curves, with m and s labeled. y 1 0.8 0.6 0.4 0..5 x −4 −2 0 2 4 6 8 10 Figure 13.5-1 The normal curve with a mean of 0 and standard deviation of 1 is called the standard normal curve. The equation of the standard normal curve is y 1 12p x 2 e 2 y 0.4 0.2 −4 −3 −2 −1 0 1 2 3 4 Figure 13.5-2 x The standard normal curve can be thought of as a parent function for all normal curves. • A change in • A change in m s results in a horizontal translation of the curve. results in a horizontal stretch and vertical compression of the curve, or vice versa, so that the resulting area is still 1. Section 13.5 Normal Distributions 891 Graphing Exploration The normal curves below have the same value of and different values of Graph the curves in the same viewing window and describe the results. m. s y 1 12p x2 e 2 y 1 12p e 2 x 3 2 1 2 y 1 12p 2 x 4 2 1 2 e The normal curves below have the same value of and different values of Graph the curves below in the same viewing window and describe the results. s. m y 1 12p x2 2 e y 1 312p x2 18 e y 3 12p 9x2 2 e Equation of Normal Curve A random variable is said to have a normal distribution with mean given by the equation if its density function is and standard deviation S M y 1 S12P (x M)2 2S2 e Most of the time, the mean and standard deviation of an entire population cannot be measured. The population mean and standard deviation are often estimated by using a sample. Example 1 Using Sample Information A paper in Animal Behavior gives 11 sample distances, in cm, from which a bat can first detect a nearby insect. The bat does this by sending out high-pitched sounds and listening for the echoes. Assume the population is normally distributed. 62, 23, 27, 56, 52, 34, 42, 40, 68, 45, 83 a. Compute the mean and standard deviation of this sample. b. Draw a normal curve to represent the distribution. Figure 13.5-3 Solution a. The mean is 48.36 and the sample standard deviation is 18.08, as shown in Figure 13.5-3. b. The normal curve is shown in Figure 13.5-4. 892 Chapter 13 Statistics and Probability 0.04 0 0 Figure 13.5-4 102 ■ The Empirical Rule Consider the intervals formed by one, two, and three standard deviations on either side of the mean, as shown below. The empirical rule describes the areas under the normal curve over these intervals. µ σ− 3 µ σ− 2 µ σ+ 3 68% 95% 99.7% Empirical Rule In a normal distribution: NOTE Each percentage in the empirical rule can also be interpreted as the probability that a data value chosen at random will lie within one, two, or three standard deviations of the mean. • about 68% of the data values are within one standard deviation of the mean. • about 95% of the data values are within two standard deviations of the mean. • about 99.7% of the data values are within three standard deviations of the mean. Example 2 Running Shoes A pair of running shoes lasts an average of 450 miles, with a standard deviation of 50 miles. Use the empirical rule to find the probability that a new pair of shoes will have the following lifespans, in miles. a. between 400 and 500 miles b. more than 550 miles Section 13.5 Normal Distributions 893 Solution Figure 13.5-5 shows the normal distribution with the intervals of one, two, and three standard deviations on either side of the mean. 300 350 400 450 500 550 600 Figure 13.5-5 x miles a. The area under the curve between 400 and 500 miles is approximately 68% of the total area. Since the area under a density function corresponds to the probability, the probability that a pair of shoes will last between 400 and 500 miles is about 0.68. b. The area under the curve between 350 and 550 miles is 95% of the total area, which leaves 5% for the area less than 350 and greater than 550. Since the normal curve is symmetric, the area greater than 550 is exactly half of this, or 2.5%. Thus, the probability that a pair of shoes will last more than 550 miles is about 0.025. ■ The Standard Normal Curve In general, determining the area under a normal curve is very difficult. Because of this, it is common to standardize data to match the normal curve with a mean of 0 and standard deviation of 1, for which these areas are known. Comparing Data Sets Sarah and Megan are high school juniors. Sarah scored 660 on the SAT, and Megan scored 29 on the ACT. Who did better? Although the scores on both tests are normally distributed, the mean and standard deviation are very different. One way to compare the distributions is to adjust the scales of the axes, as shown in Figure 13.5-6. SAT µ σ = 500 = 100 ACT = 18 µ = 6 σ 200 300 400 500 600 700 800 0 6 12 18 24 30 36 Figure 13.5-6 x x 894 Chapter 13 Statistics and Probability A more precise way to compare two scores from different data sets is to use the standard deviation as a unit of measurement. Each score is represented by the number of standard deviations above or below the mean. Example 3 Comparing Scores Use the standard deviation as a unit of measurement to compare the SAT and ACT scores for Sarah and Megan. Solution Sarah’s score of 660 is 160 points above the mean, which is 500. The stan- dard deviation is 100, so Sarah’s score is 160 100 1.6 standard deviations above the mean. Megan’s score of 29 is 11 points above the mean, which is 18. The stan- dard deviation is 6, so Megan’s score is 11 6 1.83 standard deviations above the mean. Thus, Megan’s score is better than Sarah’s score. ■ The number of standard deviations that a data value is above the mean is called the z-value. In Example 3, Sarah’s z-value is 1.6 and Megan’s z-value is 1.83. For a normal distribution, the z-values correspond to values on the standard normal curve, as shown in Figure 13.5-7. z-values: 200 −3 300 −2 400 −1 500 0 600 1 700 2 Figure 13.5-7 z-Values The z-value of the value x in a data set with mean standard deviation is S x 800 3 M and z x M S The area under a normal curve between x a and x b is equal to the area under the standard normal curve between the z-value of a and the z-value of b. Once z-values are determined, corresponding areas under the normal curve are often found using a table. Because of the symmetry of the normal curve, it is only necessary to include positive z-values in the table. Section 13.5 Normal Distributions 895 The following table gives the area under the normal curve between 0 and the given z-values, as shown in Figure 13.5-8. y −3 −2 −1 0 z 1 2 3 Figure 13.5-8 x z 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Area 0.04 0.08 0.12 0.16 0.19 0.23 0.26 0.29 0.32 0.34 Example 4 Using z-Values to Determine Area A normally distributed data set has a mean of 25 and a standard deviation of 5. Find the probability that a data value chosen at random is between 23 and 28. Solution First, find the z-values for 23 and 28, given z 23 25 5 0.4 and s 5. 0.6 m 25 z 28 25 5 0.4 Divide the area into two parts: the area from to 0, and the area from 0 to 0.6, as shown in Figure 13.5-9. The areas are found in the table above. Because the normal curve is symmetric with respect to the area from 0.4 to 0 is the same as the area from 0 to 0.4, or 0.16. The area from 0 to 0.6 is 0.23. The total area is the sum of the two areas, 0.39. Thus, the probability that a randomly chosen data value is between 23 and 28 is 0.39. x 0, Technology Tip Normalcdf under the DISTR menu of TI will calculate the area under the normal curve in a given interval. The parameters are Normalcdf (lower bound, upper ).sm, bound, 0.16 y 0.23 x −0.4 0.6 Figure 13.5-9 ■ Example 5 Response Times The EMT response time for an emergency is the difference between the time the call is received and the time the ambulance arrives on the scene. Suppose the response times for a given city have a normal distribution 896 Chapter 13 Statistics and Probability m 6 s 1.2 with estimate the probability of the following response times. minutes and minutes. For a randomly received call, a. between 6 and 7 minutes c. less than 7 minutes b. less than 5 minutes Solution a. The z-values are z 6 6 1.2 0 and z 7 6 1.2 0.8. The probability of a response time between 6 and 7 minutes is the area under the standard normal curve from 0 to 0.8, which is approximately 0.29. b. The z-value is z 5 6 1.2 0.8. The area from 0 to 0.8 is approxi- mately 0.29, so the probability of a response time under 5 minutes is (see Figure 13.5-10). about 0.5 0.29 0.21 half of total area = 0.5 y 0.21 0.29 −0.8 Figure 13.5-10 x c. The area under the curve from 0 to 6 minutes is 0.5. From part a, the area under the curve from 6 to 7 minutes is 0.29. Thus, the probability of a response time under 7 minutes is 0.5 0.29 0.79. ■ Exercises 13.5 In Exercises 1–4, refer to the normal curve below. The area of each square of the grid is 0.01. 3. Estimate the area under the curve from 16 to 24. 4. Estimate the area under the curve from 12 to 28. Graph the normal curves for the following values of M and S. 8 12 16 20 24 28 32 1. Find m. 2. Estimate s. x 5. m 10, s 12 6. m 40, s 12 7. m 500, s 100 8. m 3, s 5 Suppose the heights of adult men are normally distributed. The heights of a sample of 30 men are shown below, in inches. 65, 83, 69, 67, 69, 67, 67, 72, 85, 68, 73, 65, 67, 65, 72, 71, 67, 73, 68, 72, 61, 75, 66, 78, 65, 71, 68, 76, 67, 68 Section 13.5 Normal Distributions 897 9. Compute the mean and standard deviation of this 23. the probability that the wait time for a table is sample. between 25 and 38 minutes 10. Draw a normal curve that represents the 24. the probability that the wait time for a table is less distribution of adult male heights, based on the sample. than 22 minutes Suppose that the heights of adult women are normally inches. distributed with Use the properties of the normal curve and the Empirical Rule to find the probability that a randomly chosen woman is within the given range. inches and S 2.5 M 65 11. taller than 65 inches 12. shorter than 67.5 inches 13. between 62.5 inches and 67.5 inches 14. between 60 inches and 70 inches 15. between 57.5 inches and 67.5 inches 16. A student t
ook two national standardized tests while applying for college. On the first test, m 32 and s 6. the second, on which test did he do better? m 475 and If he scored 630 on the first test and 45 on and on the second test, s 75, 17. Four students took a national standardized test for which the mean was 500 and the standard deviation was 100. Their scores were 560, 450, 640, and 530. Determine the z-value for each student. 18. If a student’s z-value was 1.75 on the test described in problem 17, what was the student’s score? 19. A sample of restaurants in a city showed that the average cost of a glass of iced tea is $1.25 with a 7¢. standard deviation of Three of the restaurants charge z-value for each restaurant. $1.00, and $1.35. Determine the 95¢, 20. If a new restaurant charges a price for iced tea that has a z-value of what is the tea’s actual cost? 1.25 (see Exercise 19), then At a certain restaurant, the wait time for a table is norM 30 S 10 mally distributed with minutes. Use the table on page 895 to estimate the following: minutes and 21. the probability that the wait time for a table is between 30 and 35 minutes 22. the probability that the wait time for a table is between 24 and 30 minutes Daytime high temperatures in New York in February are normally distributed with an average of and a standard deviation of 30.2 8.5. 25. Estimate the probability that the temperature on a given day in February is 39° or higher. 26. Estimate the probability that the temperature on a given day in February is 22° or lower. 27. Estimate the probability that the temperature on a given day in February is between 13° and 39°. 28. Estimate the probability that the temperature on a given day in February is between 25° and 30°. 29. Estimate the probability that the temperature on a given day in February is between 27° and 38°. The quartiles of a normal distribution are the values that divide the area under the curve into fourths4 −3 −2 Q1 Q3 2 3 4 Q2 (center) 1st and The the left and right of the mean, or quartiles are approximately 3rd 0.675S to Q1 M 0.675S and Q3 M 0.675S 30. Find Q1 s 4. and Q3 for a distribution with m 20 and 31. Suppose the scores on an exam are normally m 70 Q1 s 10. Find and distributed with Q3 , and interpret the result. and 32. For the exam in Exercise 31, what exam score would place a student in the top 25% of the class? C H A P T E R 13 R E V I E W Important Concepts Section 13.1 Section 13.2 Section 13.3 Section 13.4 Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .843 Population. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 843 Sample . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 843 Frequency table . . . . . . . . . . . . . . . . . . . . . . . . . . 844 Bar graph . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 845 Pie chart . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 845 Uniform, symmetric, and skewed distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 846 Stem plot . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 847 Histogram . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 849 Mean . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 853 Median . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 854 Mode . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 855 Standard deviation . . . . . . . . . . . . . . . . . . . . . . . 857 Range . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 859 Interquartile range . . . . . . . . . . . . . . . . . . . . . . . 860 Five-number summary and box plot . . . . . . . . . 861 Experiment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 864 Sample space. . . . . . . . . . . . . . . . . . . . . . . . . . . . 864 Event . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 865 Probability. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 865 Probability distribution . . . . . . . . . . . . . . . . . . . . 865 Mutually exclusive events. . . . . . . . . . . . . . . . . . 866 Complement . . . . . . . . . . . . . . . . . . . . . . . . . . . . 866 Independent events. . . . . . . . . . . . . . . . . . . . . . . 867 Random variable. . . . . . . . . . . . . . . . . . . . . . . . . 869 Expected value . . . . . . . . . . . . . . . . . . . . . . . . . . 869 Probability density function . . . . . . . . . . . . . . . . 871 Experimental estimate of probability . . . . . . . . . 875 Probability simulation. . . . . . . . . . . . . . . . . . . . . 875 Equally likely outcomes . . . . . . . . . . . . . . . . . . . 877 Fundamental counting principle. . . . . . . . . . . . . 879 Permutations and combinations . . . . . . . . . . . . . 880 Section 13.4.A Binomial experiment . . . . . . . . . . . . . . . . . . . . . . 884 Binomial distribution . . . . . . . . . . . . . . . . . . . . . 887 898 Chapter Review 899 Section 13.5 Normal curve . . . . . . . . . . . . . . . . . . . . . . . . . . . 889 Normal distribution . . . . . . . . . . . . . . . . . . . . . . 889 Standard normal curve . . . . . . . . . . . . . . . . . . . . 890 Empirical rule . . . . . . . . . . . . . . . . . . . . . . . . . . . 892 z-value. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 894 Important Facts and Formulas mean: x ©xi n x3, p xn x1, x2, are ordered from smallest to largest, then the median If is the middle entry when n is odd and the average of the two middle entries when n is even. for n odd, the value in the n 1 2 position median • for n even, the average of the values in the n 2 and n 2 1 positions Population standard deviation s B © xi 1 2 x n 2 Sample standard deviation s © 2 x xi 1 n 1 2 B Suppose an experiment has a sample space of n outcomes, all of 1 n, which are equally likely. Then the probability of each outcome is and the probability of an event E is given by number of outcomes in E n P E 2 1 Fundamental Counting Principle Consider a set of k experiments. Suppose the first experiment has outcomes, the second has n1 number of outcomes is n1 outcomes, and so on. Then the total p nk for all k experiments. n2 n2 Permutations If r items are chosen in order without replacement from n possible items, the number of permutations is nPr n! n r 1 ! 2 Combinations If r items are chosen in any order without replacement from n possible items, the number of combinations is nCr n! n r ! 2 r! 1 900 Chapter Review nCrprqnr In a binomial experiment, where p is the probability of success, and is the probability of failure. The expected value of a binomial distribution is np, 1npq. and the standard deviation is r successes in n trials q 1 p P 2 1 A random variable is said to have a normal distribution with mean m if its density function is given by the equation and standard deviation s y 1 s12p x m 2s2 2 2 1 e Empirical Rule In a normal distribution: • about 68% of the data values are within one standard deviation of the mean. • about 95% of the data values are within two standard deviations of the mean. • about 99.7% of the data values are within three standard deviations of the mean. The z-value of the value x in a data set with mean m and standard deviation s is z x m s . Review Exercises Exercises 1–4 refer to the following description: A group of bird-watchers is trying to determine what types of birds are common to their area. The group observed 21 sparrows, 15 purple finches, 10 chickadees, 5 cardinals, and 2 blue jays. Section 13.1 1. Is the data qualitative or quantitative? 2. Create a frequency table for the data. 3. Create a bar graph for the data. 4. Create a pie chart for the data. Exercises 5–17 refer to the following description: a study is done to determine the average commuting time of employees at a company. A total of 34 employees are surveyed, with the following results (in minutes). 31.9, 34, 30.7, 39, 33.1, 35.2, 30.5, 32.7, 29.4, 33.4, 22.3, 31.9, 32.3, 29.4, 33, 18.2, 29.1, 32.5, 22.2, 36.1, 27.7, 36, 31.9, 26, 31.7, 23.2, 30.7, 24.4, 33, 28.4, 28.8, 23.3, 32.2, 22.8 5. Is the data qualitative or quantitative? 6. Is the data discrete or continuous? Chapter Review 901 7. Create a stem plot for the data. 8. Create a histogram for the data with class intervals of 5 minutes. 9. Use your histogram from Exercise 8 to describe the shape of the data. Section 13.2 10. Find the mean of the data. 11. Find the median of the data. 12. Find the mode of the data. 13. Find the sample standard deviation of the data. 14. Find the range of the data. 15. Find the first and third quartiles of the data. 16. Find the interquartile range of the data. 17. Create a box plot for the data. Section 13.3 Exercises 18–22 refer to the following probability distribution: Outcome Probability 1 0.1 2 ? 3 0.4 4 ? 5 0.1 18. List the sample space for the probability distribution. 19. Suppose the probabilities are the same for outcomes 2 and 4. Complete the probability distribution. 20. What is the probability that the outcome is an even number? 21. What is the probability that the outcome is greater than 2? 22. Suppose the experiment with the given probability distribution is repeated 3 times. Assuming the trials are independent, what is the probability of the outcome {1, 2, 3}? An experiment consists of spinning the spinner at left 3 times. A probability distribution for the outcomes is given below. W white, R red 1 2 Outcome WWW WWR WRW RWW WRR RWR RRW RRR Probability 0.512 0.128 0.128 0.128 0.032 0.032 0.032 0.008 A random variable is assigned to the number of times the spinner lands on red. 23. What is the range of the random variable? 24. What is the probability that the value of the random variable is 2? 902 Chapter Review Section 13.4 25. Create a probability distribution for the random variable. 26. Find the expected value of the random variable, and interpret the result. 27. Graph the probability density function of the random variable, and shade the area of the graph that corresponds to the pro
bability that the spinner lands on white 3 times. 28. Suppose the experiment is repeated 25 times, with the following results: WRW, RRW, WWW, WWW, WWW, WWW, WWR, RWW, WWW, WRW, WWW, WWR, WWW, RWW, WWW, RWR, WWW, WRW, WWW, RWW, WWW, WWW, WRW, WWW, WWR Write a probability distribution of the random variable, based on the experimental results. How do the probabilities compare to your results in Exercise 25? Exercises 29–33 refer to the following experiment: A sock drawer contains 6 identical black socks, 8 identical white socks, and 1 blue sock. A sock is chosen randomly from the drawer. Assume all socks are equally likely to be chosen. 29. What is the probability of choosing a black sock? a white sock? a blue sock? 30. Suppose a black sock is chosen. What is the probability that the next sock chosen will also be black? Hint: how many of each color are left in the drawer? 31. Suppose a white sock is chosen. What is the probability that the next sock chosen will also be white? Hint: how many of each color are left in the drawer? 32. Suppose a blue sock is chosen. What is the probability that the next sock chosen will also be blue? Hint: how many of each color are left in the drawer? 33. Use your results from Exercises 30 – 32 to determine the probability of choosing a pair if two socks are chosen randomly from the drawer. 34. A lottery ticket involves matching 5 numbers between 1 and 50 in any order. What is the probability of matching all 5 numbers? What is the probability of matching any 4 numbers? 35. Suppose there are 4 people on a subcommittee, and you do not know their last names. If you have a list of 10 last names of all of the people in the committee, what is the probability of correctly guessing the last names of the people in the subcommittee? Chapter Review 903 Section 13.4.A A binomial experiment consists of randomly choosing 7 tiles imprinted with letters of the alphabet. An outcome of a vowel is considered a success, and a consonant is considered a failure. The probability of success is p 0.44. 36. Complete the probability distribution below for the number of vowels. Number of vowels Probability ? 37. Find the expected value and standard deviation of the number of vowels. S 8. Suppose the scores on an exam are normally distributed with Use properties of the normal curve, the empirical rule, and the table on page 895 to answer Exercises 38–46. M 75 and Section 13.5 38. Write the equation of the normal curve for the distribution of the scores. 39. Graph the normal curve for the distribution of the scores. 40. Estimate the probability that a randomly chosen score is greater than 75. 41. Estimate the probability that a randomly chosen score is between 67 and 83. 42. Estimate the probability that a randomly chosen score is greater than 59. 43. Estimate the probability that a randomly chosen score is less than 99. 44. Estimate the probability that a randomly chosen score is between 75 and 78. 45. Estimate the probability that a randomly chosen score is between 70 and 82. 46. Estimate the probability that a randomly chosen score is less than 74. C H A P T E R 13 Area Under a Curve y h x a b Figure 13.C-1 Many applications of calculus involve finding the area under a curve. In probability, for example, it is often necessary to find areas under a normal curve or other probability density function. In other areas, such as physics, the area under a curve can be used to determine total distance traveled or the total amount of force on an object. In this section, properties of probability are used to estimate the area under a curve. Example 1 Area Model for Probability Consider the graph in Figure 13.C-1. Suppose a point in the rectangle is chosen randomly. If A is the area of the shaded region, write a formula in terms of A, a, b, and h for the probability that the point is in the shaded region. Solution If the point is chosen randomly, then all points in the sample space are equally likely to be chosen. The sample space is all points in the rectangle, and the event above can be described as the set of all points in the shaded area, both of which are infinite. Since it is impossible to divide the number of outcomes in the event by the number of outcomes in the sample space, the areas of the regions are used instead. The area of the rectangle is area of shaded region area of rectangle A b a 2 h 1 ■ A probability simulation can also be used to find the probability in Example 1. Suppose points in the rectangle are chosen randomly, and 470 of them are in the shaded region. Recall that the experimental estimate of the probability of an event is n 1000 E P 1 2 number of trials with an outcome in E n 470 1000 0.47 By setting the two probability estimates equal to each other and solving, a formula can be found for the area A in terms of a, b, and h. h 1 904 A b a 0.47 2 A 0.47 h b a 2 1 Area Under a Curve In general, the area A under a curve between the x-values a and b with a b may be estimated as follows: 6 1. Draw a rectangle with a base of length b a that contains the desired area. Let h be the height of the rectangle. 2. Randomly choose n points in the rectangle, using a calculator or computer. Determine the number of points e that lie in the desired area. 3. The area is approximated by the formula below. A e n h 1 b a 2 Using Probability Simulations To generate a large number of points and determine the number of points that lie under a curve, a calculator or computer program is usually used. A sample program is given below. The equation of the graph must be entered in and the calculator window should contain the desired region. Note: the choice of window will not affect the answer. Y1, Prompt A Prompt B Prompt H Prompt N 0 S E For (K, 1, N) B A rand A S X ) ( H rand S Y Pt-On (X, Y) Y Y1 If E 1 S E End (X) Disp (E/N)H( B A ) A is the left bound of the rectangle. B is the right bound of the rectangle. H is the height of the rectangle. N is the number of points in the rectangle. E is the number of points in the shaded region. This command generates an x-coordinate, X. This command generates a y-coordinate, Y. This command displays the point (X, Y). The point is tested to determine whether it is under the curve. This command displays the estimate of the area. Example 2 Area of a Quarter-Circle Use a probability simulation to estimate the area under the curve with equa- tion y 21 x2 from 0 to 1. Since this region is one-fourth of a circle with radius 1, the area should be p 4 0.785. 905 Solution The result of a simulation with The area estimate is 0.782, which is very close to the predicted area. points is shown in Figure 13.C-2. n 500 1.2 1.175 1.175 0.25 Figure 13.C-2 ■ This method can also be used to find areas under a normal curve. Example 3 Area Under a Normal Curve a. Use a probability simulation to estimate the area under the normal curve with m 70 and s 10 between a 85 and b 95. b. Suppose the scores on an exam are normally distributed with m 70 Estimate the probability that a randomly chosen score is s 10. and between 85 and 95. Solution a. The normal curve has the equation y 0.03989e0.005(x70)2 . Trace to find a good value of h. The closer the value of h is to the maximum value of the function over the given interval, the better the estimate will be. A good choice of h is 0.014. The result of a simulation with n 500 points is shown in Figure 13.C-3. The area estimate is 0.0616. 0.05 0 40 100 Figure 13.C-3 b. The probability that a randomly chosen score is between 85 and 95 is approximately the area under the normal curve, or about 0.06. ■ 906 Exercises In Exercises 1 – 6, estimate the area under the given curve between the given values of a and b. 8. The function y 1 1x is not defined at x 0, and 1. y x2 2. y sin x a 0, b 1 a 0, b p 3. y x2 5x 4 a 1, b 4 4. y e x 5. y 1 x 6. y 1x a 2, b 10 a 3, b 5 a 7, b 9 the graph has a vertical asymptote. However, calculus can be used to show that the area under the curve from 0 to 1 is finite. Use a probability a 0 simulation to estimate the area by letting b 1. h 3, 4, for each value of h to be sure you have a good estimate. and Compare the value of the area estimates for and 5. Run the simulation at least twice 9. Use a probability simulation to verify that the area under the standard normal curve is 1. Hint: according to the Empirical Rule, 99.7% of the area under a normal curve is within of the mean. 3s 7. Suppose the weights of apples from a certain tree are normally distributed with s 18 g. chosen apple weighs between 100 g and 150 g. Estimate the probability that a randomly and m 138 g 907 C H A P T E R 14 Limits and Continuity Terminal Velocity Parachutists have two forces acting on them, as does any free-falling body. One force is gravity, which causes the body to speed up as it falls, and the other is air resistance, which causes the body to slow down. As the body moves faster, the air resistance builds until it nearly balances the gravitational force. So the body speeds up very little after it has fallen some distance. Terminal velocity is achieved when the air resistance approaches the gravitational pull. See Exercise 43 in Section 14.5. 908 Chapter Outline 14.1 14.2 Limits of Functions Properties of Limits 14.2.A Excursion: One-Sided Limits 14.3 The Formal Definition of Limit (Optional) 14.1 > 14.2 14.4 Continuity 14.5 Limits Involving Infinity Chapter Review can do calculus Riemann Sums Interdependence of Sections > > > 14.3 14.4 14.5 The mathematics presented in previous chapters deals with static prob- lems, such as What is the size of the angle? What is the average speed of a car between t 0 and t 5 minutes? Calculus, on the other hand, deals with dynamic problems, such as At what rate is the angle increasing in size? How fast is the car going at time t 4.2 minutes? The key to dealing with such dynamic problems is the concept of limit, which is introduced in this chapter. 14.1 Limits of Functions Objectives • Use the informal definition of limit Many
mathematical problems involve the behavior of a function at a particular value: What is the value of the function f when x c? x 1 2 rather than at The underlying idea of limit, however, is the behavior of the function near x c You have dealt with limits informally in previous chapters, but this section will discuss limits in more detail and give the notation used when talking about them. x c. Suppose you want to describe the behavior of 0.1x4 0.8x3 1.6x2 2x 8 x 4 f x 1 2 909 910 Chapter 14 Limits and Continuity Technology Tip Letting the value of x that produces the hole be the center of the displayed parts of the x-axis will produce a graph that shows the hole as shown in Figure 14.1-1. When holes are at integer values, a decimal window will normally show the holes. Technology Tip Once the function is entered into the calculator, use the calculator’s table feature to generate values of the function. Set the INPUT or INDPNT to USER or ASK, and enter the values of x. To see what happens to the values of when x is very close to 4. Notice that the function is not defined when x 4. when x is very close to 4, observe the graph of the function in the viewing window shown in Figure 14.1-1. (See the Technology Tip about how to produce a graph that shows the hole.) x f 1 2 3 0 0 8 Figure 14.1-1 To further explore the behavior of the function near following Graphing Exploration. x 4, perform the Graphing Exploration in the viewing window with x f Graph and 2 1 0 y 3. Use the trace feature to move along the graph, and examwhen x takes values close to 4. Your results ine the values of should be consistent with the following table of values. x f 2 1 3.5 x 4.5 x approaches 4 from the left > x approaches 4 from the right > x 3.9 3.99 3.999 f x 2 1 1.8479 1.9841 1.9984 4 * 4.001 4.01 4.1 2.0016 2.0161 2.168 The exploration and the table suggest that as x gets closer and closer to get closer and closer 4 from either side, the corresponding values of to 2. Furthermore, by taking x close enough to 4, the corresponding values can be made as close as you want to 2. x f 1 2 For instance, 3.99999 f 1 2 1.999984 and f 4.00001 1 2 2.000016 Notation The statement above is usually expressed by saying “ The limit of f x 2 1 as x approaches 4 is 2, ” which is written symbolically as Section 14.1 Limits of Functions 911 lim xS4 f x 1 2 2 or lim xS4 0.1x4 0.8x3 1.6x2 2x 8 x 4 2 Informal Definition of Limit The following definition of “limit” in the general case is similar to the situation previously described, but now f is any function, c and L are fixed real numbers, and the phrase “arbitrarily close” means “as close as you c 4 want.” In the previous discussion, and L 2. Informal Definition of Limit Let f be a function and let c be a real number such that defined for all values of x near , except possibly at itself. Suppose that x c is f(x) x c whenever x takes values closer and closer but not equal to c (on both sides of c), the corresponding values of f(x) get very close to, and possibly equal, to the same real number L and that the values of can be made arbitrarily close to L by taking values of x close enough to c, but not equal to c. f(x) Then it is said that The limit of the function f(x) as x approaches c is the number L, which is written f(x) L lim xSc NOTE The limit in Example 1 is very important in calculus. Example 1 Limit of a Function If f x 1 2 sin x f x , find lim xS0 x . 2 1 Solution f x is not defined when Although be used with the graph of the function to examine the values of x is very close to 0. The table feature can also be used. a calculator’s trace feature can when x f 2 1 2 1 x 0, Calculator Exploration f Create a table of values for x 0. than and larger than same value as x approaches 0 from both sides? 1 Are the function values approaching the with values of x both smaller x 2 sin x x 912 Chapter 14 Limits and Continuity 2 The exploration should suggest that lim xS0 f x 1 2 1 or equivalently, lim xS0 sin x x 1, 2 2 a fact that will be proved in calculus. ■ 2 Figure 14.1-2 Example 2 Limit of a Function Find lim xS2 f , where f is the function given by the following two-part rule. x 1 2 f x 1 2 0 if x is an integer 1 if x is not an integer e Solution A calculator is not much help here, but the function is easily graphed by hand. y 3 2 1 0 −1 −1 −2 −3 −3 −2 x 1 2 3 Figure 14.1-3 When x is a number very close but not equal to 2 (either greater than 2 or less than 2), the corresponding value of is 1, and this is true no matter how close x is to 2. Thus, x f 1 2 lim xS2 f x 1 2 1. Because same as the value of the function f at 2 2 1 f 0 x 2. by definition, the limit of f as x approaches 2 is not the ■ Limits and Function Values If the limit of a function f as x approaches c exists, this limit may not be equal to f(c). In fact, f(c) may not even be defined. Very often the limit of a function as x approaches a point is equal to the value of the function at that point. Section 14.1 Limits of Functions 913 2 Example 3 Limit of a Function and a Function Value If f x 2 1 sin2 px cos px, find f 0.5 1 2 and lim xS0.5 f x . 2 1 3 3 Solution 2 Figure 14.1-4 0.5 f 1 2 f˛ 1 2b a sin2 p 2 cos p 2 12 0 1. Using the trace feature on the graph of gests that the limit is a number near 1. f x 1 2 , shown in Figure 14.1-4, sug- A much narrower viewing window is needed to determine the limit more precisely. Graphing Exploration 2 1 x f Graph 0.99 y 1.01. both sides of x 0.5, f 1 2 in a viewing window with and Use the trace feature to move along the graph on and confirm that as x gets closer and closer to x 0.5 0.49 x 0.51 gets closer and closer to 1. The Exploration suggests that lim xS0.5 f x 1 2 Thus, the value of the function at approaches 0.5. 1 f 1 x 0.5 0.5 . 2 is the same as the limit as x ■ NOTE Whenever a calculator was used in preceding examples, it was said that the information provided by the calculator suggested that the limit of the function was a particular number. Although such calculator explorations provide strong evidence, they do not constitute a proof and in some instances can be very misleading. (See Exercise 36.) Nevertheless, a calculator can help you develop an intuitive understanding for limits, which is needed for a rigorous treatment of the concept. Nonexistence of Limits Not every function has a limit at every number. Limits can fail to exist for several reasons. 914 Chapter 14 Limits and Continuity Nonexistence of Limits The limit of a function f as x approaches c may fail to exist if: 1. f(x) becomes infinitely large or infinitely small as x approaches c from either side 2. f(x) approaches L as x approaches c from the right and f(x) as x approaches c from the left approaches M, with M L, 3. f(x) oscillates infinitely many times between two numbers as x approaches c from either side The examples that follow illustrate each of these possibilities. 10,000 Example 4 A Function that Approaches Infinity Discuss the existence of lim xS0 1 x2. Solution 0.1 0.1 Figure 14.1-5 shows the graph of f x 1 2 1 x2 near x 0. As x approaches 1,000 Figure 14.1-5 y 2 1 −1 −2 −2 −1 become larger 0 from the left or right, the corresponding values of and larger without bound — rather than approaching one particular number—which can be verified by using the trace feature. Therefore, x f 1 2 lim xS0 1 x2 does not exist. ■ Example 5 A Function that Approaches Two Values x Find 0 lim xS0 x 0x , if it exists. 1 2 Solution The function f 0 x 1 2 x 0x is not defined when x 0. Figure 14.1-6a According to the definition of absolute value, x x There are two possibilities. x 6 0. x when 0 0 0 x when x 7 0 and 0 If x 7 0, then f If x 6 0, then f 0 0 x 0x x 0x . 1 Consequently, the graph of f looks like Figure 14.1-6a. A table of values for is shown in Figure 14.1-6b. x f 1 2 Figure 14.1-6b If x approaches 0 from the right, that is, through positive values, then the Section 14.1 Limits of Functions 915 f x corresponding value of that is, from negative values, then the corresponding value of always ding values of the definition of limit. Therefore, the limit does not exist. is always 1. If x approaches 0 from the left, is Thus, as x approaches 0 from both sides of 0, the correspondo not approach the same real number, as required by 1 ■ Example 6 An Oscillating Function Find lim xS0 sin p x , if it exists. Solution To understand the behavior of f x is close to 0. sin p x x 1 2 , consider what happens when As x takes values: Then p x takes values: from from from 1 2 1 4 1 6 to to to 1 4 1 6 1 8 2p to 4p 4p to 6p 6p to 8p p x completes one period of the sine function from Thus, the graph of sin x 1 x 1 2 4 , to another from x 1 4 nomenon occurs for negative values of x. Consequently, the graph of f 1 oscillates infinitely often between and 1, with the waves becoming more and more compressed as x approaches 0, as shown in Figure 14.1-7. and so on. A similar phe- x 1 6 to , y 1 0 −1 −2 −1 x 2 1 Figure 14.1-7 As x approaches 0, the function takes every value between nitely many times. In particular, p x 1 does not exist. real number. Therefore, and 1 infidoes not approach one particular sin x f 2 lim xS0 1 ■ 916 Chapter 14 Limits and Continuity Exercises 14.1 In Exercises 1–10, complete the table and use the result to estimate the given limit. 1. lim xS3 x 3 x2 2x 3 2.9 2.99 2.999 3.001 3.01 3.1 x f x 2 1 2. lim xS3 x 3 x2 9 2.9 2.99 2.999 3.001 3.01 3.1 x f x 2 1 3. lim xS0 2x 2 22 x x f x 1 2 8. lim xS1 0.9 0.99 0.999 1.001 1.01 1..1 1.01 1.001 0.999 0.99 0.9 f x 1 2 9. lim xS0 cos x 1 x 0.1 0.01 0.001 0.001 0.01 0.1 x f x 1 2 0.1 0.01 0.001 0.001 0.01 0.1 10. lim xSp 4 tan . lim xS0 2x 5 25 x 0.1 0.01 0.001 0.001 0.01 0.1 x f x 2 1 5. lim xS7 22 x 3 x 7 x 7.1 7.01 7.001 6.999 6.99 6.9 f x 1 2 6. lim xS0 21 x 1 x x f x 2 1 7. lim xS1 0.1 0.01 0.001 0.001 0.01 0..78 0.785 0.7853 0.7854 0.7855 0.786 f x 1 2 In Exercises 11–26, use a calculator to find a reasonable estimate of the limit. If the limit does not exist, explain why. 11. lim xS1 x6 1 x4 1
12. lim xS2 x5 32 x3 8 13. lim xS3 x2 x 6 x2 2x 3 14. lim xS1 x2 1 x2 x 2 15. lim xS1 x3 1 x2 1 17. lim xS0 tan x x x3 19. lim xS0 x tan x x sin x 21. lim xS3 2x 23 x 3 23. lim xS0 x ln 0 x 0 16. lim xS2 x2 5x 6 x2 x 6 18. lim xS0 tan x x sin x 20. lim xS0 x sin 2x x sin 2x 22. lim xS25 2x 5 x 25 24. lim xS0 e2x 1 x 25. lim xS0 ex 1 sin x 26. lim xS0 a x sin 1 xb 30. In Exercises 27–32, use the graph of the function f to determine the following: lim xS3 f(x) lim xS0 f(x) lim xS2 f(x) −4 −3 −2 27. 28. 291 −1 −2 −3 −4 4 3 2 1 0 −1 −1 −2 −3 −4 4 3 2 1 0 −1 −1 −2 −3 −4 −4 −3 −2 −4 −3 −2 −4 −3 − Section 14.1 Limits of Functions 917 y y y 4 3 2 1 0 −1 −1 −2 −3 −4 4 3 2 1 0 −1 −1 −2 −3 −4 4 3 2 1 0 −1 −1 −2 −3 − 31. 32. −4 −3 −2 −4 −3 −2 33. a. Graph the function f whose rule is if x 6 2 if 2 x 6 2 if x 2 if x 7 2 Use the graph in part a to evaluate the following limits. x lim xS2 d. lim xS2 lim xS1 c. Explain why value of c. lim xSc t 1 x 2 does not exist for every 36. Critical Thinking If f 1 cos x6 x12 x 1 2 , then 1 2 1 2 , x f as is shown in calculus. A calculator lim xS0 or computer, however, may indicate otherwise. f Graph 1 0.1 x 0.1, determine the values of What does this suggest that the limit is? and use the trace feature to in a viewing window with when x approaches 0. x x f 2 1 2 918 Chapter 14 Limits and Continuity 34. a. Graph the function g whose rule is g x 1 2 µ x2 x 2 2 3 x if x 6 1 if 1 x 6 1 if x 1 if x 7 1 Use the graph in part a to evaluate the following limits. x lim xS1 lim xS0 lim xS1 d. c. 35. Critical Thinking Consider the function t whose rule is x t 1 2 0 if x is rational 1 if x is irrational e 14.2 Properties of Limits Objectives • Find the limit of the constant function the identity function • Use the properties of limits • Find the limit of polynomial functions rational functions • Use the Limit Theorem Most of the functions that appear in calculus are combinations of simpler functions, such as sums, products, quotients, and compositions. This section presents rules for finding limits of combinations of functions without a table or a graph. There are two easy, but important, cases where the limit of a function may be found by evaluating the function. The first occurs with constant funcnote that as x lim tions, such as xS3 is always the number 5, so 5. The same 5. 2 approaches 3, the corresponding value of 5 5. that that is, x is the number To find Thus, lim xS3 5 lim xS3 lim xS3 2 thing is true for any constant function. 1 1 2 lim xS3 1 2 Limit of a Constant If d is a constant, then lim xSc d d. The same phenomenon occurs with the identity function, which is given If c is any real number, then the statement “x by the rule approaches c” is exactly the same as the statement “I(x) approaches c” because for every x. Thus, 2 1 x x. x x I I 1 2 Limit of the Identity Function For every real number c, lim xSc x c. Section 14.2 Properties of Limits 919 Properties of Limits There are a number of facts that greatly simplify the computation of limits. Rigorous proofs of these properties will not be given, knowing the central idea is more important now. For instance, suppose that as x approaches c, the values of a function f approach a number L and the values of a function g approach a number M. Then it is plausible that as x L M approaches c, the values of the function and that Following is a formal the values of the function statement in terms of limits. approach approach L M. f g f g Properties of Limits If f and g are functions and c, L, and M are numbers such that lim xSc f(x) L and lim xSc g(x) M , In Property 4, NOTE f gb1 x lim 2 xSc a M 0 and L 0, and may not exist. does not exist if L 0. M 0 If the limit may or then 1. lim xSc ( f g)(x) lim xSc L M [f(x) g(x)] 2. lim xSc ( f g)(x) lim xSc L M [f(x) g(x)] 3. lim xSc (f g)(x) lim xSc L M [f(x) g(x)] 4. lim xSc a f gb (x) lim xSc a f(x) g(x)b L M 2f (x) 2L 5. lim xSc (for M 0 ) (provided f(x) 0 for all x near c) x f L lim xSc These properties are often stated somewhat differently. For instance, M, because 2 lim xSc and similarly for the other properties. Property 1 can be written as lim xSc f g lim xSc lim xSc and 1 x g g 21 Limits of Polynomial Functions Properties 1–3, together with the facts about limits of constants and the identity function presented at the beginning of the section, now make it easy to find the limit of any polynomial function. Example 1 Limit of a Polynomial Function If f x 1 2 x2 2x 3, find lim xS4 f x 1 2 lim xS4 1 x2 2x 3 . 2 920 Chapter 14 Limits and Continuity Solution lim xS4 1 2x lim 3 xS4 x lim xS4 lim xS4 x lim xS4 x 2 lim xS4 4 x2 lim xS4 2 lim xS4 x 3 3 2 1 2 x2 2x 3 lim xS4 lim xS4 4 21 1 16 27 2 x lim xS4 x lim xS4 4 2 8 2 1 3 Properties 1 and 2 3 Property 3 Limit of a constant Limit of x Note that the limit of value of the function at x f 2 x 4, x2 2x 3 namely, f 1 at 4 x 4 4 1 2 1 is the same as the 3 27. 2 2 ■ 4 2 1 2 Because any polynomial function consists of sums and products of constants and x, the same argument used in Example 1 works for any polynomial function and leads to the following conclusion. Limits of Polynomial Functions If f(x) is a polynomial function and c is any real number, then f(x) f(c). lim xSc In other words, the limit is the value of the polynomial function f at x c. Limits of Rational Functions The fact in the previous box and Property 4 of limits make it easy to compute limits of many rational functions. Example 2 Limit of a Rational Function If f x 1 2 x3 3x2 10 x2 6x 1 , find lim xS2 f x . 2 1 3.1 Solution 4.7 4.7 3.1 Figure 14.2-1 0.86, The graph of f near as shown in Figure 14.2-1. You can determine the limit exactly by noting that suggests that the limit is a number near x 2 x f 1 2 is the quotient of the two functions x3 3x2 10 and h x g 1 2 x2 6x 1. x 1 2 As x approaches 2, the limit of each can be found by evaluation of the functions at Therefore, x 2. Section 14.2 Properties of Limits 921 lim xS2 f x 1 2 2 lim xS2 lim 1 xS2 lim xS2 x3 3x2 10 x2 6x 1 b Q x3 3x2 10 x2 6x 1 1 23 3 22 10 22 6 2 1 6 7 6 7 0.857 2 limit property 4 limits of polynomial functions Note that the limit of f x 1 2 as x approaches 2 is the number f 2 . 2 1 ■ The procedure in Example 2 works for other rational functions as well. Limits of Rational Functions Let f(x) be a rational function and let c be a real number such that f(c) is defined. Then f(x) f(c). lim xSc In other words, the limit of a rational function as x approaches c is the value of the function at if the function is defined there. x c, When a rational function is not defined at a number, different techniques must be used to find its limit there—if it exists. Example 3 Limit of a Rational Function If f x 2 1 x2 2x 3 x 3 , find lim xS3 f x . 2 1 Solution f x is not defined when x 3, Because the limit cannot be found by evaluating the function. Although it could be estimated graphically, it can be found by using algebraic methods. Begin by factoring the numerator. 2 1 x2 2x 3 x 3 1 x 1 x 3 21 x 3 2 Because the numerator and denominator have a common factor, the rational expression may be reduced. x 1 x 3 21 x 3 1 2 x 1, for all x 3. The definition of the limit as x approaches 3 involves only the behavior of the function near The preceding equation shows have exactly the same that both x 3. and not at g x and the function 922 Chapter 14 Limits and Continuity Figure 14.2-2 Limit Theorem NOTE Do not confuse the variable h with the function h. The meaning should be clear in context. values at all numbers except x approaches 3. x 3. So they must have the same limit as x2 2x 3 x 3 lim xS3 lim xS3 1 x 1 2 3 1 4, as illustrated in Figure 14.2-2. ■ The technique illustrated in Example 3 applies in many cases. When two functions have identical behavior, except possibly at they will have the same limit as x approaches c. x c, If f and g are functions that have limits as x approaches c and f(x) g(x) for all x c , then lim xSc f(x) lim xSc g(x). Recall that the difference quotient of a function f is given by The difference quotient can be evaluated for a specific value of x, say x c, to obtain a new form Limits of the difference quotient of a function f play an important role in calculus. When computing such limits, the variable is often the quantity h, as in the following example. Example 4 Limit of a Difference Quotient If f x 2 1 x2, find lim hS0 . Solution Using algebra, write the difference quotient as follows 52 5 h 2 h 25 10h h2 1 10h h2 h h 25 2 Section 14.2 Properties of Limits 923 When the difference quotient is written this way, it is easy to see that it is a function of h, and the function is not defined when Find the limit of the difference quotient as h approaches 0. h 0 lim hS0 2 lim hS0 10h h2 h 10 h h 10 h h 1 lim hS0 lim hS0 2 1 10 0 10 2 Factor the numerator. Limit Theorem limit of a polynomial function ■ Exercises 14.2 In Exercises 1 – 8, use the following facts about the functions f, g, and h to find the required limit. h(x) 2 f(x) 5 lim xS4 g(x) 0 lim xS4 lim xS4 1. lim xS4 1 f g x 2 21 3. lim xS4 f g 1 x 2 2. lim xS4 1 g h x 2 21 4. lim xS4 g h 1 x 2 5. lim xS4 f g x 2 21 1 6. lim xS4 3 7. lim xS4 3h 2f gb 1 x 2 a 8. lim xS4 a 4h b 1 2 h x 2 4 1 f 2g 20. lim xS2 22. lim xS0 In Exercises 24–27, find 21. lim xS0 22 x 22 x 23. 0 lim xS3 x 3 x 3 0 f(2 h) f(2) h . lim hS0 x 2 24. 26. f f x x 1 1 2 2 x2 x2 x 1 25. 27. f f x x 1 1 2 2 x3 2x In Exercises 9–23, find the limit, if it exists. If the limit does not exist, explain why. 9. lim xS2 1 6x3 2x2 5x 3 2 10. lim xS11 x7 2x5 x4 3x 4 2 11. lim xS2 3x 1 2x 3 12. lim xS3 x2 x 1 x2 2x 13. lim xS1 2x3 6x2 2x 5 14. lim xS2 2x2 x 3 15. ° lim xS0 ¢ 16. ° lim xS0 17. lim xS1 a 1 x 1 2 x2 1b 18. lim xS2 c x2 x 2 2x x 2 d 19. lim xS0 x2 x 0 0 In Exercises 28–29, use a unit circle diagram to explain why the given statement is true. 28. sin t 1 lim tSp 2 29. cos t 0 lim tSp 2 Exercises 30–34 involve the greatest integer function, f(x) [x], which is defined to be the greatest integer that is
less than or equal to a given number x. See Section 3.1. Use a calculator as an aid in analyzing these problems. 30. For x h 1 exists. 2 31. For x g 1 exists. 2 x 3 4 x 4 3 , find lim xS2 h 1 x 2 , if the limit x x 3 4 , find lim xS2 g 1 x 2 , if the limit 3 x 4 2 x x 3 4 , find lim xS3 r 1 x 2 , if the limit 32. For r x 1 exists. 924 Chapter 14 Limits and Continuity 33. For k x 1 exists. 2 x x 3 4 x 3 4 , find lim xS1 k 1 x 2 , if the limit 34. For f x 1 2 x 0 x 0 , find lim hS0 . 36. Critical Thinking Give an example of functions f x and g and a number c such that neither f lim xSc 1 2 nor lim xSc g 1 x 2 exists, but lim xSc f g x 2 21 1 does exist. 35. Critical Thinking Give an example of functions f x and g and a number c such that neither f nor lim xSc g 1 x 2 exists, but lim xSc 1 f g x 2 2 1 lim 2 xSc does exist. 1 14.2.A Excursion: One-Sided Limits Objectives • Find one-sided limits The function whose graph is shown in Figure 14.2.A-1 is defined for all values of x except x 4. y 4 2 −2 0 2 −2 −4 Figure 14.2.A-1 x 6 As x approaches 4 from the right, that is, takes values larger than but close f to 4, the graph shows that the corresponding values of get very close 1 as x approaches 4 from the right to 2. Consequently, “the limit of is 2.” x x f 2 1 2 xS4 f lim x 1 2 2 x 7 4 The small plus sign on 4 indicates that only the values of x with are considered. Similarly, as x approaches 4 from the left, the graph Conshows that the corresponding values of sequently, “the limit of get very close to as x approaches 4 from the left is 1. 1. x x f f 2 1 1 2 lim xS4 f x 1 2 1 The small minus sign indicates that only values of x with considered. x 6 4 are Section 14.2.A Excursion: One-Sided Limits 925 These “one-sided” limits are a bit different than the “two sided” limits discussed in Section 14.2. When x approaches 4 from the left and from the right, the corresponding values of do not approach a single number, so the limit as x approaches 4, as defined in Section 14.1, does not exist. x f 2 1 The same notation and terminology are used in the general case, where f is any function and c and L are real numbers. The definition of the “righthand limit,” xSc f lim x 2 1 L, ” in place of the phrase “on both sides of is obtained by inserting “ c” in the definition of limit in Section 14.2. The function f need not be defined when x 6 c. x 7 c Similarly, inserting “ definition produces the definition of the “left-hand limit.” ” in place of “on both sides of c” in the same x 6 c Again, the function f need not be defined when x 7 c. f lim xSc x 2 1 L Example 1 One-Sided Limit Find xS3 lim 1x 3 1 A B . Solution f 2x 3 1 The function x 3 and values of x to the right of 3. The graph of ure 14.2.A-2a, and a table of values is shown in 14.2.A-2b. is defined only when x x f 2 1 2 1 x 3, that is, for is shown in Fig- 2.2 2.9 0 4 Figure 14.2.A-2a Figure 14.2.A-2b The values of approach 1 as x approaches 3 from the right. There- 2 x f 1 1x 3 1 A fore, xS3 lim 1. B ■ Computing One-Sided Limits The computation of one-sided limits is greatly facilitated by the following fact. 926 Chapter 14 Limits and Continuity All the results about limits in Section 14.2, such as the properties of limits, limits of polynomial functions, and the Limit Theorem, x S c. ” remain valid if “ ” is replaced by either “ x S c x S c ” or “ Example 2 Using Properties of Limits Find xS319 x2. lim Solution 29 x2 1 2 f x The function because the quantity under the radical is negative for other values of x. Compute as x approaches 3 from the left by using the properties the limit of of limits: is defined only when x f 2 1 3 x 3, lim xS3 9 x2 29 x2 2 lim 2 1 xS3 9 lim 2 lim xS3 xS3 2 lim xS3 9 lim xS3 x2 x lim xS3 x B BA A property 5 property 2 property 3 29 32 0 limit of a constant limit of the identity function ■ Three Types of Limits Notice that there are three kinds of limits: left-hand limits, right-hand limits, and “two-sided” limits as defined in Section 14.1. Example 1 exhibits a function that has a right-hand limit at but no left-hand or two sided limit. Even when a function has both a left-hand and a right-hand limit at these limits may not be the same, as was shown in the introduction of this Excursion. x 3, x c, f x x c It is clear, however, that a function which has a two-sided limit L at x c. necessarily has L as both its left-hand and right-hand limit at If the can be made arbitrarily close to L by taking x close enough values of to c on both sides of c, then the same thing is true if you take only values of x on the left of c or on the right of c. Conversely, if a function has the same left-hand and right-hand limits at then this number must be a two-sided limit as well. In summary, x c, 1 2 Two-sided Limits Let f be a function and let c and L be real numbers. Then f(x) L lim xSc exactly when xSc f(x) L lim and lim xSc f(x) L. Section 14.2.A Excursion: One-Sided Limits 927 Example 3 Limits From the following graph, find a. d. lim xS2 f x 1 2 lim xS6 f x 1 2 b. e. lim xS2 f x 1 2 lim xS6 f x 1 2 c. lim xS2 2 − Figure 14.2.A-3 Solution The graph shows that a. d. lim xS2 f x 1 2 lim xS6 f x 2 1 3 5 b. e. lim xS2 f x 1 2 lim xS6 f x 1 2 3 4 c. lim xS2 f x 1 2 3 ■ Exercises 14.2.A In Exercises 1–6, use a calculator to find a reasonable estimate of the limit. 7. 1. lim xSp sin x 1 cos x 3. xS0 2x lim 1 ln x 2 5. xS0 lim sin 6x x 2. lim 1 xSp 2 sec x tan x 2 4. lim xS0 a x 1 xb 6. lim xS0 sin 3x 1 sin 4x In Exercises 7–10, use the graph of the function f to determine the required limits. a. c. xS2 f lim 1 x xS3 f lim 1 x 2 2 b. d. xS0 f lim xS3 f lim 1 −1 −2 −3 −4 −4 −3 −2 x 1 2 3 4 928 8. 9. 10. Chapter 14 Limits and Continuity y y y 4 3 2 1 0 −1 −1 −2 −3 −4 4 3 2 1 0 −1 −1 −2 −3 −4 4 3 2 1 0 −1 −1 −2 −3 −4 −4 −3 −2 −4 −3 −2 −4 −3 −. a. b. xS2 f lim 1 x xS2 f lim 11. In Exercise 33a of Section 14.1, find xS2 f lim 1 x xS2 f lim 12. In Exercise 34a of Section 14.1, find xS1 g lim 1 x xS1 g lim xS1 g lim 1 x xS1 g lim d. d. b. c. a In Exercises 13–22, find the limit. 13. xS1 lim A 2x 1 3 B 15. xS4 lim x 4 x2 16 17. xS3 lim 3 x2 9 19. 20. 25 2x x B 2x 3 23x lim xS2.5A xS3 lim A 14. 16. xS3 23 x lim x 2 x 2 xS2 0 lim 0 18. xS2 lim x 1 x2 x 2 B 21. xS3 lim 0 a x 3 x 3 0 2x 3 1 b 22. xS4 lim 2 0 A x 0 4 x2 B Exercises 23 and 24 involve the greatest integer function, See Exercise 30 in Section 14.2. f(x) [x]. 23. Find xS2 lim 24. Find lim xS3 3 1 x 4 x xS2 and lim x . 4 3 x 3 4 2 and lim xS3 1 x x . 4 2 3 Section 14.3 The Formal Definition of Limit 929 14.3 The Formal Definition of Limit (Optional) Objectives • Use the formal definition of limit The informal definition of limit in Section 14.1 is quite adequate for understanding the basic properties of limits and for calculating the limits of many familiar functions. This definition, or one very much like it, was used for more than a century and played a crucial role in the development of calculus. Nevertheless, the informal definition is not entirely satisfactory; it is based on ideas that have been illustrated by examples but not precisely defined. Mathematical intuition, as exemplified in the informal definition of limit, is a valuable guide; but it is not infallible. On several occasions in the history of mathematics, what first seemed intuitively plausible has turned out to be false. In the long run, the only way to guarantee the accuracy of mathematical results is to base them on rigorously precise definitions and theorems. This section takes the first step in building this rigorous foundation by developing a formal definition of limit. In order to keep the discussion as concrete as possible, suppose f is a funcYou do not need to know the rule of f or tion such that 12. f x lim xS5 1 2 anything else about it to understand the following discussion. The informal definition of the statement in Section 14.1 has two components: –lim xS5 f x 1 2 12– that was given A. As x takes values very close to but not equal to 5, the f get very close—and possibly are x corresponding values of equal—to 12. B. The value of x f 1 2 1 2 want) to 12 by taking x sufficiently close (but not equal) to 5. can be made arbitrarily close (as close as you In a sense, Component A is unnecessary because it is included in Component B: If the values of can be made arbitrarily close to 12 by taking x close enough to 5, then presumably can be made to get very close 2 to 12 by taking values of x very close to 5. x x f f 1 1 2 Consequently, to obtain the formal definition of limit, begin with Component B of the informal definition. f(x) 12 means that lim xS5 as close as you want to 12 by taking x close enough to 5. the value of f(x) can be made [1] The definition above, referred to as Definition [1], will be modified step by step until the formal definition of limit is reached. Definition [1] says, in effect, that there is a two-step process involved in finding a limit, if it exists: 1. Know how close you want 2. Determine how close x must be to 5 to guarantee this. to be to 12, x f 1 2 Definition [1] can now be restated as Definition [2]. 930 Chapter 14 Limits and Continuity y 12.01 11.99 5 11.99 6 f x is in this interval when x 1 Figure 14.3-1 6 12.01 2 y ε ε 12 δ δ 5 Figure 14.3-2 x x f(x) 12 means that lim xS5 f(x) should be to 12, you know how close x must be to 5 to guarantee it. whenever you specify how close [2] For example, if you want to be within 0.01 of 12, that is, between 11.99 and 12.01, you can find how close x must be to 5 to guarantee that x f 1 2 2 1 Any value of x in the interval around 5 shown in Figure 14.3-1 will produce 11.99 6 f 6 12.01. x 11.99 6 f x 6 12.01. 1 2 But “arbitrarily close” implies much more. You must be able to find how close x is to 5 regardless of how close you want to be to 12. If you want to be within 0.002 of 12, or 0.0001 of 12, or within any distance of 12, you must be able to find how close x must be to 5 in each case to accomplish this. So, Definition [2] can be restated as follows(x) 12 means that
no matter what positive number lim xS5 you specify in measuring how close you want f(x) to be to 12, you must be able to find how close x must be to 5 in order to guarantee that f(x) is that close to 12. [3] 2 1 e x d should be to 12 will be denoted by the Greek letter Hereafter, the small positive number you specify in measuring how close f (epsilon). When you know how close x should be to 5 to accomplish this, denote the (delta). Presumably the number which number by the Greek letter measures how close x must be to 5, will depend on the number which x measures how close you want to be to 12. Using this language, Def1 inition [3] becomes Definition [4]. f(x) 12 means that lim xS5 there is a positive number this property: If x is within of 5 but not equal to 5, then f(x) is within of 12, and possibly equal to 12. for every positive number that depends on with d, e, D E [4] D E E f 2 Although Definition [4] is essentially the formal definition, somewhat briefer notation is usually used. If you think of ment f x 1 2 means that and 12 as numbers on the number line, then the state- is within of 12 e f x 1 2 the distance from f x to 12 is less than e. 2 1 Because distance on the number line is measured by absolute value (see Sections 2.4 and 2.5.A), the last statement can be written as f x 1 2 0 12 0 6 e. Similarly, saying that x is within of 5 and not equal to 5 means that the distance from x to 5 is less than but greater than 0, that is 0 6 d d x 5 6 d. 0 0 e, NOTE d, Epsilon, and are Greek letters delta, that are often used to represent small amounts. Section 14.3 The Formal Definition of Limit 931 Using this notation, Definition [4] becomes the desired formal definition. f(x) 12 means that lim xS5 there is a number x 5 if 0 66 D 66 D, then 00 00 00 for each positive number E, 66 E. that depends on f(x) 12 such that 00 E, [5] Definition [5] is sufficiently rigorous because the imprecise terms “arbitrarily close” and “close enough” in the informal definition have been replaced by a precise statement using inequalities that can be verified in specific cases, as will be shown in the examples that follow. There is nothing special about 5 and 12 in the preceding discussion; the entire analysis applies equally well in the general case and leads to the formal definition of limit, which is just Definition [5] with c in place of 5, L in place of 12, and f for any function. Definition of Limit Let f be a function and let c be a real number such that f (x) is defined for all x, except possibly in some open interval containing c. The limit of f (x) as x approaches c is L, which is written x c, f(x) L, lim xSc provided that for each positive number number E that depends on with the property E, there is a positive D if 0 66 x c 00 00 66 D, then f(x) L 66 E. 00 00 This definition is often called the e-d definition of limits. Example 1 Proving a Limit Let f x 1 2 4x 8 and prove that lim xS5 f x 1 2 12. Solution c 5, L 12, Apply the definition of limit with e ure 14.3-3 illustrates the situation. Suppose that d Find a positive number with the property 6 d, then x 5 If 0 6 f f 4x 8. and Figis any positive number. x 2 1 x 1 12 0 2 6 e. 0 Let be the number d , and show that this will work. For now, do not d was found; just verify that the following argu- 0 0 e 4 d e 4 y 12 + ε 12 12 − ε x 5 5 + δ 5 − δ worry about how ment is valid. Figure 14.3-3 932 Chapter 14 Limits and Continuity If 0 x 5 0 6 d, then 4x 20 6 e 0 12 6 e 0 12 6 e 0 e 7 0, 6 d, then 12 lim xS5 1 2 0 1 4x 8 2 x f Multiply both sides by 4. 4 a 0 0 b ab 4 0 0 Rewrite 20 as 8 12. 4x 8 This verifies that for each x 5 If 0 6 This completes the proof that d there is a positive with the property 12 6 e. x f 0 1 2 0 ■ Proofs like the one in Example 1 often seem mysterious to beginners. d Although they can follow the argument after has been found, they do d . Example 2 gives a fuller picture of the processes not see how to find used in proving statements about limits. Example 2 proves that for any positive number d there exists such that x f ˛1 2 0 9 0 6 e whenever 0 6 0 6 d. 0 e, x 1 Example 2 Use the E-D Definition of Limit Prove that lim xS1 1 2x 7 9. 2 Solution and L 9. Let be a positive number e 2x 7, c 1, x f˛ 1 In this case, 2 d and find a with the property If 0 6 x 1 0 0 6 d, then 2x 7 9 0 2 0 1 6 e. In order to get some idea which might have this property, work backwards from the desired conclusion, namely, d The last statement is equivalent to 2x 7 9 0 2 0 1 6 e. 2x 2 6 e, 0 0 which in turn is equivalent to each of the following statements Section 14.3 The Formal Definition of Limit 933 When the conclusion is written this way, it suggests that the number e 2 would be a good choice for d. Everything up to here has been “scratch work.” Now give the actual proof, written forwards. Given a positive number e, let be the number d e 2 If 0 6 . x 1 0 0 6 d, then 2x 2x 7 f ˛ Multiply both sides by 2. 2 a 0 0 b ab 2 0 0 0 0 0 0 Rewrite 2 as 7 9. 2x 7 f ˛1 x 2 Figure 14.3-4 Therefore, d e 2 has the required property, and the proof is complete. ■ Proving Limit Properties Once the algebraic scratch work was done in Examples 1 and 2, the limit proofs were relatively easy. In most cases, however, a more involved argument is required. In fact, it can be quite difficult to prove directly from the definition, for example, that lim xS3 such complicated calculations can often be avoided by using the various limit properties given in Section 14.2. Of course, these properties must first be proved using the definition. Surprisingly, the proofs are comparatively easy. Fortunately, . x2 4x 1 x3 2x2 x 1 3 Example 3 Proving Limit Properties Let f and g be functions such that Prove that lim xSc x f ˛1 2 1 2 x f˛ 1 lim xSc g˛1 x 22 L and lim xSc L M. M. x g˛1 2 Solution Scratch Work: If erty e is any positive number, find a positive with the prop- d If 0 6 x c 0 0 6 d, then x f˛ 1 2 0 1 g˛1 x 22 1 L M 2 0 6 e. 934 Chapter 14 Limits and Continuity Note the following result of the triangle inequality. (See Section 2.4.) x f ˛1 2 0 1 g˛1 x 22 1 L M 2 0 x f˛ 1 0 1 x f1 1 g˛ If there is a with the property d If 0 6 x c 0 0 6 d, then x 2 f ˛1 0 g ˛1 x d then the smaller quantity 6 d. e x c when 0 6 x f ˛1 0 1 1 Such a can be found as follows. 2 0 22 2 0 0 L g ˛1 0 0 L M x M 6 e, 0 2 will also be less than Proof Let be any positive number. Because e inition of limit with e 2 in place of e: lim xSc f ˛1 x 2 L, apply the def- there is a positive number d1 with the property 6 d1, then 0 6 e 2 If 0 6 x c L f ˛1 x . 2 0 Similarly, because lim xSc 0 g˛1 x 2 0 M, there is a positive number with the property d2 6 d2, then 1 If 0 6 x c 0 0 d Then Now let be the smaller of the two numbers d d2. x c x c and it must be true that x c and 0 6 if 0 6 0 6 6 d, 6 d1 d1 0 0 0 0 6 d2. 0 d2, so that d d1 and Therefore, Consequently, if L 6 e 2 6 d˛ 1 2 0 6 g˛1 x 0 22 2 0 x f ˛1 2 0 1 0 1 and x g ˛˛1 1 x g˛1 2 M 2 M 0 2 0 0 then x f ˛1 0 1 x f ˛1 0 2 6 e e 2 2 e It has been shown that for any x c If 0 6 0 lim xSc f ˛1 1 6 d, then g˛1 L M 22 x 0 1 . 0 x 2 Therefore, e 7 0, there is a g˛1 f ˛1 x 2 d 7 0 x 22 1 with the property: L M 6 e. 2 0 ■ The proofs of the other limit properties and theorems of Section 14.2 are introduced in calculus. Section 14.3 The Formal Definition of Limit 935 One-Sided Limits The formal definition of limit may easily be carried over to one-sided limits, as defined informally in Excursion 14.2.A, by using the following fact: x c 6 d exactly when d 6 x c 6 d, 0 which is equivalent to 0 x c 6 d exactly when c d 6 x 6 c d. 0 0 Thus, the numbers between c and tance of c, and the numbers between distance of c. d d c d lie to the right of c, within disand c lie to the left of c, within c d Consequently, the formal definition of right-hand limits can be obtained 6 d from the definition above by replacing the phrase “if ” c 6 x 6 c d. ” For a formal definition of the left-hand limits, with “ 0 6 replace the phrase “if c d 6 x 6 c. ” with “if Exercises 14.3 In Exercises 1–12, use the formal definition of limit to prove the given statement, as in Example 1. 1. 2. 3. 4. 5. 6. 7. 8. 9. lim xS1 lim xS5 lim xS0 lim xS2 lim xS7 lim xS1 lim xS2 lim xS4 10. lim xS1 3x 2 lim xS3 1 4x 6x 3 1 1 2 15 2 5 2 2x 19 1 4 4 p p x 6 2 2x 7 1 1 2 5 2 11. 12. lim xS2 lim xS3 1 1 2x 5 2x 4 2 2 1 10 In Exercises 13 and 14, use the formal definition of limit to prove the statement. 13. 14. lim xS0 lim xS0 x2 0 x3 0 In Exercises 15 and 16, let f and g be functions such that lim xSc f˛(x) L and lim xSc g˛(x) M. 15. Critical Thinking Prove that L M. g˛1 lim xSc f ˛1 22 x x 1 2 Hint: see Example 3. 16. Critical Thinking If k is a constant, prove that lim xSc k f x 2 1 kL. 936 Chapter 14 Limits and Continuity 14.4 Continuity Objectives • Determine if a function is continuous at a point • Determine if a function is continuous on an interval • Apply properties of continuous functions Calculus deals in large part with continuous functions, and the properties of continuous functions are essential for understanding many of the key theorems in calculus. This section presents the intuitive idea of continuity, its formal definition, and the various properties of continuity—which were used in the work with graphs in earlier chapters. Continuity Informally Let c be a real number in the domain of a function f. Informally, the function f is continuous at if you can draw the graph of f at and near c, f ˛1 the point without lifting your pencil from the paper. For example, 1 each of the four graphs in Figure 14.4-1 is the graph of a function that is continuous at x c. x c 22 c (c, f(c)) c y y x (c, f(c)) x c y y (c, f(c)) c (c, f(c)) c x x Figure 14.4-1 Thus, a function is continuous at is connected and unbroken. x c if its graph around the point 1 c, f ˛1 c 22 On the other hand, none of the functions whose graphs are shown in Figure 14.4-2 is continuous at Try to draw one of these graphs at c and near without lifting your pencil from the paper. x
c. c, f ˛1 1 22 (c, f(c)) y c a. y c. Section 14.4 Continuity 937 (c, f(c)) c y b. y x x x x c c Figure 14.4-2 d. Figure 14.4-2 shows that a function is discontinuous, that is, not continx c. uous, at if the graph has a break, gap, hole, or jump when x c Calculators and Discontinuity When a calculator uses “connected” mode to graph a function, it plots a number of points and then connects them with line segments to produce a curve. Thus, the calculator assumes that the function is continuous at any point it plots. For example, a calculator may not show the hole in graph d of Figure 14.4-2, or it may insert a vertical line segment where the graph jumps in graph b of Figure 14.4-2. Consequently, a calculator may present misleading information about the continuity of a function. Analytic Description of Continuity The goal is to find a mathematical description of continuity at a point that does not depend on having the graph given in advance. This is done by expressing in analytic terms the intuitive geometric idea of continuity given above. 938 Chapter 14 Limits and Continuity If the graph of f can be drawn at and near a point pencil from paper, then there are two possibilities: c, f ˛1 1 c 22 without lifting (c, f(c)) is an interior point of the graph. (c, f(c)) (c, f(c)) (c, f(c)) is an endpoint of the graph. Figure 14.4-3 Continuity at an Interior Point c c, f˛ 1 If the point 1 c, f ˛1 drawn around the point 1 the very least, the following two statments are true. is an interior point of the graph, and the graph can be without lifting pencil from paper, then at 22 22 c x c x t, 2 x x • • must be defined for must be defined for f ˛1 f ˛1 2 f ˛1 is not defined for some t near c, there will be a hole in the graph For if at the point which would require lifting the pencil when drawing the graph. See graphs c and d of Figure 14.4-2 for functions that are not The situation can be described more precisely by saying: defined at when t is any number near c 1 1 x c. t, f 22 t t , 2 f(x) is defined for all x in some open interval containing c. In other words, there are numbers a and b with such that f(x) is defined for all x with a 66 x 66 b. In particular, f(c) is defined. a 66 c 66 b [1] x c, Although condition [1] is necessary in order for f to be continuous at x c this condition by itself does not guarantee continuity. For instance, and the graphs a and b in Figure 14.4-2 show functions whose graphs are defined for all values of x near c, but are not continuous at x c. is defined, there are two conditions that can prevent a function from c f ˛1 If being continuous at 2 x c: • There is a jump at does not exist. x c, that is, the limit of x f ˛1 2 as x approaches c • There is a hole in the graph at c f ˛1 but it is not equal to x c, 2 x c, . that is, limit of x f˛ 1 2 exists at The conditions that prevent a function from being continuous at a point are shown in Figure 14.4-4. Notice the reason that the graph is not continuous at and at x c. x b, x a, at Section 14.4 Continuity 939 y lim f(x) does not exist. x→b f(a) is not defined. lim f(x) ≠ f(c) x→c x ba c Figure 14.4-4 The preceding analysis leads to the following formal definition of continuity for interior points. Definition of Continuity Let f be a function that is defined for all x in some open interval containing c. Then f is said to be continuous at under the following conditions: x c 1. f(c) is defined 2. lim xSc 3. lim xSc exists f˛(x) f˛(x) f˛(c) Example 1 Continuity at a Point Without graphing, show that the function 2x2 x 1 x 5 x f ˛1 2 is continuous at x 3. Solution To show continuity of f at x 3, 3 f˛ 1 3 f ˛1 2 2 show that 2x2 x 1 x 5 . lim xS3 213 8 940 Chapter 14 Limits and Continuity By the properties of limits given in Section 14.2, lim xS3 f x 2 1 lim xS3 2x2 x 1 x 5 2x2 x 1 x 5 2 1 x2 lim xS3 lim xS3 2 lim xS3 lim xS3 3 2 1 1 213 8 3 f˛ 1 2 x 2 limit of a quotient limit of a root 1 2 limit of a polynomial Therefore, lim xS3 f ˛1 and f is continuous at x 3. ■ The facts about limits presented in Sections 14.1 and 14.2 and the definition of continuity provide justification for several assumptions about graphs that were made earlier in this book. Every polynomial function is continuous at every real number. Every rational function is continuous at every real number in its domain. Every exponential function is continuous at every real number. Every logarithmic function is continuous at every positive real number. f ˛(x) sin x number. h(x) tan x and g ˛(x) cos x are continuous at every real is continuous at every real number in its domain. Continuity on an Interval Consider continuity at an endpoint of the graph of a function f, such as a, f shown in Figure 14.4-5. b, f or b a 1 1 22 1 1 22 Continuity of Special Functions NOTE One-sided limits, which were discussed in Section 14.2.A, are a prerequisite for the material on continuity on an interval. Section 14.4 Continuity 941 (b, f(b)) (a, f(a)) Figure 14.4-5 x a is that the graph of f can be The intuitive idea of continuity at drawn at without lifting the pencil from the paper. Essentially the same analysis that was given above can be made here if we consider only values of x to the right of In An short, continuity at the endpoint and to the right of the point x a. a . a, f a, f a, f 22 22 22 means that b, f b , 1 22 f xSa f lim which leads to the 2 2 1 1 analogous discussion applies to the endpoint formal definition. 1 Continuity from the Left and Right A function f is continuous from the right at x a provided that xSa f(x) f(a). lim A function f is continuous from the left at x b provided that xSb f˛(x) f(b). lim Example 2 Continuity at an Endpoint Show that 2x x f ˛1 2 is continuous from the right at x 0. Solution x The function 2 from the right at f ˛1 2x, x 0 which is not defined when and because 20 0, f 0 2 1 xS0 1x 0 f lim xS0 f lim x 1 2 x 6 0, is continuous 0 . 2 1 ■ The most useful functions are those that are continuous at every point in an interval. Consider the following three examples. • • • h 1 g ˛1 f ˛1 x tan x 2 is continuous at every number in the interval ln x sin x is continuous at every number in 1 is continuous at every number in 0, q 2 q, q 1 2 942 Chapter 14 Limits and Continuity Intuitively, this means that their graphs can be drawn over the entire interval without lifting the pencil from the paper. Most of the functions in this book are of this type. Continuity on an Interval A function f is said to be continuous on an open interval (a, b) provided that f is continuous at every value in the interval. A function f is said to be continuous on a closed interval [a, b] provided that f is continuous from the right at x b, continuous from the left at value in the open interval (a, b). and continuous at every x a, Analogous definitions may be given for continuity on intervals of the form a, b q, q q, b q, b a, q a, q and a Example 3 Continuity of a Function Discuss the continuity of the function shown in Figure 14.4-6. y 2 −4 −2 0 2 4 6 x −2 Figure 14.4-6 Solution x 3 The function is discontinuous at 3, 2 5, 3 on each of the intervals , and , 2, q x 2, . 1 2 3 4 1 2 but it is continuous ■ Properties of Continuous Functions Using the definition is not always the most convenient way to show that a particular function is continuous. It is often easier to establish continuity by using the following facts. Section 14.4 Continuity 943 Properties of Continuous Functions If the functions f and g are continuous at the following functions is also continuous at x c: x c, then each of 1. the sum function f g 2. the difference function f g 3. the product function fg f g, 4. the quotient function g (c) 0 Proof By the definition of the sum function, Because f and g are continuous at Therefore, by the first property of limits, lim xSc x c 2 1 2 f lim xSc 1 f g x 2 21 x c, and lim xSc lim xSc c 22 g 2 x 2 1 f x lim xSc lim xSc This says that f g 1 is continuous at The remaining statements are proved similarly, using limit properties 2, 3, and 4. Example 4 Continuity of Functions x f 1 sin x Assume that and Prove that the following functions are continuous at x3 5x 2 x3 5x 2 sin x x3 5x 2 g 1 b. a. x 2 2 1 x3 5x 2 2 d. 2 sin x 1 sin x x3 5x 2 x 0. 2 are continuous at x 0. c. sin x 1 2 1 Solution Because f an g are continuous at are continuous at , each of the following functions by the listed property of continuous functions. x 0 1 2 x 0 x3 5x 2 x3 5x 2 1 x3 5x 2 f gb a 1 2 1 1 2 fg 1 x 2 sin x sin x sin x 1 1 2 sin x x3 5x 2 a. b. c. d sum of continuous functions difference of continuous functions product of continuous functions quotient of continuous functions ■ 944 Chapter 14 Limits and Continuity Composite Functions Composition of functions is often used to construct new functions from given ones. Continuity of Composite Functions If the function f is continuous at continuous at continuous at x c. x f(c), x c and the function g is then the composite function g f is Example 5 Continuity of Composite Functions 2x3 3x2 x 7 is continuous at x 2. Show that h x 2 1 Solution The polynomial function 23 3 and at 5 because by limit property 5 x f ˛1 2 2 7 5. 2 2 f 2 2 1 2 1 x3 3x2 x 7 g˛1 The function is continuous at x x 2 is continuous 2x 2 lim xS5 By the box above, with which is given by g g f 21 is also continuous at x x f 1 1 1 2 g 1 22 x 2. 2x 2lim xS5 c 2 f and x 25 g c 5, 5 . 2 1 1 2 the composite function x3 3x2 x 7 2 2x3 3x2 x 7 g f, ■ The Intermediate Value Theorem This section’s introduction to continuity will close by mentioning, without proof, a very important property of continuous functions. The Intermediate Value Theorem If the function f is continuous on the closed interval [a, b] and k is any number between f(a) and f(b), then there exists at least one number c between a and b such that f(c) k. The truth of the Intermediate Value Theorem can be understood geometrically by remembering that since f is continuous on the graph 4 b, f of f can be drawn from the point without lift1 ing the pe
ncil from the paper. As suggested in Figure 14.4-7, there is no way that this can be done unless the graph crosses the horizontal line y k, to the point a, b 3 b 1 6 k 6 f where f a, f 22 22 Section 14.4 Continuity 945 (b, f(b)) (c, f(c)) = (c, k) y f(b) k f(a) (a, f(a)) x a c b Figure 14.4-7 The first coordinate of the point where the graph crosses this line is some number c between a and b, and its second coordinate is because the point is on the graph of f. But c, f is also on the line so its sec k. ond coordinate must be k; that is, 22 c 1 The Intermediate Value Theorem further explains why the graph of a continuous function is connected and unbroken. If the function f is continuous on the interval to b a the point f and . a then its graph cannot go from the point 22 without moving through all the y values between f f c 2 1 y k, c 1 f a, b 3 22 b, f a The graphical method of solving equations that has been used throughout this book is based on the Intermediate Value Theorem, as are some root-finding features on calculators. If f is continuous on the interval a, b 4 3 f a have opposite signs, then 0 is a number between and 2 1 k 0, and there is at least one number c between a and b such that In other is a solution of the equation words, and Consequently, by the Intermediate Value Theorem, with 2 2 x c 0. 0. f So when a calculator shows that the graph of a continuous function f has points above and below the x-axis, there really is an x-intercept between x these points, that is, a solution of Zoom-in uses this fact by look2 ing at smaller and smaller viewing windows that contain points of the graph on both sides of the x-axis. The closer together the points are horizontally, the better the approximation of the x-intercept, or solution, that can be read from the graph. 0. f 1 946 Chapter 14 Limits and Continuity Exercises 14.4 In Exercises 1 and 2, use the graph to find all the numbers at which the function is not continuous. 4. y 1. y 2. x x 1 3 y 1 1 In Exercises 3 – 6, determine whether the function x 0, whose graph is given is continuous at and at x 2, x 3. at 3. y −1 1 3 x −1 1 3 5. y −1 1 3 6. y −1 1 3 x x x In Exercises 7–12, use the definition of continuity and the properties of limits to show that the function is continuous at the given number. x2 2x2 5x 4 , x 1 2 7. 8. 9. x x f 1 g˛1 f x 1 2 2 2 10. h x 2 1 11. f x 1 2 12. k x 1 2 1 x2 5 x 2 7, x 3 2 1 x2 3x 10 2 1 x2 9 1 x2 x 6 2 1 x 3 x2 x 1 1 x2x x 6 1 28 x2 2x2 5 2 2, x 36 , x 2 , x 2 x2 6x 9 2 x2 1 2 1 , x 2 2 In Exercises 13–18, explain why the function is not continuous at the given number. 3, x 3 13. f x 1 2 14 x2 4 x2 x 2 , x 2 15. f x 1 2 x2 4x 3 x2 x 2 , x 1 16. x g˛1 2 sin • 1 p x if x 0 if x 0 , x 0 17. f x 1 2 x2 if x 0 if x 0 1 e , x 0 18. f x 1 2 22 x 22 x , x 0 In Exercises 19–24, determine whether or not the function is continuous at the given number. 19. f x 1 2 2x 4 if x 2 2x 4 if x 7 2 e , x 2 20. x g˛1 2 2x 5 if x 6 1 2x 1 if x 1 e , x 1 21. f x 1 2 x2 x if x 0 2x2 if x 7 0 e , x 0 22. g x 2 1 x3 x 1 if x 6 2 3x2 2x 1 if x 2 e , x 2 Section 14.4 Continuity 947 26. x g˛1 2 µ if x 2 x2 x 6 x2 4 if x 2 4 5 27. f x 2 1 • x2 1 if x 6 0 x if 0 6 x 2 2x 3 if x 7 2 28. h x 2 1 • if x 6 1 and x 0 1 x x2 if x 1 29. Critical Thinking For what values of b is the following function continuous at x 3? f x 2 1 bx 4 if x 3 bx2 2 if x 7 3 e 30. Critical Thinking Show that continuous at x 0. f x 2 1 2 0 x 0 is x c A function f that is not defined at is said to have x c if there is a funca removable discontinuity at x c, tion g such that g(c) is defined, g is continuous at and In Exercises 31–34, show that the function f has a removable discontinuity by finding an appropriate function g. g ˛(x) f˛(x) x c. for 31. x f˛ 1 2 x 1 x2 1 32. f x 1 2 33. f x 1 2 x2 x 0 0 2 2x 4 x x 3 , x 3 23. 24, x 2 0 34. f x 1 2 sin x x Hint: See Example 1 of Section 14.1. In Exercises 25–28, determine all numbers at which the function is continuous. 35. Show that the function 25. x f˛ 1 2 µ if x 1 discontinuity at x 0 x2 x 2 x2 4x 3 3 2 if x 1 x 0 x 0x f˛ 1 that is not removable. has a 2 948 Chapter 14 Limits and Continuity 14.5 Limits Involving Infinity Objectives In the discussion that follows, it is important to remember that • Define limits involving infinity • Use properties of limits at infinity • Use the Limit Theorem NOTE Excursion 14.2.A is a prerequisite for some of the material that follows. q, There is no real number called “infinity,” and the symbol usually read “infinity,” does not represent any real number. which is q Nevertheless, the word “infinity” and the symbol are often used as a convenient shorthand to describe the way some functions behave under certain circumstances. Generally speaking, “infinity” indicates a situation in which some numerical quantity gets larger and larger without bound, meaning that it can be made larger than any given number. Similarly, “negative infinity,” indicates a situation in which a numerical quantity gets smaller and smaller without bound, meaning that it can be made smaller than any given negative number. q, In Section 14.1, several ways were discussed in which a function might fail to have a limit as x approaches a number c. The word “infinity” is often used to describe one such situation. Consider the function f whose graph is shown in Figure 14.5-1. y 1 3 5 x Figure 14.5-1 The graph shows that as x approaches 3 from the left or right, the corresponding values of do not get closer and closer to a particular number. Instead, they become larger and larger without bound. Although there is no limit as defined in Section 14.1, it is convenient to describe this situation symbolically by writing x f 2 1 q, x f˛ 1 as x approaches 3 is infinity.” Similarly, f which is read “the limit of does not have a limit as x approaches 1 from the left or right, because the f˛ 1 corresponding values of get smaller and smaller without bound. We x as x approaches 1 is negative infinity” and write say that “the limit of lim xS3 x f˛ 1 x 2 2 2 f˛ 1 2 lim xS1 f˛ 1 x 2 q. Section 14.5 Limits Involving Infinity 949 x 5, Near small on the right side of 5, so we write the values of f˛ 1 x 2 get very large on the left side of 5 and very xS5 f lim x 1 2 q and xS5 f lim x 1 2 q, which are read “The limit as x approaches 5 from the left is infinity” and “the limit as x approaches 5 from the right is negative infinity.” There are many cases like the ones illustrated above in which the language of limits and the word “infinity” can be useful for describing the behavior of a function that actually does not have a limit in the sense of Section 14.1. Example 1 Infinite Limits Describe the behavior of f x 1 2 5 x4 near x 0. Solution The graph of f is shown in Figure 14.5-2. The trace feature indicates that get small without bound as x approaches 0 from the the values of left or from the right. x f 1 2 0.5 0.5 0 500,000 Figure 14.5-2 Therefore, lim xS0 5 x4 q. Example 2 Infinite Limits ■ Describe the behavior of the function g x 1 2 8 x2 2x 8 near x 2. Solution As shown in Figure 14.5-3, the graph of g is not continuous at the left of x ues of x 2 get small without bound. To get large without bound, and the val- x 2, to the right of g 1 x 2 the values of g x 2. 1 2 950 Chapter 14 Limits and Continuity 10 3 5 10 Figure 14.5-3 Therefore, xS2 lim 8 x2 2x 8 q and xS2 lim 8 x2 2x 8 q. ■ Infinite Limits and Vertical Asymptotes The “infinite limits” considered in Figure 14.5-1 and in Examples 1 and 2 can be interpreted geometrically: Each such limit corresponds to a vertical asymptote of the graph. Vertical Asymptotes is a vertical asymptote of the graph of the x c The vertical line function f if at least one of the following is true. xSc f(x) lim xSc f(x) lim xSc f(x) lim xSc f(x) lim lim xSc lim xSc f(x) f(x) 7 Limits at Infinity 80 80 Whenever the word “limit” has been used up to now, it referred to the behavior of a function when x was near a particular number c. Now, the behavior of a function when x takes very large or very small values will be considered. That is, the end behavior of a function will be discussed. 1 The graph of f Figure 14.5-4 x 1 2 5 1 24e x 4 1 is shown in Figure 14.5-4. Graphing Exploration Produce the graph shown in Figure 14.5-4 and use the trace feature as x gets larger and larger. Are the values to find values of approaching a single value? x f 1 2 As you move to the right, the graph gets very close to the horizontal In other words, as x gets larger and larger, the corresponding line y 6. NOTE Due to rounding, the trace feature on most calculators will display y 1 when x is smaller 60. than approximately However, the value of the function is always greater than 1. Why? Limits at Infinity Section 14.5 Limits Involving Infinity 951 f values of bolically as 1 x 2 get closer and closer to 6, which can be expressed sym- 2 The last statement is read “the limit of 1 xSq f lim x 6. f x 1 2 as x approaches infinity is 6.” Toward the left the graph gets very close to the horizontal line is, as x gets smaller and smaller, the corresponding values of x and closer to 1. (See note.) It is said that “the limit of negative infinity is 1,” which is written x f 1 1 2 y 1; that get closer as x approaches f 2 xSq f lim x 1. 2 1 The types of limits when x gets large or small without bound are similar to those in Section 14.1 in that the values of the function do approach a fixed number. The definition in the general case is similar: f is any function, L is a real number, and the phrase “arbitrarily close” means “as close as you want.” Let f be a function that is defined for all number a. If x 77 a for some as x takes larger and larger positive values, increasing without bound, the corresponding values of f(x) get very close, and possibly are equal, to a single real number L and the values of f(x) can be made arbitrarily close (as close as you want) to L by taking large enough values of x, then the limit of f(x) as x approaches infinity is L, which is
written xS f(x) L. lim x 6 a x 7 a ” with “ Limits as x approaches negative infinity are defined analogously by replacing “ ”, “increasing” with “decreasing”, and “larger and larger positive” with “smaller and smaller negative” in the preceding definition. These definitions are informal because such phrases as “arbitrarily close” have not been precisely defined. Rigorous definitions, similar to those in Section 14.3 for ordinary limits, are discussed in Exercises 49–50. Horizontal Asymptotes and Limits at Infinity Limits as x approaches infinity or negative infinity correspond to horizontal asymptotes of the graph of the function. 952 Chapter 14 Limits and Continuity Horizontal Asymptotes y L The line function f if either is a horizontal asymptote of the graph of the xSqq f(x) L lim or xSqq f(x) L. lim Example 3 Limits at Infinity Discuss the behavior of 1 approaches negative infinity. f x 1 x 2 as x approaches infinity and as x Solution 1 When x is a very large positive number, 1 x is a positive number that is very close to 0. Similarly, when x is a very small negative number—which 20 20 is large in absolute value—such as 5,000,000, that is very close to 0. These facts suggest that tote, or 1 x y 0 is a negative number is a horizontal asymp- 1 Figure 14.5-5 as confirmed in Figure 14.5-5. ■ xSq f lim x 1 2 0 and xSq f lim x 1 2 0, Example 4 Limits at Positive and Negative Infinity Discuss the behavior of as x approaches negative infinity. x f 1 2 x3 10x 5 as x approaches infinity and Solution As shown in Figure 14.5-6, approaches infinity or as x approaches negative infinity. does not approach a single value as x x f 2 1 100,000,000,000 5,000 0 0 0 0 5,000 Figure 14.5-6 100,000,000,000 Section 14.5 Limits Involving Infinity 953 Thus, xSq f lim x and xSq f lim 1 However, the situations are often described by writing 2 1 2 x , as defined in Section 14.1, do not exist. xSq f lim x 2 1 q and xSq f lim x 1 2 q ■ In fact, no polynomial graph has a horizontal asymptote. That is, no polynomial function has a limit as x approaches infinity or negative infinity. Limit of a Constant Function The limits of constant functions are easily found. Consider, for example, As x approaches infinity or negative infinity, the the function corresponding value of and 5. xSq f lim 5 is always the number 5, so A similar argument works for any constant function. xSq f lim 5 Limit of a Constant If c is a constant, then xSqq c c lim and lim xSqq c c. Properties of Limits Infinite limits have the same useful properties that ordinary limits have. For instance, suppose that as x approaches infinity, the values of a function f approach a number L, and the values of a function g approach a number M. Then it is plausible that the values of approach L M, L M, x fg and so forth. Similar remarks 1 apply when x approaches negative infinity. the values of approach f g 21 21 x 2 1 2 Properties of Limits at Infinity If f and g are functions and L and M are numbers such that xSqq f(x) L lim and xSqq g(x) M, lim then 1. 2. 3. 4. 5. xSqq( f g)(x) lim lim xSqq( f g)(x) lim lim xSqq( fg)(x) lim lim xSqq lim a f gb (x) lim xSqq xSqq( f(x) g(x)) lim xSqq( f(x) g(x)) lim xSqq( f (x) g(x)) lim xSqq f(x) B A A xSqq f(x) lim f(x) L g(x)b M xSqq g(x) lim f(x) 0 xSqq f(x) lim xSqq f(x) lim xSqq g(x) lim B M 0 for all large x provided a , 2f(x) 2L, lim xSqq xSqq g(x) L M xSqq g(x) L M L M Properties 1–5 also hold with for all small x. erty 0 5, f x 1 2 q in place of q, provided that for prop- 954 Chapter 14 Limits and Continuity Limit of c x n If c is a constant, then Property 3 and Example 3 show that lim xSq c x lim xSqa c 1 xb lim xSq c b Q lim xSq a 1 xb c 0 0. Repeatedly using Property 3 with this fact and Example 3, note that the result holds for any integer n 2 . p 1 xb lim xSq c x 1 x 1 x c xn lim xSqa lim xSq 0 0 0 p 0 0 c xba lim xSq a 1 xba lim xSq 1 xb p lim xSq a 1 xb A similar argument works with and produces this useful result, which is essentially a formal statement of half of the Big-Little Concept discussed in Section 4.4. in place of q q Limit Theorem If c is a constant, then for each positive integer n, lim xSq c x n 0 and xSq lim c x n 0 The Limit Theorem and the limit properties now make it possible to determine the limit, if it exists, of any rational function as x approaches infinity or negative infinity. Example 5 End Behavior of a Rational Function 5 Describe the end behavior of 3x2 2x 1 2x2 4x 5 f x 2 1 and justify your 10 conclusion. 20 Solution 5 Figure 14.5-7 f If you graph 3x2 2x 1 2x2 4x 5 that there appears to be a horizontal asymptote close to in Figure 14.5-7. This can be confirmed algebraically by computing y 1.5, to the right of the y-axis you will see as shown x , 2 1 lim xSq 3x2 2x 1 2x2 4x 5 . Property 4 cannot be used directly because neither the numerator nor denominator have a finite limit as x approaches infinity, as discussed in Example 4. To rewrite the expression in an equivalent form, divide both Section 14.5 Limits Involving Infinity 955 numerator and denominator by the highest power of x that appears, namely x2. Dividing both by x2 is the same as multiplying by form of 1, so the value of the fraction is not changed. 1 x2 1 x2 , a lim xSq 3x2 2x 1 2x2 4x 5 lim xSq 1 x2 5 x2 3x2 2x 1 x2 2x2 4x 5 x2 2x x2 4x x2 1 x2 5 x2 1 3x2 x2 2x2 x2 3 2 x 2 4 x 3 2 x x2b lim xSq lim xSq lim xSqa lim xSqa 2 4 x 5 x2b lim xSq 3 lim xSq 2 x lim xSq lim xSq 2 lim xSq 4 x lim xSq 1 x2 5 x2 3 lim xSq 2 x lim xSq 1 x2 5 x2 lim xSq 4 2 lim x xSq 3 0 0 2 0 0 3 2 property 4 property 1, 2 limit of constant limit theorem ■ A slight variation on the last example can be used to compute certain limits involving square roots. Example 6 Limits at Infinity Find each limit. a. lim xSq 23x2 1 2x 3 b. lim xSq 23x2 1 2x 3 956 Chapter 14 Limits and Continuity Solution a. Only positive values of x need to be considered when finding the limit as x approaches infinity. When x is positive, Therefore, 2x2 x. 23x2 1 2x 3 lim xSq lim xSq lim xSq lim xSq lim xSq B 23x2 1 x 2x 3 x 23x2 1 2x2 2x 3 x 3x2 1 x2 2x 3 x 3 1 B x2 2 3 x 3 1 x2 2 3 xb 3 1 lim xSqB lim xSqa B lim xSqa x2b lim xSqa 2 3 xb lim xSq B 3 lim xSq 1 x2 3 x 2 lim xSq lim xSq 23 0 2 0 multiply by 1 x 1 x 2x2 x, for x 7 0 2a 2b a b B property 4 property 5 property 1 constant limit and limit theorem 23 2 b. To compute the limit as x approaches negative infinity, you need only consider negative values of x and use the fact that when x is negative, Then an argument similar to the one in part a shows that x 2x2. 2 2 For instance, 2 24. 2 2 1 lim xSq 23x2 1 2x 3 23 2 ■ Section 14.5 Limits Involving Infinity 957 Although the properties of limits and some algebraic ingenuity can often by used to compute limits, as in the preceding examples, more sophisticated techniques are needed to determine certain limits. This is the case, for example, with the proof that n lim nSqa 1 1 nb exists and is the number e. Exercises 14.5 In Exercises 1–8, use a calculator to estimate the limit. 10. 2x2 1 x 1 1 2D 2x2 x 1 x D 1. 2. lim xSq lim xSq C C 3. lim xSq x 4 3 2 3 x x3 5. lim xSq sin 1 x 7. lim xSq ln x x 8. lim xSq 1 5 1.1 1 x 20 2 4. lim xSq 5 4 x x 2x x 5 4 6. lim xSq sin x x 11. In Exercises 9–14, list the vertical asymptotes of the graph, if any exist. Then use the graph of the function to find xSq f(x) lim and xSq f(x). lim 9. y 3 2 1 −10 −5 5 10 x 12. y y y 3 2 1 −20 −10 −1 3 2 1 −20 −1 −2 −3 15 10 5 10 20 30 20 40 60 x x x −40 −20 −5 20 40 60 −10 −15 958 13. 14. Chapter 14 Limits and Continuity 15 10 5 −20 −10 −5 −10 −15 y y 12 8 4 −20 −4 −8 −60 −40 x x 10 20 30 20 40 60 In Exercises 15–20, use the limit theorem and the properties of limits to find the horizontal asymptotes of the graph of the given function. 15. f x 1 2 3x2 5 4x2 6x 2 16. g x 2 1 x2 x2 2x 1 17. h x 2 1 2x2 6x 1 2 x x2 18. x k 1 2 3x x2 4 2x x3 x2 19. f x 1 2 3x4 2x3 5x2 x 1 7x3 4x2 6x 12 20. g x 2 1 2x5 x3 2x 9 5 x5 In Exercises 21–39, use the limit theorem and the properties of limits to find the limit. 21. lim xSq 1 x 3 x 2 21 2x2 x 1 2 22. 1 lim xSq 3x 2 2x 1 21 3x2 2x 5 2 23. lim xSq 3x 1 x2b a 24. lim xSq 1 3x2 1 2 2 25. lim xSq 3x x 2 a 2x x 1b 26. lim xSq a x x2 1 2x2 x3 xb 27. lim xSq 2x 2x2 2x 29. lim xSq 3x 2 22x2 1 28. lim xSq x 2x2 1 30. lim xSq 3x 2 22x2 1 31. lim xSq 22x2 1 3x 5 32. lim xSq 22x2 1 3x 5 33. lim xSq 23x2 3 x 3 34. lim xSq 23x2 2x 2x 1 35. lim xSq x2 2x 1 2x4 2x 36. lim xSq 2x6 x2 2x3 Hint: Rationalize the denominator. 37. lim xSq 38. lim xSq 1 2x 1 2x 2x 2 2x 3 39. lim xSq A 2x2 1 x Hint: Multiply by B 2x2 1 x 2x2 1 x . In Exercises 40–42, find the limit by adapting the hint from Exercise 39. 40. 41. 42. lim xSq lim xSq A lim xSq x 2x2 4 A 2x2 1 2x2 1 B B 2x2 5x 5 x 1 A B 43. A free-falling body has two forces acting on it: gravity, which causes the body to speed up as it falls, and air resistance, which causes the body to slow down. Assuming that a free-falling body has an initial velocity of zero and that its velocity is proportional to the force due to air resistance, the velocity of a falling object can be written as a function of the amount of elapsed time. velocity f t 2 1 mg 1 e k m t R k Q The variable m represents the mass of the body, g is acceleration due to gravity ( 32 feet per second per second), and k is the contant of proportionality. For a fall with a parachute, k 1.6m; without a parachute, k 0.18m. When the air resistance has built until it nearly balances the gravitational force, the body speeds up very little. Upon reaching this condition, the body continues to move downward with a constant maximum speed called terminal velocity. Find the terminal velocity of falling bodies with a parachute and without a parachute by finding the limit as t approaches infinity. In Exercises 44–45, find the limit. 44. Critical Thinking lim xSq x x 0 0 45. Critical Thinking lim xSq 46. Critical Thinking Let function and find denote the greatest integer a. 3 lim xSq x x 4 b. lim xSq 3 x x 4 47. Critical Thinking F
ind lim xSq 4x 3x 2x 2x 1 . Section 14.5 Limits Involving Infinity 959 48. Critical Thinking Let f x 1 2 be a nonzero polynomial x g with leading coefficient a, and let nonzero polynomial with leading coefficient c. Prove that be a 2 1 a. if degree f b. if degree f x 1 2 x 2 1 6 degree g x 1 2 , then lim xSq 0. degree g x 2 1 , then lim xSq f g a c . c. if degree f x 1 2 7 degree g x 1 2 , then lim xSq does not exist Formal definitions of limits at infinity and negative infinity are given in Exercises 49 and 50. Adapt the discussion in Section 14.3 to explain how these definitions are derived from the informal definitions in this section. 49. Critical Thinking Let f be a function and L be a real number. Then the statement that for each positive number real number k that depends on with the following property: lim xSq e, e 2 1 x means f there is a positive L If x 7 k, then 0 f x 2 1 L 0 6 e Hint: Concentrate on the second part of the informal definition. The number k measures “large enough,” that is, how large the values of x must be in order to guarantee that you want to L. is as close as x f 2 1 50. Critical Thinking Let f be a function and L be a real means number. Then the statement that for each positive number negative real number n that depends on with the following property: L x 2 there is a e xSq f lim e, 1 If x 6 n, then 14 R E V I E W Important Concepts Section 14.1 Section 14.2 Section 14.2.A Section 14.3 Section 14.4 Section 14.5 Limit notation. . . . . . . . . . . . . . . . . . . . . . . . . . . . 910 Informal definition of limit. . . . . . . . . . . . . . . . . . 911 Limits and function values. . . . . . . . . . . . . . . . . . 912 Nonexistence of limits . . . . . . . . . . . . . . . . . . . . . 914 Limit of a constant . . . . . . . . . . . . . . . . . . . . . . . . 918 Limit of the identity function . . . . . . . . . . . . . . . . 918 Properties of limits . . . . . . . . . . . . . . . . . . . . . . . . 919 Limits of polynomial functions . . . . . . . . . . . . . . 920 Limits of rational functions . . . . . . . . . . . . . . . . . 921 The limit theorem . . . . . . . . . . . . . . . . . . . . . . . . . 922 Limit of a function from the right . . . . . . . . . . . . 924 Limit of a function from the left. . . . . . . . . . . . . . 924 Computing one-sided limits. . . . . . . . . . . . . . . . . 925 Two-sided limits . . . . . . . . . . . . . . . . . . . . . . . . . . 926 The formal definition of limit. . . . . . . . . . . . . . . . 931 Proving limit properties . . . . . . . . . . . . . . . . . . . . 933 x c Continuity at . . . . . . . . . . . . . . . . . . . . . . . . 936 Continuity on an interval . . . . . . . . . . . . . . . . . . . 942 Properties of continuous functions. . . . . . . . . . . . 943 Continuity of composite functions . . . . . . . . . . . . 944 The Intermediate Value Theorem . . . . . . . . . . . . . 944 Infinite limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . 948 Vertical asymptotes. . . . . . . . . . . . . . . . . . . . . . . . 950 Limit of a function as x approaches infinity or negative infinity . . . . . . . . . . . . . . . . . . . . . . . . 951 Horizontal asymptotes . . . . . . . . . . . . . . . . . . . . . 952 Properties of limits at infinity . . . . . . . . . . . . . . . 953 960 Chapter Review 961 Review Exercises In Exercises 1–2, use a calculator to estimate the limit. Section 14.1 1. lim xS0 3x sin x x 2. lim xSp 2 1 sin x 1 cos 2x In Exercises 3–4, use the graph of the function to determine the limit. 3. lim xS2 f x 1 2 y 3 2 1 −1 −1 −2 −3 −3 −2 x 1 2 3 4. lim xS1 f x 1 2 y 3 2 1 −1 −1 −2 −3 −3 −2 x 1 2 3 Section 14.2 In Exercises 5–6, assume that lim xS3 f(x) 5 and g(x) 2. lim xS3 5. Find lim xS3 2f 2 x 2 4 x 2 1 6. Find lim xS3 2f f 1 2 x 1 x 2 2g g x 1 1 2 x 2 In Exercises 7–10, find the limit if it exists. If the limit does not exist, explain why. 7. lim xS1 x2 1 x2 3x 2 9. lim xS0 21 x 1 x 8. lim xS2 x2 x 6 x2 x 2 10. lim xS2 x2 2x 3 x2 6x 9 962 Chapter Review 11. If f x 1 2 x2 1, find lim hS0 . 12. If f x 1 2 3x 2 and c is a constant, find lim hS0 . Section 14.2.A 13. 0 lim xS5 x 5 x 5 0 14. lim xS7 A 27 x2 6x 2 B Section 14.3 In Exercises 15–16, use the formal definition of limit to prove the statement. 15. lim xS3 1 2x 1 7 2 16. lim xS2 1 2 a x 3 b 4 Section 14.4 In Exercises 17–18, determine whether the function whose graph is given is continuous at x 3 x 2. and 17. 18. y y 3 2 1 −1 −1 −2 −3 3 2 1 −1 −1 −2 −3 −3 −2 −3 −2 19. Show that f x 1 2 x2 x 6 x2 9 a. continuous at has the given traits. b. discontinuous at x 3 Chapter Review 963 20. Is the function given by f x 1 2 3x 2 if x 3 10 x if x 7 3 e continuous at x 3? Justify your answer. Section 14.5 In Exercises 21–22, find the vertical asymptotes of the graph of the given function, and state whether the graph moves upward or downward on each side of each asymptote. 21. f x 1 2 x2 1 x2 x 2 22. g x 2 1 x2 1 x2 3x 2 In Exercises 23–26, find the limit. 23. lim xSq 2x3 3x2 5x 1 4x3 2x2 x 10 24. lim xSq 4 3x 2x2 x3 2x 5 25. lim xSq a 2x 1 x 3 4x 1 3x b 26. lim xSq 23x2 2 4x 1 In Exercises 27–28, find the horizontal asymptotes of the graph of the given function algebraically, and verify your results graphically with a calculator. 27. f x 1 2 x2 x 7 2x2 5x 7 28. f x 1 2 x 9 24x2 3x 2 C H A P T E R 14 Riemann Sums Calculus deals with rates of change, such as the speed of a car, and problems such as the following: If you know the continuously changing speed of a car at any instant, can you determine how far the car has traveled? A special case of this question will be answered in this section by using Riemann sums. Example 1 Total Distance Given Velocities Suppose a racecar is moving with increasing velocity. The velocity of the car at various times is given in the table. Estimate the total distance traveled in the 4-second interval. Solution Because the velocity is increasing, the car has gone at least 14 feet during the first second, at least 29 feet during the second, at least 61 feet during the third second, and at least 128 feet during the fourth second. During the four-second interval, the car has traveled at least 14 29 61 128 232 feet underestimate Therefore, 232 feet is an underestimate of the total distance traveled. An overestimate can be found by noting that the car travels no more than 29 feet in the first second, no more than 61 feet in the next second, no more than 128 in the third second, and no more than 268 feet in the last second. Altogether, the car traveled no more than 29 61 128 268 486 feet overestimate Therefore, the total distance traveled is between 232 feet and 486 feet. ■ The lower and upper estimates can be represented on a graph, where the velocity is shown as a smooth curve passing through each point given in the table, and the estimates of the distance traveled each second are represented by the area of rectangles. See Figure 14.C-1. The darker rectangles represent the underestimate for each second and the darker and lighter rectangles stacked together represent the overestimate. Because the time interval between each measurement is 1 second, each rectangle is 1 unit wide. Each height corresponds to how far the car could have traveled during each time interval. Therefore, the areas of the darker rectangles are 14, 29, 61, and 128, and the sum of the areas represents the total underestimate of 232. Time (sec) Velocity (ft/sec) 0 1 2 3 4 14 29 61 128 268 velocity 250 200 150 100 50 time 0 1 2 3 4 Figure 14.C-1 964 Similarly, the sum of the areas of the darker and lighter rectangles, which represents the overestimate, is 486. There is a difference of feet between the estimates. This difference can also be found by adding the areas of the lighter rectangles. 486 232 254 A Better Estimate Time (sec) Velocity (ft/sec) To get a better estimate of how far the car traveled during the 4-second interval, the velocity is measured for each half second, as shown in the table. The graph shown in Figure 14.C-2 displays the new data. 0 0.5 1 1.5 2 2.5 3 3.5 4 14 20 29 42 61 88 128 185 268 velocity Overestimate 250 200 150 100 50 time 0 1 2 3 4 Figure 14.C-2 In the first half second, the car travels at least 14 1 2b a 7 feet and at most 1 2b 10 20 a for both the underestimate and the overestimate as feet. The distance traveled in each half second is calculated vi where val and represents the velocity as measured at one end of the time inter¢t represents length of each the time interval. vi1 ¢t 2 Underestimate 14 a 1 2b 1 2b 20 a 1 2b 1 2b 29 a 1 2b 88 a 1 2b 128 a 1 2b 29 a 1 2b 128 a 42 a 1 2b 185 a 1 2b 42 a 1 2b 61 a 1 2b 185 a 1 2b 61 a 1 2b 20 a 88 a 1 2b 283.5 feet 268 a 1 2b 410.5 feet The difference between the estimates is again shown as the area of the lighter rectangles. This difference is 410.5 283.5 127. Notice that the difference between the better estimates is half what it was in Example 1. By halving the intervals of measurement, the difference between the estimates is halved. Similarly, if the interval of measurement was given for every tenth of a second, the estimates would differ by 254 25.4 feet, and if the interval of measurement was every thousandth of a second, the difference between the estimates would be 254 0.254. 0.001 0.1 2 1 1 2 965 Example 2 Accuracy of Estimates How frequently must the velocity be measured to ensure that the estimated distances traveled by the race car are within 5 feet of each other? Solution The difference between the velocity at the beginning and end of the meas¢t, urements is then the difference between the estimates is For the differences of the estimates to be within 5 feet, If the time between each measurement is 268 14 254. 254 ¢t . 2 1 254 ¢t 2 1 6 5 or ¢t 6 0.0196850394 seconds . ■ As the length of the intervals of measurement become smaller and smaller, the underestimate and overestimate approach the same number—the area under the curve—which represents the total distance traveled. Riemann Sums . Suppose that velocity is given as an incr
easing function of time: 2 To find the total distance traveled by a moving object over the time individe the interval into n equally spaced times terval a 6 t 6 b, 1 t v f t0 a, t1, p , tn b, where each time interval is ¢t b a n in duration. ti2 , The velocity at the beginning or end of each time interval is given by so the estimated distance traveled during each interval is velocity times the length of each time interval. f 1 ¢t f ti2 Both the underestimate and the overestimate of the total distance traveled can be written as the sum of the individual distances. The t02 underestimate sum begins with and the overestimate sum begins with distance at each ¢t ¢t ¢t f f 1 1 t12 1 f t12 1 ¢t f t22 1 ¢t p f tn2 1 f t02 1 ¢t f t12 1 ¢t p f 1 tn12 n f ¢t a i1 n1 ¢t a i0 t22 ¢t f 1 ¢t ti2 1 overestimate underestimate ¢t f ti2 1 ¢t p f ¢t tn2 1 lim nSq f t12 1 n nSq a i1 lim ¢t f ti2 1 As Examples 1 and 2 show, these estimates will be very close to each other when n is very large and the correspoding is very small. In fact, the actual total distance can be found by taking the limit of one of these sums as n gets very large without bound, written n S q. ¢t velocity f(t4) = f(b) Total distance traveled f(t0) = f(a) f(t3) f(t2) f(t1) t0 = a t1 t2 t3 t4 = b Figure 14.C-3 966 This limit can be interpreted geometrically as the area between the graph of f and the horizontal axis from t a to t b. The sum a Riemann sum. The limit of the Riemann sums n a i1 ¢t ˛f ti2 1 is called n lim nSq a i1 ¢t f ti2 1 is called the definite integral of f from a to b and is denoted by b dt . f t 1 2 a Definite integrals are studied fully in calculus. They have a wide variety of other applications, including determining the lengths of curves and the volumes of irregularly shaped solids, finding the amount of work done by a force, and making sophisticated probability calculations. Exercises 1. A driver slams on the brakes and comes to a stop 3. a. For the diagram below, estimate the shaded in five seconds. The following velocities are recorded after the brakes are applied. area with an error of at most 0.1. b. How can the shaded area be approximated to any desired degree of accuracy? Time (sec) 0 1 2 3 4 Velocity (ft/sec) 102 70 46 29 12 5 0 a. Find an upper and lower estimate of the distance traveled by the car after the brakes were applied. b. Sketch the graph of velocity versus time, and show the upper and lower estimates and the difference between them. c. How often would the velocity need to be measured to assure that the estimates differ by less than 5 feet? by less than 1 foot? 2. Use the grid to estimate the area of the region x ± 4. bounded by the curve, the horizontal axis, and the lines that are within 4 square units of one another. Explain your procedure. Get an upper and a lower estimate y 4 2 −4 −2 0 2 4 t y − x2 3 y = e −1 0 t 1 4. Estimate the total distance an object travels t 10 between represents the velocity v of the object in ft/sec. if the graph below t 0 and 10 t t cos t 5. Suppose the velocity of an object is given by 0 t 1.5. v Estimate the distance traveled during the 1.5-second interval, accurate to one decimal place. for 1 2 6. A snail is crawling at a velocity given by v 1 t 2 where Estimate the distance that the snail crawls during the second hour. hours and v is in feet per hour. 1 t , t 2 1 967 APPENDIX Algebra Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 969 A.1 Integral Exponents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 969 A.2 Arithmetic of Algebraic Expressions . . . . . . . . . . . . . . . . . . . . . . . . . 973 A.3 Factoring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 978 A.4 Fractional Expressions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 981 A.5 The Coordinate Plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 987 Advanced Topics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 994 B.1 The Binomial Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 994 B.2 Mathematical Induction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1002 Geometry Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1011 G.1 Geometry Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1011 Technology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1018 T.1 Graphs and Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1018 T.2 Lists, Statistics, Plots, and Regression . . . . . . . . . . . . . . . . . . . . . . . 1027 T.3 Programs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1033 Glossary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1035 Selected Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1054 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1148 968 Appendix ALGEBRA REVIEW This appendix reviews the fundamental algebraic facts that are used frequently in this book. You must be able to handle these algebraic manipulations in order to succeed in this course and in calculus. A.1 Integral Exponents Exponents provide a convenient shorthand for certain products. If c is a c3 real number, then denotes cc and denotes ccc. More generally, for any positive integer n c2 cn denotes the product ccc p c (n factors). In this notation c1 is just c, so we usually omit the exponent 1. Example 1 34 3 3 3 3 81 2 1 and 5 2 1 2 For every positive integer n, Example 2 2 21 21 0n 0 2 p 2 21 21 0 0. 2 32. 2 To find 2.4 1 9 2 use the ^ (or ) key on your calculator:* or ab xy 2.4 ^ 9 ENTER* which produces the (approximate) answer 2641.80754. ■ ■ Because exponents are just shorthand for multiplication, it is easy to determine the rules they obey. For instance, c3c5 1 ccccccc cccc 21 ccccc ccc 2 ccccccc cccc c8, that is, c3c5 c35. ccc c3, that is, c7 c4 c74. c7 c4 The ENTER key is labeled EXE on Casio calculators. Section A.1 Integral Exponents 969 Similar arguments work in the general case: To multiply cm by cn, add the exponents: cmcn cmn. To divide cm by cn, subtract the exponents: cm cn cmn. Example 3 42 47 427 49 and 28 23 283 25. ■ The notation as follows: cn can be extended to the cases when n is zero or negative If If c 0, c 0 then c0 is defined to be the number 1. and n is a positive integer, then cn is defined to be the number 1 cn . 00 and negative powers of 0 are not defined (negative powers Note that of 0 would involve division by 0). The reason for choosing these definifor nonzero c is that the multiplication and division rules for tions of exponents remain valid. For instance, n c c5 c0 c5 1 c5, so that c5c0 c50. 7 c7 1 c0, so that c7c 7 c77. c7c 1 c7b a Example 4 6 1 216 3 1 2 2 63 12 3,201,969.857. and 0.287 1 5 1 2 1 1 2 1 32 . 5 2 A calculator shows that ■ If c and d are nonzero real numbers and m and n are integers (positive, negative, or zero), then we have these Exponent Laws 1. 3. 5. cmcn cmn (cm)n cmn cn n c dn dR Q 2. 4. 6. cm n cm cn (cd)n cndn cn 1 cn 970 Algebra Review * means “approximately equal to.” Example 5 Here are examples of each of the six exponent laws. 1. 2. 3. p5p2 p52 p3 1 p3. x9 x4 x94 x5. 3 5 1 2 2 51 3 2 5 2 6. 4. 5. 6. 5 25x5 32x5. 2x 1 2 10 7 3b a 710 310 . 1 5 x 1 1 x5b a x5. ■ The exponent laws can often be used to simplify complicated expressions. Example 6 CAUTION 5 is not the same as Part 4 of Example 5 5 32x5 2x 2x 2 1 2x 5. shows that 1 and not 2x 5. 2 1 1 a. b. c. 2x2y3z 2 4 24 c Law (4) x2 4 2 1 1 y3 2 4z4 16x8y12z4. c Law (3) 4 r6 s4 . s2 1 2 2 r6s c Law (3) 3s2 r 2 2 3 2 r 1 2 c Law (4) x5y6 3 x2 y 2 1 c x5y6 2 2y2 x2 2 2 1 2 c Law (3) Law (4) 2 x5 y2 1 x2y 1 Law (3) Law (2) x5y6 x4y2 c x54y62 xy4. c ■ ■ It is usually more efficient to use the exponent laws with the negative exponents rather than first converting to positive exponents. If positive exponents are required, the conversion can be made in the last step. Example 7 Simplify and express without negative exponents 2 a 1 3b a b2c3 5 1 2 2c 2 2 . Solution 2 a 1 3b a b2c3 5 2 2 2c 2 1 a 2 b2 2 1 3 2 a 1 2 2 2 c3 1 5 b 2 2c 2 2b a 6b a 4c 6 10c 1 c Law (4) c Law (3) 6 1 4 2b 10 1 61 a4b6c 2c 7 a4b6 c7 . 2 a c Law (2) Section A.1 Integral Exponents ■ 971 1 1 1 1, Since be equal to 1. Every odd power of 1 1 1 any even power of 1 12, will 2 for instance Consequently, for every positive number c such as is equal to 2 1 1 4 2 1; 21 1 1. 1, 1 1 1 5 or c)n [(1)c]n (1)ncn cn cn e if n is even . if n is odd Example 8 3 1 2 4 34 81 and 5 1 2 3 53 125. ■ CAUTION 4, 12 1 Be careful with negative bases. For instance, if you want to compute ^ 4 ENTER the calculator will interpret this as a negative answer. To get the correct answer, you must key in the parentheses: which is a positive number, but you key in 12 2 1 and produce 124 1 2 2 ( ( ) 12 ) ^ 4 ENTER. Exercises A.1 In Exercises 1–18, evaluate the expression. 19. x2 x3 x5 20. y y4 y6 2. 62 3 2 4 4 2 2 1 1 4. 1 2 4 2 6. 8. 22 42 4b 10. 2 5 7b a 2 2 7b a 12. 33 3 7 14. 16. 1 33 3 2 1 43 5 2 2 42 5 1 1. 3. 2 6 2 1 5 4 3 5. 1 32 23 1 2 2 2 22 1 1 7. 9. 3 5 4 b a 3 1 3b a 3 2 3b a 11. 24 27 2 2 2 2 2 1 22 3 3 32 2 3 13. 15. 17. 1 23 1 4 2 18. 32 1 3 a 1 3 2 b In Exercises 19–38, simplify the expression. Each letter represents a nonzero real number and should appear at most once in your answer. 972 Algebra Review 21. 23. 25. 27. 29. 1 1 1 1 1 31. a 33. 35. 37. 1 1 1 0.03 y2 y7 2 33x 2x2 2 3x2y 2 2 a2 7a 21 21 3a3 2 3w 4w 2 2 21 3 2w 1 2 2b3a3 1 2 2x 2 3a4 3 2y 1 2 9x 3 4x 2 1 2 1 2 2 0 2x2y 3xy 2 2 1 22. 24. 26. 28. 30. 32. 34. 36. 38. 1 1 1 1 1 1 1 z3 z5 1.3 2
3y3 2 2xy3 b3 21 45y2 3 2 b2 3b 2 21 2 2 1 5d 2 4 2d 3d 1 2 1 3 c4d5c 2x 2 2 1 2 2y 2 1 3y2 2 3 3x 2 2y3 3 1 2 3x 2y4 0 2 In Exercises 39–42, express the given number as a power of 2. 39. 41. 1 1 64 2 2 24 16 2 3 2 40. 3 1 8b a 42 16 b In Exercises 43–60, simplify and write the given expression without negative exponents. All letters represent nonzero real numbers. 43. x4 3 x2 1 x3 2 45. 2 e6 c4b a 3 c3 e b a 47. 2 ab2c3d4 abc2d b a 49. 2 a6 4 b b a 51. 53. a a 2 c5 3 b d 3 3x y2 b 2 x 2y3b a 55. 1 1 3b2c a 2c3 ab 2 1 3 2 2 2 57. 1d c 1 2 44. 46. 4 z2 t3 b a 2 x7 y6b a 5 z3 t b a 4 y2 x b a 3x 48. 1 2 y2 2 1 2xy2 3x2 2 3 2 1 2 2 2 b 50. x y a 52. 54. 1 1 b a x 2y a 2 2y x b 2 5u2v 2uv2 b a 3uv 2u2v b a 3 56. 1 2cd2e 3de 5c 1 1 3 2 2 2 58. 3 1 x2y 1 2 3 4 2 b2 a2 59. a2 1 1 a a 3 2 60. 2 2 a b In Exercises 61–66, determine the sign of the given number without calculating the product. 2.6 3 2 1 4.3 2 2 6.7 5 2 61. 63. 65. 66.1 45.8 3 2 7 2 1 4.6 2 1 7.9 6 7.2 2 8.5 1 9 2 1 4 2 62. 64. 1 1 7 2.5 3 2 2 1 1269 4.1 2 4 2 In Exercises 67–72, r, s, and t are positive integers and a, b, and c are nonzero real numbers. Simplify and write the given expression without negative exponents. 67. r 3 sr 3 t c 6b s 2 70. 1 68. 71. 1 t1 4 4 2t 2 69. t a6 4 b b a rbs c s ctb t r 2 2 1 1 72. 1 1 s arb btcr t 2 s 2 In Exercises 73–80, give an example to show that the statement may be false for some numbers. 73. ar br 75. arbs ab 1 r 2 a b 1 rs 2 74. aras ars 76. r cr c 77. r s c cr cs 79. a 2 1 2 a2 78. 80. 1 1 a 1 b 1 2 21 ab 1 a b 2 21 ab A.2 Arithmetic of Algebraic Expressions Expressions such as b 3c2, 3x2 5x 4, 2x3 z, x3 4xy p x2 xy are called algebraic expressions. Each expression represents a number that is obtained by performing various algebraic operations (such as addition or taking roots) on one or more numbers, some of which may be denoted by letters. A letter that denotes a particular real number is called a constant; its value remains unchanged throughout the discussion. For example, the Greek . letter Sometimes a constant is a fixed but unspecified real number, as in “an angle of k degrees” or “a triangle with base of length b.” has long been used to denote the number 3.14159 p p A letter that can represent any real number is called a variable. In for example, the variable x can be any real the expression 2x 5, Section A.2 Arithmetic of Algebraic Expressions 973 x 3, then 2x 5 2 3 5 11. If x 1 2 , then If number. 2x 5 2 1 2 5 6, and so on.* Constants are usually denoted by letters near the beginning of the alphabet and variables by letters near the end of the alphabet. Consequently, it is understood that c and d in expressions such as are constants and x and y are variables. cy2 dy, cx d and The usual rules of arithmetic are valid for algebraic expressions: Commutative Laws: a b b a and ab ba Associative Laws: a b c a 1 Distributive Laws: 2 b c 1 2 and ab 1 2 c a bc 1 2 b c 2 a 1 ab ac and b c 1 2 a ba ca. Example 1 Use the distributive law to combine like terms; for instance, 3x 5x 4x 3 5 4 x 12x. 2 1 In practice, you do the middle part in your head and simply write 3x 5x 4x 12x. ■ Example 2 In more complicated expressions, eliminate parentheses, use the commutative law to group like terms together, and then combine them. a2b 31c 5ab 71c ■ B 7a2b a2b 31c 5ab 71c 7a2b a2b 7a2b 31c 71c 5ab ⎧⎪⎨⎪⎩ ⎧⎪⎪⎪⎨⎪⎪⎪⎩ 41c 8a2b 5ab. Combine like terms: A B A Regroup: ■ CAUTION Be careful when parentheses are preceded by a minus sign: b 3 means , so that by the distributive law, b 3 b 3 Here’s the reason: b 3. and not 1 1 b 3 2 1 Similarly, 2 7 y 1 1 21 . 1 2 3 b 3. 2 2 1 1 21 b 3 2 974 Algebra Review *We assume any conditions on the constants and variables necessary to guarantee that an z 0 algebraic expression does represent a real number. For instance, in and in we assume c 0. we assume 1z 1 c The examples in the Caution Box illustrate the following. Rules for Eliminating Parentheses Parentheses preceded by a plus sign (or no sign) may be deleted. Parentheses preceded by a minus sign may be deleted if the sign of every term within the parentheses is changed. The usual method of multiplying algebraic expressions is to use the distributive laws repeatedly, as shown in the following examples. The net result is to multiply every term in the first sum by every term in the second sum. Example 3 , we first apply the distributive law, 3y2 7y 4 2 as a single number: 1 To compute treating y 2 21 3y2 7y 4 1 y 2 Distributive law: 21 1 2 3y2 7y 4 2 1 2 3y2 7y 4 3y2 7y 4 y 2 3y3 7y2 4y 6y2 14y 8 3y3 7y2 6y2 4y 14y 8 3y3 8. ⎧⎪⎪⎨⎪⎪⎩ ⎧⎪⎪⎨⎪⎪⎩ 13y2 18y 1 2 Regroup: Combine like terms: Example 4 1 2x 5y We follow the same procedure with 3x 4y 3x 4y 2x 3x 4y 1 5y 2x 3x 2x 4y 6x2 8xy 15xy 20y2 6x2 20y2. 7xy 2x 5y 5y ⎧⎪⎪⎨⎪⎪⎩ 21 21 1 1 2 1 2 2 3x 4y : 2 2 3x 5y 1 2 4y ■ ■ Observe the pattern in the second line of Example 4 and its relationship to the terms being multiplied: 2x 5y 3x 4y 2 21 1 2x 3x 2x 4y > > 5y 3x > 2 5y 4y > 2 1 1 First terms 3x 4y 2x 5y 2x 5y 2x 5y 1 1 1 21 21 21 3x 4y 3x 4y Outside terms 2 2 2 Inside terms Last terms Section A.2 Arithmetic of Algebraic Expressions 975 CAUTION The FOIL method can be used only when multiplying two expressions that each have two terms. This pattern is easy to remember by using the acronym FOIL (First, Outside, Inside, Last). The FOIL method makes it easy to find products such as this one mentally, without the necessity of writing out the intermediate steps. Example 5 3x 2 x 5 2 21 1 3x2 15x 2x 10 3x2 17x 10. c c c c First Outside Inside Last ■ Exercises A.2 In Exercises 1–54, perform the indicated operations and simplify your answer. 1. x 7x 2. 5w 7w 3w 3. 6a2b 8b a2 2 1 4. 6x31t 7x31t 15x31t 1 3 1 A 1 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 16. x2 2x 1 1 2 u3 u 2 3 2 x3 3x2 4 2 1 T u4 2u3 5 u 2 R Q 3 u3 u 2 2 1 T u4 2u3 5 u 2 R Q u4 u4 S S 1 1 6a2b 3a1c 5ab1c B A 4z 6z2w 2 1 2 z3w2 6ab2 3ab 6ab1c B A 8 6z2w zw3 4z3w2 2 x5y 2x 3xy3 2 9x x3 1 1 x 1y z 2 3 1 2x3 2 1 x 1y z 1 4 2x x5y 2xy3 xy 1 1 2 2 x2 xy 1y z x 1 2 x2 3xy 2 x2 2 2x 1 x2y 2 xy 6xy2 1 2 4ax 2a2y 2ay 17. 3ax 19. 6z3 1 2z 5 1 2 4a 6b 2a2b 2 x 2 2 21 21. 3ab 1 x 1 23. 1 976 Algebra Review 15. 5y 1 21 3y2 1 2 2 18. 20. 22. 24. x2 3xy 2y2 2x 1 3x2 1 3ay 21 x 2 1 1 12x6 7x5 2 4ay 5y 2 2x 5 2 21 B 2 2y 2 2 3w 1 2 a 2 21 y 8 2 2 3x y 21 2 y 6 21 w 2 21 ab 1 y 8 21 3x y x 6 2 2 2 2x 3y 2 2s2 9y 21 4x3 5y2 2s2 9y 2 4x3 5y2 2 21 26. 28. 30. 32. 34. 36. 38. 40. 42 50. 3y 1 y 2 3y 1 2 21 25. 27. 29. 31. 33. 35. 37. 39. 41. 43. 44. 45. 46. 47. 48. 49. 51. 52. 53. 54 2x 4 x 3 2 21 y 4 2 2x 5 21 3y2 4 21 x 4 2 2 4a 5b 2 2 y 3 21 3x 7 y 3 x 4 21 4a 5b y 11 21 2 2 2 5x b 2 4x3 y4 2 2 3x2 2y4 2 2 2c2 3c 1 2 y2 3y 1 2 2x2 xy y2 21 3w2 4w 3 2 x2 2xy 3y2 2 4x 2 21 x 2 21 3y 2 2 x 3 2 y 2 21 2y x 21 3x 2y 2 3x y 2 y x 21 2 c 2 21 2y 3 x 2y 21 5w 6 21 5x 2y 21 3x 1 2x 1 x 1 21 y 2 21 x 4y 21 2x y 21 In Exercises 55–64, find the coefficient of in the given product. Avoid doing any more multiplying than necessary. x2 x2 1 21 1 13 x x 1 2 13 x A BA x2 x 1 1 21 B x 1 55. 57. 59. 61. 62. 63. 64. 1 1 1 1 1 1 1 x2 3x 1 21 x3 2x 6 2x 3 2 x2 1 2 21 x 2 3 2 x2 x 1 x2 x 1 2 21 56. 58. 60. 2x2 1 2x 1 21 2x2 1 2 21 x2 3x 2 1 2x 2 1 4x2 x 1 2 2 In Exercises 65–70, perform the indicated multiplication and simplify your answer if possible. 65. 66. 67. 69. A A A A 1x 5 1x 5 B BA 21x 12y 21x 12y BA B 68. 3 1y 2 1 13x 2 BA x 13 B 70. A 7w 12x A 2y 13 BA 2 B 15y 1 B In Exercises 71–76, compute the product and arrange the terms of your answer according to decreasing powers of x, with each power of x appearing at most once. Example: (ax b)(4x c) 4ax2 (4b ac)x bc. 71. 73. 75. 1 1 1 ax b 3x 2 2 21 ax b bx a 21 x b 2 x c 21 x a 21 72. 74. 76. 2 4x c dx c 21 3rx 1 rx 1 2dx c 2 4x r 2 21 3cx d 1 1 21 79. 81. 82. 1 1 1 xm 2 21 2xn 5 xn 3 2 x3n 4xn 1 21 80. yr 1 ys 4 2 21 1 2 3y 2k yk 1 yk 3 2 21 In Exercises 83–92, find a numerical example to show that the given statement is false. Then find the mistake in the statement and correct it. 2 b 5, is false Example: The statement The when mistake is the sign on the 2. The correct statement is (b 2) b 2 but (b 2) b 2. (5 2) 7 5 2 3. since 83. 84. 85. 87. 89. 91. 92. 1 1 1 x 3y 4 y 2 3y 2 3 1 x 2 3y 4 1 x y 2 2 x y2 2 7xy 7x 7y 21 2 1 y y y y3 x 3 a b x 2 2 21 a2 b2 21 x2 5x 6 a3 b3 2 3 2x3 86. 88. 90. 1 1 1 2x 2 x y a b 2 2 x2 y2 2 2 a2 b2 In Exercises 93 and 94, explain algebraically why each of these parlor tricks always works. 93. Critical Thinking Write down a nonzero number. Add 1 to it and square the result. Subtract 1 from the original number and square the result. Subtract this second square from the first one. Divide by the number with which you started. The answer is 4. 2 94. Critical Thinking Write down a positive number. In Exercises 77–82, assume that all exponents are nonnegative integers and find the product. Example: 2xk(3x xn1) (2xk)(3x) (2xk)(xn1) 77. 3r343t 6xk1 2xkn1. 78. 2xn 8xk 2 21 1 Add 4 to it. Multiply the result by the original number. Add 4 to this result and then take the square root. Subtract the number with which you started. The answer is 2. 95. Critical Thinking Invent a similar parlor trick in which the answer is always the number with which you started. Section A.2 Arithmetic of Algebraic Expressions 977 A.3 Factoring Factoring is the reverse of multiplication: We begin with a product and find the factors that multiply together to produce this product. Factoring skills are necessary to simplify expressions, to do arithmetic with fractional expressions, and to solve equations and inequalities. The first general rule for factoring is Common Factors If there is a common factor in every term of the expression, factor out the common factor of highest degree. Example 1 4x6 8x, for example, each term contains a factor of In 4x6 8x 4x x5 2 x3y2 2xy3 3x2y4 so that Similarly, the common factor of highest degree in xy2 . 2 is 4x, 1 and x3y2 2xy3 3x2y4 xy2 x2 2y 3xy2 . 2 1 ■ You can greatly increase your factoring proficiency b
y learning to recognize multiplication patterns that appear frequently. Here are the most common ones. Quadratic Factoring Patterns Difference of Squares Perfect Squares u2 v2 (u v)(u v) u2 2uv v2 (u v)2 u2 2uv v2 (u v)2 Example 2 a. b. c. 2, 3y x2 9y2 x2 9y2 y2 7 y2 36r2 64s2 can be written x 3y x 3y 21 2 17 B A A 2 8s 6r 1 2 1 2 3r 4s 2 2 x2 . 2 y 17 2 1 3r 4s 1 1 2 1 2 1 y 17 BA 6r 8s 21 4 1 2 .* B 6r 8s 2 3r 4s 21 a difference of squares. Therefore, 3r 4s . 2 ■ When a polynomial has integer coefficients, we normally look only for factors with integer coefficients. But when it is easy to find other factors, as here, we shall do so. 978 Algebra Review Example 3 Since the first and last terms of to use the perfect square pattern with 4x2 36x 81 4x2 36x 81 u 2x 2 36x 92 2 2 2x 9 92 2x 2x 2 1 1 2 are perfect squares, we try and v 9: 2x 9 2. 2 1 ■ Cubic Factoring Patterns Difference of Cubes Sum of Cubes Perfect Cubes u3 v3 (u v)(u2 uv v2) u3 v3 (u v)(u2 uv v2) u3 3u2v 3uv2 v3 (u v)3 u3 3u2v 3uv2 v3 (u v)3 Example 4 a. b. c. x 5 x 5 x3 125 x3 53 x3 8y3 x3 1 1 3 x3 12x2 48x 64 x3 12x2 48x 43 21 21 x 2y x 2y x2 x 2y 1 x2 2xy 4y2 x2 5x 52 2 x2 5x 25 2 2 3 21 2y 1 1 . 1 2 2 2 4 2y . 2 x3 3x2 4 3x 42 43 x 4 3. 1 2 ■ When none of the multiplication patterns applies, use trial and error to factor quadratic polynomials. If a quadratic has two first-degree factors, then the factors must be of the form for some constants a, b, c, d. The product of such factors is ax b cx d cx d ax b and 1 21 2 acx2 adx bcx bd acx2 x bd. ad bc 1 2 Note that ac is the coefficient of and bd is the constant term of the product polynomial. This pattern can be used to factor quadratics by reversing the FOIL process. x2 Example 5 x2 9x 18 If ficient of only integer factors of 1). The only possibilities for b and d are then we must have a ± 1 and cx d ax b (constant term). Thus, factors as 1 bd 18 ) and x 2 21 , 2 ac 1 c ± 1 (coef(the ± 1, ± 18 or ± 2, ± 9 or ± 3, ± 6. We mentally try the various possibilities, using FOIL as our guide. For x 9 x 2 . example, we try 2 so this prodThe sum of the outside and inside terms is and check this factorization: 1 9x 2x 11x, b 2, d 9 21 Section A.3 Factoring 979 uct can’t be d 6 x2 9x 18. By trying other possibilities we find that leads to the correct factorization: x2 9x 18 x 3 b 3, x 6 . 2 ■ 21 1 Example 6 6x2 11x 4 To factor c whose product is 6, the coefficient of product is the constant term 4. Some possibilities are cx d , 2 x2, ax b as 21 1 we must find numbers a and and numbers b and d whose ac 6 a c ±1 ±6 ±2 ±3 ±3 ± 2 ±6 ±1 bd 4 b d ±1 ±4 ±2 ±2 ±4 ±1 Trial and error shows that 2x 1 3x 4 2 21 1 6x2 11x 4. ■ Occasionally the patterns above can be used to factor expressions involving larger exponents than 2. Example 7 a. b. x6 y6 x8 1 1 Example 8 x3 y3 x3 y3 2 1 2 x3 2 1 x y 1 x4 2 y3 1 x2 xy y2 x4 1 x4 1 x4 1 21 2 1 2 21 x y 21 x4 1 x2 1 x2 1 2 21 21 21 21 21 1 1 1 2 x2 xy y2 . 2 21 x2 1 x 1 2 21 x 1 . 2 To factor x4 2x2 3, Then, x4 2x2 3 1 x2 u x2. let 2 2x2 3 u2 2u 3 2 u 1 21 1 x2 1 21 1 x2 1 BA A u 3 2 x2 3 2 x 13 ■ ■ ■ x 13 BA . B ■ ■ Example 9 can be factored by regrouping and using the distrib- 3x3 3x2 2x 2 utive law to factor out a common factor: 3x2 3x3 3x2 2x 2 1 2 2 1 x 1 1 3x2 2 2 1 2 x 1 21 1 x 1 . 2 2 ■ ■ ■ 980 Algebra Review Exercises A.3 In Exercises 1–58, factor the expression. 37. x3 125 38. y3 64 1. x2 4 3. 9y2 25 2. x2 6x 9 4. y2 4y 4 39. x3 6x2 12x 8 40. y3 3y2 3y 1 41. 8 x3 42. z3 9z2 27z 27 5. 81x2 36x 4 6. 4x2 12x 9 43. x3 15x2 75x 125 7. 5 x2 8. 1 36u2 9. 49 28z 4z2 10. 25u2 20uv 4v2 44. 27 t3 46. x3 1 11. x4 y4 13. x2 x 6 15. z2 4z 3 12. x2 1 9 14. y2 11y 30 16. x2 8x 15 17. y2 5y 36 18. z2 9z 14 19. x2 6x 9 20. 4y2 81 21. x2 7x 10 22. w2 6w 16 23. x2 11x 18 24. x2 3xy 28y2 25. 3x2 4x 1 26. 4y2 4y 1 27. 2z2 11z 12 28. 10x2 17x 3 29. 9x2 72x 30. 4x2 4x 3 31. 10x2 8x 2 32. 7z2 23z 6 33. 8u2 6u 9 34. 2y2 4y 2 35. 4x2 20xy 25y2 36. 63u2 46uv 8v2 45. x3 1 47. 8x3 y3 49. x6 64 51. y4 7y2 10 48. 50. 3 1 x 1 2 1 x5 8x2 52. z4 5z2 6 53. 81 y4 54. x6 16x3 64 55. z6 1 56. y6 26y3 27 57. x4 2x2y 3y2 58. x8 17x4 16 In Exercises 59–64, factor by regrouping and using the distributive law (as in Example 9). 59. x2 yz xz xy 60. x6 2x4 8x2 16 61. a3 2b2 2a2b ab 62. u2v 2w2 2uvw uw 63. x3 4x2 8x 32 64. z8 5z7 2z 10 65. Critical Thinking Show that there do not exist real numbers c and d such that x c . x d 1 21 2 x2 1 A.4 Fractional Expressions Quotients of algebraic expressions are called fractional expressions. A quotient of two polynomials is sometimes called a rational expression. The basic rules for dealing with fractional expressions are essentially the same as those for ordinary numerical fractions. For instance, the “cross products” are equal: 2 6 4 3. In the general case we have 2 4 3 6 and Section A.4 Fractional Expressions 981 Properties of Fractions 1. Equality rule: a b c d exactly when ad bc. 2. Cancellation property: If k 0, then ka kb a b . The cancellation property follows directly from the equality rule because ka b kb a. 1 2 1 2 Example 1 Here are examples of the two properties: x x 1 x2 2x 1. 2. x2 2x x2 x 2 x4 1 x2 1 1 because the cross products are equal: 1 x2 1 21 x2 1 21 x2 1 1 2 x 1 x3 x2 2x 2 x2 1 1 x2 1. 2 x2 x 2 1 x. 2 ■ A fraction is in lowest terms if its numerator (top) and denominator (bottom) have no common factors except To express a fraction in lowest terms, factor numerator and denominator and cancel common factors. ± 1. Example 2 x2 x 6 x2 3x 21 21 1 1 x 3 x 1 . ■ To add two fractions with the same denominator, simply add the numera b ators as in ordinary arithmetic: Subtraction is done a c b c b . similarly. Example 3 7x2 2 x2 3 4x2 2x 5 x2 3 7x2 2 2 1 4x2 2x 5 1 x2 3 2 7x2 2 4x2 2x 5 x2 3 3x2 2x 7 x2 3 . ■ *Throughout this section we assume that all denominators are nonzero. 982 Algebra Review To add or subtract fractions with different denominators, you must first find a common denominator. One common denominator for a/b and c/d is the product of the two denominators bd because both fractions can be expressed with this denominator: a b ad bd and c d bc bd . Consequently, a b c d ad bd bc bd ad bc bd and a b c d ad bd bc bd ad bc bd . Example 4 2x 1 3x x2 2 x 1 x2 2 3x 1 x 1 3x 1 x2 2 3x 2 2 2 1 x 1 2 21 x 1 2x 1 3x 2x 1 1 1 21 2 x 1 2 x 1 1 3x 2x2 x 1 3x3 6x 3x2 3x 3x3 2x2 5x 1 3x2 3x . 1 2 ■ Although the product of the denominators can always be used as a common denominator, it’s often more efficient to use the least common denominator. The least common denominator can be found by factoring each denominator completely (with integer coefficients) and then taking the product of the highest power of each of the distinct factors. Example 5 1 120 1 In the sum 100 120 23 3 5. The distinct factors are 2, 3, 5. The highest exponent of 2 in either denominator is 3, the highest of 3 is 1, and the highest of 5 is 2. So the least common denominator is the denominators are 23 3 52 600 100 22 52 and , 1 100 1 120 6 600 5 600 11 600 . ■ Example 6 To find the least common denominator of 3x 7 x4 x3 , factor each of the denominators completely: 1 x2 2x 1 , 5x x2 x , and x 1 1 2 2, x x 1 1 2 , x3 x 1 , 2 1 Section A.4 Fractional Expressions 983 x 1, The distinct factors are nator is determined by the highest power of each factor: x 1. and x, The least common denomi- x3 ■ To express one of several fractions in terms of the least common denominator, multiply its numerator and denominator by those factors in the common denominator that don’t appear in the denominator of the fraction. Example 7 The preceding example shows the least common denominator (LCD) of 1 x 1 2, 2 x 1 5x x 1 2 1 3x 7 x 1 x3 , and 1 2 2 1 1 x 1 5x x 1 1 1 x 1 5x x 1 x 1 3x 7 x 1 x3 1 2 x 1 2 3x 7 x 1 x3 1 1 1 2 to be x3 x 1 2 2 1 x 1 . Therefore, 2 2 2 x3 x3 x2 x2 21 21 x3 2 x 1 2 2 x 1 x3 1 x 1 5x3 1 x 1 x3 1 3x 7 x3 21 . 21 x 1 Example 8 To find 1 z 3z z 1 z2 z 1 1 2 1 z 3z z 1 z2 z 1 1 2 2 z3 z 1 we use the LCD z 2 z 1 2 3z2 3z2 z 1 z3 1 1 z 1 2 z z2 2z 1 3z3 3z2 z3 z 1 2 1 2z3 4z2 2z ■ Multiplication of fractions is easy: Multiply corresponding numerators and denominators, then simplify your answer. Example 9 x2 1 x2 2 3x 4 x 1 1 1 3x 4 x2 1 21 x 1 x2 2 21 1 x 1 x 1 2 2 3x 4 21 x2 2 21 x 1 21 2 984 Algebra Review 1 2 1 x 1 3x 4 21 x2 2 2 . ■ Division of fractions is given by the rule: Invert the divisor and multiply: a b c d a b d c ad bc . Example 10 x2 x 2 x2 6x 9 x2 1 x 3 x 3 x2 1 x2 x 2 x2 6x 9 x 2 x 1 2 21 21 1 1 x 3 x 1 x 1 21 1 2 2 . 2 ■ Division problems can also be written as fractions. For instance, means 8 2 8 2 4. Similarly, the compound fraction a b c d means a b c d . So, the basic rule for simplifying compound fractions is: Invert the denominator and multiply it by the numerator. Example 11 16y2z 8yz2 yz 6y3z3 16y2z 8yz2 6y3z3 yz 16 6 y5z4 8y2z3 2 6 y52z43 12y3z. y2 y 2 y3 y y2 y 2 1 y3 y y 2 1 y2 21 a. b. 2 y3 y y2 y 2 y 1 y2 1 2 y 2 1 1 y 2 21 y2 1 . 2 ■ Exercises A.4 In Exercises 1–10, express the fraction in lowest terms. 1. 4. 63 49 x2 4 x 2 2. 5. 121 33 x2 x 2 x2 2x 1 3. 6. 13 27 22 10 6 4 11 12 z 1 z3 1 7. a2 b2 a3 b3 8. x4 3x2 x3 x c 9. 1 x2 cx c2 21 x4 c3x 2 10. x4 y4 x2 y2 1 21 x2 xy 2 Section A.4 Fractional Expressions 985 In Exercises 11–28, perform the indicated operations. In Exercises 43–60, compute the quotient and express in lowest terms. 5 6 3c e 13. 19 7 a 1 2b 1 3 16. r s s t t r 2a b2 3a b3 19. 1 x 1 1 x 11. 14. 17. 3 7 1 a b c 2 5 2a b c b 12. 15. 18. 7 8 c d a b 20. 1 2x 1 1 2x 1 21 x2 8x 16 22. 1 x 1 xy 1 xy2 23. 1 x 1 3x 4 24. 3 x 1 4 x 1 25. 1 x y x y x3 y3 26. 27. 28 21 x 2 2 2 1 x 1 4x 1 x 2 3 2 21 x y x2 xy x y 2 2 21 1 x 1 3 1 2 6x 2 x 1 2 1 4 3 2 x2 y2 1 2 2 In Exercises 29–42, express in lowest terms. 29. 3 4 12 5 10 9 31. 3a2c 4ac 8ac3 9a2c4 7x 11y 66y2 14x3 30. 10 45 6 14 1 2 32. 6x2y 2x y 21xy 34. ab c2 cd a2b ad bc2 43. 5 12 4 14 46. 3x2y 2 xy 2 1 3xyz x2y 44. 47. 100 52 27 26 x 3 x 4 2x x 4 45. uv v2w uv u2v 48 x2 2x x2 4 49. x y x 2y x y 2 a xy b
50. u3 v3 u2 v2 u2 uv v2 u v 51. c d 2 2 1 c2 d2 cd 52 53. 1 x2 1 x 1 y2 55. 3 6 y 1 1 y 1 1 3x 5 6x2 1 4y 1 y 1 x2 57. 59 54. 56. 58. 60. 33. 35. 37. 39. 40. 41. 42. 3x 9 2x 8x2 x2 9 36. 4x 16 3x 15 2x 10 x 4 5y 25 3 y2 y2 25 38. 6x 12 6x 8x2 x 2 u u 1 u2 1 u2 t2 t 6 t2 6t 9 t2 4t 5 t2 25 2u2 uv v2 4u2 4uv v2 8u2 6uv 9v2 4u2 9v2 2x2 3xy 2y2 6x2 5xy 4y2 6x2 6xy x2 xy 2y2 In Exercises 61–67, find a numerical example to show that the given statement is false. Then find the mistake in the statement and correct it. 61. 1 a 1 b 1 a b 62. x2 x2 x6 1 x3 63. 2 1 1a 1bb a 1 a b 64. r s r t 1 s t 65. u v v u 1 66. 1 x 1 y 1 xy 1x 1y 67. A 1 1x 1y B x y 986 Algebra Review A.5 The Coordinate Plane The Distance Formula We shall often identify a point with its coordinates and refer, for example, to the point (2, 3). When dealing with several points simultaneously, it is customary to label the coordinates of the first point the sec1 x3, y32 and so on.* Once the plane is ond point coordinatized, it’s easy to compute the distance between any two points: the third point x1, y12 x2, y22 , , , 1 1 The Distance Formula The distance between points (x1, y1) x2)2 (y1 and y2)2 . (x2, y2) is 2(x1 y Before proving the distance formula, we shall see how it is used. −2 2 (−1, −3) −2 −4 10 Figure A.5-1 x Example 1 To find the distance between the points 1, 3 A.5–1, substitute and formula. x1, y12 for 1 2 1 1 1, 3 1 2, 4 2 for 2 and 1 x2, y22 1 2, 4 in Figure in the distance 2 (2, −4) Distance formula: Substitute: Simplify: 1 2 x22 distance 2 x1 1 2 2 2 3 2 2 2 19 1 110 2 1 1 1 2 1 y22 y1 3 1 2 1 3 4 2 4 2 22 The order in which the points are used in the distance formula doesn’t make a difference. If we substitute for x2, y22 1, 3 x1, y12 2, 4 and for , 1 2 1 2 1 1 we get the same answer 232 1 2 110. 1 2 4 1 2 ■ CAUTION cannot be 2a 2 4b 2 simplified. In particular, it is not equal to a 2b. Example 2 To find the distance from (a, b) to b for numbers, substitute a for tance formula: x1, 2a, b 1 y1, , 2a for 2 where a and b are fixed real x2, in the disfor and b y2 2 x1 1 x22 2 y1 1 y22 1 a 2a 2 2 2 2 a 2 1 2a2 4b2 22 2 2a2 2 2b 1 2 2 ■ x1 * “ ” is read “x-one” or “x-sub-one”; it is a single symbol denoting the first coordinate of the first point, just as c denotes the first coordinate of (c, d). Analogous remarks apply to y1, x2, and so on. Section A.5 The Coordinate Plane 987 Proof of the Distance Formula and Q in the plane. We must find length d of line segment PQ. Figure A.5-2 shows typical points P y y1 y2 ⎜y1 − y2 ⎜ P(x1, y1) d R x1 Q(x2, y2) x x2 ⎜x1 − x2 ⎜ Figure A.5-2 x2 to As shown in Figure A.5-2, the length of RQ is the same as the distance from x1 Similarly, the length y2 0 of PR is the same as the distance from y1 . According to the Pythagorean Theorem* the length d of PQ is given by: on the x-axis (number line), namely, y2 . to on the y-axis, namely, 0 x2 0 x1 y1 x1 Since Length PQ 2 d2 0 c2 length PR 1 y2 0 y1 0 c2 0 , 2 y22 Since the length d is nonnegative, we must have d 2 y22 length RQ 2 2 x2 0 (because x22 c2 0 d2 2 2 y1 y1 x1 x1 x22 The distance formula can be used to prove the following useful fact (see Exercise 54). this equation becomes: The Midpoint Formula The midpoint of the line segment from y2 2 x2 2 y1 x1 , a b (x1, y1) to (x2, y2) is y (−1, 4) 2 −1 Example 3 (1, 5 2) (3, 1) x To find the midpoint of the segment joining 1 3, x1 mula in the box with x2 2 x2 1 3 2 1, y2 2 4, x1 y1 y1 , , b a a 4 1 2 3 as shown in Figure A.5-3. 1, 4 and 2 y2 and (3, 1), use the for 1. The midpoint is 5 2b 1, a b ■ Figure A.5-3 *See the Geometry Review Appendix. 988 Algebra Review y y = x2 − 2x − 1 4 3 2 1 (−1, 2) x 1 2 3 (1.5, −1.75) −2 −1 (0, −1) −2 Figure A.5-4 (x, y) r (c, d) Figure A.5-5 Circle Equation Graphs A graph is a set of points in the plane. Some graphs are based on data points. Other graphs arise from equations, as follows. A solution of an equation in variables x and y is a pair of numbers such that the substitution of the first number for x and the second for y produces a true statement. For instance, 5x 7y 1 because 3, 2 1 2 5 3 7 is a solution of 2 1 1 is not a solution because 2 2 2, 3 2 1 The graph of an and equation in two variables is the set of points in the plane whose coordinates are solutions of the equation. Thus the graph is a geometric picture of the solutions. 5 2 1 7 3 1. Example 4 y x2 2x 1 The graph of is shown in Figure A.5-4. You can readily verify that each of the points whose coordinates are labeled is a soluis a solution because tion of the equation. For instance, 1 02 2 0, 1 1. 0 1 2 1 2 ■ Circles If (c, d) is a point in the plane and r a positive number, then the circle with center (c, d) and radius r consists of all points (x, y) that lie r units from (c, d), as shown in Figure A.5-5. According to the distance formula, the statement that “the distance from (x, y) to (c, d) is r units” is equivalent to: Squaring both sides shows that (x, y) satisfies this equation r2 1 Reversing the procedure shows that any solution (x, y) of this equation is a point on the circle. Therefore, 2 1 2 The circle with center (c, d) and radius r is the graph of (x c)2 (y d)2 r 2. x c 2 y d 2 r2 We say that ter (c, d) and radius r. If the center is at the origin, then the equation has a simpler form: is the equation of the circle with cenand 0, 0 c Circle at the Origin The circle with center (0, 0) and radius r is the graph of x2 y2 r2. Section A.5 The Coordinate Plane 989 Example 5 a. Letting r 1 shows that the graph of is the circle of radius 1 centered at the origin, as shown in Figure A.5-6. This circle is called the unit circle. x2 y2 1 b. The circle with center 3, 2 1 and radius 2, shown in Figure A.5-7, is 2 the graph of the equation x 11 2 3 1 22 1 y 2 2 2 22 or equivalently. ■ y 2 1 −2 −1 −1 −2 x 1 2 (−3, 2) 2 y 4 3 2 1 x Figure A.5-6 Figure A.5-7 −5 −4 −3 −2 −1 1 Example 6 Find the equation of the circle with center (2, 4). 1 3, 1 2 that passes through x Solution We must first find the radius. Since (2, 4) is on the circle, the radius is the distance from (2, 4) to 1 2 3 as shown in Figure A.5-8, namely, 2 11 25 126 3 22 The equation of the circle with center at 3, 1 and radius 26 2 x2 6x 9 y2 2y 1 26 x2 y2 6x 2y 16 0. 1 1 22 1 y 1 2 2 1 1 126 2 B 126 is ■ The equation of any circle can always be written in the form x2 y2 Bx Cy D 0 for some constants B, C, D, as in Example 6 (where D 16 determined. C 2, Conversely, the graph of such an equation can always be B 6, . 2 y (2, 4) 26 (3, −1) Figure A.5-8 990 Algebra Review Example 7 To find the graph of sides by 3 and rewrite the equation as 3x2 3y2 12x 30y 45 0, we divide both x2 4x 1 2 1 y2 10y 2 15. Next we complete the square in both expressions in parentheses (see page x2 4x, we add 4 (the square of half the 91). To complete the square in coefficient of x) and to complete the square in we add 25 (why?). In order to have an equivalent equation we must add these numbers to both sides: y2 10y x2 4x 4 1 1 x 2 y2 10y 25 y 5 2 15 4 25 2 2 14 2 1 2 2 1 Since radius 14 1 114. 114 2, 2 this is the equation of the circle with center (2, 5) and ■ Exercises A.5 In Exercises 1–8, find the distance between the two points and the midpoint of the segment joining them. In Exercises 11–14, find the equation of the circle with given center and radius r. 1. 3. 5. 7. 3, 5 1 1, 5 1 12, 7 2 2, 1 2 13, 2 B a, b , 2 1 1 b, a 2 2. 4. 6. 8. 1, 5 2 3, 2 1 1 , 2, 4 2 2, 3 2 1 1, 15 , 1 A , A B 2 12, 13 s, t , 2 1 1 0, 0 2 9. According to the Information Technology Industry Council, there were about 12 million personal computers sold in the United States in 1992 and about 36 million in 1998. a. Represent the data graphically by two points. b. Find the midpoint of the line segment joining these points. c. How might this midpoint be interpreted? What assumptions, if any, are needed to make this interpretation? 10. A standard baseball diamond (which is actually a square) is shown in the figure at right. Suppose it is placed on a coordinate plane with home plate at the origin, first base on the positive x-axis, and third base on the positive y-axis. The unit of measurement is feet. a. Find the coordinates of first, second, and third base. b. If the left fielder is at the point (50, 325), how far is he from first base? c. How far is the left fielder in part b from the right fielder, who is at the point (280, 20)? 11. 13. 1 1 3, 4 ; r 2 2 ; r 12 0, 0 2 12. 14. 1 1 2, 2 2 B In Exercises 15–18, sketch the graph of the equation. 15. 16. 17. 18 y2 2nd base 3rd base 90 ft 90 ft Pitcher's mound 60.5 ft 1st base 90 ft 90 ft Home plate Section A.5 The Coordinate Plane 991 In Exercises 19–24, find the center and radius of the circle whose equation is given. 19. x2 y2 8x 6y 15 0 20. 15x2 15y2 10 21. x2 y2 6x 4y 15 0 22. x2 y2 10x 75 0 23. x2 y2 25x 10y 12 24. 3x2 3y2 12x 12 18y In Exercises 25–27, show that the three points are the vertices of a right triangle, and state the length of the hypotenuse. [You may assume that a triangle with sides of lengths a, b, c is a right triangle with hypotenuse c provided that a2 b2 c2. ] 25. 1 1, 1 , 1 0, 0 2 12 2 2, 2 , 1 2 12 2 2 12 2 b , 26. a , 0 , b a 0, 0 , 1 2 27. 3, 2 1 0, 4 , 1 2 , 2 1 2, 3 2 28. What is the perimeter of the triangle with vertices (1, 1), (5, 4), and 1 2, 5 ? 2 40. Find the three points that divide the line segment 4, 7 to 1 2 10, 9 into four parts of equal 2 from 1 length. 41. Find all points P on the x-axis that are 5 units from (3, 4). Hint: P must have coordinates (x, 0) for some x and the distance from P to (3, 4) is 5. 42. Find all points on the y-axis that are 8 units from 2, 4 . 2 1 43. Find all points with first coordinate 3 that are 6 units from 1 2, 5 . 2 44. Find all points with second coordinate 1 that are 4 units from (2, 3). 45. Find a number x such that (0, 0), (3, 2), and (x, 0) are the vertices of an isosceles triangle, neither of whose two equal sides lie on the x-axis. 46. Do Exercise 45 if one of the two equal sides
lies on the positive x-axis. 47. Show that the midpoint M of the hypotenuse of a right triangle is equidistant from the vertices of the triangle. Hint: Place the triangle in the first quadrant of the plane, with right angle at the origin so that the situation looks like the figure. In Exercises 29–36, find the equation of the circle. 29. Center (2, 2); passes through the origin. y (0, r) 30. Center 1, 3 ; 2 1 passes through 31. Center (1, 2); intersects x-axis at 32. Center (3, 1); diameter 2. 4, 2 . 2 and 3. 1 1 33. Center 5, 4 1 the x-axis. 34. Center 2, 6 1 ; tangent (touching at one point) to ; tangent to the y-axis. 2 2 35. Endpoints of diameter are (3, 3) and 1 36. Endpoints of diameter are 3, 5 1 and 1 2 37. One diagonal of a square has endpoints 1, 1 . 2 7, 5 . 2 3, 1 1 Find the endpoints of the other 2, 4 and 1 diagonal. . 2 38. Find the vertices of all possible squares with this property: Two of the vertices are (2, 1) and (2, 5). Hint: There are three such squares. 39. Do Exercise 38 with (c, d) and (c, k) in place of (2, 1) and (2, 5). 992 Algebra Review M x (s, 0) 48. Show that the diagonals of a parallelogram bisect each other. Hint: Place the parallelogram in the first quadrant with a vertex at the origin and one side along the x-axis, so that the situation looks like the figure. 2 y (a, b) c (a + c, b) c (c, 0) x 49. Show that the diagonals of a rectangle have the same length. Hint: Place the rectangle in the first quadrant of the plane and label its vertices appropriately, as in Exercises 47–48. 50. If the diagonals of a parallelogram have the same length, show that the parallelogram is actually a rectangle. Hint: See Exercise 48. 51. Critical Thinking For each nonzero real number k, the graph of all possible such circles. 1 2 x k 2 y2 k2 is a circle. Describe 52. Critical Thinking Suppose every point in the coordinate plane is moved 5 units straight up. a. To what point does each of these points go: (2, 2), (5, 0), (5, 5), (4, 1)? 0, 5 , b. Which points go to each of the points in 1 2 part a? c. To what point does (a, b) go? a, b 5 d. To what point does 2 1 4a, b e. What point goes to ? 1 f. What points go to themselves? 2 go? 53. Critical Thinking Let (c, d) be any point in the c 0. Prove that (c, d) and plane with 2 lie on the same straight line through the origin, on opposite sides of the origin, the same distance from the origin. Hint: Find the midpoint of the line segment joining (c, d) and c, d . 1 c, d 1 2 54. Critical Thinking Proof of the Midpoint Formula Let P and Q be the points x2, y22 respectively and let M be the point with coordinates x1, y12 and 1 1 x1 a x2 2 , y1 y2 2 . b Use the distance formula to compute the following: a. the distance d from P to Q; b. the distance from M to P; d1 c. the distance from M to Q. d2 d2. d. Verify that d1 d2 e. Show that d1 1 2 Hint: Verify that 1 2 d. and d2 d1 d. d f. Explain why parts d and e show that M is the midpoint of PQ. Section A.5 The Coordinate Plane 993 ADVANCED TOPICS B.1 The Binomial Theorem The Binomial Theorem provides a formula for calculating the product x y for any positive integer n. Before we state the theorem, some pre1 liminaries are needed. 2 n Let n be a positive integer. The symbol product of all the integers from 1 to n. For example, n! (read n factorial) denotes the 2! 1 2 2, 3! 1 2 3 6, 4! 1 2 3 4 24, 5! 1 2 3 4 5 120, 10 10 3,628,800 . In general, we have this result: n Factorial Let n be a positive integer. Then p n! 1 2 3 4 (n 2)(n 1)n. 0! is defined to be the number 1. Learn to use your calculator to compute factorials. You will find ! in the PROB (or PRB) submenu of the MATH or OPTN menu. Calculator Exploration 15! is such a large number your calculator will switch to scientific notation to express it. Many calculators cannot compute factorials larger than 69! If yours does compute larger ones, what is the largest factorial that you can compute without getting an error message? 994 Advanced Topics If r and n are integers with 0 r n, then Binomial Coefficients Either of the symbols n r b a or nCr denotes the number n! r!(n r)! . n r b a is called a binomial coefficient. For example, 5C3 4C2 a 5 3b 4 2b a 5! 5 3 3! 1 4! 4 2 2! 1 ! 2 ! 2 5! 3!2! 4! 2!2 21 21 1 6. 10 nCr Binomial coefficients can be computed on a calculator by using Comb in the PROB (or PRB) submenu of the MATH or OPTN menu. or Calculator Exploration Compute 56C47 56 47b a . Although calculators cannot compute 475!, they can compute many binomial coefficients, such as 475 400b a , because most of the factors cancel out (as in the previous example). Check yours. Will it also compute 475 20 b a ? The preceding examples illustrate a fact whose proof will be omitted: Every binomial coefficient is an integer. Furthermore, for every nonnegative integer n, n 0b a 1 and n nb a 1 because n 0b a n nb a n! n 0 n! n n 2 0! 1 n! 1 n! 0!n! n! n! ! 1 and n! n!0! n! n! 1. ! 2 If we list the binomial coefficients for each value of n in this manner, we find that they form a rectangular array. Section B.1 The Binomial Theorem 995 o 3 0b a 4 0b a 0 0b a 2 1b a 4 2b a 1 1b a 3 2b a 1 0b a 3 1b a 2 0b a 4 1b a 2 2b a 4 3b a 3 3b a 4 4b a ∞ Calculating each binomial coefficient, we obtain the following array of numbers: row 0 row 1 row 2 row 3 row 4 1 † ∞ This array is called Pascal’s triangle. Its pattern is easy to remember. Each entry (except the 1’s at the beginning or end of a row) is the sum of the two closest entries in the row above it. In the fourth row, for instance, 6 is the sum of the two 3’s above it, and each 4 is the sum of the 1 and 3 above it. See Exercise 47 for a proof. In order to develop a formula for calculating we first calculate these products for small values of n to see if we can find some kind of pattern: 1 2 x y n, () 1x 1y 1x2 2xy 1y2 1x3 3x2y 3xy2 1y3 1x4 4x3y 6x2y2 4xy3 1y4 One pattern is immediately obvious: the coefficients here (shown in color) for example, this are the top part of Pascal’s triangle! In the case means that the coefficients are the numbers n 4, 1 4 6 4 1 4 0b a , 4 1b a , 4 2b a , 4 3b , a 4 4b . a 996 Advanced Topics If this pattern holds for larger n, then the coefficients in the expansion of x y are n 1 2 n 0b a , n 1b a , n 2b a , n 3b a , p , n n1b a , n nb a . As for the xy-terms associated with each of these coefficients, look at the pattern in () above: the exponent of x goes down by 1 and the exponent of y goes up by 1 as you go from term to term, which suggests that the terms of the expansion of (without the coefficients) are: x y n 1 2 xn, xn1y, xn2y2, xn3y3, . . . , xyn1, yn. Combining the patterns of coefficients and xy-terms and using the fact that n 0b a 1 and the expansion of n nb a x y 1 1 n. 2 suggests that the following result is true about The Binomial Theorem For each positive integer n, (x y)n xn n 1b a xn1y xn2 y2 n 2b a xn3y3 p n 3b a n n 1b a xyn1 yn. Using summation notation and the fact that the Binomial Theorem compactly as n 0b a 1 n nb a , we can write x y 2 1 n n a j0 a n jb xnjy j. The Binomial Theorem will be proved in Section B.2 by means of mathematical induction. We shall assume its truth for now and illustrate some of its uses. Example 1 Expand x y 8. 2 1 Solution We apply the Binomial Theorem in the case n 8: x y 1 2 8 x8 8 1b a x7y a 8 2b 8 4b a x6y2 x4y4 8 3b a 8 5b a x 5y3 x3y5 8 6b a x2y6 8 7b a xy7 y8. The coefficients can be computed individually by hand or by using (or COMB) on a calculator; for instance, nCr 8C2 8 2b a 8! 2!6! 28 or 8C3 8 3b a 8! 3!5! 56. Section B.1 The Binomial Theorem 997 Alternatively, you can display all the coefficients at once by making a table 8Cx, of values for the function as shown in Figure B.1-1 at left. x f 1 2 Substituting these values in the preceding expansion, we have x y 8 x8 8x7y 28x6y2 1 2 Example 2 Expand 1 z 6. 2 1 Solution Note that y z, 1 z 1 n 6: and Figure B.1-1 56x5y3 70x4y4 56x3y5 28x2y6 8xy7 y8. ■ z 2 1 and apply the Binomial Theorem with x 1, 1 z 2 1 6 16 6 1b a z 15 1 2 6 2b a z 14 1 2 2 6 3b a 13 1 z 3 2 1 6 1b a z 6 2b a z2 6 3b a z3 12 6 4b a 6 4b a 4 z 2 1 6 5b z4 6 5b a z5 z6 1 6z 15z2 20z3 15z4 6z5 z6. ■ Example 3 2 x The unemployment rate in the United States can be modeled by .0051x4 .1813x3 2.2202x2 10.8212x 12.1394 f 1 where x 4 corresponds to 1994.* Write and simplify the rule of a function g(x) that provides the same information as f but has x 0 corresponding to 1994. 4 x 14 1 2 *Source: Bureau of Labor Statistics Solution f The graph of g will be the graph of f shifted 4 units to the left, which x g means that 2 1 x4 .0051 2 1 x4 4x3 .0051 x 4 . 2 1 4 .1813 210.8212 12.1394 x4 1813 32.2202 x4 1 2 1 2 4x 6x2 4 4 4 1 2 1 2 1 2 3x x3 3x2 4 4 1 2 1 3 x2 2x 2.2202 2 4 4 2 1 3 x 4 10.8212 12.1394 2 1 2 2 1 4 x4 0051x4 .099x3 .534x2 .455x 5.92 g x 1 2 ■ 998 Advanced Topics Example 4 Show that 1.001 1 2 1000 7 2 without using a calculator. Solution 1 .001 and apply the Binomial Theorem with x 1, We write 1.001 as y .001, and n 1000: 1000 1.001 1 2 1 .001 1 11000 1000 2 1000 1 b a 1 1000 a 1999 1 .001 2 other positive terms .001 other positive terms. But 1000 1 b 1000! 1!999! 1,000 and 2 other positive terms. a .001 1 1 2 1 b 1 1000 999! 1000. 999! 1000 1 1 1.001 1 2 Therefore, 1 b 1 other positive terms 1000 a .001 2 2 Hence, 1.001 1 2 1000 7 2. ■ n. Sometimes we need to know only one term in the expansion of If you examine the expansion given by the Binomial Theorem, you will see that in the second term y has exponent 1, in the third term y has exponent 2, and so on. Thus, 1 2 x y Properties of the Binomial Expansion In the binomial expansion of (x y)n, The exponent of y is always one less than the number of the term. Furthermore, in each of the middle terms of the expansion, The coefficient of the term containing yr is n . r b a The sum of the x exponent and the y exponent is n. For instance, in the ninth term of the expansion of x y 13, 2 1 y has expo- nent 8, the coefficient is 13 8 b a , and x must have exponent 5 (
since 8 5 13 2 . Thus, the ninth term is 13 8 b a x5y8. Section B.1 The Binomial Theorem 999 Example 5 Find the ninth term of the expansion of 2x2 a 13 41y 16 b . Solution We shall use the Binomial Theorem with n 13 and with 2x2 in place of x and in place of y. The remarks on the previous page show that the 41y 16 ninth term is Since 41y y 13 8 b 1 a 2x2 5 2 a 1 4 and 41y 16 b 8 13 8 b 1 a 2x2 5 2 a 8 41y 16 b . 1 2, 1 4 16 1312 3 1 2 2 we can simplify as follows 25 x2 B 2 y2 34 24 34 x10y2 13 12 11 10 9 5 4 3 2 13 8 b 25x10 34 x10y2 13 8 b a 13 8 b a 286 9 x10y2. ■ 20. a 1c 1 1cb 7 22. 1 3x 2 x2 13 13 1 B i2 1 where 19. 1 c 10 2 1 21. 23. 24. 25. x 3 x 1 1 13 A 13 1 A 1 i 6, 2 1 26. The median income of U.S. households (in thousands of dollars) from 1992 through 2002 can be modeled by x f 1 2 .001x4 .003x3 .211x2 .322x 30.345 2 x 12 1 2 where x 2 corresponds to 1992.* Write and simplify the rule of a function g(x) that provides the same information as f but has x 0 corresponding to 2000. *Source: U.S. Census Bureau Exercises B.1 In Exercises 1–10, evaluate the expression. 1. 6! 2. 11! 8! 3. 12! 9!3! 4. 9! 8! 7! 5. 7. 8. 9. 5 3b a 5 2b a 6 a 3b 6. 12 11b a 11 10b a 7 0 b a 6 0b a 6 0b a 6 1b a 6 1b a 6 2b a 6 2b a 6 3b a 6 3b a 6 4b a 6 4b a 6 5b a 6 5b a 6 6b a 6 6b a 100 96 b a 10. 75 72b a In Exercises 11–16, expand the expression. 11. 14 12. 15. a b 7 2 2x y2 1 1 5 2 13. 16. a b 5 2 3u v3 1 1 6 2 In Exercises 17–26, use the Binomial Theorem to expand and (where possible) simplify the expression. 17. A 1x 1 6 B 18. A 2 1y 5 B 1000 Advanced Topics In Exercises 27–32, find the indicated term of the expansion of the given expression. 27. third, 29. fifth, 1 5 2 x y 1 c d 28. fourth 30. third, 1 31. fourth, 7 2 u u 2 b a 32. fifth, 1x 12 A B 7 33. Find the coefficient of 9. 2x y2 34. Find the coefficient of 10. x3 3y 2 2 1 1 x5y8 in the expansion of x12y6 in the expansion of 35. Find the coefficient of 1 x3 in the expansion of 2x 1 x2b a 6 . 36. Find the constant term in the expansion of y 1 2yb a 10 . 37. a. Verify that 9 1b a 9 and 9 8b a 9. b. Prove that for each positive integer n, n 1b a n and a when n n 1b n 9 n. and Note: Part a is just the case n 1 8. 38. a. Verify that 7 2b a b. Let r and n be integers with 7 . 5b a 0 r n. Prove n rb that a case when n n rb a n 7 and r 2. . Note: Part a is just the 39. Prove that for any positive integer n, 2n n 0b a n 1b a n 2b a p n nb a . Hint: 2 1 1. 40. Prove that for any positive integer n, n 0b a n 1b a n 2b a n 3b a n 4b a p 1 k 2 1 n kb a p 1 n 2 1 n nb a 0. 42. a. Use DeMoivre’s Theorem on page 441 to find cos u i sin u 4. 2 1 b. Use the fact that the two expressions obtained in part a and in Exercise 41 must be equal to express cos 4 and sin cos in terms of sin and 4u u. u u 43. a. Let f be the function given by f x5. Let x 1 2 h be a nonzero number and compute f (but leave all binomial x h f x 2 1 1 coefficients in the form 2 5 rb a here and below). b. Use part a to show that h is a factor of and find 1 x h 2 h c. If h is very close to 0, find a simple approximation of the quantity f 1 See part b 44. Do Exercise 43 with x5. x f 45. Do Exercise 43 with x5. x f 1 1 2 2 x8 in place of x12 in place of f x 1 2 f x 2 1 46. Let n be a fixed positive integer. Do Exercise 43 in place of x5. xn with x x f f 1 2 1 2 47. Let r and n be integers such that a. Verify that b. Verify that c. Prove that 1 1 2 2 ! ! 0 r n 1b 1b a Hint: Write out the terms on the 2 4 ! 2 4 for any ! r n 1. left side and use parts a and b to express each of them as a fraction with denominator r 1 1 simplify the numerator, and compare the result Then add these two fractions, n r !. ! 1 2 2 with n 1 r 1b a . d. Use part c to explain why each entry in Pascal’s triangle (except the 1’s at the beginning or end of a row) is the sum of the two closest entries in the row above it. 48. a. Find these numbers and write them one below 112, the next: 113, 110, 114. 111, b. Compare the list in part a with rows 0 to 4 of Pascal’s triangle. What’s the explanation? and row 5 of c. What can be said about 115 41. Use the Binomial Theorem with to find cos u i sin u x sin u where 4 1 2 y cos u i2 1. and Pascal’s triangle? d. Calculate all integer powers of 101 from 1010 1018, list the results one under the other, and compare the list with rows 0 to 8 of Pascal’s triangle. What’s the explanation? What happens with 1019? to Section B.1 The Binomial Theorem 1001 B.2 Mathematical Induction Mathematical induction is a method of proof that can be used to prove a wide variety of mathematical facts, including the Binomial Theorem, DeMoivre’s Theorem, and statements such as: The sum of the first n positive integers is the number 2n 7 n for every positive integer n. For each positive integer n, 4 is a factor of 7n 3n. n n 1 2 1 2 . All of the preceding statements have a common property. For example, a statement such as The sum of the first n positive integers is the number n(n 1) 2 or, in symbols is really an infinite sequence of statements, one for each possible value of n: n 1: n 2: n 3 and so on. Obviously, there isn’t time enough to verify every one of the statements on this list, one at a time. But we can find a workable method of proof by examining how each statement on the list is related to the next statement on the list. For example, for n 50, the statement is 1 2 3 p 50 50 51 1 2 2 . At the moment, we don’t know whether or not this statement is true. But just suppose that it were true. What could then be said about the next statement, the one for n 51: 1 2 3 p 50 51 51 52 1 2 2 ? Well, if it is true that 1 2 3 p 50 50 51 1 2 2 then adding 51 to both sides and simplifying the right side would yield these equalities: 1002 Advanced Topics 1 2 3 p 50 51 50 51 1 2 3 p 50 51 50 2 50 1 51 2 1 51 2 2 2 2 1 51 2 1 2 3 p 50 51 1 51 2 1 2 3 p 50 51 51 2 2 51 1 2 51 1 2 50 2 2 52 1 2 . 2 Since this last equality is just the original statement for clude that n 51, we con- If the statement is true for n 50, then it is also true for n 51. We have not proved that the statement actually is true for that if it is, then it is also true for n 51. n 50, but only We claim that this same conditional relationship holds for any two consecutive values of n. In other words, we claim that for any positive integer k, 1 If the statement is true for n k 1. n k, then it is also true for The proof of this claim is the same argument used earlier (with k and k 1 in place of 50 and 51): If it is true that Original statement for n k ] then adding these equalities: k 1 to both sides and simplifying the right side produces 21 Original statement for n k 1 ] 1 is valid for each positive integer k. We We have proved that claim have not proved that the original statement is true for any value of n, but n k, only that if it is true for Applyk 1, 2, 3, . . . , we see that a recursive pattern emerges. ing this fact when Beginning with the smallest positive integer, 1, then it is also true for n k 1. Section B.2 Mathematical Induction 1003 ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ...2 ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ If the statement is true for n 1 1 2; If the statement is true for n 2 1 3; If the statement is true for n 3 1 4; If the statement is true for n 50 1 51; If the statement is true for n 51 1 52; ... and so on. n 1, then it is also true for n 2, then it is also true for n 3, then it is also true for n 50, then it is also true for n 51, then it is also true for 2 n 1 in a position /2. We are finally , the statement is true for n 4, for and hence for the original statement is true for every positive integer n. the original statement: n 1 since 2 . Since and hence it must also be true for and so on, for every value of n. Therefore, Now apply in turn each of the propositions on list n 2, to prove Obviously, it is true for 2 1 n 3, /2. 2 The preceding proof is an illustration of the following principle: Suppose there is given a statement involving the positive integer n and that: (i) The statement is true for n 1. n k (ii) If the statement is true for (where k is any Principle of Mathematical Induction positive integer), then the statement is also true for n k 1. Then the statement is true for every positive integer n. Property (i) is simply a statement of fact. To verify that it holds, you must This is usually easy, as in the prove the given statement is true for preceding example. n 1. Property (ii) is a conditional property. It does not assert that the given statement is true for then it is also true for So to verify that property (ii) holds, you need only prove this conditional proposition: but only that if it is true for n k 1. n k, n k, If the statement is true for n k, then it is also true for n k 1. In order to prove this, or any conditional proposition, you must proceed as in the previous example: Assume the “if” part and use this assumption to prove the “then” part. As we saw earlier, the same argument will usually work for any possible k. Once this conditional proposition has been proved, you can use it together with property (i) to conclude that the 1004 Advanced Topics given statement is necessarily true for every n, just as in the preceding example. Thus proof by mathematical induction reduces to two steps: Step 1 Prove that the given statement is true for n 1. Step 2 Let k be a positive integer. Assume that the given statement is true for n k. Use this assumption to prove that the statement is true for n k 1. Step 2 may be performed before step 1 if you wish. Step 2 is sometimes referred to as the inductive step. The assumption that the given statein this inductive step is called the induction ment is true for hypothesis. n k Example 1 Prove that 2n 7 n for every positive integer n. Solution Here the statement involving n is 2n 7 n. Step 1 When n 1, we have the statement 21 7 1. This is obviously true. Step 2 Let k be any positive integer. We assume that the statement is true for n k, 2k 7 k. We shall use this assumption to 2k1 7 k 1. prove that the statement is true for t
hat is, we assume that n k 1, that is, that We begin with the induction hypothesis:* of this inequality by 2 yields: 2k 7 k. Multiplying both sides 3 2 2k 7 2k 2k1 7 2k. Since k is a positive integer, we know that of the inequality we have k 1, k 1. Adding k to each side k k k 1 2k k 1. Combining this result with inequality 3 , we see that 2k1 7 2k k 1. *This is the point at which you usually must do some work. Remember that what follows is the “finished proof.” It does not include all the thought, scratch work, false starts, and so on that were done before this proof was actually found. Section B.2 Mathematical Induction 1005 2k1 7 k 1. ThereThe first and last terms of this inequality show that n k 1. This argument works for any fore, the statement is true for positive integer k. Thus, we have completed the inductive step. By the Principle of Mathematical Induction, we conclude that for every positive integer n. 2n 7 n ■ Example 2 Simple arithmetic shows that 72 32 49 9 40 4 10 and 73 33 343 27 316 4 79. In each case, 4 is a factor. These examples suggest that 7n 3n. This conjecture can be proved by induction as follows. For each positive integer n, 4 is a factor of Step 1 n 1, When 71 31 4 4 1, the statement is “4 is a factor of the statement is true for n 1. 71 31.” Since Step 2 7k1 3k1. 7k 3k 4D. n k, Let us denote the other factor by D, so We must use this assumpthat is, that 4 is a Let k be a positive integer and assume that the statement is true for 7k 3k. that is, that 4 is a factor of that the induction hypothesis is: tion to prove that the statement is true for factor of 7k1 3k1 7k1 7 3k 7 3k 3k1 7k 3k 1 2 7 3 4D 1 2 4 3k 4D 2 7D 3k . [Induction hypothesis] 7 3 4 [ [Factor out 4] 7 7 7 4 7 3k 7 3k 0 ] Here is the proof: 7 3 3k n k 1, [Factor] [Since 3k ] 2 2 7k1 3k1. From this last line, we see that 4 is a factor of Thus, the staten k 1, and the inductive step is complete. Therefore, ment is true for by the Principle of Mathematical Induction the conjecture is actually true for every positive integer n. 1 1 1 1 2 ■ Another example of mathematical induction, the proof of the Binomial Theorem, is given at the end of this section. Sometimes a statement involving the integer n may be false for and (possibly) other small values of n, but true for all values of n beyond a n 1, 2n 7 n2 particular number. For instance, the statement and all larger values of n. A variation on 2, 3, 4. But it is true for the Principle of Mathematical Induction can be used to prove this fact and similar statements. See Exercise 28 for details. is false for n 5 n 1 1006 Advanced Topics A Common Mistake with Induction It is sometimes tempting to omit step 2 of an inductive proof when the given statement can easily be verified for small values of n, especially if a clear pattern seems to be developing. As the next example shows, however, omitting step 2 may lead to error. Example 3 7 1 1 2 is said to be prime if its only positive integer factors are An integer itself and 1. For instance, 11 is prime since its only positive integer factors are 11 and 1. But 15 is not prime because it has factors other than 15 and 1 (namely, 3 and 5). For each positive integer n, consider the number n2 n 11. f n 2 1 You can readily verify that 11, f 2 1 f 13, f 1 2 1 2 3 2 1 17, f 4 1 2 23, f 31 5 1 2 and that each of these numbers is prime. Furthermore, there is a clear pattern: The first two numbers (11 and 13) differ by 2; the next two (13 and 17) differ by 4; the next two (17 and 23) differ by 6; and so on. On the basis of this evidence, we might conjecture: For each positive integer n, the number 2 We have seen that this conjecture is true for nately, however, it is false for some values of n. For instance, when 2, 3, 4, 5. Unfortun 11, 1 n2 n 11 n 1, is prime. n f 112 11 11 112 121. 11 f 1 2 But 121 is obviously not prime since it has a factor other than 121 and 1, namely, 11. You can verify that the statement is also false for but true for n 13. n 12 ■ In the preceding example, the proposition n k, k 1 11. If the statement is true for then it is true for n k 1 k 10 and is false when If you were not aware of this and tried to complete step 2 of an inductive proof, you would not have been able to find a valid proof for it. Of course, the fact that you can’t find a proof of a proposition doesn’t always mean that no proof exists. But when you are unable to complete step 2, you are warned that there is a possibility that the given statement may be false for some values of n. This warning should prevent you from drawing any wrong conclusions. Proof of the Binomial Theorem We shall use induction to prove that for every positive integer n, x y 1 2 n xn n 1b a n 2b a xn1y xn2y2 n 3b a xn3y3 p n n 1b a xyn1 yn. Section B.2 Mathematical Induction 1007 This theorem was discussed and its notation explained in Section B.1. Step 1 n 1, When equation, and the statement reads true. there are only two terms on the right side of the preceding This is certainly x1 y1. x y 1 2 1 Step 2 Let k be any positive integer and assume that the theorem is true for that is, that x y xk2 y2 p k xk xk1y k 1b a k 2b a 1 2 n k, k rb a xkryr p k k 1b a xyk1 yk. [On the right side of this equation, we have included a typical middle term k rb a xkryr. The sum of the exponents is k, and the bottom part of the binomial coefficient is the same as the y exponent.] We shall use this assumption to prove that the theorem is true for that is, that k 1 1 b a xk1y2 p k1 xk1 n k 1, xky 1b a xkryr1 p k 1 k b a xyk yk1. We have simplified some of the terms on the right side; for instance, k 1 But this is the correct state- and 1 1 n k 1: ment for , term is and the bottom part of each binomial coefficient is the same as the y exponent. and so on; the sum of the exponents of each middle The coefficients of the middle terms are k 1 1 b a , , 2 In order to prove the theorem for binomial coefficients: For any integers r and k with we shall need this fact about 0 r 6 k, n k 1, 4 k r 1b a k rb a k 1 r 1b a . A proof of this fact is outlined in Exercise 47 on page 1001. n k 1, k1 we first note that x y x y To prove the theorem for x y k. 1 2 1 Applying the induction hypothesis to 2 21 x y 1 we see that k, 2 x y 1 2 k1 x y 1 2 xk xk2y2 p a xk1y k k 2b 1b a k 2 yr1 p r 1b xk1y p yk yxk xk r1 a 1 k rb a xkryr xyk1 yk k k 1b xk1y p yk. a k 1b a 1008 Advanced Topics xxk k 1b a Next we multiply out the right-hand side. Remember that multiplying by x increases the x exponent by 1 and multiplying by y increases the y exponent by 1. k1 xk1 x y 1 2 k 1b a xky k 2b a k r 1b a xky k 1b a xk1y2 k 2b a xkr1yr k rb xk1y2 p a xkryr1 p k k 1b a k xk2y3 p rb a xkryr1 x2 yk1 xyk xyk yk1 1 k xk r 1b a 1xky k 1b a r1 2 yr2 p k k 1b a xk1y2 p k 2b a xkryr1 p 1 k 1b a k k 1b a xyk yk1. xk1 k r 1b a k rb a 4 Now apply statement to each of the coefficients of the middle terms. k 2b a k 1 , 1 b a k 1b a k 1 2 b a . and so on. Then For instance, with r 1, statement 4 shows that Similarly, with k 1b a the expression above for r 0, 1 x y a k1 x y 1 2 k1 xk1 k k 0b a 1b becomes k 1 2 b a xk1y2 p 1 k 1 1 b a 2 xky k 1 r 1b a xkryr1 p xyk yk1. k 1 k b a n k 1, Since this last statement says the theorem is true for the inductive step is complete. By the Principle of Mathematical Induction the theorem is true for every positive integer n. Exercises B.2 In Exercises 1–18, use mathematical induction to prove that each of the given statements is true for every positive integer n. 1. 1 2 22 23 24 p 2n1 2n 1 6. 7. 1 2 1 4 1 8 p 1 2n 1 1 2n 32 33 34 p 3n1 3n 1 2 8. a 1 1 1 1 1ba 1 1 3b 2ba p 1 n 1 n 1 1 1 nb a n n 1 2 n 1 2. 3. 4. 1 3 5 7 p n2 2 4 6 8 p 2n n2 n 2n 1 2 1 5. 12 22 32 p n2 n 1 n 1 2n 1 2 21 6 9. n 2 7 n 11. 3n 3n 13. 3n 7 n 1 10. 2n 2 7 n 12. 3n 1 2n 14. n 3 2b a 7 n 15. 3 is a factor of 22n1 1 Section B.2 Mathematical Induction 1009 16. 5 is a factor of 24n2 1 17. 64 is a factor of 32n2 8n 9 18. 64 is a factor of 9n 8n 1 19. Let c and d be fixed real numbers. Prove that c c d 2 1 p 1 c 2d c 3d 2c 3 n 1 1 2 d 2 4 20. Let r be a fixed real number with r 1. Prove that 1 r r2 r3 p rn1 rn 1 r 1 . Remember: reduces to 1 r0; r0 1. so when n 1 the left side State the two steps necessary to use this principle to prove that a given statement is true for all (See discussion on page 1005.) n q. In Exercises 29–34, use the Extended Principle of Mathematical Induction (Exercise 28) to prove the given statement. 29. 2n 4 7 n for every n 5. (Use 5 for q here.) 30. Let r be a fixed real number with r 7 1. Then for every integer n 2. (Use 2 n 7 1 nr 1 r 2 1 for q here.) 31. n2 7 n for all n 2 32. 2n 7 n2 for all n 5 as a 33. 3n 7 2n 10n for all n 4 21. a. Write each of x2 y2, product of x y b. Make a conjecture as to how x y x3 y3, and another factor. xn yn and another written as a product of factor. Use induction to prove your conjecture. x4 y4 can be and 22. Let x2 12; x1 32 22 12; 6 2 x3 xn 22 12; for every positive integer n. and so on. Prove that In Exercises 23–27, if the given statement is true, prove it. If it is false, give a counterexample. 23. Every odd positive integer is prime. 24. The number n2 n 17 is prime for every positive integer n. 25. n 1 1 2 2 7 n2 1 for every positive integer n. 26. 3 is a factor of the number n3 n 3 for every positive integer n. 27. 4 is a factor of the number n4 n 4 for every positive integer n. 28. Let q be a fixed integer. Suppose a statement involving the integer n has these two properties: i. The statement is true for ii. If the statement is true for n q. n k (where k is any integer with also true for k q 2 n k 1. , then the statement is Then we claim that the statement is true for every integer n greater than or equal to q. a. Give an informal explanation that shows why this claim should be valid. Note that when q 1, Mathematical Induction. this claim is precisely the Principle of b. The claim made before part a will be called the Extended Principle of Mathematical Induction. 1010 Advanced Topics
34. 2n 6 n! for all n 4 35. Let n be a positive integer. Suppose that there are three pegs and on one of them n rings are stacked, with each ring being smaller in diameter than the one below it (see the figure). We want to transfer the stack of rings to another peg according to these rules: (i) Only one ring may be moved at a time; (ii) a ring can be moved to any peg, provided it is never placed on top of a smaller ring; (iii) the final order of the rings on the new peg must be the same as the original order on the first peg. a. What is the smallest possible number of moves when n 2? n 3? n 4? b. Make a conjecture as to the smallest possible number of moves required for any n. Prove your conjecture by induction. 36. The basic formula for compound interest was discussed in Chapter 5. Prove by induction that the formula is valid whenever x is a positive integer. [Note: P and r are assumed to be constant.] 37. Use induction to prove DeMoivre’s Theorem: For any complex number any positive integer n, z r cos u i sin u 1 and 2 zn rn cos nu 2 1 3 i sin nu . 2 4 1 GEOMETRY REVIEW G.1 Geometry Concepts An angle consists of two half-lines that begin at the same point P, as in Figure G.1-1. The point P is called the vertex of the angle and the halflines the sides of the angle. M θ P a. Q P b. Figure G.1-1 An angle may be labeled by a Greek letter, such as angle in Figure G.1-1a, or by listing three points (a point on one side, the vertex, a point on the other side), such as angle QPM in Figure G.1-1b. u In order to measure the size of an angle, we must assign a number to each angle. Here is the classical method for doing this: 1. Construct a circle whose center is the vertex of the angle. 2. Divide the circumference of the circle into 360 equal parts (called degrees) by marking 360 points on the circumference, beginning with the point where one side of the angle intersects the circle. Label these points and so on. 3. The label of the point where the second side of the angle intersects 1°, 2°, 3°, 0°, the circle is the degree measure of the angle. For example, Figure G.1-2 on the next page shows an angle of measure 25 degrees (in symbols, u and an angle of measure 135°. 25° b 2 Section G.1 Geometry Concepts 1011 30˚ 20˚ 10˚ 0˚ θ 1 2 0˚ 130˚ 140˚ 9 0 ˚ ˚ 80˚ 0 0 1 0 ˚ 1 1 70˚60˚50˚ 4 0 Figure G.1-2 An acute angle is an angle whose measure is strictly between such as angle 90°. and 90°, in Figure G.1-2. A right angle is an angle that measures 90° An obtuse angle is an angle whose measure is strictly between 180°, in Figure G.1-2. such as angle and 0° b u A triangle has three sides (straight line segments) and three angles, formed at the points where the various sides meet. When angles are measured in degrees, the sum of the measures of all three angles of a triangle is always 180. For instance, see Figure G.1-3. A B 60° 90° 45° 30° C C 90° 45° A B Figure G.1-3 17° 143° 20° 49° 71° 60° A right triangle is a triangle, one of whose angles is a right angle, such as the first two triangles shown in Figure G.1-3. The side of a right triangle that lies opposite the right angle is called the hypotenuse. In each of the right triangles in Figure G.1-3, side AC is the hypotenuse. Pythagorean Theorem If the sides of a right triangle have lengths a and b and the hypotenuse has length c, then c2 a2 b2. a c b 1012 Geometry Review Example 1 Consider the right triangle with sides of lengths 5 and 12, as shown in Figure G.1-4. 5 c 12 Figure G.1-4 According to the Pythagorean Theorem the length c of the hypotenuse 169 132, satisfies the equation: we see that c must be 13. c2 52 122 25 144 169. Since ■ Theorem I If two angles of a triangle are equal, then the two sides opposite these angles have the same length. Example 2 Suppose the hypotenuse of the right triangle shown in Figure G.1-5 has length 1 and that angles B and C measure each. 45° B x A 45° 1 45° C Figure G.1-5 Then by Theorem I, sides AB and AC have the same length. If x is the length of side AB, then by the Pythagorean Theorem: x2 x2 12 2x2 1 x2 1 2 x 1 2 1 12 B 12 2 . Section G.1 Geometry Concepts 1013 (We ignore the other solution of this equation, namely, x A 1 2 , since x represents a length here and thus must be nonnegative.) Therefore, the sides of a 90°45°45° triangle with hypotenuse 1 are each of length 12 2 . ■ In a right triangle that has an angle of 30 side opposite the hypotenuse. angle is one-half the length of the 30, the length of the Example 3 Suppose that in the right triangle shown in Figure G.1-6 angle B is and the length of hypotenuse BC is 2. 30° C A 2 x 30° B Figure G.1-6 angle, namely, side AC, has length 30° By Theorem II the side opposite the 1. If x denotes the length of side AB, then by the Pythagorean Theorem: 12 x2 22 x2 3 x 13. ■ Example 4 The right triangle shown in Figure G.1-7 has a 30° angle at C, and side AC has length 13 2 . A 3 2 B 30° C x Figure G.1-7 Theorem II 1014 Geometry Review Let x denote the length of the hypotenuse BC. By Theorem II, side AB has length 1 2 x. By the Pythagorean Theorem: 2 1 2 a x b a x2 4 2 x2 x2 13 2 b 3 4 3 3 4 4 x2 1 x 1. x2 Therefore, the triangle has hypotenuse of length 1 and sides of lengths 1 2 13 2 and . ■ Two triangles, as in Figure G.1-8, are said to be similar if their corre. sponding angles are equal (that is, 2 Thus, similar triangles have the same shape but not necessarily the same size. A D; B E; C F and F C A B D E Figure G.1-8 Section G.1 Geometry Concepts 1015 Theorem III 1016 Geometry Review Suppose triangle ABC with sides a, b, c is similar to triangle C F DEF with sides d, e, f (that is, A D; B E; ). Then a d b e c f . These equalities are equivalent to . The equivalence of the equalities in the conclusion of the theorem is easily verified. For example, since we have a d b e ae db . Dividing both sides of this equation by be yields: ae be a b db be d e . The other equivalences are proved similarly. Example 5 Suppose the triangles in Figure G.1-9 are similar and that the sides have the lengths indicated 10 F Figure G.1-9 Then by Theorem III, In other words, so that length AC length DF length BC length EF 18 s 3 10 3s 1018 s 10 3 b a 18. Similarly, by Theorem III, length AB length DE length BC length EF so that 3 1 r 10 3r 10 r 10 3 . Therefore the sides of triangle DEF are of lengths 10, 10 3 , and 10 3 18. ■ Section G.1 Geometry Concepts 1017 Technology This appendix describes calculator features used throughout the book. The first section focuses on graphing functions and sequences; the second section presents procedures for creating lists, statistics, statistical plots, and regression equations. Students who are unfamiliar with a graphing calculator should complete the entire appendix; all students may use it as a reference to specific features seen in the text. T.1 Graphs and Tables Graphing calculators have a MODE or SET UP feature that allows you to set different modes for number format, type of angle measure, graph type, and drawing mode. Select the following modes by using the procedures explained in detail below for TI-84 Plus/TI-83 and Casio 9850GB-Plus. Number format: Float or Standard Angle measure: Degree Graph type: Function or Rect Drawing mode: Connected or Connect TI-84/TI-83 Press MODE, then use the up/down arrow keys to move from one row to another and the left/right arrow keys to move to a desired selection. When the cursor is on a desired selection, press ENTER to choose that option. After all selections have been made, press 2nd QUIT to return to the home screen. Casio 9850 From the MAIN MENU, select RUN EXE. Then press SHIFT SET UP. Use the up/down arrow key to select the different mode types. Use the F1 through F6 buttons to select the desired setting for each mode. 1018 Technology Appendix Solutions and Graphs The graph of an equation in two variables is the set of points in the plane whose coordinates are solutions of the equation. Thus, the graph is a geometric picture of the solutions. y x2 consists of all points where x is a real numThe graph of ber. A table of values for this function is shown below along with the graphical representation of the points. See Figure T1-1a. The points suggest that the graph looks like Figure T1-1b, which is obtained by connecting the plotted points and extending the graph. , 1 2 x, x2 x 2 1.5 1 0.5 0 0.5 1 1.5 2 y x2 4 2.25 1 0.25 0 0.25 1 2.25 2 0 −1−1 x 1 2 −2 0 −1−1 x 1 2 Figure T1-1a Figure T1-1b Using Technology to Graph Functions Technology improves graphing speed and accuracy by plotting a large number of points quickly. A graphing calculator graphs in the same way you would graph by hand, but it uses many more points. Viewing Windows The first step in graphing with technology is to choose a preliminary viewing window, which is the portion of the coordinate plane that will appear on the screen. Using an inappropriate viewing window may display no portion of a graph or may misrepresent its important characteristics. The viewing window for the graph in Figure T1-2a is the rectangular x, y region indicated by the dashed blue lines. It includes all points 2 1 4 x 5 To display whose coordinates satisfy this viewing window on a calculator, press the WINDOW (or SHIFT V-Window) key, and enter the appropriate numbers, as shown in Figure T1-2b for the TI-84/TI-83. Other calculators are similar. The settings Xscl 1 put the tick marks 1 unit apart on the respective axes. This is usually the best setting for small viewing windows but not for large ones. 3 y 6. Yscl 1 and and Technology Appendix 1019 (−4, 6) −4 −2 (−4, −3) y 6 4 2 −1 −3 (5, 6) x (5, −3) 1 3 5 Figure T1-2a Figure T1-2b Resolution Xres sets the pixel resolution for function graphs. When Xres is 1, functions are evaluated and graphed at each pixel (point) along the x-axis. At 10, functions are evaluated and graphed at every 10th pixel along the x-axis. Some calculators do not have a Xres setting, and on those that do, it should normally be set at 1. Graphing Functions The follow
ing example outlines the procedure for graphing an equation. Example 1 Using Technology to Graph a Function Use technology to graph the relation 2u3 8u 2v 4 0. Solution Step 1 Choose a preliminary viewing window. If it is unknown where the graph lies in the plane, start with a viewing . This window setting is window with called the standard window on most calculators. The window may be adjusted to fit the functions after graphing. 10 x 10 10 y 10 and Figure T1-3a To display the viewing window editor on a calculator, press WINDOW (or SHIFT V-Window) and enter the appropriate numbers, as shown in Figure T1-3a. Xscl and Yscl determine where the tick marks will be displayed on the axes. Setting both to 1 is usually best for small viewing windows but not for large ones. Step 2 Solve for the output variable, if necessary. Calculators can only graph functions in a form where the output variable is expressed as a function of the input variable. In this example, assume that the output variable is v and solve the given equation for v: 1020 Technology Appendix 2u3 8u 2v 4 0 2v 2u3 8u 4 v u3 4u 2 Rearrange terms. Divide by .2 Figure T1-3b 10 10 10 10 Figure T1-3c 6 −6 −3 10 10 10 Figure T1-4 , replace v Because the calculator graphs functions of the form with y and replace u with x. The function can be represented as the following. x 1 2 y f y x3 4x 2 Step 3 Enter the function by selecting the following. TI-84/TI-83 Casio 9850 Y GRAPH from the MAIN MENU Key in the function, as shown in Figure T1-3b. Display the graph by pressing GRAPH (or DRAW). The graph of the function is shown in Figure T1-3c. Step 4 If necessary, adjust the viewing window for a better view. Notice that the point where the graph crosses the y-axis is not clear in the standard window. Changing the viewing window and displaying the graph again shows that the graph crosses the y-axis at 2. Figure T1-3d shows the graph after TRACE—which is discussed below—was pressed and the arrow keys were used to move the cursor to the y-intercept. ■ 3 Graphing Relations Some relations must be written as two separate functions before they can be graphed. The method used to graph the function in Example 1 can be used to graph any equation that can be solved for y. Figure T1-3d Example 2 Graphing a Relation Graph the relation given by the equation 12x2 4y2 16x 12 0 . 10 Solution Solve the equation for y: 12x2 4y2 16x 12 0 4y2 12x2 16x 12 y2 3x2 4x 3 y ± 23x2 4x 3 The graph is shown in Figure T1-4, where the upper portion of the graph and lower portion. y 23x2 4x 3 y 23x2 4x 3 represents represents the ■ Other Graphing Calculator Features Trace The trace feature allows you to display the coordinates of points on a graph of a function by using the left/right arrow keys, as illustrated in Technology Appendix 1021 Technology Tip When you get a blank screen, press TRACE and use the left/right arrow keys. The coordinates of the points on the graph will be displayed at the bottom of the screen, even though they are not in the current window. Use these coordinates as a guide for selecting a viewing window in which the graph does appear. Figure T1-3d. The TRACE feature is located on the different calculators as follows. TI-84/TI-83 Casio 9850 Trace is on the keyboard. Trace is the 2nd function above F1. Zoom Using the ZOOM feature of a graphing calculator will change the viewing window. ZOOM IN may be used to obtain better approximations for the coordinates of a point, while ZOOM OUT may be used to view a larger portion of the graph. TI-84/TI-83 Casio 9850 ZOOM is on the keyboard. ZOOM is the 2nd function above F2. To use the ZOOM feature, select Zoom In (or IN) from the ZOOM menu, move the cursor to the desired point, and press ENTER. The coordinates of the cursor’s position are shown at the bottom of the screen. Repeatedly using Zoom In will give better approximations of the coordinates of points. The scaling factors used to Zoom In or Zoom Out may be adjusted on most calculators. To set the ZOOM factors, look for Set Factors or FACT in the Zoom menu or in the MEMORY submenu of ZOOM of TI-84/TI-83. Maximums and Minimums There are several ways to approximate the coordinates of a high point or a low point: using the TRACE feature as discussed in Example 1, using the ZOOM feature, and using the maximum finder. The maximum/ minimum finder automatically finds the highest/lowest point with the calculator’s greatest degree of accuracy. On some calculators, left and right bounds and an initial guess must be indicated. On other calculators, the cursor must be placed near the maximum point. The maximum finder is referenced by the term in the second column of the chart following this paragraph; it can be found in the menu referred to in the third column. The minimum finder works in a similar way. Minimum finders are found in the same menu. Model TI-84/TI-83 Casio 9850 Reference maximum MAX Menu(s) CALC (2nd TRACE) G-Solv 1022 Technology Appendix Technology Tip To use a square viewing window, use ZSquare, or SQR in the ZOOM menu. This will adjust the x scale while keeping the y scale fixed. Technology Tip For a decimal window that is also square, use ZDecimal in the TI-84/TI-83 ZOOM menu and INIT in the Casio 9850 V-Window menu. Graphing Exploration In a standard viewing window, graph the relation 12x2 4y2 16x 12 0, which was given in Example 2. Use TRACE, ZOOM, and the maximum and minimum finders to find the coordinates of the maximum and minimum points of the relation. Use the up/down arrow keys to switch between the functions that represent the relation. Compare the values you get from each feature and determine which is most accurate, which is fastest, and which is easiest to use. The relative maximum point’s exact coordinates are and the relative minimum point’s exact coordinates are , , . Square Windows Another useful screen is one in which one-unit segments on the x-axis are the same one-unit segments on the y-axis. This type of window is called a square window. Because calculator screens are wider than they are high, the y-axis in a square window must be shorter than the x-axis. On many calculators, such as the TI-84/TI-83, the ratio of height to width is about 2 3 when square. Therefore, a square window with could have 6 y 7 height-to-width ratio. Check your calculator’s height-to-width ratio by looking at the scales after producing a square window. A square window should be used to display circles and perpendicular lines. 20 2 10 x 10 2 : 3 , so an x-axis with 20 units will have on its y-axis about , on a screen that has a 0 y 13 13 2 3 1 units , or Graphing Exploration and y 2x 2 y 0.5x are perpendicular because the The lines product of their slopes is . Graph both in the standard viewing window, and then graph them in a square window. Describe the different appearances of the lines in both window types. 1 Decimal Window When using the TRACE feature, a calculator typically displays points as 2.34042519, for example, rather than as 2.3 or 2.34. A screen in which the points represent values with one or two decimal places is called a decimal window. A decimal window is appropriate when accuracy to one or two decimal places is desired. Technology Appendix 1023 Technology Tip The screen widths, in pixels, of commonly used calculators: TI-84/TI-83 and Casio 9800: 95 TI-86, Sharp 9600, Casio 9850: 127 HP-38: 131 TI-89: 159 TI-92: 239 Figure T1-5a Choosing a viewing window carefully can make the trace feature much more convenient, as the following Exploration demonstrates. Graphing Exploration • Graph y 5 9 1 x 32 2 , which relates the temperature x in degrees Fahrenheit and the temperature y in degrees Celsius. • Use a viewing window with 40 y 40 and 0 x k , where k is one less than the number of pixels, or points, on your calculator screen divided by 10. On a TI-84/TI-83, for example, k 95 1 . (See Technology Tip at left.) 9.4 10 • Use the TRACE feature to determine the Celsius temperatures corresponding to 20°F and 77°F. • Set the window to 32 k 20 x 32 k 20 , where k is the same as above. This is a decimal window with (32, 0) at its center. • Graph the equation again, and use TRACE to determine the Celsius equivalent of 33.8°F. Function Tables Table of Function Values The table feature of a calculator is a convenient way to display points and evaluate functions. By setting the initial input value and the increment value, a table of values may be displayed for functions stored in the memory, as shown in Figure T1-5a. If more than one function is in the function memory, the output values for each are stored in separate columns of a table. To access the table setup screen, select the following. TI-84/TI-83 Casio 9850 TBLSET on the keyboard RANG in the TABLE menu The increment is labeled Tbl on TI-84/TI-83 and Pitch on Casio 9850. y x3 4x 2 Figure T1-5b shows a table of values for the function which is stored as Y1. , ¢ Figure T1-5b Select the following to view the table of values. TI-84/TI-83 Casio 9850 TABLE on the keyboard Press TABL. Graphs On and Off An equation stays in the equation memory until you delete it. If there are several equations in memory and you want to graph only some of them, turn those equations “on’’ and the others “off.’’ On most calculators an equation is “on’’ if its equal sign is shaded and “off’’ if its equal sign is clear. 1024 Technology Appendix Graphing Conventions 1. Unless directed otherwise, use a calculator for graphing. 2. Complete graphs are required unless a viewing window is specified or the context of a problem indicates that a partial graph is acceptable. 3. If the directions say “obtain the graph,” “find the graph,” or “graph the equation,” you need not actually draw the graph on paper. For review purposes, however, it may be helpful to record the viewing window used. 4. The directions “sketch the graph’’ mean to draw the graph on paper, indicating the scale on each axis. This may involve simply copying the display that is on the calculator scre
en, or it may require graphing if the calculator display is misleading. Calculator Exploration 1. Tick Marks a. Set Xscl 1 so that adjacent tick marks on the x-axis are one unit apart. Find the largest range of x values such that the tick marks on the x-axis are clearly distinguishable and appear to be equally spaced. b. Do part a with y in place of x. 2. Viewing Window Look in the ZOOM menu to find out how many built-in viewing windows your calculator has. Take a look at each one. 3. Maximum/Minimum Finders Use your minimum finder to approximate the x-coordinates of the lowest point on the graph of 3 y 8 y x3 2x 5 . in the window with 0 x 5 and The correct answer is x 2 3 A 0.81649658 . How good is your approximation? 4. Square Windows Find a square viewing window on your cal- culator that has 4 x 4 . 5. Dot Graphing Mode To see the points plotted by a calculator without the connecting segments, set your calculator to DOT mode, then graph Try some other equations as well. y 0.5x3 2x2 1 in the standard window. Technology Appendix 1025 Sequence Graphing Many calculators can produce a table of values and graph functions that are defined by recursive and nonrecursive sequences. In a recursive sequence, the nth term is defined in relation to a previous term or terms, such as n 2 3u u . In a nonrecursive 2n 5. sequence, the nth term is a function of the variable n, such as or To enter, display a table of values for, and graph a sequence, select the following. In general, first set the mode to sequence and enter the sequence rule. • Set the window parameters, set the drawing mode (dot or connected), and graph the sequence. • Choose the table settings, and display the table. TI-84/TI-83 MODE Y access the sequence editor screen SEQ nMin u(n) u(nMin) the minimum n value evaluated the rule of the sequence the value of the sequence at the minimum n value (recursive sequence only) WINDOW nMin the smallest n value evaluated; must be an integer the largest n value evaluated nMax PlotStart first term plotted PlotStep incremental n value (graphing only); designates which points are plotted; does not affect sequence evaluation Xmin, Xmax, Xscl, Ymin, Ymax, Yscl same as function graphing values displays the graph of sequences stored in memory table settings TblStart smallest n value ¢ Tbl increment GRAPH TBLSET TABLE displays a table of values of sequences stored in memory Casio 9850 RECUR TYPE a general term of the sequence { n a a linear recursion between two terms linear recursion between three terms un } n1 n2 RANG displays table range settings Start starting value of n End ending value of n a , b value of first term 1 1 (the value of n increments by 1) TABL displays a table of values G-PLT displays a graph of the data in the table Press EXIT to return to the table and EXIT again to return to the sequence editor. NOTE In the TI-84/ TI-83, nMax, PlotStart, and PlotStep must be integers greater than 1. NOTE In the TI-84/ TI-83, both Indpnt (independent) and Depend (dependent) values may be set by the user or the calculator. Input can be set Sharp 9600 by the user or the calculator. Auto: Calculator enters values. Ask: User enters values. 1026 Technology Appendix T.2 Lists, Statistics, Plots, and Regression NOTE Words printed in all caps refer to buttons to be pressed or menu choices to be selected. For example, STAT EDIT directs you to press the STAT button and then to select EDIT from the options shown. Most calculators allow you to store one or more lists that can be used with statistical operations and graphs. Although the procedures needed to create the statistical values seen in this book are shown below, refer to your calculator’s manual for a complete guide. This section outlines the procedures to create the following items: • statistical lists for 1-variable and 2-variable data • statistics for 1-variable and 2-variable data • histograms and box plots for 1-variable data • scatter plots of 2-variable data • regression equations of 2-variable data Refer to Sections 1.5, 4.3.A, 5.7, and 13.1 for examples that involve lists and regression. Lists Call up the list editor to enter data into lists by using the commands below. TI-84/TI-83 STAT Lists are L1, L2, . . ., L6. EDIT Casio 9850 STAT Lists are List 1, List 2, . . ., List 6. Press ENTER (or EXE) after each entry in a list to proceed to the next position. Use the left/right arrow keys to move among lists and the up/down arrow keys to move within a list. One-Variable Statistics Data for 1-variable statistics can be entered into the list editor in two ways: • Each value is entered into a single list. • Each value is entered into one list and its frequency into the corresponding position of a second list to create a frequency table. The symbols commonly used to represent the statistics for 1-variable data are as follows: n x x sn1 Sx, sx x sn , ©x ©x2 minX Q1 Med Q3 maxX sample number mean standard deviation from the sample population standard deviation from the sample sum of the data sum of squares of the data smallest value of the data value of the first quartile median of the data value of the third quartile largest value of the data Technology Appendix 1027 NOTE Directions for Casio 9850 begin from the main menu screen. To compute 1-variable statistics, use the following procedures. TI-84/TI-83 STAT CALC 1: 1-Var Stats • If the data is contained in one list, enter the list name by pressing the list name shown above the number keys 1 through 6. • If the data was entered as a frequency table, enter the list name of the data, press [ , ] (comma), and enter the name of the list that contains the frequency. Casio 9850 STAT EXE CALC SET • If the data is contained in one list: 1. Set XList to the list that contains the data by pressing the key below the appropriate list name. 2. Use the down arrow key to highlight 1Var Freq and set 1Var Freq to 1 by pressing F1. • If the data is contained in a frequency table. 1. Set 1Var XList to the list that contains the data by pressing the key below the appropriate list name. 2. Set 1Var Freq to the list that contains the frequency of the data. Press EXIT to return to the list editor. Press 1VAR to display the statistics. Histograms and Box Plots (1-Variable Graphs) A histogram is a bar graph that displays 1-variable data values on the x-axis and represents each corresponding frequency as the height of the box above the x-axis. A box plot and a modified box plot display the values Q1 to Q3 as a box with a vertical line at the median. A box plot displays the values of xMin to Q1 and of Q3 to xMax as whiskers at either end of the box, and a modified box plot displays outliers (values at least 1.5 * (Q3–Q1) below Q1 and above Q3) as points beyond the whiskers. To create a graph of 1-variable data, enter the data into a single list or into a frequency table, and use the procedures to create a histogram or a box plot. TI-84/TI-83 • STAT PLOT • Highlight On and press ENTER. Plot1 • Highlight the desired graph type by using the arrow keys, and press ENTER. • Use the down arrow key to select Xlist. Enter the name of the list that contains the data by using the keys L1 through L6, which are the 2nd functions above the number keys 1 through 6. • Move the cursor to Freq and enter 1 if the data is contained in a single list, or enter the name of the list that contains the frequency. 1028 Technology Appendix • If a modified box plot is selected, select the type of symbol that will denote outliers. • Set the window values for xMin and xMax, and if creating a histogram, set the yMin and yMax values. Box plots ignore yMin and yMax values. • GRAPH Casio 9850 STAT EXE • GRPH SET • Select the StatGraph area desired by choosing GPH1, GPH2, or GPH3. • Use the down arrow key to highlight Graph Type. Press to display the graph type options. Choose Hist, Box (MedBox), or Box (MeanBox). MedBox shows the distribution of the data items that are grouped within Q1, Med, and Q3. MeanBox shows the distribution of the data around the mean when there is a large number of data items, and a vertical line is drawn at the mean. • Select XList and enter the name of the list that contains the data. • Select Frequency and enter 1 if the data is in one list or the name of the list that contains the frequency of the data. • Choose the graph color. • Press EXIT. • Press GPH1 (GRPH2 or GRPH3) to display the StatGraph area. Press EXIT to return to the previous screen. Two-Variable Statistics and Graphs Two-variable statistics for both the x-variable and the y-variable include all the statistics listed for 1-variable statistics; that is, the mean of the y-variable data is denoted as , the minimum value of the y-variable data is denoted by minY or yMin, and so forth. Additionally, xy, the sum of the product of corresponding data pairs, is also given. g y To compute 2-variable statistics, use the following procedure after entering the data pairs into two lists. TI-84/TI-83 STAT CALC 2–Var Stats • Enter the list names that contain the data pairs separated by a comma, and press ENTER. The first named list is the x-list, and the second named list is the y-list. Casio 9850 STAT EXE CALC SET • Use the down arrow key to select 2Var XList, and enter the name of the list that contains the x-data. Similarly, enter the name of the list that contains the y-data. 2Var Freq specifies the list where paired-variable frequency values are located. Enter 1 if the data pairs are entered separately in the XList, and set 2Var Freq to 1 if the data pairs are entered separately in the XList and the Ylist. • EXIT 2VAR Technology Appendix 1029 Scatter Plots Scatter plots graph the data points from Xlist and Ylist as coordinate pairs. The general procedure for graphing 2-variable data contained in two lists is as follows. • Define the statistical plot. • Define the viewing window. • Display the graph. TI-84/TI-83 • STAT PLOT, select the desired stat plot editor, highlight On, and press ENTER. • Select the ic
on that represents the type of graph desired and press ENTER. • Enter the names of the lists that contain the x-data and the y-data. Choose the graph type and Mark symbol. Option GS.D. represents a scatter plot. • GRAPH Casio 9850 STAT EXE GPH SET • Select the graph number: GPH1, GPH2, or GPH3. • Highlight Graph Type and choose Scat. • Enter the names of the lists that contain the x-data and the y-data, enter the frequency of the data pairs, choose the mark type and the graph color. EXIT • V-Window, enter the viewing window values, EXIT • GPH, select the graph number that contains the desired settings Regression Equations Most graphing calculators can compute the following types of regression equations. Regression Type Model linear quadratic cubic quartic logarithmic exponential power y ax b y ax2 bx c y ax3 bx2 cx d y ax4 bx3 cx2 dx e y a ln x y aebx y axn y c 1 ae y a sin bx bx c 1 d 2 logistic sinusoidal 1030 Technology Appendix Reference LinReg or LinearReg QuadReg CubicReg QuartReg LnReg or LogReg ExpReg PwrReg or PowerReg Logistic or LogisticReg SinReg The procedure to compute regression equations follows. Enter the data into two lists. TI-84/TI-83 STAT CALC Select the regression model by using the up/down arrow keys. Enter the names of the lists separated by a comma by using the L1 through L6 keys, which are the 2nd functions for the numerals 1 through 6. The default is L1, L2. Casio 9850 While viewing the lists, press CALC REG. Select the regression model by pressing the corresponding F key. Additional models are accessed by pressing F6. Displaying a Regression Equation’s Graph Set the bounds of the display window by selecting the following. TI-84/TI-83 WINDOW V-Window Casio 9850 TI-84/TI-83 There are two methods for entering a regression equation into the Y editor for display: automatic and manual. Note that each time a regression equation is found, the contents of RegEQ are overwritten with the new regression equation. • Automatic The calculator can automatically place the regression equation into the Y editor as Y3 when computing a regression equation. The following entry places the linear regression equation for the lists L1 and L2 into Y3 in the Y editor. LinReg L1,L2,Y3 • Manual Whenever a regression model is found, the calculator places the regression equation into the variable RegEQ (or RegEqn), which can be entered into the Y editor. Press VARS Statistics EQ RegEQ from the Y editor screen. A regression equation is displayed using the procedure to graph any type of equation. Casio 9850 • While viewing the data lists, press GRPH. • Select GPH1, GPH2, or GPH3 to view the desired graph type. (Press SET to change the options.) • Choose the type of regression model desired (X represents linear regression.) COPY lets you store the displayed regression equation as a function in the Y editor. Use the up/down arrow keys to select the desired Y variable. DRAW graphs the displayed regression equation Technology Appendix 1031 Regression Coefficients r2 R2 r , the correlation When some regression models are created, values of ), the coefficient of determination, are computed coefficient, and (or and stored as values. Values of and are computed for the following regression models: linear, logarithmic, power, and exponential. The value is computed for the following regression models: quadratic, cubic, of and quartic. No correlation coefficient or coefficient of determination is given for logistic and sinusoidal regression models. R2 r2 r TI-84/TI-83 r r2 , and are displayed when a R2 The variables , regression equation is computed by executing the DiagnosticOn instruction, which is found in the CATALOG menu. When DiagnosticOff is set, the values for , are not displayed. , and R2 r2 r Casio 9850 Neither form of a coefficient of determination is computed on a Casio. Scatter Plots of Residuals Residuals, also called relative error, are stored as a list of values in a variable named Resid or RelErr on some calculators. Scatter plots of the residuals can be graphed but may not be visible in the same viewing window as the data. Adjust the window to view the scatter plot of the residuals. TI-84/TI-83 From a STAT PLOT menu set the following: ON Type Scatter Plot XList Ylist Mark Ln (the list that contains the input data) RESID (which is found in LIST NAMES) as desired GRAPH Casio 9850 Residuals are not automatically computed when a regression equation is found. 1032 Technology Appendix T.3 Programs The programs listed here are of two types: programs to give older calculators some of the features that are built-in on newer ones and programs to do specific tasks discussed in this book (such as synthetic division). Each program is preceded by a Description, which describes, in general terms, how the program operates and what it does. Some programs require that certain things be done before the program is run (such as entering a function in the function memory); these requirements are listed as Preliminaries. Occasionally, italic remarks appear in brackets after a program step; they are not part of the program, but are intended to provide assistance when you are entering the program into your calculator. A remark such as “[MATH NUM menu]” means that the symbols or commands needed for that step of the program are in the NUM submenu of the MATH menu. Fraction Conversion (Built-in on TI-82/83/84/85/86) Description: Enter a repeating decimal; the program converts it into a fraction. The denominator is displayed on the last line and the numerator on the line above. Casio 9850 Fix 7 ”N ”? S N 0 S D Lbl 1 D 1 S D N D TI-82/83/84 Quadratic Formula (Built-in on other calculators) [MATH NUM menu] Rnd 2 [MATH NUM menu] 0 1 Goto 1 S N [DISP menu] 2 1 Frac Ans Ans .5 1 Norm (Int N) D Description: Enter ax2 bx c 0; the coefficients of the program finds all real solutions. the quadratic equation :ClrHome [Optional] :Disp ”AX2 BX C 0” [Optional] :Prompt A :Prompt B :Prompt C B2 4AC : 1 :If S 6 0 2 S S :Goto 1 B 1S B 1S /2A 2 /2A 2 1 1 :Disp :Disp :Stop :Lbl 1 :Disp “NO REAL ROOTS” Technology Appendix 1033 Synthetic Division (Built-in on TI-89/92) Preliminaries: Enter the coefficients of the dividend (in order of decreasing powers of x, putting in zeros for missing coefficients) as list L1 (or List 1). If the coefficients are 1, 2, 3, for example, key in {1, 2, 3,} The symbols { } are on the keyboard or in the LIST menu. and store it in The list name is on the keyboard of TI-82/83 and Sharp 9600. On TI-85/86 and HP-38, type in L1. On Casio 9850, use “List” in the LIST submenu of the OPTN menu to type in List 1. L1. L1 F x 1 2 x a and enter a. The proDescription: Write the divisor in the form x , gram displays the degree of the quotient (in order of decreasing powers of x), and the remainder. If the program pauses before it has displayed all these items, you can use the arrow keys to scroll through the display; then press ENTER (or OK) to continue. the coefficients of Q Q x 2 1 2 1 TI-82/83/84/85/86 :ClrHome [ClLCD on TI-85/86] :Disp “DIVISOR IS X A ” :Prompt A S L2 [See Preliminaries for how to enter list names] :L1 S N :dim L1 3 :For (K, 2, N) L11 : 1 :End A L21 on TI-85/86] S L21 dimL L1 K 1 K K 22 2 2 [LIST OPS menu] S S dim L2 :round(L2(N),9) N 1 : 1 :Disp “DEGREE OF QUOTIENT” 2 R [MATH NUM menu] 1 :Disp dim L2 :Disp “COEFFICIENTS” :Pause L2 :Disp “REMAINDER” :Disp R Casio 9850 [OPTN MATH NUM menu] Rnd Ans S R [See Preliminaries for how to enter list names] Norm [SETUP DISP menu] [OPTN LIST menu] Seq(List 2 [X], X, 1, N 1, 1) S List 2 “DEGREE OF QUOTIENT” [OPTN LIST menu] A List 2 K 1 3 4 2 S List 2 K 4 3 dim List 2 1 “COEFFICIENTS” List 2 “REMAINDER” R X A ” “DIVISOR IS A ”? S A “ List 1 S List 2 List 1 S N dim 2 S K Lbl 1 List Goto 1 Fix 9 K 4 [SETUP DISP menu] List 2 [N] 1034 Technology Appendix Glossary A absolute value (of a complex number) a bi 2a2 b2 (p. 638) 0 0 absolute value of a number For any real number c, if braic definition). number line (geometric definition). (p. 107, 108) and if (alge, then is the distance from c to 0 on the c c c 6 0 c c 0 , then c c 0 0 0 0 0 0 absolute-value inequalities For a positive number k and any real number r, r k r k (p. 129) k is equivalent to is equivalent to r k k and or r k. r 0 0 0 0 acute angle an angle with a degree measure of less than (p. 414) 90° addition and subtraction identities trigonometric identities involving a function of the sum or difference of two angle measures (p. 582) adjacency matrix a matrix used to represent the connections between vertices in a directed network (p. 809) adjacent side (of a right triangle) abbreviated adj, the side of a given acute angle in a right triangle that is not the hypotenuse (p. 415) ambiguous case When the measures of two sides of a triangle and the angle opposite one of them are known, there may be one, two, or no triangles that satisfy the given measures. (p. 627) amplitude The amplitude of a sinusoidal function is one-half of the difference between the maximum and minimum function values and is always positive. (p. 497) See also sinusoidal function. analytic geometry the study of geometric properties of objects using a coordinate system (p. 691) angle a figure formed by two rays and a common endpoint (p. 413) angle of depression the angle formed by a horizontal line and a line below it (p. 427) angle of elevation the angle formed by a horizontal line and a line above it (p. 426) angle of inclination If L is a nonhorizontal straight line in a coordinate plane, then the angle of inclination of L is the angle formed by the part of L above the x-axis and the x-axis in the positive direction. (p. 589) Angle of Inclination Theorem If L is a nonvertical line with angle of inclination (p. 589) then tan slope of L. u u, u Angle Theorem If is the angle between nonzero u v u v cos u vectors u and v, then cos u u v u v. (p. 672) and angular speed a measure of speed of a point rotating at a constant rate around the center of a circle, given as
the angle through which the point rotates over time (p. 440) approach infinity Output values of a function that get larger and larger without bound as input values increase are said to approach infinity. (p. 201) arc an unbroken part of a circle (p. 434) arc length the length of an arc, which is equal to the radius times the radian measure of the central angle of the arc (p. 435, 439) arccosine function the inverse cosine function, denoted by arccos x (p. 533) x g 1 2 arcsine function the inverse sine function, denoted by arcsin x (p. 530) x g 1 2 arctangent function the inverse tangent function, denoted by arctan x (p. 535) g x 1 2 area of a triangle formula The area of any triangle ABC in standard notation is 1 2 ab sin C. (p. 632) Glossary 1035 argument (p. 639) the angle u in a trigonometric expression arithmetic progression See arithmetic sequence. arithmetic sequence a sequence in which the difference between each term and the preceding term is always constant (p. 22) asymptotes of a hyperbola two lines intersecting at the center of a hyperbola which the hyperbola approaches but never touches (p. 701) augmented matrix a matrix in which each row represents an equation of a system and contains the coefficients of the variables in the equation (p. 795) average See mean. average rate of change For any function f, the average rate of change of with respect to x as x x f 1 2 changes from a to b is the value change in f x 1 change in p. 216) axes (singular: axis) nate system (p. 5) the number lines in a coordi- axis of a parabola the line through the focus of a parabola perpendicular to the directrix of the parabola (p. 709) B bar graph a visual display of qualitative data in which categories are displayed on the horizontal axis and frequencies or relative frequencies on the vertical axis (p. 845) Basic Properties of Logarithms only for for all real k; and logb 1 0 blogb x x x 7 0; for all and logb b 1; x 7 0. logb x is defined logb bk k (p. 364, 372) bell curve See normal curve. Bernoulli experiment See binomial experiment. If c is a number far from 0, then Big-Little Concept 1 c is a number close to 0. Conversely, if c is a number close to 0, then 1 c is a number far from 0. (p. 280) binomial distribution the probability distribution for a binomial experiment (p. 888) binomial experiment A probability experiment that can be described in terms of just two outcomes is a binomial experiment, also known as a Bernoulli experiment. It must meet the following conditions: the experiment consists of n trials whose outcomes are either 1036 Glossary successes or failures, and the trials are identical and independent with a constant probability of success, p, and a constant probability of failure, q 1 p. (p. 885) bounds test a test used to determine the lower and upper bounds for the real zeros of a polynomial function (p. 256) C Cartesian coordinate system a two-dimensional coordinate system that corresponds ordered pairs of real numbers with locations in a coordinate plane (p. 5) center of a hyperbola the midpoint of the segment that has the foci of the hyperbola as endpoints (p. 701) center of an ellipse the midpoint of the segment that has the foci of the ellipse as endpoints (p. 692) central angle an angle whose vertex is at the center of a circle (p. 434) central tendency a value that is used to represent the center of an entire data set (p. 853) change in x the horizontal distance moved from one point to another point in a coordinate plane; sometimes denoted and read “delta x” (p. 31) ¢ x, change in y the vertical distance moved from one point to another point in a coordinate plane; sometimes denoted and read “delta y” (p. 31) ¢ y, Change-of-Base Formula For any positive number logb x ln x ln b logb x log x log b (p. 374) and x, . closed interval an interval of numbers in which both endpoints of the interval are included in the set; denoted with two brackets (p. 118) coefficient nomial (p. 239) the numerical factor of a term in a poly- coefficient of determination a statistical measure, that is the proportion of variation often denoted by in y that can be attributed to a linear relationship between x and y in a data set (p. 47) r2, Cofunction Identities relate the sine, secant, and tangent functions to the cosine, cosecant, and cotangent functions, respectively (p. 586) trigonometric identities that combination an arrangement of objects in which order is not important; a collection of objects (p. 880) common difference the constant number, usually denoted by d, that is the difference between each term and the preceding term in an arithmetic sequence (p. 22) common logarithm (of x) at the number x, denoted log x (p. 356) the value of g x 1 2 log x common logarithmic function the inverse of the log x exponential function (p. 356) denoted 10x, x x g f 1 1 2 2 common ratio the constant value, denoted by r, given by the quotient of consecutive terms in a geometric sequence (p. 58) complement of an event are not contained in the event (p. 866) the set of all outcomes that complete graph a graph that shows all of its important features, including all peaks and valleys, points where it touches an axis, and suggests the general shape of the portions of the graph that are not in view (p. 82) completely factored (over the set of real numbers) a polynomial written as the product of irreducible factors with real coefficients (p. 253) completing the square a process used to change an x2 bx expression of the form into a perfect square by adding a suitable constant (p. 92) complex number system the number system that consists of real and nonreal numbers (p. 293) complex numbers numbers of the form where a is a real number and bi is an imaginary number (p. 294) a bi, complex plane a coordinate plane with the horizontal axis labeled for real numbers and the vertical axis labeled for imaginary numbers (p. 301, 638) components (of a vector) u the vector point at the origin and terminal point at (a, b) (p. 655) the numbers a and b in where u is a vector with initial a, b , I H composite functions the composite function of f and g is 2 read “g circle f” or “f followed by g.” (p. 193) If f and g are functions, then g f x g 21 x f 1 1 1 concave up a description of the way a curve bends if for any two points in a given interval that lie on the curve, the segment that connects them is above the curve (p. 154) concavity a description of the way that a curve bends, such as concave up or concave down (p. 154) conic section a curve that is formed by the intersection of a plane and a double-napped right circular cone (p. 691); Let L be a fixed line, called a directrix, P a fixed point not on L, and e a positive constant. The set of all points X in the plane such that distance between X and the fixed point XP XL distance between X and the fixed line conic section with P as one of its foci. (p. 747) e is a conjugate The conjugate of the complex number a bi a bi, is is the number (p. 296, 309) and the conjugate of a bi. a bi conjugate pairs (p. 296) complex numbers a bi and a bi Conjugate Zero Theorem For every polynomial function f, if the complex number z is a zero of f, then its conjugate, is also a zero of f. (p. 309) z, consistent system a system of equations with at least one solution (p. 781) constant function A function is said to be constant on an interval if its graph is a horizontal line over the interval; that is, if its output values are always constant as the input values are increasing. (p. 152) constant polynomial a polynomial that consists of only a constant term (p. 240) constant term the coefficient is written in the form (p. 239) in a polynomial that anxn an1xn1 p a1x a0 a0 , 22 constraints that exist in linear programming (p. 829) restrictions, represented by inequalities, composition of functions a way of combining functions in which the output of one function is used as the input of another function (p. 193) compound inequality an inequality that compares more than two quantities and contains more than one inequality symbol (p. 118) compound interest This interest on an investment is compounded, or becomes part of the investment, at a particular interest rate per time period. If P dollars is invested at annual interest rate r per time period (expressed as a decimal), then the amount A after t 1 r time periods is A P (p. 345) t. 1 2 concave down a description of the way a curve bends if for any two points in a given interval that lie on the curve, the segment that connects them is below the curve (p. 154) continuous compounding This compound interest is compounded infinitely many times during a time period. If P dollars is invested at annual interest rate r and compounded continuously, then the amount A after t years is A Pert. (p. 348) continuous function a function whose graph is an unbroken curve with no jumps, gaps, or holes (p. 261, 939) S2, S1, S3, p convergent series a geometric series in which the terms of the sequence of partial sums get closer and closer to a particular real number S in such Sk a way that the partial sum when k is large enough; a geometric series with com6 1 mon ratio r such that is arbitrarily close to S (p. 77) r 0 0 coordinate plane See Cartesian coordinate system. Glossary 1037 coordinate system a system of locating points in a plane or in space by using ordered pairs or ordered triples, respectively, of real numbers (p. 5) any point of a feasible region where at corner point least two of the graphs of the constraints intersect (p. 829) correlation coefficient a statistical measure, often denoted by r, of how well the least squares regression line fits the data points that it models (p. 47) cosecant ratio For a given acute angle u triangle, the cosecant of is written as csc and is equal to the reciprocal of the sine ratio of the given angle. (p. 416) u in a right u u cosine ratio For a given acute angle in a right triu u is written as cos and is equal to angle, the cosine of the ratio of the ad
jacent side length to the length of the hypotenuse. (p. 416) cotangent ratio For a given acute angle u triangle, the cotangent of equal to the reciprocal of the tangent ratio of the given angle. (p. 416) is written as cot and is u in a right u coterminal angles angles formed by different rotations that have the same initial and terminal sides (p. 434) cubic function a third-degree polynomial function (p. 240) cycle (of a periodic function) a portion of the graph of a periodic function in which the function goes through one period (p. 493) D data information gathered in a statistical experiment (p. 843) decreasing function A function is said to be decreasing on an interval if its graph always falls as you move from left to right over the interval; that is, its output values are always decreasing as the input values are increasing. (p. 152) degenerate conic section a point, line, or intersecting lines formed by the intersection of a plane and a double-napped right circular cone (p. 691) degree (measure) a unit of angle measure that of a circle, denoted with the degree symbol equals 1 360 (p. 414) ° 2 1 degree (of a polynomial) the exponent of the highest power of the variable that appears with a nonzero coefficient in a polynomial (p. 240) 1038 Glossary delta See change in x or change in y. DeMoivre’s Theorem For any complex number, z r zn rn and for any positive integer n, cos n u i sin n u cos u i sin u (p. 644) . , 1 2 1 denominator below the fraction bar (p. 4) 2 the expression in a fraction that lies deviation (of a data value) value from the mean of the data set (p. 857) the difference of a data x difference function For any functions 1 their difference function is the new function f g x (p. 191) g f x x f . 1 21 2 1 2 1 2 and g x , 2 1 2 difference quotient the quantity a function f (p. 219 for differential calculus a method of calculating the changes in one variable produced by changes in a related variable (p. 138) dimensions of a matrix used to indicate the number of rows and columns in a matrix (Example: an m n matrix has m rows and n columns) (p. 804) directed network a finite set of connected points in which permissible directions of travel between the points are indicated (p. 809) direction (of a vector) line segment representing a vector positive x-axis (p. 662) the angle that the directed a, b I H makes with the directrix of a parabola the line in the formation of a parabola such that its distance from any point on the parabola is equal to the distance from that point to the focus of the parabola (p. 709) discontinuous function a function that has one or more jumps, gaps, or holes (p. 937) the expression discriminant formula, used to determine the number of real solu(p. 93) tions of a quadratic equation ax2 bx c 0 in the quadratic b2 4ac . (p. 108) distance (between real numbers) The distance on the number line between real numbers c and d is c d 0 distance difference the constant difference between the distances from each focus of a hyperbola to a point on the hyperbola (p. 700) 0 distribution an arrangement of numerical data in order (usually ascending) (p. 846) divergent series a geometric series that is not convergent; a geometric series with common ratio r such that (p. 77) 1 r 0 0 f x h Division Algorithm If a polynomial by a nonzero polynomial q x polynomial 2 h 0 x x f that 2 x r has a degree less than the degree of the divisor, 2 1 x h (p. 243) . 2 1 x is divided 2 then there is a quotient such or r and a remainder polynomial q x where either DMS form a form of degree measure expression which includes degrees, minutes, and seconds (p. 414) domain the set of first numbers in the ordered pairs of a relation (p. 6, 142) domain convention Unless information to the contrary is given, the domain of a function f includes every real number input for which the function rule produces a real number output. (p. 145) dot product a real number produced by multiplying corresponding components of two vectors and adding the products (p. 670) Double-Angle Identities involving a function of an angle multiplied by 2 (p. 593) trigonometric identities E eccentricity (of an ellipse or hyperbola) e distance between the foci where , distance between the vertices e 7 1 all ellipses and for all hyperbolas (p. 745) the ratio 0 6 e 6 1 for elementary row operations operations used on an augmented matrix that produce an augmented matrix of an equivalent system (p. 796) eliminating the parameter expressing a curve that is given by parametric equations as part of the graph of an equation in x and y (p. 757) ellipse For any points P and Q in the plane and any number k greater than the distance from P to Q, an ellipse, with foci P and Q, is the set of all points (x, y) such that the sum of the distance from (x, y) to P and the distance from (x, y) to Q is k. (p. 692) a conic section with eccentricity between 0 and 1, not inclusive (p. 747) empirical rule a rule that describes the areas under the normal curve over intervals of one, two, and three standard deviations on either side of the mean in terms of percentages of the number of data values (p. 892) 0 end behavior the far left and far right of the coordinate plane when x the shape of the graph of a function at is large (p. 262, 287) 0 endpoints (of an interval) set of numbers that can be expressed as an interval from c to d (p. 118) the numbers c and d in a equal matrices same dimensions and equal corresponding entries (p. 804) two or more matrices that have the equivalent equations solutions (p. 81) equivalent inequalities same solutions (p. 119) equations that have the same inequalities that have the equivalent systems same solutions (p. 795) systems of equations with the equivalent vectors vectors, such as u and v, that have the same magnitude and direction, denoted u v (p. 654) Euler’s Formula the identity (p. 688) eix cos x i sin x even function a function f for which for all x in its domain, its graph symmetric with respect to the y-axis (p. 188, 482) x f 2 2 1 1 x f any outcome or collection of outcomes in the event sample space of an experiment (p. 865) eventually fixed point orbit of c for a given function eventually produces constant output values (p. 202) the number c for which the eventually periodic point the number c for which the orbit of c for a given function eventually produces repeating output values (p. 202) expected value of a random variable the average value of the outcome values for a random variable (p. 869) experiment one or more observable outcomes (p. 864) in probability, any process that generates explicit form of a geometric sequence In a geometric sequence n 1. for all with common ratio r, un6 5 (p. 60) rn1u1 un explicit form of an arithmetic sequence In an arithmetic sequence d un with common difference d, n 2. un6 5 for all n 1 (p. 23) u1 1 2 exponential decay decay that can be represented by x f a function of the form 1 quantity at time x, P is the initial quantity when and decreases when x increases by 1 (p. 351) is the factor by which the quantity 0 6 a 6 1 Pax, x 0, where is the x f 1 2 2 exponential function (with base a) a function where a is any positive real whose rule is number, and whose domain is all real numbers (p. 336) ax, x f 2 1 exponential growth growth that can be represented x f is the by a function of the form 2 x 0, quantity at time x, P is the initial quantity when and is the factor by which the quantity increases when x increases by 1 (p. 349) Pax, a 7 1 where x f 2 1 1 Glossary 1039 exponential model represent the trend in a data set (p. 389) an exponential function used to parabola is equal to the distance from that point to the directrix (p. 709) extraneous roots See extraneous solutions. extraneous solutions tion that are not solutions of the original equation (p. 110) solutions of a derived equa- F Factor Theorem A polynomial 0. if and only if tor x a a f x f 2 (p. 245) 1 1 2 1 factorial The notation n! is read “n factorial” and describes the product of all the integers from 1 through n. (p. 520, 880) 2 has a linear fac- Formula for Roots of Unity For each positive integer n, the n distinct nth roots of unity are i sin 2ku n cos 2ku n (p. 648) for k 0, 1, 2, p , and n 1. fractional equation an equation formed by a fractional expression equal to 0 (p. 114) fractional expression the quotient f 1 g 1 g are algebraic expressions and x 2 x 2 x and g x 1 2 , f x where 0 2 1 (p. 114) 1 2 the reciprocal of the frequency (of a sound wave) period of the sinusoidal function that represents a sound wave (p. 558) feasible region the region of the coordinate plane that is the intersection of the graphs of the constraints in linear programming (p. 829) Fibonacci sequence the sequence by Leonardo Fibonacci in the thirteenth century a2 in which sum of the two preceding terms; (p. 21) 5 n 3, an an an1 and for is the an2 1, 1, an6 a1 discovered finite differences the differences between each y-value and the preceding one in a table of values (p. 43) the octant of a three-dimensional coordi- first octant nate system in which all coordinates are positive (p. 790) first quartile See quartiles. five-number summary (of a data set) ing list of values: minimum, first quartile, second quartile, third quartile, and maximum (p. 861) the follow- fixed point (of an orbit) the number c for which the orbit of c for a given function produces constant output values (p. 202) focal axis of a hyperbola the line through the foci of a hyperbola (p. 701) foci (singular, focus) of a hyperbola the points in the formation of hyperbola such that the difference of the distances from each focus to a point on the hyperbola is constant (p. 701) foci (singular, focus) of an ellipse the points in the formation of an ellipse such that the sum of the distances from each focus to a point on the ellipse is constant (p. 692) focus of a parabola the point in the formation of a parabola such that its distance from any point on the 1040 Glossary function a special
type of relation in which each member of the domain corresponds to one and only one member of the range (p. 7) A function consists of a set of inputs called the domain, a rule by which each input determines one and only one output, and a set of outputs called the range. (p. 142) function notation There is a customary method of denoting a function in abbreviated form. If f denotes a function and a denotes a number in the domain, then f(a) denotes the output of the function f produced by input a. (p. 9) function rule a set of operations that defines a function (p. 7) Fundamental Counting Principle In a set of k experiments, if the first experiment has n1 p , outcomes, the second has the kth has nk of outcomes for all k experiments is (p. 879) and outcomes, then the total number outcomes, n2 n1 n2 p nk. Fundamental Theorem of Algebra Every nonconstant polynomial has a zero in the complex number system. (p. 307) Fundamental Theorem of Linear Programming The maximum or minimum of the objective function (if it exists) always occurs at one or more of the corner points of the feasible region. (p. 829) G Gauss-Jordan elimination the method of using elementary row operations on an augmented matrix to produce a matrix in reduced row-echelon form that represents an equivalent system (p. 797) general form (of a line) a linear equation in the where A and B are not both form equal to zero (p. 39) Ax By C 0, geometric progression See geometric sequence. geometric sequence a sequence in which terms are formed by multiplying a preceding term by a nonzero constant (p. 58) graph a visual display of a set of points (p. 30) graph of an equation the set of points whose coordinates are solutions of the equation (p. 30) the calculation performed by graphical zero finder a graphics calculator in which the x-intercepts of the graph of a function are identified; labeled ROOT, ZERO, or X-INCPT in the TI-83 and Sharp 9600 CALC menu, the Casio 9850 G-SOLVE menu, the MATH submenu of TI-86/89 GRAPH menu, and the FCN submenu of the HP-38 PLOT menu (p. 84) greatest integer function a piecewise-defined function denoted as that converts a real number 4 x into the largest integer that is less than or equal to x (p. 147) x x f 2 1 3 H Half-Angle Identities involving a function of an angle divided by 2 (p. 596) trigonometric identities half-open interval an interval of numbers in which one endpoint of the interval is included in the set and the other endpoint of the interval is not included; denoted with one bracket and one parenthesis, respectively (p. 118) Heron’s Formula The area of any triangle ABC in s a standard notation is where s 1 a b c 2 (p. 633) s b s c 1s 21 21 . , 1 2 1 2 Hertz the unit of measure for the frequency of a sound wave, where one Hertz is one cycle per second (p. 559) histogram a display of quantitative data in which the data is divided into classes of equal size and displayed along the horizontal axis and frequencies or relative frequencies on the vertical axis (p. 848) f x g h rational function hole in a graph A point is omitted in the graph of a function and is not contained by an asymptote. For any x 1 x 1 duces both a zero numerator and a zero denominator, if the multiplicity of d as a zero of the related function g is greater than or equal to the multiplicity of d as a zero of the related function h, then the graph of f has a hole at and number d that pro- x d. (p. 283) 2 2 1 2 horizontal asymptote a horizontal line that the graph of a function approaches but never touches or crosses as gets large (p. 284, 952) x 0 0 horizontal compression For any positive number c 7 1, cx is the graph of f com1 pressed horizontally, toward the y-axis, by a factor the graph of y f 2 of 1 c . (p. 179) horizontal line a line that has a slope of zero and an equation of the form (p. 37) where b is the y-intercept y b, horizontal shifts For any positive number c, the y f graph of the left, and the graph of shifted c units to the right. (p. 175) is the graph of f shifted c units to is the graph of horizontal stretch For any positive number the graph of c 6 1, is the graph of f stretched hori- y f 2 zontally, away from the y-axis, by a factor of cx 1 1 c . (p. 179) horizontal line test A function f is one-to-one if and only if no horizontal line intersects the graph of f more than once. (p. 209) hyperbola For any points P and Q in the plane and any positive number k, a hyperbola with foci P and Q is the set of all points (x, y) such that the absolute value of the difference of the distance from (x, y) to P and the distance from (x, y) to Q is k. (p. 700) a conic section with eccentricity greater than 1 (p. 700) hypotenuse abbreviated hyp, the side of a right triangle that is across from the right angle (p. 415 cos 1 tan t cos t, 2 (p. 459) identities involving sin tan sin t, p t 1 2 identity an equation that is true for all values of the variable for which every term of the equation is defined (p. 454) n n identity matrix The on the diagonal from the top left to the bottom right and 0s in all other entries. For any AIn matrix, denoted InA A. matrix A, (p. 815) n n In, has 1s imaginary axis plane where each imaginary number sponds to the point (0, b) (p. 638) the vertical axis in the complex corre- 0 bi imaginary numbers a number of the form bi, where b is a real number and i is the imaginary unit (p. 294) The number . (p. 294, 297) i 11 inconsistent system a system of equations with no solutions (p. 781) Glossary 1041 increasing function A function is said to be increasing on an interval if its graph always rises as you move from left to right over the interval, that is, if its output values are always increasing as the input values are increasing. (p. 152) independent events rence or non-occurrence of one event has no effect on the probability of the other event (p. 867) two events such that the occur- infinite geometric series p an6 a1 where 5 with common ratio r (p. 77) a3 a2 , the infinite series is a geometric sequence infinite limit a limit of infinity as x approaches some constant c; corresponds to a vertical asymptote (p. 949) infinite sequence a sequence with an infinite number of terms (p. 13) infinite series continues without end, or an infinite sequence; an a1 expression of the form the sum of terms of a sequence that in which p a3 a2 , an is q a real number; also denoted by an (p. 76) a n1 inflection point tion changes concavity (p. 154, 266) a point where the graph of a func- initial point (of a vector) that extends from point P to point Q (p. 653) the point P in a vector initial side the starting position of a ray that is rotated around its vertex (p. 433) input (of a relation) numbers in the ordered pairs of a relation (p. 7) the values denoted by the first instantaneous rate of change the rate of change of a function at a particular point (p. 234) integers numbers and their opposites: 3, the set of numbers that consists of whole p , 0, 1, 2, (p. 3) 1, 2, 3, p one number c between a and b such that (p. 944) f c 1 2 k. interquartile range a measure of variability resistant to outliers; the difference between the first and third quartiles (p. 860) intersection method a method of solving an equax tion of the form and x y2 and finding the x-coordinate of each point of intersection (p. 82) on the same screen of a graphics calculator by graphing g g f y1 interval (of numbers) between two fixed numbers (p. 118) the set of all numbers lying c 6 d: interval notation There is a customary method of denoting an interval of numbers. For real numbers c and d with that that that such that (c, d) denotes all real numbers x such [c, d) denotes all real numbers x such and (c, d] denotes all real numbers x c x d; c 6 x 6 d; c x 6 d; [c, d] denotes all real numbers x such c 6 x d. (p. 118) interval of convergence the set of values of x for which an infinite series converges to a function (p. 520) inverse cosine function the inverse of the cosine function with a domain restricted to by (p. 533) cos 1, 1 1 x denoted x g , 3 4 1 2 inverse function an inverse relation that is a function (p. 205) inverse of a matrix For an inverse of A is an AB In such that 1 In AA A and n n and 1A In. BA In, (p. 816) matrix A, the A 1, matrix B, also denoted or equivalently, n n inverse relation the result of exchanging the input and output values of a function or relation (p. 205) inverse sine function the inverse of the sine func- tion with a domain restricted to p 2 , p 2 T S , denoted by integral calculus a method of calculating quantities such as distance, area, and volume (p. 138) sin 1 x (p. 530) g x 1 2 intercepts (of a rational function) rational function, f, has a y-intercept, it occurs at f(0), and the x-intercepts occur at the numbers that are zeros of the numerator and not zeros of the denominator. (p. 279) If the graph of a intercepts (of polynomial functions) The graph of a polynomial function of degree n has one y-intercept, which is equal to the constant term, and at most n x-intercepts. (p. 264) interest a fee paid for the use of borrowed money; calculated as a percentage of the principal (p. 100) Intermediate Value Theorem If the function f is continuous on the closed interval [a, b] and k is any number between then there exists at least and b a f f , 1 2 1 2 1042 Glossary inverse tangent function the inverse of the tangent function with a domain restricted to by g x 1 2 tan 1 x (p. 535) p 2 , p 2 T S , denoted invertible matrix an exists an inverse matrix (p. 815) n n matrix for which there irrational number not be expressed as a fraction of integers (p. 4) the set of real numbers that can- irreducible (polynomial) be written as the product of polynomials of lesser degree (p. 253) a polynomial that cannot iterations (of a function) tions of a function with itself (p. 199) the repeated composi- K kth partial sum the sum of the first k terms of a where k is a positive integer (p. 26) sequence , un6 5 L Law of Cosines For any triangle ABC in a2 b2 c
2 2bc cos A, standard notation, b2 a2 c2 2ac cos B, (p. 617) and c2 a2 b2 2ab cos C. Law of Sines For any triangle ABC in standard nota- tion, a sin A b sin B c sin C . (p. 625) Laws of Exponents For any nonnegative real numbers c and d and any rational numbers r and s, crcs crs, c 0 cr dr 2 2 1 r 1 c cr cr cs d 0 r crdr, crs s crs, (p. 330) c 0 c db and cd cr leading coefficient highest power of the variable in a polynomial (p. 240) the nonzero coefficient of the least-squares regression line the one and only one line for which the sum of the squares of the residuals for a set of data is as small as possible (p. 47) length (of a vector) point Q in a vector that extends from point P to point Q, denoted the distance from point P to ! PQ (p. 653) limit a number (or infinity) that a function value approaches but never reaches as the domain values of that function approach a particular value or infinity (p. 909, 931) limit at infinity a real number limit as x gets large or small without bound; corresponds to a horizontal asymptote (p. 951) 1 x g f (p. 922, 954) Limit Theorem If f and g are functions that have limits as x approaches c and x then lim f lim g 2 1 xSc xSc linear combination The vector be a linear combination of i and j, where j are unit vectors. (p. 662) v ai bj i is said to and 1, 0 I x c, for all , 1 H I linear function a first-degree polynomial function (p. 240) linear programming a process that involves finding the maximum or minimum output of a linear function, called the objective function, subject to certain restrictions called constraints (p. 829) linear regression the computational process for finding the least-squares regression line for a set of data (p. 47) linear speed a measure of speed of a point rotating at a constant rate around the center of a circle, given as the distance that the point travels over time (p. 440) linear system a system of equations in which all equations are linear (p. 779) local extremum (plural: extrema) maximum or a local minimum (p. 266) either a local local maximum (plural: local maxima) A function is said to have a local maximum of graph of f has a peak at the point 1 for all x near c. (p. 153) f x c that is, f 1 c, f at ; if the 6 f c 2 c 22 x c 1 1 2 1 2 local minimum (plural: local minima) A function is said to have a local minimum of if the graph of f has a valley at the point for all x near d. (p. 153) f x d that is, ; d f 2 1 d, f at d 7 f 22 x d 1 1 1 2 1 2 logarithm to the base b (of x) g at the number x, denoted logb x x the value of logb x (p. 371) 1 2 logarithmic function to the base b the inverse of x f the exponential function x g 2 1 b 7 1 bx, , where b is a fixed positive number and logb x (p. 371) denoted 2 1 limit of a constant (p. 918) If d is a constant, then lim d d. xSc logarithmic model represent the trend in a data set (p. 389) a logarithmic function used to limit of a constant at infinity If c is a constant, xSqlim c c. then xSqlim c c (p. 953) and is a polyno- 1 f c x x f limit of a polynomial function If mial function and c is any real number, then lim f xSc limit of a rational function If function and c is any real number such that defined, then (p. 920) (p. 921 is a rational f is c 1 2 2 1 limit of the identity function For every real number c, (p. 918) 2 1 lim x c. xSc lim f xSc logarithmic scale a scale of numbers, such as the Richter scale, that is determined by a logarithmic function to measure logarithmic growth, which is very gradual and slow (p. 368) logistic model y a 1 be kx, a logistic function of the form where a, b, and k are constant, used to represent the trend in a data set (p. 389) lower bound (for the real zeros of a polynomial function) a polynomial function r and s are real numbers and the number r such that all the real zeros of are between r and s, where (p. 255) r 6 s x f 1 2 Glossary 1043 M Multiplication Principle See Fundamental Counting Principle. magnitude (of a vector) The length of a vector v v 2a2 b2. (p. 653) is a, b I H major axis of an ellipse the segment connecting the vertices of an ellipse (p. 692) Mandelbrot Set that the orbit of 0 under the function does not approach infinity (p. 304) z2 c f x 1 2 the set of complex numbers c such mathematical model a mathematical description or structure, such as an equation or graph, which illustrates a relationship between real-world quantities and which is often used to predict the likely value of an unknown quantity (p. 43) matrix an array of numbers often used to represent a system of equations (p. 795) multiplicity (of a zero) If occurs m times in the complete factorization of a polynomial, then a is called a zero with multiplicity m of the related polynomial function. (p. 265) is a factor that x a mutually exclusive events two events in a sample space that do not have outcomes in common (p. 866) N natural logarithm (of x) the number x, denoted ln x (p. 358) the value of g x 1 2 ln x at natural logarithmic function the inverse of the nat ln x x f ural exponential function (p. 358) denoted ex, x g 1 2 2 1 matrix addition addition of corresponding entries of matrices that have the same dimensions (p. 804) natural numbers The set of numbers that consists of p counting numbers: 1, 2, 3, (p. 3) matrix equation a matrix equation in the form AX B that represents a system of equations, where A contains the coefficients of the equations in the system, X contains the variables of the system, and B contains the constants of the equations (p. 814) matrix multiplication a method of multiplying two compatible matrices to produce a product matrix (p. 806) matrix subtraction subtraction of corresponding entries of matrices that have the same dimensions (p. 804) mean a measure of central tendency—also known as the average, denoted by and read “x bar”—that is calculated by adding the data values and then dividing the sum by the number of data values (p. 853) x mean of a random variable See expected value of a random variable. median a measure of central tendency that is, or indicates, the middle of a data set when the data values are arranged in ascending order (p. 855) minor axis of an ellipse the segment through the center of the ellipse, perpendicular to the major axis, and with points of the ellipse as endpoints (p. 692) Negative Angle Identities sin t, sin tan t t cos 1 tan t t cos t, 2 (p. 459) 1 1 2 2 negative correlation the relationship between two real-world quantities when the slope of the leastsquares regression line that represents the relationship is negative, that is, as one quantity increases, the other quantity decreases (p. 52) no correlation the relationship between two realworld quantities when the correlation coefficient, r, for a least-squares regression line that represents the relationship is close to zero; no apparent trend in the data (p. 52) nonlinear system a system of equations in which at least one equation is nonlinear (p. 779) nonnegative integers (p. 3) 1, 2, 3, p the set of whole numbers: 0, nonsingular matrix See invertible matrix. norm (of a vector) See magnitude. normal curve the graph of a probability density function that corresponds to a normal distribution: bell-shaped and symmetric about the mean, with the x-axis as a horizontal asymptote (p. 889) minute a unit of degree measure equal to degree (p. 414) 1 60 of a normal distribution a distribution of data that varies about the mean in such a way that the graph of its probability density function is a normal curve (p. 889) mode a measure of central tendency that is given by the data value that occurs most frequently in a data set (p. 855) modulus See absolute value of a complex number. 1044 Glossary nth root of a complex number The nth root of a bi zn a bi. is any of the n solutions of the equation (p. 645) nth roots For any real number c and any positive or2n integer n, the nth root of c is denoted by either c and is defined to be the solution of 1 c n odd or the nonnegative solution of even and nonnegative. (p. 328) xn c when n is xn c when n is numerator above the fraction bar (p. 4) the expression in a fraction that lies numerical derivative A calculator term for the instantaneous rate of change at a given input value; denoted nDeriv, nDer, d/dx, dY/dX, or (p. 237) 0. O objective function a linear function of which a minimum or maximum is obtained in linear programming (p. 829) oblique asymptote See slant asymptote. oblique triangle a triangle that does not contain a right angle (p. 617) octant one of eight regions into which a threedimensional coordinate system is divided by the intersection of the three coordinate planes (p. 790) odd function a function f for which for all x in its domain, its graph symmetric with respect to the origin (p. 189, 483) f 1 2 x f x 1 2 one-stage path (of a network) a direct path from one vertex to another in a directed network (p. 809) one-to-one function a function f in which a b; f relation is a function (p. 208) only when f b a 2 1 1 2 a function whose inverse open interval ther endpoint of the interval is included in the set; denoted with two parentheses (p. 118) an interval of numbers in which nei- opposite side (of a right triangle) abbreviated opp, the side of a right triangle that is across from a given acute angle of the triangle (p. 415) the sequence of output values orbit (of a number) produced by iterating a given function with that number; the orbit of a number c for a given function is c, f (p. 200 ordered pair A pair of real numbers in parentheses, separated by a comma, is used to locate or represent a point in a coordinate plane. The first number represents the horizontal distance from the origin and the second number represents the vertical distance from the x-axis. (p. 5) ordered set of numbers a set of numbers such that for any two numbers a and b in the set, exactly one of a 7 b the following statements is true: (p. 118) a 6 b, a b, or orientation (of a parametric curve) that a parametric curve is traced out (p. 757) the direc
tion origin the point of intersection of the axes in a coordinate system (p. 5) origin symmetry A graph is symmetric with respect x, y to the origin if whenever x, y is also on it. (p. 187) 1 orthogonal vectors perpendicular vectors; vectors u and v such that is on the graph, then u v 0 (p. 673) 2 2 1 outlier from the general trend of the distribution (p. 847) a data value that shows a strong deviation output (of a relation) second numbers in the ordered pairs of a relation (p. 7) the values denoted by the P parabola the shape of the graph of a quadratic function (p. 165) For any line L in the plane and any point P not on line L, a parabola with focus P and directrix L is the set of points such that the distance from X to P is equal to the distance from X to line L. (p. 709) a conic section with eccentricity equal to 1 (p. 709) parabolic asymptote a parabolic curve that the graph of a function approaches as gets large (p. 286) x 0 In a plane, these lines have the same parallel lines slope. All vertical lines are also parallel. (p. 38) 0 parallel vectors vectors that are scalar multiples of each other (p. 671) the third variable used as input for the parameter two functions that form a pair of parametric equations (p. 157, 785) parameterization a pair of parametric equations that describe a given curve (p. 755) parametric equations a pair of continuous functions that define the x- and y-coordinates of points in a coordinate plane in terms of a third variable, the parameter (p. 157) parametric graphing graphing parametric equations (p. 157) parent function a function with a certain shape that has the simplest algebraic rule for that shape (p. 172) partial fraction decomposition See partial fractions. partial fractions When a fraction is decomposed (broken down) and written as the sum of fractions, the terms of the sum are called partial fractions, and the sum is called the partial fraction decomposition of the original fraction. (p. 838) Glossary 1045 partial sums of a geometric sequence For each positive integer k, the kth partial sum of a geometric sequence with common ratio r 1 is un6 5 1 rk u1a 1 r b k a n1 un . (p. 61) partial sums of an arithmetic sequence For each positive integer k, the kth partial sum of an arithmetic sequence 5 ku1 un6 k with common difference d is k a n1 k 1 2 k 2 1 or un u1 d 2 1 uk2 k a n1 un . (p. 27) period (of a function) the smallest value of k in a function f for which there exists some constant k such that f (p. 457, 498) See also sinusoidal function. for every number t in the domain of t k f t f 2 2 1 1 periodic function a nonconstant function that repeats its values at regular intervals; a function f for which there exists some constant k such that f f 2 (p. 457) for every number t in the domain of f t k t 1 2 1 periodic orbit duces repeating output values (p. 202) an orbit for a given function that pro- periodic point (of an orbit) the orbit of c for a given function produces repeating output values (p. 202) the number c for which periodicity identities cos t cos t ± 2p , sin t sin t ± 2p , 1 t ± p 2 (p. 458) tan t tan 1 2 1 2 permutation an arrangement of objects in a specific order (p. 880) In a plane, two lines are per- perpendicular lines pendicular when their slopes are negative reciprocals 1 (having a product of ). Vertical lines and horizontal lines are perpendicular to each other. (p. 38) phase shift a number representing the horizontal translation of a sinusoidal graph (p. 502) See also sinusoidal function. pie chart a visual display of qualitative data in which categories are displayed in sectors of a circle, where the central angle of each sector is the product of (p. 845) the relative frequency of that category and 360° piecewise-defined function a function whose rule includes several formulas, each of which is applied to certain values of the domain, as specified in the definition of the function (p. 146) plane curve the set of all points (x, y) such that x f t 1 functions of t on an interval I (p. 755) and that f and g are continuous y g and t 1 2 2 point-slope form a linear equation in the form y y1 x1, y12 1 m is a point on the line (p. 36) where m represents the slope and x x12 , 1 1046 Glossary polar axis from the pole in a polar coordinate system (p. 734) the horizontal ray extending to the right r, u polar coordinates coordinates polar coordinate system, where r gives the distance from u the point to the pole, and with the polar axis as its initial side and the segment from the pole to the point as its terminal side (p. 734) is the measure of the angle of a point in the 1 2 polar form of a complex number For the complex number where the polar form is a r cos u, cos u i sin u b r sin u. r 1 and a bi, a bi (p. 639) r , , 2 0 0 pole the origin of a polar coordinate system (p. 734) an algebraic expression that can be anxn an1xn1 p polynomial written in the form a3x3 a2x2 a1x a0, ger, x is a variable, and each of stant (p. 239) where n is a nonnegative intea1, p , is a con- a0, an polynomial equation (of degree n) an equation that can be written in the form 0, a1x a0 variable, and each of where n is a nonnegative integer, x is a an is a constant (p. 94) anxn an1xn1 p a1, p , a0, polynomial form of a quadratic function a quadwhere a, x f ratic function in the form 2 b, and c are real numbers and ax2 bx c, a 0 (p. 164) 1 polynomial function a function whose rule is given by a polynomial (p. 240) polynomial model represent the trend in a data set (p. 273) a polynomial function used to population a group of individuals or objects studied in a statistical experiment (p. 843) positive correlation the relationship between two real-world quantities when the slope of the leastsquares regression line that represents the relationship is positive, that is, as one quantity increases, the other quantity increases (p. 52) positive integers 3, (p. 3) p the set of natural numbers: 1, 2, Power Law of Logarithms For all positive b and v, vk and all k, and (p. 366, 373) b 1, k logb v . logb1 2 1 2 y axr, power model where a and r are constant, used to represent the trend in a data set (p. 389) a power function of the form If both sides of an equation are Power Principal raised to the same positive integer power, then every solution of the original equation is a solution of the derived equation. However, the derived equation may have solutions that are not solutions of the original equation. (p. 112) power-reducing identities that relate second-degree expressions to first-degree expressions (p. 595) trigonometric identities prime number an integer greater than 1 whose only factors are itself and 1 (p. 20) principal borrowed (p. 100) an amount of money that is deposited or a number from 0 to 1 (or probability (of an event) 0% to 100%) that indicates how likely an event is to occur; calculated by dividing the number of elements in the event by the number of elements in the sample space (p. 865) probability density function A function with the property that the area under the graph corresponds to a probability distribution. (p. 871) probability distribution a table that describes the rule of a function P(E) that gives the probability of an event, where the domain of the function is the sample space and the range of the function is the closed interval [0, 1] (p. 865) probability of a binomial experiment P(r successes in n trials) success, and (p. 886) where p is the probability of is the probability of failure. nCrprqnr, q 1 p probability of a complement probability p, then the complement of the event has probability If an event E has 1 p. (p. 866) quadratic equation an equation that can be written where a, b, and c are real in the form constants and ax2 bx c 0, a 0 (p. 88) quadratic formula The solutions of a quadratic equation ax2 bx c 0 are x (p. 92) b ± 2b2 4ac 2a . quadratic function a function whose rule is a second-degree polynomial (p. 163, 240) qualitative data data that is categorical in nature, such as “liberal,” “moderate,” and “conservative” (p. 843) quantitative data numerical data (p. 843) quartic function a fourth-degree polynomial function (p. 240) quartiles These values divide a data set into fourths. The median, or second quartile divides the data into a lower half and an upper half; the first quartile Q1 is the Q3 median of the lower half; and the third quartile is the median of the upper half. (p. 860) Q2, quotient function For any functions x 2 their quotient function is the new function f 1 and g x , 2 1 f gb1 p. 192) and g x , 2 1 2 quotient identities tan t sin t cos t and cot t cos t sin t product function For any functions x their product function is the new function . (p. 192) f g fg 21 1 Product Law of Logarithms For all positive b, v, and w, and (p. 365, 373) logb v logb w. b 1, logb1 vw 2 product-to-sum trigonometric identities involving the product of two functions (p. 599) ! OQ , projection vector A vector called the projection of u on v, determined by constructing a segment from the terminal point of a vector u perpendicular to another vector v at a point Q on the vector, where point O is the initial point of both vectors, denoted projv u. (p. 674) Pythagorean identities sin2 t cos2 t 1 (p. 456) the identity and the identities derived from it Pythagorean Theorem In a right triangle with legs a and b and hypotenuse c, a2 b2 c2. (p. 421) Q (p. 455) Quotient Law of Logarithms For all positive b, v, and w, and b 1, logba v wb logb v logb w. (p. 366, 373) R radian measure The radian measure of an angle in standard position is the length of the arc along the unit circle from the point (1, 0) on the initial side to the point P where the terminal side intersects the unit circle. (p. 436) radical equations as square roots, cube roots, etc.) of expressions that contain variables (p. 111) equations that contain roots (such radioactive decay decay in the amount of a radioactive substance that can be modeled by the function P x f 2 stance, and h is the half-life
of the substance (p. 352) x h, corresponds to when the decay began, where P is the initial amount of the sub- 0.5 2 1 x 0 1 the four regions into which a coordinate quadrants plane is divided by its axes, usually indicated by Roman numerals I, II, III, and IV (p. 5) radiocarbon dating a process of determining the age of an organic object by using the amount of carbon-14 remaining in the object (p. 352) Glossary 1047 random sample a sample in which all members of the population and all groups of members of a given size have an equal chance of being in the sample (p. 843) random variable a function that assigns a number to each element in the sample space of an experiment (p. 869) range (of a data set) imum and minimum data values in a data set (p. 859) the difference between the max- range (of a function) the ordered pairs of a relation (p. 6, 142, 447) the set of second numbers in rational exponent A rational exponent is a rational number with a nonzero denominator; for any positive real number c and rational number with positive denominator, (p. 330) t k c 1 k ct 2 1 1 c k t 2 1 or c ct 2k c t. B A t k k 2k t rational function a function whose rule is the quotient of two polynomials, defined only for input values that produce a nonzero denominator (p. 278) rational number the set of real numbers that can be expressed as a fraction of an integer numerator and an integer denominator, where the denominator is not equal to zero (p. 4) Rational Zero Test If a rational number written in r s, lowest terms, is a zero of the polynomial function anxn p a1x a0, x f 2 1 an, p , an are integers with a0, factor of the constant term, (p. 251) leading coefficient, where the coefficients 0 0, and s is a factor of the and a1 a0 an. then r is a rational zeros numbers (p. 250) zeros of a function that are rational rationalizing (a denominator) writing equivalent fractions with no radicals in the denominator (p. 332) real axis where each real number point (a, 0) (p. 638) the horizontal axis in the complex plane a 0i corresponds to the real numbers rational numbers and irrational numbers (p. 3) the set of numbers that consists of real solutions numbers (p. 88) solutions of an equation that are real real zeros 0 f x 1 2 solutions of an equation of the form that are real numbers (p. 245) reciprocal identities metric functions and their reciprocals (p. 455) identities that relate trigono- rectangular coordinate system See Cartesian coordinate system. recursive form of a geometric sequence In a geo run1 for some nonzero conmetric sequence un , un6 5 1048 Glossary stant r and all n 2. (p. 59) recursive form of a sequence a method of defining a sequence when given the first term and the procedure for determining each term by using the preceding term (p. 15) recursive form of an arithmetic sequence In an un6 un arithmetic sequence , 5 n 2. stant d and all (p. 22) un1 for some con- d reduced row-echelon form This form of an augmented matrix satisfies the following conditions: all rows consisting entirely of zeros (if any) are at the bottom; the first nonzero entry in each nonzero row is a 1 (called leading 1); any column containing a leading 1 has zeros in all other entries; and each leading 1 appears to the right of leading 1s in any preceding row. (p. 797) reference angle the positive acute angle formed by the terminal side of x-axis (p. 449) in standard position and the u reflections The graph of reflected across the x-axis, and the graph of 1 is the graph of f reflected across the y-axis. (p. 177) is the graph of relation a correspondence between two sets; a set of ordered pairs (p. 6) 2 , x is divided Remainder Theorem If a polynomial by then the remainder is x c f 1 (p. 244) 1 residual a measure of the error between an actual data value and the corresponding value given by a model; point contained by the model (p. 44) where (x, r) is a data point and (x, y) is a r y, c f . 2 1 2 Richter scale a logarithmic scale used to measure the magnitude of an earthquake (p. 368) right angle an angle with a degree measure of (p. 414) 90° solutions of an equation of the form roots and equal to the zeros of f (p. 83) f x 1 2 0 roots of unity the n distinct nth roots of 1 (the solutions of ) (p. 648) zn 1 rotation equations a coordinate plane to the corresponding point in the plane after a rotation (p. 729) equations that relate a point in rule of the function See function rule. S sample a subset of the population studied in a statistical experiment, whose information is used to draw conclusions about the population (p. 843) sample space the set of all possible outcomes in an experiment (p. 864) scalar a real number, often denoted by k, used in scalar multiplication (p. 655) scalar multiplication (with vectors) an operation in which a scalar k is multiplied by a vector v to produce another vector, denoted by kv (p. 655) scalar multiplication with matrices multiplication of each entry in a matrix by a real number (p. 805) scatter plot a graphical display of statistical data plotted as points on a coordinate plane to show the relationship between two quantities (p. 5) 0 (p. 673) Schwartz Inequality For any vectors u and v, u v u v. 0 secant line (of a function) the straight line determined by two points on the graph of a function; the slope of the secant line joining points b, f 1 rate of change of the function from a to b (p. 218) on the graph of a function, equal to the average and a, f 22 22 b a 1 1 1 u secant ratio For a given acute angle in a right triu u angle, the secant of is written as sec and is equal to the reciprocal of the cosine ratio of the given angle. (p. 416) second a unit of degree measure equal to 1 60 of a minute, or 1 3600 of a degree (p. 414) second quartile See quartiles. second-degree equation See quadratic equation. self-similar the property possessed by fractals that every subdivision of the fractal has a structure similar to the structure of the whole (p. 305) sequence an ordered list of numbers (p. 13) sequence notation a customary method of denoting a sequence or terms of a sequence in abbreviated form: u1, u3, p , (p. 14) u2, un series the sum of the terms of a sequence (p. 76) sides (of an angle) that form an angle (p. 413) the two rays, segments, or lines Sigma notation See summation notation. simple harmonic motion motion that can be described by a function of the form a sin or x f 1 bt c 2 d a cos bt c d x f (p. 549) 1 2 1 2 1 2 simple interest interest that is generally used when a loan or a bank balance is less than 1 year; calculated where I is the simple interest, P is the prinby cipal, r is the annual interest, and t is time in years (p. 100) I Prt, sine ratio For a given acute angle gle, the sine of ratio of the opposite side length to the length of the hypotenuse. (p. 416) in a right trianis written as sin and is equal to the u u u sinusoid the wave shape of the graph of a sine or cosine function (p. 510, 548) sinusoidal function A function whose graph is the shape of a sinusoid and can be expressed in the form f x f or 2 1 2p b 2 is the amplitude, 2 where is the period, a cos 2 c b a sin bt c bt c d, d x a 1 1 1 0 0 is the phase shift, and d is the vertical shift. (p. 548) skewed distribution a type of distribution in which the right or left side of its display indicate frequencies that are much greater than those of the other side (p. 846) slant asymptote a nonvertical and nonhorizontal line that the graph of a function approaches as large (p. 286) x 0 0 gets slope of a line The value of the ratio y1 x1 are points contained by the y2 x2 ¢y ¢x , where line and x1, y12 x1 1 x2. and x2, y22 (p. 32) 1 slope-intercept form a linear equation in the form y mx b, sents the y-intercept (p. 33) where m represents the slope and b repre- solution of a system of equations a set of values that satisfy all the equations in the system (p. 780) solution of an equation the value(s) of the variable(s) that make the equation true (p. 30) solution to an inequality (in two variables) the region in the coordinate plane consisting of all points whose coordinates satisfy the inequality (p. 827) solving a triangle finding the lengths of all three sides and the measures of all three angles in a triangle when only some of these measures are known (p. 424) sound waves periodic air pressure waves created by vibrations (p. 558) special angles angles of degree measure 60° (p. 418) 30°, 45°, or square root of squares For every real number c, 2c2 (p. 109) c . 0 0 square system a system of equations that has the same number of equations as variables (p. 814) standard deviation a measure of variability that describes the average distance of data values from the mean, given by the square root of the variance (p. 857) standard equation of a hyperbola For any point (h, k) in the plane and positive real numbers a and b, Glossary 1049 2 1 x h a2 2 1 2 y k b2 2 1 or 1 2 x h b2 2 1 2 y k a2 2 1. (p. 701, 720) standard equation of a parabola For any point in the plane and nonzero real number p, h, k 2 1 x h . 1 720 or 2 1 2 2 2 1 1 (p. 710, standard equation of an ellipse For any point in the plane and real numbers a and b with x h a2 x h b2 y k b2 y k a2 h, k a 7 b 7 0, 1. 1 1 1 or p. 693, 720) standard form (of a line) used to display the equation of a line without fractions, a linear equation in the form to zero (p. 39) where A and B are not both equal Ax By C, standard normal curve a normal curve with a mean of 0 and a standard deviation of 1 (p. 890) standard notation (of triangles) a method of labeling triangles in which each vertex is labeled with a capital letter to denote the angle at that vertex, and the length of the side opposite that vertex is denoted by the same letter in lower case (p. 617) standard position (of an angle) an angle in the coordinate plane with its vertex at the origin and its initial side on the positive x-axis (p. 434) standard viewing window the window or screen of a graphics calculator that displays 10 y 10, graphi
cs calculators (p. 84) 10 x 10 listed in the ZOOM menu of most and standardize (data) normally distributed in order to match the standard normal curve (p. 893) to adjust the scale of data that is stem plot a display of quantitative data in a tabular format consisting of the initial digit(s) of the data values, called stems, on the left and the remaining digits of the data values, called leaves, on the right; commonly used to display small data sets (p. 847) step function a function, such as the greatest-integer function, whose graph consists of horizontal line segments resembling steps (p. 157) sum function For any functions sum function is the new function f (p. 191 and x 21 2 , their x g 1 2 sum of an infinite geometric series The sum S a1 1 r vergent geometric series with common ratio r such that r is a con- r3a1 r2a1 ra1 where p (p. 77) 6 1. a1 , 0 0 1050 Glossary sum-to-product sin x ± sin y or cos x ± cos y (p. 599) trigonometric identities involving summation notation a customary method of denoting the sum of terms by using the Greek letter Sigma c2 p cm ) as follows: (p. 25) c1 c3 ck © ( m a k1 supplementary angle identity For any acute angle u, (p. 628) sin u sin 180° u . 1 2 symmetric distribution a type of distribution in which the right and left sides of its display indicate frequencies that are mirror images of each other (p. 846) synthetic division an abbreviated notation for performing polynomial division when the divisor is a first-degree polynomial (p. 241) system of equations a set of two or more equations in two or more variables (p. 779) T tangent line (of a function) a line that touches the graph of a function at exactly one point; the slope of the tangent line to a curve at a point, equal to the instantaneous rate of change of the function at that point (p. 235) u tangent ratio For a given acute angle u angle, the tangent of is written as tan and is equal to the ratio of the opposite side length to the adjacent side length. (p. 416) in a right tri- u term (of a sequence) a number in a sequence (p. 13) terminal point (of a vector) that extends from point P to point Q (p. 653) the point Q in a vector terminal side the final position of a ray that is rotated around its vertex (p. 433) third quartile See quartiles. transformation form of a quadratic function a x h a 2 k, quadratic function in the form 2 1 a 0 where a, h, and k are real numbers and (p. 164) x f 2 1 Triangle Sum Theorem The sum of the measures of 180°. the angles in a triangle is (p. 421) trigonometric form of a complex number See polar form of a complex number. trigonometric functions of a real variable a function whose rule is a trigonometric ratio in the coordinate plane with domain values in radian measure (p. 445) trigonometric ratios (in a triangle) combinations of side length ratios of a right triangle (p. 415) the six possible trigonometric ratios (in the coordinate plane) six trigonometric ratios defined in terms of a triangle determined by the coordinates of a point on the terminal side of an angle in standard position and the origin (p. 444) the two-stage path (of a network) a path in a directed network from one vertex to another with exactly one intermediate vertex (p. 810) U uniform distribution a type of distribution in which all of the data values have approximately the same frequency; its display is level (p. 846) unit circle the circle of radius 1 centered at the origin of the coordinate plane (p. 435) unit vector a vector with a length of 1 (p. 661) upper bound (for the real zeros of a polynomial function) the number s such that all the real zeros of a polynomial function f(x) are between r and s, where r and s are real numbers and (p. 255) r 6 s V variability the spread of a data set (p. 857) variance a measure of variability given by the average of squared deviations (p. 858) vector a quantity that involves both magnitude and direction; represented geometrically by a directed line segment or arrow and denoted by using its endpoints, or by a boldface lowercase letter, such as u such as (p. 653) ! PQ , u . I a, b I H (p. 657) u (p. 658) H vector addition If a c, b d u v H and v c, d I H , then and v a, b I c, d H , I vector subtraction If u v a c, b d H . I vertex (of an angle) two rays, segments, or lines that form an angle (p. 413) the common endpoint of the vertex of a parabola the intersection of the axis of a parabola and the parabola; the midpoint of the segment from the focus of the parabola to the directrix (p. 709) vertical asymptotes a vertical line that the graph of a function approaches but never touches or crosses because it is not defined there (p. 284, 950) the graph of vertical compression For any positive number y c f c 6 1, compressed vertically, toward the x-axis, by a factor of c. (p. 179) is the graph of f x 1 2 vertical line a line that has an undefined slope and x c, an equation of the form real number (p. 37) where c is a constant y f vertical shifts For any positive number c, the graph is the graph of f shifted upward c units, x of 1 c is the graph of f shifted and the graph of downward c units. (p. 174) See also sinusoidal function. y f c x 1 2 2 vertical stretch For any positive number graph of cally, away from the x-axis, by a factor of c. (p. 179) the is the graph of f stretched verti, vertical line test A graph in a coordinate plane represents a function if and only if no vertical line intersects the graph more than once. (p. 151) vertices of a hyperbola the points where the line through the foci intercepts the hyperbola (p. 701) vertices of a network the points that are connected in a network (p. 809) vertices of an ellipse the points where the line through the foci intercepts the ellipse (p. 692) W whole numbers natural numbers and zero: 0, 1, 2, the set of numbers that consists of p (p. 3) work The work W done by a constant force F as its point of application moves along the vector d is W F d. (p. 678) X then x-axis often the name of the horizontal axis of a coordinate plane with the positive direction to the right and the negative direction to the left (p. 5) x-axis symmetry A graph is symmetric with respect to the x-axis if whenever x, y x, y is on the graph, then is also on it. (p. 185) 1 x-coordinate usually the first real number in an ordered pair (p. 5) 2 2 1 the x-coordinate of a point where a graph x-intercept crosses the x-axis; the x-intercepts of the graph of f, equal to the zeros of f and the solutions of (p. 83) 0 x f 1 2 x-intercept form of a quadratic function a quadx s a f where ratic function in the form 1 a 0 a, x, s, and t are real numbers and x t , 2 21 (p. 164) x 2 1 Glossary 1051 x-intercept method a method of solving an equation of the form finding the x-intercepts (p. 84) by graphing y f 0 and -axis nate system with positive direction upward (p. 790) the vertical axis in a three-dimensional coordi- zero polynomial the constant polynomial 0 (p. 240) y-axis often the name of the vertical axis of a coordinate plane with the positive direction up and the negative direction down (p. 5) Zero Product Property If a product of real numbers ab 0, is zero, then at least one of the factors is zero; if (p. 89) then b 0. a 0 or y-axis symmetry A graph is symmetric with respect x, y to the y-axis if whenever x, y is on the graph, then is also on it. (p. 184) 1 y-coordinate usually the second real number in an ordered pair (p. 5) 2 2 1 zeros of a function inputs of the function that produce an output of zero (p. 83) z-value a value that gives the number of standard deviations that a data value in a normal distribution is located above the mean (p. 894) 1052 Glossary Acknowledgments Photo Credits: page 2, Ron Behrmann/International Stock Photography; page 80, Orion Press/Natural Selection; page 140, Peter Van Steen/HRW Photo; page 326, Laurence Parent; page 412, Todd Gipstein/National Geographic Society Image Collection; page 472, Peter Van Steen/HRW Photo/Courtesy Jim Reese at KVET, Austin, Tx; page 522, Coco McCoy/Rainbow/PictureQuest; page 570, Telegraph Color Library/FPG International; page 616, Scott Barrow/International Stock Photography; page 690, Stone; page 778, Tom McHugh/Photo Researchers, Inc.; page 842, SuperStock; page 908, Tom Sanders/Photri/The Stock Market Illustration Credits: Abbreviations used: (t) top, (c) center, (b) bottom, (l) left, (r) right, (bkgd) background All work, unless otherwise noted, contributed by Holt, Rinehart and Winston. Chapter Four: page 306 (tc), Pronk&Associates; Chapter Six: page 470 (cl), NETS; page 471 (br), NETS. Chapter Eleven: page 713 (b), NETS. Chapter Thirteen: page 882 (br), NETS. Acknowledgments 1053 Answers to Selected Exercises 2.5, 0 1 H ; 3, 1 2 ; I D 1 3, 1 1.5, 3 1 5. 2 P 1 4, 2 1 b. y (–5, 4) ; 2 (4, 1) x 2 (3, –2) (–2, –3) Chapter 1 1. Section 1.1, page 10 3, 3 ; 2 1 F ; 0, 2 1 2 6, 3 B 1 0, 0 A E 1.5, 3 G ; 3. P 1 2 1 ; 2 2, 0 C ; 2 1 2 7. y 500 400 300 200 100 2 x 0 987654321 10 9. a. About $0.94 in 1987 and $1.19 in 1995. b. About 26.6% c. In the first third of 1985 and from 1989 onward. 11. a. Quadrant IV b. Quadrants III or IV 13. a. y (–2, 3) (–5, –4) (3, 2) x (4, –1) 1054 Answers to Selected Exercises c. They are mirror images of each other, with the x-axis being the mirror. In other words, they lie on the same vertical line, on opposite sides of the x-axis, the same distance from the axis. 15. Yes. Each input produces only one output. 17. No. The value 5 produces two outputs. 19. (500, 0); (1509, 0); (3754, 35.08); (6783, 119.15); (12500, 405); (55342, 2547.10) 21. Each input (income) yields only one output (tax). 23. Postage is a function of weight since each weight determines one and only one postage amount. But weight is not a function of postage since a given postage amount may apply to several different weights. For instance, all letters under 1 oz use just one first-class stamp. 25. Domain: all real numbers between 3 inclusive; range: all real numbers between 3, inclusive and 3, 4 and 27. 2 is
output of 1 2 ; 0 of 5 2 ; and 3 of 5 2 . 2; 3 of 0; 2 of 1; 1 of 2.5; and 0 29. 1 of is output of 1.5. 31. 1 is output of 2; 1; 1 of 0; 1 of 3 1 2 2, 3 (approximately) of ; and 1.5 of 1. 33. a. 3, 4 4 3 d. 0.5 b. 3 e. 1 4 2 c. f. 1 Section 1.2, page 19 1. 20 0 3. 20 0 5. u1 6 and un un1 2 10 0 −10 10 10 10 7. u1 6 and un un1 5 30 0 10 0 4 2 4 3 11 2 11 3 25 2 25 3 53 2 53 3 109 400; 0.8un1; un u1 u2 u3 u4 u5 u1 u2 u3 u4 u5 u0 9. 11. 13. 15. For 2 rays: 6; un u4 1; u2 un1 for 3 rays: n 1 for 3; u3 n 3 for 4 rays: 325 0 25 17. 19. 21. 23. u25 300 12 , u1 seats un un1 2 for 2 n 35 ; u30 70 for n 1 ; 6500, for n 1 ; 1.06un1 0.75un1 u0 u10 u0 u8 u0 u20 30,000 un , $53,725.43 35,000 un , 26,901 students 50,000 un , 41,216 u35 ; 4000 0.9un1 40,250 . for n 1 ; 25. a. The items listed are the first ten primes. Every other number less than 29 can be factored into a product of smaller integers. b. 59, 61, 67, 71 27. 4, 9, 25, 49, 121 29. 3, 7, 13, 19, 23 31. 1, 1, 2, 3, 5, 8, 13, 21, 34, 55 33. n 1: 5 1 n 2: 5 1 n 3: 5 1 n 4: 5 1 n 5: 5 1 n 6: 5 1 n 7: 5 1 n 8: 5 1 n 9: 5 1 n 10 13 21 34 12 2 9 32 3 16 42 4 49 72 5 121 112 6 324 182 7 841 292 8 2209 472 9 5776 762 10 15,129 1232 2 2 2 55 1 2 1 2 Section 1.3, page 29 1. 13; un 2n 3 15 0 0 10 u6 104.858 cm Answers to Selected Exercises 1055 3. 5; un n 4 15 4 6 0 2 5. 8; un n 2 21 2 15 Section 1.4, page 40 1. a. C b. B c. B d. D 3. Slope, 2; y-intercept, b 5 5. Slope, 3 7 ; y-intercept, b 11 7 7. Slope, 5 2 15. y 9. Slope, 4 11. t 22 13. t 12 5 10 L P(1, C) x (0, 0) (1, 0) Slope of 10 9. 224 17. 30 3 2 1 3 n 1 2 4 2; 0 5 7. 45 11. 13. 87 15. 21 4 3 2n 19. un1 un arithmetic with 1 2 d 2 21. un un1 5 3n 2 d 3 2 c 2n 2 d 2 2n 4 7n 10 3n 2 15 2 arithmetic with 23. 25. 27. 1 un1 un arithmetic with 14; un 25; un u5 u5 29. u5 0; un 5 3 n 1 1 2 2 3 2 ; 49; 17. 23. y x 2 y 7 3 x 34 9 19. y x 8 21. y x 5 25. Perpendicular 27. Parallel 29. Parallel 31. Perpendicular 33. Yes. The slopes are , , 2 5 and 9 8 perpendicular result in a right triangle. y 3 2 y 3x 7 5 2 39. 37. x . 35. Two sides y x 5 43. k 11 3 41. y x 2 45. un un u1 1 2 1 4n 6 n 1 d 2 n 1 4 2 or 47. The common difference is y 4x 6 6 . d 6 2 or y 6x 13 un un 2 21 n 1 u1 1 n 1 7 1 6n 13 u1 1 2 1 8n 10 un The slope is n 1 d 2 n 1 8 2 or m 8 un y 8x 10 and the y-intercept is b 10. 51. Both have slope and different y-intercepts. A B y 449x 9287 x 9; y $13,328 1 x 24; y $20,063 53. a. b. 1990 2 2005 1 2 31. 710 33. 156 2 3 35. 2550 37. 20,100 39. $77,500 in tenth year; $437,500 over ten years 41. 428 43. 23.25, 22.5, 21.75, 21, 20.25, 19.5, 18.75 1056 Answers to Selected Exercises 57. a. b. 55. $375,000; $60,000 y 5x 150 x 5, y 125 pounds x 7, y 185 pounds 50x 110,000 22x 110,000 59. a. c ft ; 2 1 6 ft 2 r 72x b. x 2 d. x 5000 1 61. a. 63. a. y 8.50x 50,000 x 10 b. $11, $9.50, $9 per hat b. x 30 15. No High School Diploma 12.31x 238 y2 1. a. Section 1.5, page 53 x 5 y 3 4 4 Sum of squares 3 Model B still has least error. Slope 1.0564054 y 1.0564054x 21.077892 b. 3. a. b. c. Line described in b predicts a higher number of workers. 5. negative correlation 7. very little correlation 9. a. 0.09, 0.17, 1.22, 3.13, 5.14, 8.26 b. not linear; 35 0 0 60 11. a. 4.6, 8.3, 14.3, 23.5, 37.2, 56.9, 84.3, 121.4, 170.7, 234.2 b. not linear; 1000 0 0 110 13. a. 446.9, 405.2, 515.8, 785.3, 298, 413 b. linear; positive correlation 7100 0 0 1000 0 0 High School Graduate 15.17x 354 y4 1000 0 0 Some College y1 20.74x 392 1000 0 0 College Graduate 34.86x 543 y3 1000 0 0 12 12 12 12 10 y 0.0292x 4.0149 y 0.4078x 16.8494 17. a. b. c. The income of the rich is increasing faster than the income of the poor is decreasing. d. The income gap will increase. Answers to Selected Exercises 1057 19. a. y 4x 82 b. The amount of federal money in loans is increasing and the amount in grants and workstudy is decreasing. 13. u6 5 16 ; un 1 1 n1 5 2 2n2 15. 315 32 19. un un1 21. un un1 n 1 a 2b n1 1 a 2b 5n2 2 n1 51 2 17. 381 1 2 ; geometric with r 1 2 5; geometric with r 5 20 23. u5 1; un 1 1 n164 2 4n2 1 n1 1 2 4n5 25. u5 1 16 ; un 1 4n3 27. u5 8 25 ; un 2n2 5n3 29. 254 31. 4921 19,683 33. 665 8 100 0 0 Loan data Grant/work-study data c. 1983 40 21. a. 35. a. Since for all n, the ratio r is un1 un 1.71 1 1.71 1.191n1 1.191n 1 2 2 1.71 the sequence is geometric. b. $217.47 1.191n 1 1.71 21 1.191n 1.191 1 2 2 1.191, 39. 37. 23.75 ft 31 a n1 41. $1898.44 2n1 1 231 1 2 231 1 1 2 cents $21,474,836.47 1 1 43. un log u1 log log u1 u1rn1 u1rn1. log un 2 rn1 n 1 log 1 2 2n1 r 2. 5 and the sum of the preceding k 1 2 1 2n1 1 2k1 1 2 45. The sequence is 2k1, kth term is terms is the k1 a n1 2k1 1. log r and 6 2 th partial sum of the sequence, So for any k, the 47. 37 payments Chapter 1 Review, page 67 is an irrational number. 1. 3. 13 e 2.718 is an irrational number. 5. 0 is a whole number. 7. 1121 is a natural number. 9. 5 is a natural number. 11. Answers may vary; for example, numbers that are not rational. p and 12 are real 13. This is not a function; the input 2 has more than one output. 15. This is a function; for each input there is exactly one output. 17. all real numbers greater than or equal to 2 and not equal to 3 −1 0 17 y 1.27x 23.8318 b. c. The model gives a negative median time for approval in 2009, and it will not be useful for data extrapolation. y 0.08586x 22.62286 14.72 23. a. b. for 1992, a terrific prediction. c. This model may not remain valid for any future dates. The rate of improvement seems to be slowing down. r 0.9797 d. e. The data appears to be linear because the residuals do not form a pattern. 2 0 2 Section 1.6, page 63 1. Arithmetic 9. u6 5. Arithmetic 160; un 1 256 11. u6 ; un 2n1 5 1 4n2 100 3. Geometric 7. Geometric 1058 Answers to Selected Exercises 19. r 4 f g 21. a. b. c. radius: 150 meters; area: 70,685.83 square 502t 2500pt t 1 t 1 2 2 meters d. 12.73 hr x 3 4 x 5 23. 27. 33. false 39. a. 1.618 25. x 0 and 29. 3 35. $1862.96. 4 x 5 31. false 37. 12.5 cm b. 1.618033989 . . . 4, 82, un 10, 28, 6n 11 244 43. un 9 6n 47. 78 41. 45. 49. 8 0 0 7 b. 4 5 51. a. 1 53. y 3x 7 55. a. y-intercept is 0, a 4 3b ; b. y 5 3 x 4 3 57. x 5y 29 61. true 59. m 5 63. false 65. true 67. (d) 69. (e) 71. 5 3 81. a. 13 −10 −10 10 10 0 Managerial 8 0 Females 17 −10 10 0 Males 73. a. 75. c; 79. a. y 0.25x 62.9 m 75 100 b. x 40, y 72.9 yrs. 77. d; m 20 83. a. (negative slope, y-intercept at y1 (negative slope, y-intercept at (positive slope, y-intercept at b. Managerial, y2 11.74); Female, y3 7.34); Male, 15.48) y 0.3654x 10.6741 b. $11.77, $14.33; Both answers are close to actual amounts. c. $15.79 85. un 2 3n1 20 0 70 89. 55 87. un 384 n1 1 2b a 91. 315 32 b. Nonlinear c. 20.10 19.58 0.52 25.23 20.79 4.44 48.55 34.89 13.66 98.92 69.17 29.75 The finite differences show that the data is not linear. 20.79 20.10 0.69 34.89 25.23 9.66 69.17 48.55 20.62 93. Second method is better. Answers to Selected Exercises 1059 Chapter 1 can do calculus, page 79 1. 1 4. 2 3 2. Diverges 5. 500 0.6 833 1 3 7. 4 212 8. Diverges 10. 13. 2 9 8428 99 11. 14. 37 99 10,702 4995 3. 3 7 6. 5.7058 9. 1 2 12. 15. 597 110 18,564 4995 16. Because there are infinitely many terms that are getting larger and larger, the sum cannot converge to a finite number. 17. The sum approaches 1 3 . 0 0 30 −1 Chapter 2 Section 2.1, page 87 1. 3 3. 3 5. 2 7. 11. 15. 19. 21. 25. 29. x 2.426 x 1.475, x 1.379, x 2.115, x 2.102 x 0.951 x 2.390 9. 13. 17. 1.237 1.603 0.254, 1.861 23. 27. 33. x 7.033 35. 31. x 2 3 39. x 13 41. 2004 1.164 x 1.750, x 1.453, x 0, x 1.601, 0.507, 1.329 1.192 x 1.752 x 0, x 0.651, x 2.207 1.151 37. x 1 12 43. 1999 3. 1. 5. or 3 Section 2.2, page 95 x 3 or 5 y 1 2 u 1 or 4 3 x ± 3 x ± 140 ± 6.325 x2 9; x2 40; 11. 13. 15. 9. 7. x 2 or 7 t 2 or 1 4 or 4 3 x 1 4 1060 Answers to Selected Exercises 17. 19. 21. 23. 25. 27. x2 4; x ± 2 3x2 12; 5s2 30; 25x2 4 0; s ± 16 ± 2.449 s2 6; x2 4 25 w2 28 3 ; x ± 2 5 3w2 8 20; w ± ; 28 3 A ± 3.055 x 1 ± 113 1 15 2 w A B or A 1 15 2 B 29. x 2 ± 13 31. x 3 ± 12 33. No real number solutions 35. 39. ± 12 x 1 2 4 ± 16 5 u 47. x 3 or 6 51. x 5 or 3 2 37. x 2 ± 13 2 41. 2 43. 2 45. 1 49. x 1 ± 12 2 53. No real number solutions 55. No real number solutions 57. 61. 65. x 1.824 or 0.470 y ± 1 or ± 16 y ± 2 or ± 1 12 59. 63. 67. x 13.79 x ± 17 x ± 1 15 69. k 10 or 10 71. k 16 73. k 4 Section 2.3, page 105 1. The two numbers: x and y; their sum is 15: the difference of their squares is 5: x y 15; x2 y2 5. 3. English Language Length of rectangle Width of rectangle Perimeter is 45 Area is 112.5 Mathematical Language x y x y x y 45 or 2x 2y 45 xy 112.5 5. Let x be the old salary. Then the raise is 8% of x. Hence, old salary raise $1600 1600 8% of x 2 1 x 0.08x 1600. x 7. The circle has radius pr2 p 82 64p. r 16 2 8, so its area is Let x be the amount by which the radius is to be reduced. Then p and the new area is 1 48p less than the original area, that is, 2 64p 48p, p 2 16p. p or equivalently, 8 x 8 x 8 x 2, 2 1 1 2 2 r 8 x which must be 9. $366.67 at 12% and $733.33 at 6% 11. 2 2 3 qt 13. 65 mph 15. 34.75 and 48 17. Approximately 1.75 ft 19. Red Riding Hood, 54 mph; wolf, 48 mph 21. a. 6.3 sec b. 4.9 sec 23. a. Approximately 4.4 sec 25. 23 cm by 24 cm by 25 cm b. After 50 sec r 4.658 27. 29. x 2.234 31. 2.2 by 4.4 by 4 ft high Section 2.4, page 116 12 y 2, 3w 2 0 −1 6 8 0 3. −6 −5 −4 −3 −2 −1 109876543210 Since 3w 6 or 10, dividing by 3, w 2, 3 5. 0 2x Since 2x 1 or 9, dividing by 2, x 1 2 , 9 2 . 3x 7. 0 2 1 2 0 5 or 0 3x 2 5 0 −7 −6 −5 −4 −3 −2 −1 0 1 2 3 Since 3x 7 or 3, dividing by 9. x 6 or 3 15. x 3 2 3, x 7 3 1. , 13. x 2 11. x 3 2 17. x 5 or 1 or 3 or 1 19. x 1 or 4 or 5 133 2 or 5 133 2 21. For any real number x, the distance between 2x2 cannot be a negative number. Therefore, 12 has no real number solution. 3x and 2x2 3x 0 23. Let x 0 Joan’s ideal body weight. The difference Therefore, 0 x 120 0.05x or x 120 0.05x 0 0.05x. x 120 0.95x 120 x 126.3 1.05x 120 x 114.3 To the nearest pound, Joan’s ideal weight is either 114 pounds or 126 pounds. 25. If the true w
ind speed differs by 5 feet per second true wind speed from the measured speed, let x 5 and x 20 0 0 x 20 5 or x 20 5 x 25 x 15 or The true wind speed is between 15 and 25 feet per second. CL 0.0097 0 x 4 0.0497 33. (in practice) and x 2 31. 27. 29. 35. x 1 or 2 37. x 9 39. x ± 3 x 1 2 47. x 1.658 53. No solutions 61. x 1 63. x 1 43. x ± 0.73 or 2.59 or x 3 51. or 1.40 x 1 or 7 57. x 7 41. 45. 49. 55. 59. or 4 x 1 2 x 1.17 x 6 x 0.457 x 3 ± 141 4 65. u x2 1 K2 B 67. b a2 A2 1 B 69. a. I x x2 1024 1 3 2 2 Section 2.5, page 124 b. 22.63 ft 1. 5. x 0 −2 3. −3 −2 −1 0 1 32 9. −4 −2 0 1 11. [5, 8] 17. q, a 3 2 T 23. 1, q 1 2 29. q, a 4 7b 35. 5, q 3 2 13. 19. 3, 14 2, q 2 2 1 1 25. (2, 4) 31. 37. , q 7 b 17 S x 6 b c a 15. 21. 27. 33. 8, q 2 q, 8 5 T 3 a 3, 1, 5 2b 1 8b S S between Joan’s actual weight and her ideal weight is x 120. 39. c 6 x 6 a c 41. 1 x 3 Answers to Selected Exercises 1061 43. x 9 121 2 x 1 133 2 or or x 9 121 2 x 1 133 2 45. 47. 49. 51. 53. or x 1 or or 2.26 x 0.76 1 6 x 6 2 x 3.51 or 55. 0.5 6 x 6 0.84 57. x 6 1 3 or x 7 2 59. 2 6 x 6 1 or 1 6 x 6 3 61. x 7 1 63. x 9 2 or x 7 3 65. 67. 3 6 x 6 1 or 17 6 x 6 17 x 5 or x 7 5.34 69. x 6 3 or 1 2 6 x 6 5 71. x 7 1.43 73. x 3.79 or x 0.79 75. Approximately 8.608 cents per kwh 81. 77. More than $12,500 1 6 x 6 19 10 6 x 6 35 and 83. y 20 x 85. 1 t 4 79. Between $4000 and $5400 87. 2 6 t 6 2.25 Section 2.5.A, page 131 1. 4 3 x 0 5. x 6 2 or x 7 1 9. x 6 53 40 or x 7 43 40 3. 7. 11. 6 x 6 11 6 7 6 x 11 20 x 7 2 13. x 6 5 or 5 6 x 6 4 3 or x 7 6 5x 15 5x 4 0 0 1 0 0 1 11 5x 4 2 6 E. 0 11 0 . Thus, 2 Chapter 2 Review, page 135 1. 5. x 2.7644 x 3.2843 9. No real solutions 3. 7. x 3.2678 x 1.6511 11. z 3 ± 2111 5 13. x 3 or 3 or 12 or 12 15. 2 17. gold, 3 11 19. 2 2 9 hrs 25. 25 oz; silver, 8 11 oz 21. 9.6 ft 23. 4 ft 27. b 1 2 2 1 29. x 1 2 or 11 2 31. 33. 0 4 3x 1 0 3x 1 4 or 3x 1 4 3x 3 3x 5 x 5 3 x 1 2x2 x 2 0 Squaring both sides, x2 x 2 0 x 1 0 2 x 2, 1 x 2 21 1 Both of these check in the original equation. x2 6x 8 x 1 0 35. Set the numerator equal to 0. or x 1 4 or x 5 4 1 x 2 x2 6x 8 0 x 4 0 2 x 2, 4 21 1 7 6 x 6 3 37. 15. 17. 19. 21. 23. 25. 1 6 x 6 13 x 1 or x 7 16 or 1 x 0 13 6 x 6 1 x 6 16 x 2 or 0 6 x 6 2 3 1.43 6 x 6 1.24 or 2 6 x 6 8 3 or 27. x 6 0.89 or x 7 1.56 29. x 2 or x 14 3 31. 1.13 6 x 6 1.35 or 1.35 6 x 6 1.67 then multiplying both sides by 5 33. If 0 x 3 0 shows that . 0 x 3 But 5 x 3 2 0 0 1 0 1062 Answers to Selected Exercises Neither solution causes the denominator to be 0. x 5 15 39. No solutions 2 41. a. 8, q 1 2 b. 1 q, 5 4 43. 7 4 S , q b 45. 47. 51. 55. 57. 59. 61. b 49. e 53. x 7 or x 7 4 , q and 1 a 3 0 x 1 q, 2 2 or 1 x 1 4, 5 8b S x 6 213 or x 6 1 113 6 3 6 x 6 213 x 7 1 113 or 6 y 17 x 4 3 or y 13 or x 0 Chapter 2 can do calculus, page 139 1. Numerical Method a. Each base must be greater than 0 and less than 10 yards. The nr in the chart indicates that no rectangle can be formed with a base length of 5 yards or more because the opposite bases of a rectangle are the same length, and, since 5 5 10, make the sides of a rectangle. there would be no wire left to b. length 1 yd 1.5 yd height 4 yd 3.5 yd 2 yd 3 yd 2.5 yd 2.5 yd 3 yd 2 yd 3.5 yd 1.5 yd 4 yd 1 yd area 4 yd2 5.25 yd2 6 yd2 6.25 yd2 6 yd2 5.25 yd2 4 yd2 2.25 yd2 4.5 yd 5 yd 5.5 yd 0.5 yd nr — nr — 6 nr b. x y 1 2 3 4 5 5.92 5.66 5.20 4.47 3.32 area 5.92 11.32 15.6 17.88 16.6 — 6 c. 20 c. 8 0 0 d. The maximum area 6.25 yd2 appears to occur when the base length is 2.5 yd. Analytical and Graphical Method 2l 2w 10 2w 10 2l 5l l2 A l2 5l 2 8 0 0 6 Using the maximum finder on a graphing calculator indicates that the maximum area of 6.25 occurs at approximately 2.5 yd. yd2 2. Numerical Method a. The base must be greater than 0 and less than 6 units. The nr in the table indicates that no rectangle can be formed with a base of 6 or more units because exist if x 7 6. y 236 x2 would not 0 0 7 d. A maximum area 17.88 square units appears to occur when x is 4. Analytical and Graphical Method y 236 x2 A x y A x 236 x2 20 0 0 7 Using the maximum finder on a graphing calculator indicates that the maximum area of 18 square units occurs when x is approximately 4.24. 3. Numerical Method a. The base must be greater than 0 and less than 4 units. The nr in the table indicates that no rectangle with a base of 4 or more units if x 4. because y 0 Answers to Selected Exercises 1063 c. 3 0 0 b. x y 0.5 1.0 1.5 2.0 2.5 3.0 3.5 1.75 1.5 1.25 1 0.75 0.5 0.25 4.0 nr area 0.875 1.5 1.875 2.0 1.875 1.5 0.875 — 11. y is not a function of x. 13. 23 1 2.73 15. 322 3 22 1 1.69 17. 4 19. 34 3 21. 59 12 a k 2 2 1 a k 2 23. 25. 1 1 2 x 2 27. 8 33. t2 1 2 1 2 6 4x x2 1 2 x 29. 1 35. 1 2 x 2 1 s2 2s s 1 31. 1 37. 3 2 4 39. 2x h 1 41. 1 2x h 2x 43. All real numbers 45. All real numbers d. A maximum area of 2 square units appears to occur at x 2. Analytical and Graphing Method 4x x2 2 3 0 0 4 Using the maximum finder on a graphing calculator indicates that the maximum area is 2 when x is approximately 2. Chapter 3 Section 3.1, page 148 47. All nonnegative real numbers 49. All nonzero real numbers 51. All real numbers 53. All real numbers except 2 and 3 55. [6, 12] 57. a. the greatest integer less than or 4 2 1 ; 0 0 f 3 equal to 0 is 0. f f f 1.6 2 1 1 1 1 4 2.3 3 1.6 3 2 2.3 5 2p b. c. 4 5 2p d. e. The domain of f is all real numbers. 5.76 3 1.283 0 2 4 4 3 3 2 f f 59. a. 1 0.69 c. 1 e. The domain of f is all real 0 2 2.3 b. d. f f 1 1 2 5 f f 61. a. c. e. The domain of f is all real numbers. 1.6 2 5 2p 0 2 2.3 7.6 b. d.920 1.6 2 5 2p 2 numbers 20. 3.4 5.566 63. a. A pr2 65. V 4x3 b. A 1 4 ˛pd2 1 x C 67. a. 5.75x 45,000 x x 7 0. 69. a. Let t be the number of hours since he started b. The domain is 2 1. The set of inputs is the number of hours you work 3t in each pay period. The set of outputs is the amounts of your paychecks. The function rule is the amount of pay for each hour worked multiplied by the number of hours worked. 3. The set of inputs is temperatures of gas. The set of outputs is pressures of gas. The function rule is the formula P k T . 5. y is a function of x. 7. y is not a function of x. 9. y is a function of x. 1064 Answers to Selected Exercises d t 2 1 2.25 5 t 3 4b a 8.5 8.5 3 b. The domain is t such that t 2..5 2.5 6 t 4 0 t 4. of annual income and of tax. 71. Let x f 1 2 f x 1 2 x amount amount 0 0.02 1 80 0.05 u The domain is x 0. x 2000 2 x 6000 1 2 if x 6 2000 if 2000 x 6000 if x 7 6000 Section 3.2, page 160 1. 5. The domain is 7. The domain is 6, 9 7, 8 . . 4 4 3 3 9. This is not a function; for example, there are three output values for an input value of 4. 11. This is not a function; for example, there are three output values for an input value of 2. 13. This is a function. 1 and 15. Increasing on 6, 2.5 17. Constant on 1, 1 . 1 1 2 2.5, 0 2 0, 1.7 2 q, 1 1 4 1 2 19. Increasing on q, 5.8 1 and 5.8, 0.5 0.5, q 1 2 21. Increasing on (0, 0.867) and 1 and (0.867, 2.883) q, 0 on 1 2 2 1 2 and 1 1.7, 4 2 ; decreasing on and 1, q 3 2 ; decreasing on ; decreasing on 2.883, q ; decreasing 2 23. Minimum at x 0.57735; maximum at x 0.57735 25. Minimum at 27. Minimum at x 0.7633; maximum at x 0.7633 29. a. maximum at 50x x2 x 1; x 1 A x b. To maximize area each side should be 25 in. long. 1 2 31. a. SA x 1 2 2x2 3468 x b. Base is approximately 9.5354 in. 9.5354 in.; height is same (that is, 9.5354 in.). 33. 3.1 −4.7 4.7 Therefore, the points of inflection are approximately at and 0.6, 0.6 1 . 2 1 0.6, 0.6 10 2 37. a. 10 10 10 b. This function is increasing over the interval and decreasing over the interval c. There is a local minimum at the point (1, 0). d. This function is concave up over the interval q, 1 . 2 1 1, q 2 1 1 q, q . 2 e. There is no point of inflection. 39. a. 10 10 10 10 b. This function is increasing over the intervals and decreasing over the 2, q q, 0 and 2 1 1 interval (0, 2). 2 c. There is a local maximum at the point (0, 2) and a local minimum at the point 2, 2 . 2 1 d. This function is concave upward over the interval interval 1, q 2 q, 1 1 1 . e. There is a point of inflection at (1, 0). 2 and concave downward over the −3.1 41. y This function is concave up over the interval q, 0 . and concave down over the interval Therefore, the point of inflection is at (0, 0). 2 1 35. 2.5 −2.5 2.5 −2.5 0, q 1 2 4 3 2 1 −1 −2 −3 −4 x 1 2 The function is concave up over the approximate intervals and down over the approximate interval and concave 0.6, 0.6 q, 0.6 0.6, q . 2 1 2 1 1 2 Answers to Selected Exercises 1065 43. y 51. y 2 x 6 0 55. When 0 x 2, x is positive and 6 4 2 −5 −3 1 x 45. a. y b 2 x 2 x 0 47. a. y 2 (−4, 0) (4, 0) x (0, −2) 49. a. y b (0, 5) (10, 5) x (5, 0) 1066 Answers to Selected Exercises y = [-x] x 1 1 53. y y = 2[x] x 1 1 negative, so 0 x Therefore and x 2 0 0 x 1 0 x 2. when 0 0 x 2 is x 2 . 2 1 x 2 2 57. Domain: all real numbers x such that x 2 or x 2; range: all nonnegative real numbers 59. Domain: all real numbers; range: all real numbers 61. Many correct answers, including 6 4 2 −2 −1 2 4 −2 −4 −6 63. Entire graph: 75 69. 10 2 10 near the origin: 10 2 10 65. Entire graph: 60 16 near the origin: 2 62 2 32 5 2 2 10 10 10 10 x t, y t4 3t3 t2 71. 10 10 10 x t4 3t2 5, y t Section 3.3, page 170 3. (1, 2), downward 5. 1. (5, 2), upward y-intercept 3 y-intercept 5 9. The x-intercepts are 7. The parabola opens upward. The parabola opens downward. x 2 and x 3. The parabola opens upward. x 3 4 11. The x-intercepts are and x 1 2 ˛. The 13. Vertex is parabola opens upward. 3, 4 1 y-intercept 14. The x-intercepts are or x 1.586, 4.414. . 2 x 3 22, x 3 22 4 67. 20 10 45 5 20 5 10 10 Answers to Selected Exercises 1067 15. Vertex is 1, 4 y-intercept 5 there are no x-intercepts. . 2 1 10 25. 27. 29. 31 10 10 33x2 4x 13 2x 1 x 7 21 1 x 1 3x 1 21 x 1 2 2 2 1 2 x 5 2b a 2x2 1 49 2 2 2 10 17. Vertex is (4, 14) y-intercept 2. The x-intercepts are at approximately 0.258 and 7.742. 15 5 10 19. Vertex is 1, 4 . 1 y-intercept 3. The x-intercepts are 2 10 1 and 3. 10 10 10 35. 39. f
x 2 1 b 0 37. 41. b 4, c 8 a 1 2 43. Minimum product is 4; numbers are 2 and 2 45. Two 50-ft sides and one 100-ft side 47. $3.50 49. 30 salespeople 51. 1 second; 22 ft 53. The maximum height of 35,156.25 feet is reached 46.875 seconds after the bullet is fired. Section 3.4, page 182 5. x2 h x 2 1 5 1. 7. f x 1 2 h x 2 1 x3 x 4 3 11. 5 3 10 13. 10 10 10 21. Vertex is 0.5, 24.5 24. 1 . 2 y-intercept is The x-intercepts are 3 and 4. 10 10 10 10 23 x, h h x 2 1 23 x 1 x 2 5 15. 5 5 30 2x2 14x 20 2 Answers to Selected Exercises 5 g x 1 2 x 4 3 1, g x 1 2 x 4 3 x 23. g 1 1068 17. 10 35. 3 10 10 5 5 10 19 21. 5 5 5 5 h x 2 1 23. g 25 3x ˛, h 1 2x 3 2 1 2 0 x 3 0 27. g x 1 2 1.5 1 x 3 2 2 29. Shift the graph 4 units to the right and 1 unit upward; reflect the graph across the x-axis, and stretch it vertically by a factor of 3. 31. Reflect the graph across the y-axis, shift it 2 units to the left, stretch vertically by a factor of 4, and shift 3 units downward. 33. Compress the graph horizontally by a factor of 1 , 1.3 units to the right, and shift 0.4 units 3.23 shift it upward. 3 g x 2 1 1 4 23 x 3 1, g 23 x x 1 2 37. 10 10 10 10 , y x 39. 10 9.4 9.4 10 f x 1 2 3 2 x 4, f 1 x x 1 2 41. 10 10 10 10 ˛, g x 1 2 x3 Answers to Selected Exercises 1069 43. y 51. y 45. 47. 49. y y y g x 1 2 3f 53. y 55 57. 5.1 −4.7 4.7 −1.1 21 x2 2x 2 1070 Answers to Selected Exercises 59. 3.1 37. − 4.7 4.7 −3.1 321 x2 g x 2 1 61. 6.2 −9.4 9.4 −6. 63. a. Shifts upward by 28 units b. The graph is stretched vertically by a factor of 1.00012. Section 3.4.A, page 189 1. Symmetric with respect to the y-axis 3. The graph does not have symmetry with respect to the x-axis, y-axis, or origin. However, it is symmetric with respect to the point (0, 2). 5. y 23 x x2 7. Yes 13. No Symmetric with respect to the origin. 9. Yes 15. Yes 11. Yes 17. Yes 19. Origin 21. Origin 23. y-axis 25. Odd 27. Even 29. Even 31. Even 33. Neither 35. 39. Many correct graphs, including the one shown here: (−7, f(−7)) (−5, f(−5)) −7 −5 −3 3 1 −1 −2 (−3, f(−3)) −4 (−2, f(−2)) (4, f(4)) (1, f(1)) 1 3 5 (6, f(6)) 7 1 2 2 1 x, y x, y on the graph implies that 41. Suppose the graph is symmetric to the x-axis and is 2 x, y the y-axis. If (x, y) is on the graph, then 1 on the graph by x-axis symmetry. Hence, is on the graph by y-axis symmetry. Therefore, x, y is on the 1 graph, so the graph is symmetric with respect to the origin. Next suppose that the graph is symmetric to the y-axis and the origin. If on the graph, then axis symmetry. Hence, the graph by origin symmetry. Therefore, the graph implies that is on the graph, so the graph is symmetric with respect to the x-axis. The proof of the third case is similar to that of the second case. is on the graph by y- x 1 1 x, y 1 x, y x, y 2 2 x, y , y x, y is on on is Section 3.5, page 196 1. 3. 5. f g 21 1 f g 21 1 g f 21 1 real numbers x x x 2 2 2 x3 3x 2; x3 3x 2; x3 3x 2 ; domain for each is all 1 x x 21 x2 2x 5; 2 x2 2x 5; f g 1 1 g f x domain for each is all real numbers except 0 x 21 x2 2x 5 1 x f g 21 x 2 1 1 2 ; 21 fg f gb1 1 a x 2 x 2 3x4 2x3; 3x 2 x3 ˛; a g f b1 x 2 x3 3x 2 Answers to Selected Exercises 1071 7. fg 1 2 1 f g b1 21 x 1 x2 1 21 b1 x 2 a 2 21 21 x 1 2 2x2 1 2x 1 2x2 1 2x 1 B x2 1 x 1 2x 1 x 1 x2 1 B 2x 1 2x2 1 1 x 1 B 9. Domain of fg: all real numbers except 2; domain of f g : all real numbers except 2 11. Domain of fg: all real numbers; domain of f g : all real numbers except 3 4 10 −10 1 4 1 1 −3 x g f x 2 21 x f f x 2 21 1 1 39. 41. 43 25; f g 1 f g 1; 4 x 2 21 and 1 g f x 3 15. 1 g f 2; is all real numbers. 17. 30 x2 3; 21 x 1 2 2 domain of g f x 2 21 1 1x ; domain of f g f x 1 2 45. 0.5x2 5.5x2 5 0 13. 19. 21. 23. 25. 27. 29. 31. 33. 35. 37 and 2 21 g f x ; 1 1x 1 0, q 2 f f is 1 x6; . ff ff x x 2 2 21 21 1 1 x2 ; 1 f g g f x 21 x 21 2 2 f a g f g g f 21 21 x7 1 1 x B A x9 x 2 x x 2 21 f f 21 x 2 9 9x 2 b 1 x 2 13 x 2 2 3 23 2 2 where 1 A where A x 21 x 21 , 9x 2 9 x 2 2 1 13 and and x2 2, B 7x3 10x 17, x 13 x 1 2 x 2 21 , where A x 1 2 3x2 5x 7, 49 By the definition of absolute 0 f 1 47. a. x x 3, g f 2 g f 21 1 value. When if if 0 x f 1 6 0 x f 1 the graph of f 2 2 g f 1 2 the graph of f. When below the x-axis, but the graph of reflection of f across the x-axis; therefore, the graph will be above the x-axis. is the 6 0, f x b 1 2 is the same as the graph of f is 1072 Answers to Selected Exercises y1 x3 3, y2 51. 53. 55. 2x 1, y2 y1 21 15,600 19.5n n gives the unit price as a function of n, the number of telephones produced. 57. V 256pt3 3 ; 17,157.28 cm3 59. s 10t 3 61. One such function is f x 1 x ˛. x 1 2 Section 3.5.A, page 203 1. 2, 2.16, 2.3328, 2.5194, 2.7210, 2.9387, 3.1737, 3.4276 3. 0.2, 0.64, 0.9216, 0.2890, 0.8219, 0.5854, 0.9708, 0.1133 5. 0.5, 0.375, 0.3223, 0.2888, 0.2647, 0.2462, 0.2312, 0.2189 7. approaches infinity 9. converges to 0 11. converges to 0 13. The fixed points are 3 and 2. 15. The fixed points are 0, 1, and 2 28 2 17. The fixed points are 1. and 2 28 2 ˛. 3. y 1 1 x 3 sample points on the inverse: 8, 6 1 2 2, 2 1 0, 3 , 2 1 , 2 5. 10 −10 10 −10 t t3 3t2 2 t3 3t2 2 t x1 y1 x2 y2 7. 10 19. Any real number greater than or equal to 0 is a fixed point. Any negative number is an eventually fixed point. −10 10 21. a. b. x 0.5; x 1 0 1 2 f x x terms of orbit: 0.5, x 1 terms of orbit: 1.5, 0.5, 0.5, . . . ; 0 x 0.5; is an eventually fixed point. f and 0.5, 0.5, . . . ; 0.5 is a fixed point. and 0.5 x f 0 2 1 0, . . . ; 0 is a periodic point. x 1; periodic point. x 1 terms of orbit: 0, 1, 0, 1, . . . ; 1 is a terms of orbit: 1, 0, 1, and x 0; x 1 and 10 t t4 3t2 t4 3t2 t x1 y1 x2 y2 x 11. x 5y2 4, y ± x 4 5 B Section 3.6, page 212 1. y 4 2 3 6 1 f(y) 1 2 3 4 5 9. g 13. g 17. g 21 23. No 3 5 x B 2 1 x 3 B 15. g x 1 2 19. x x2 7 4 , 1 x 0 1 2y2 1 ˛, y ± B 2 1 x 2x 5x 1 1 x 25. Yes 27. Yes 29. No Answers to Selected Exercises 1073 31. 10 −5 5 49. 1 g f 25 x x 21 25 x5 x A B 25 x A −10 3 51. 33. 35. −10 10 −3 5 − and g f 1 x 2 21 B f f x 1 22 1 x 2f 2 1 x 3f 2 1 2x 1 2 1 3 1 2x 1 1 2 2 1 3x 2 2 2 1 3x 2 1 2 3x 2 2 3x x5 21 2x 1 3x 2 d 2x 1 3x 2 d 7x 3x 2 7 3x 2 c x 53. Let y f x 1 2 mx b. Since m 0, we can solve for x and obtain Hence, the rule of the x y b m . inverse function g is g 1 and we have 21 x. f and b a 1 b a 2 1. g 2 f x 1 22 1 mx 22 1 b x x 2 21 x b m b a mx b 2 Slope a b b a y x 1. b. The line has slope 1, is Length PR 2 c. 55. a. has slope 1 and by (a), line PQ Since the product of their slopes the lines are perpendicular 2a2 2ac c2 b2 2bc c2 2a2 b2 2c2 2ac 2bc; 2 1 2c2 2bc b2 c2 2ac a2 2a2 b2 2c2 2ac 2bc is the Since the two lengths are the same, perpendicular bisector of segment PQ. Length RQ 2 d. 93 1 3 ft/sec 3. a. 0.709 gal/in. b. 2.036 gal/in. 5. a. 250 ties/mo c. 500 ties/mo e. g. ties/mo ties/mo 188 1500 55.5 9. 7. a. 2 2x h 17. b. 11. 92.5 1 b. 438 ties/mo d. 563 ties/mo f. h. ties/mo ties/mo 750 375 c. 462.5 13. 1.5858 19. 2t h 8000 15. 1 21. 2pr ph −5 37. One restricted function is x 39. One restricted function is that x h ; 2 1 2 h inverse function x 1 2 with x. x 0 0 x g 2 1 x2 x 0 (so x h 1 1x. with 2 x 0; with Another restricted inverse function x 0; inverse function function is g h 1 1x. x x 2 x g 2 1 x2 1 2 41. One restricted function is h 1 x 0; inverse function g x 2 1 x2 6 2 x 2 12x 6. with 43. One restricted function is f x 0; inverse function x 1 2 1 x2 1 with 21 21 f g x 2 x 2 45. 47. 1 1 1 f x 2 21 f 1 g x g 22 1 x f 22 and and 2 1 2 1 1 x x x 2 21 a g f 1 b g 1 x 1b a 1074 Answers to Selected Exercises . 1. a. 14 ft/sec b. 54 ft/sec c. 112 ft/sec Section 3.7, page 220 23. a. Average rate of change is 7979.9, which b. The graph represents a function of x because it means that water is leaving the tank at a rate of 7979.9 gal/min. 7979.99 gal/min. 6.5p 7980 6.1p gal/min. 6.2p d. c. c. 6p b. 25. a. b. e. It’s the same. 27. a. C, 62.5 ft/sec; D, 75 ft/sec t 9.8 sec b. Approximately c. The average speed of car D from t 4 to t 4 to t 10 sec is the slope of the secant line joining the (approximate) points (4, 100) and (10, 600), namely, 600 100 10 4 83.33 ft/sec. The average speed of car C is the slope of the secant line joining the (approximate) points (4, 475) and 800 475 10 4 54.17 ft/sec. (10, 800), namely, 29. a. From day 0 until any day up to day 94, the average growth rate is positive. b. From day 0 to day 95 c. 27, meaning that the population is decreasing at a rate of 27 chipmunks per day d. 20, 10, and 0 chipmunks per day Chapter 3 Review, page 226 11 3 7 2t 9 1 2b 2 2 7 5 3 3 x3 4 2 2 x 2 x 2b a 1 5 2b 7 2x 2h passes the vertical line test. 21. y x 23. −6 −4 1 −1 6 4 2 −2 −2 −4 −6 4 y x = t2 − 4 y = 2t + 1 (−3 ≤ t ≤ 3) x 2 4 6 25 20 5. All real numbers 2 except for 3. 7. a. For 20 miles, it costs $150. For 30 miles, it costs $202.50 b. 39 miles −10 10 1 2 1 1, f 1 2b a 1, f a 3 2b 2 −10 9. f 0 0, f 2 1 3, 3.5 11. 3 15. No local maxima; minimum at 13. 4 3 x 0.5; 0.5, q increasing on ; 2 This function is concave up for all values of x; there are no points of inflection. decreasing on 1 1 q, 0.5 17. Maximum at 1 minimum at q, 5.0704 x 5.0704; x 0.2630. Increasing on and (0.2630, ); decreasing on (5.0704, 0.2630) This function is concave up on the interval 8 a b 3 q, 8 3b and concave down on the interval There is a point of inflection at , q . 2 a x 8 3 . 19. a. The graph does not pass the vertical line test; therefore, it does not represent a function of x. Vertex is (4, 1). The y-intercept is 17. There are no x-intercepts. x2 2x 7 x f 1 2 2 27. 5 −15 10 −15 1, 6 Vertex is There are no x-intercepts. . 1 2 The y-intercept is 7. Answers to Selected Exercises 1075 29. 10 −10 10 −10 Vertex is 1 The y-intercept is The x-intercepts are 0.35, 4.2025 2 4.08. 2.4 . and 1.7. 39. Compress the graph of g toward the x-axis by a factor of 0.25, then shift the graph vertically 2 units upward. 41. Shift the graph of g horizontally 7 units to the right; then stretch it away from the x-axis by a factor of 3; then reflect it across the x-axis; finally, shift the graph
vertically 2 units upward. 43. e 45. 36 31 2b x 4 25 4 x 1 2 1 1 33. parent function: 21 2 x 2 f 1 10 (transformation form) (x-intercept form) 1x 0 0 125 Note: the right endpoint of each segment is a part of the graph; the left endpoint is not a part of the graph. −10 10 47. x-axis, y-axis, origin 49. Even 1 53. a. b. 1 −10 55. 4x 57. 51. Odd c. 2 1 x3 3 35. parent function: g x 1 2 x 0 0 −10 10 −10 10 37. parent function: f 1 10 x2 x 2 59. 82 27 x 3, 2 6 x 6 3 f 61. For and x 2 the same. For g f is the graph of of f across the x-axis. the graphs of f and , the graph of g f are , a reflection of the graph f x 63. x2, g 65. 2, 1, 0, 1, 0, 1, 0, 1 x 2 1 1 2 2x 1 67. y Inverse f x −10 10 −10 69. 71. 1076 Answers to Selected Exercises 73. x 25 y3 1, f x 1 y2, y ± 21 x x 1 2 x 2y 1, f 1 x 2 1 1 23 x5 1 x 2 1 75. The graph of f passes the horizontal line test and hence has an inverse function. It is easy to verify either geometrically [by reflecting the graph of f across the line f calculating function. ] or algebraically [by ] that f is its own inverse y x x 22 f 1 1 y 3 f(x) = −3 −3 1 x 3 x 2. 0.625 seconds The instantaneous velocity of the ball is 0 when the ball reaches its maximum height. Thus, the maximum height of the ball will be 81.25 feet. 3. s t 1 4. 14 2 16t2 300; 5. 96 0.111111 feet per second 6. 3 7. 2a 8. instantaneous rate of t 4: tangent line at change 16 y 16t 76 ; equation of 100 −2 8 −100 77. 79 21 21 0.25 4 3 g f x 2 21 1.5 x 1.5 1.5 x 6 x 6 6 x 7x x 3 1 1 2 1 4x 6 2 0.25x 1.5 2x 1 x 3 b a 2x 1 x 3 b 2x 1 2 1 2 x 3 2x 3x 1 x 2 b a 3x 1 x 2 b 3x 1 3 1 2 x 2 3x 1 3 1 x 2 x 2 7 1 7x 7x x 3 7 x 3 7x 21 1 89. a. For example, from 3 to 1 b. For example, from 1 to 2 c. For example, from 6 to 8 d. Both intervals are portions of the same line, so their slopes are the same. 91. a. $290/ton b. $230/ton c. $212/ton Chapter 3 can do calculus, page 237 1. t s 2 1 44 16t2 20t 75 feet per second change 25.132741 9. instantaneous rate of r 1, for each change of 1 unit in the When radius, the surface area of the sphere will increase by approximately 25.132741 square units. 10. instantaneous rate of change 100 dollars per phone When sold, the profit increases by approximately $100. for every additional phone x 1000, Chapter 4 Section 4.1, page 248 1. Polynomial of degree 3; leading coefficient 1; constant term 1 3. Polynomial of degree 3; leading coefficient 1; constant term 1 5. Polynomial of degree 2; leading coefficient 1; constant term 3 7. Not a polynomial 9 3x3 2x2 4x 1; 9 8 1 7 8 33 0 3 2 9 11 5 6 1 2x3 x2 3x 11; 3 25 2 quotient remainder 25 13. 7 5 5 0 35 3 245 4 1,694 6 11,830 242 35 1,690 5x3 35x2 242x 1690; 11,836 quotient remainder 11,836 Answers to Selected Exercises 1077 1 3 81. a. 83. 6 b. 5 8 85. 3 87. 2x h 11. quotient remainder 7 3 2 3 2 9. No 11. Degree 3, no; degree 4, no; degree 5, yes 15. 2 1 1 quotient remainder 17. Quotient 19. Quotient 21. Quotient 23. No 29. 222, 1 30 41. No 35. 4 8 4 6 2 4 x3 4x2 4x 6; 2 8 6 7 12 19 19 3x3 3x2 5x 11; x2 2x 6; 5x2 5x 5; remainder remainder 0 remainder 12 7x 7 25. Yes 31. 2 27. 0, 2 33. 6 37. 170,802 39. 5,935,832 43. No 45. Yes 1 1 x 21 1 f 49. 51. 47. 2x 7 x 3 21 x 2 3x 5 2 21 2x 1 2 21 x 1 x 4 x 3 x 3 21 2 x5 3x4 5x3 15x2 4x 12 x x 1 2 x5 5x4 5x3 5x2 6x 55. Many correct answers, including 53. 21 21 21 21 21 21 21 57. Many correct answers, including 21 17 100 1 21 x 5 21 1 2 x 8 x 2 59. 61. f x 2 1 k 1 k 1 63. x4 x2 1, 65. If x c were a factor of then c would be a solution of c4 c2 1. satisfy is impossible. Hence, x4 x2 1 0, c4 0 But x c that is, c would c2 0, and is not a factor. so that 67. a. Many possible answers, including: if n 3 and then since c 1, x3 1 x 1 x 1 b. Since n is odd solution of is a factor of 1 1 is not a solution of c 2 xn cn 0. xn cn and hence c x by the Factor Theorem. is not a factor of x3 1 0. c is a x c n cn Thus, 2 2 1 1 69. k 5 71. d 5 Section 4.2, page 258 1. x ± 1 or 3 3. x ± 1 or 5 5. x 4, 0, 1 or 1 2 7. x 3 or 2 9. 13. x 2 x 2 2x2 1 x 1 x 2 2 21 5; 19. Lower 17. 21 2 1 1 2 x2 3 11. 15. x 5, x2 3 x3 2, or 3 x 2 21 2 1 1 upper 2 2 21. Lower 7; upper 3 23. x 1, 2, or 1 2 25. x 1, 1 2 , or 1 3 1078 Answers to Selected Exercises 27. x 2 or 5 ± 137 2 31. x 1, 5, or ± 13 29. 33. x 1 2 x 1 3 or ± 12 or ± 13 or 1.8393 35. x 2.2470 or 0.5550 or 0.8019 or 50 37. a. The only possible rational zeros of 1 2 x are x2 2 f zero of f(x) and irrational. 23 rational zeros are 13 ± 1 is a zero of ± 3. or b. ± 2 ± 1 or 12 ± 1 (why?). But ± 2. Hence, or 12 12 is a is x2 3 ± 1 whose only possible ± 3 (why?). But or 39. a. 8.6378 people per 100,000 b. 1995 41. 2 by 2 in. c. 1991 43. a. 6° /day at the beginning; b. Day 2.0330 and day 10.7069 c. Day 5.0768 and day 9.6126 d. Day 7.6813 6.6435° /day at the end Section 4.3, page 269 1. Yes 3. Yes 5. No 7. Degree 3, yes; degree 4, no; degree 5, yes 13. The graphs have the same shape in the window but and 1000 y 5000 40 x 40 with don’t look identical. 2 2 zero of even multiplicity. and 1 15. 17. is a zero of odd multiplicity, as are 1 and 3 are zeros of odd multiplicity; 2 is a 19. (e) 21. (f) 23. (c) 25. The graph in the standard viewing window does not rise at the far right as does the graph of the highest degree term so it is not complete. x3, 27. The graph in the standard viewing window does not rise at the far left and far right as does the graph of the highest degree term not complete 20 y 40 60 y 320 35 y 20 0.005x4, so it is and and and 33. 31. 29. 33 x 2 right half: 35. Left half: 250,000; 90 x 120 37. 39. Overall: y-axis: and 3 x 3 0.1 x 0.2 20 y 30 50,000 y and and 2 x 3 15,000 y 5000 20 y 20; and near 4.997 y 5.001 and 41. a. The graph of a cubic polynomial (degree 3) has at most 3 1 2 local extrema. When is large, x 0 0 ax3, the graph resembles the graph of that is, one end shoots upward and the other end downward. If the graph had only one local extremum, both ends of the graph would go in the same direction (both up or down). Thus, the graph of a cubic polynomial has either two local extrema or none. b. These are the only possible shapes for a graph that has 0 or 2 local extrema, 1 point of inflection, is large. and resembles the graph of when x ax3 0 2, 0, 4, and 6 0 43. a. Odd b. Positive c. 45. (d) 47. y (0, 4) 4 2 −2 (−1, 0) −4 −3 −2 f(x) = x3 − 3x2 + 4 x 1 2 3 4 (2, 0) 49. y (2, 4) 4 2 h(x) = 0.25x4 − 2x3 + 4x2 x −4 −3 −2 −1 (0, 0) −2 1 2 3 4 5 (4, 0) 53. y (−1.30, 2.58) (−0.34, 0.77) (−1.69, 0) 4 2 f(x) = x5 − 3x3 + x + 1 (0.34, 1.23) (1.51, 0) x −4 −3 −2 −1 −2 1 2 3 4 (1.30, −0.58) d. 5 55. (−0.71, 174.27) y 175 150 125 100 75 50 (−3, 0) (−2.75, 0) −5 −4 −3 −2 −1 1 2 3 (−2.88, −2.34) h(x) = 8x4 + 22.8x3 − 50.6x2 − 94.8x + 138.6 (1.4, 0) (1.5, 0) 1 2 x (1.45, −0.37) 57. a. The solutions are zeros of 2 1 x and 3.99 y 4.01 4 0.01x3 0.06x2 0.12x 0.08. g This polynomial has degree 3 and hence has at most 3 zeros. 1 x 3 b. c. Suppose f(x) has degree n. If the graph of f(x) y k had a horizontal segment lying on the line k x for some constant k, then the equation would have infinitely many solutions (why?). x has degree n (why?) But the polynomial 2 and thus has at most n roots. Hence the equation f the graph cannot have a horizontal segment. has at most n solutions, which means 51. y 25 −8 −7 −6 −5 −4 −3 −2 −1 −25 (−.12, −44.73) −50 (6.72, 0 10 g(x) = 3x3 − 18.5x2 − 4.5x − 45 −75 −100 −125 −150 −175 (4.23, −167.99) Answers to Selected Exercises 1079 59. a. The general shape of the graph should be as b. shown here. The graph should cross the x-axis at the points specified, and bounce off the axis at 2. y 0.5179820180x2 20.88711289 80°; 9 A.M.: 69°; c. Noon: 2 P.M.: 14.65684316x 9. a. 50,000 1 0 83° 20 b. On the TI-83, only 3 of the roots are shown. 2 is skipped over entirely. b. Quartic c. y 1.595348011x4 58.04379735x3 630.033381x2 2131.441153x 36466.9811 d. $42,545.95 e. According to this model, income will drop steeply after 2002. 11. a. 24,000 0 0 22 b. y 0.084189248x4 0.528069153x3 66.26642628x2 397.2751554x 3965.686061 c. 1996: $19,606; The estimate is lower. Section 4.4, page 290 1. All real numbers except 3. All real numbers except 5. All real numbers except 5 2 3 15 12, 7. Vertical asymptotes x 1 x 0; 9. Hole at vertical asymptote 11. Hole at x 2; vertical asymptotes 13. any window with and 3 15 12 and 1, and x 6 x 1 x 2 115 x 110 31 x 35 any window with any window with 40 x 42 y 3; y 1; y 5 2 ; 15. 17. 19. Asymptote: y x; window: 14 x 14 and 15 y 15 21. Asymptote: y x2 x; window: 15 x 6 and 40 y 240 c. Again, d. Try the windows x 2 is missed. 20 x 3, 5,000,000 y 1,000,000. 3 x 2, part 5 x 11, 5000 y 60,000. 5000 y 5000 100,000 y 100,000 1 x 5, Then try For the third and lastly Section 4.3.A, page 276 1. Cubic 3. Quadratic 5. a. y 0.634335011x3 11.79490831x2 51.88599279x 4900.867065 b. 1987: 4898.0 per 100,000; 1995: 4635.6 per 100,000 c. 3138.2 d. Answers may vary. 7. a. 100 5 0 20 1080 Answers to Selected Exercises 23. y 29. y −6 −4 −2 2 −2 x 2 2 −1 −2 1 2 4 x vertical asymptote horizontal asymptote x 5 y 0 25. y 2 −2 −3 −2 31. x 2 vertical asymptote horizontal asymptote x 2.5 y 0 27. y 6 4 2 −1 −2 2 x 33. vertical asymptote horizontal asymptote y 3 x 1 vertical asymptote horizontal asymptote x 3 y 1 y 6 4 2 −3 −2 −1 1 2 3 x −2 −4 −6 −8 −10 vertical asymptotes horizontal asymptote x 11 −2 −5 −3 x 1 2 3 vertical asymptotes horizontal asymptote y 0 x 2, x 1 Answers to Selected Exercises 1081 35. y 431 −2 −3 x 1 2 3 vertical asymptotes horizontal asymptote y 0 x 5, x 1 37. y x −1 −2 −3 −4 vertical asymptote horizontal asymptote y 4 x 0 39. y 4 2 −2 1 3 5 −2 −4 vertical asymptotes horizontal asymptote x 1, x 5 y 1 y 41. 4 2 −2 −4 −6 −4 −2 vertical asymptotes x 3 hole at horizontal asymptote y 0 x 4, x 5 1082 Answers to Selected Exercises vertical asymptote x 2 x 3 1 −2 1 3 4 −3 oblique asymptote y x 1 45. y 28 24 20 16 12 8 4 x −4 −2 −4 −8 −12 y = 2x + 7 x 1 32 4 vertical asymptote oblique asymptote x 5 2 y 2x 7 47. y x 2 4 6 8 y = x2 + x + 1 −2 12 10 8 6 4 1 −2 −4 −6 x 1 2 3 vertic
al asymptote parabolic asymptote y x2 x 1 x 1 49. y 36 30 24 18 12 6 y = x2 + 2x + 4 −5 −3 −1 1 3 5 x −9 y x2 2x 4 and 8 y 4; hidden 2 x 2 x 5: and 0.5 y 0.5; 15 x 3 and 4 y 4; there is a hole at vertical asymptote parabolic asymptote 5 x 4.4 51. Overall: x 2 53. area near origin: hidden area near 0.07 y 0.02 9.4 x 9.4 x 2. 4.7 x 4.7 55. Overall: x 1; hole at 0.65 x 0.75 and and 2 y 2; there is a 73. a. to see the vertical asymptote, use and 3 y 3. 57. For vertical asymptotes and x-intercepts: 4.7 x 4.7 and close to the horizontal asymptote: and 8 y 8; 2 y 3 to see graph get 40 x 35 and 3 x 15 2 y 2; and hidden 0.02 y 0.01 61. 63. 59. Overall: and 4.7 x 4.7 x 4: area near 15.5 x 8.5 4.7 x 4.7 and 13 x 7 area near the origin: 0.02 y 0.02 65. Overall: 16 y 8 12 y 8 20 y 20; and 2.5 x 1 and hidden b. 20 e. x p 1 2 4x 10 x 3 f. Shift the graph of f(x) horizontally s units (to 0 0 s 6 0 ); stretch (or r to the right if (away from s 7 0; r the left if shrink) the graph by a factor of 7 1, the x-axis if 0 6 1 0 6 r ); also if the x-axis; then shift vertically downward if if r ts 0 t 7 0; t 6 0 ). 0 0 0 t 0 tx 2 1 x s 0 toward the x-axis if r 6 0, reflect the graph in units (upward 0 g. 69. a. b. c.2 1 4.002 change 1 9.3 1 9.003 0.2381; 1 4.02 0.2488; 0.2499; instantaneous rate of 0.25 1 4 0.1075; 1 9.03 0.1107; 0.1111; instantaneous rate of change 1 9 0.1111 p d. They are the same. 71. a. y x 1 x 2 b. y x 1 x 2 c. Graph (a) has a vertical asymptote at graph (b) has a vertical asymptote at and x 2 x 2. 2x2 1.25 100 a 4x 1000 x2 b 2 0.06x2 3 C x 2 100 1 1 50 x x 16.85 in. 20 x 50 x x c 1 2 b. 75. a. b. between 25 gallons and 100 gallons c. x 50 gallons c a x x 1 x 2 1 2 40,000 2.60x x 77. a. 67. b. Stretch the graph of f(x) away from the x-axis by a factor of 2. c. The graph of h(x) is the graph of f(x) shifted vertically 4 units upward; the graph of k(x) is the graph of f(x) shifted horizontally 3 units to the right; the graph of t(x) is the graph of f(x) shifted horizontally 2 units to the left. d. Shift the graph of f(x) horizontally 3 units to the right, stretch vertically by a factor of 2, then shift vertically 4 units upward. c. 79. a. y 2.60; $2.60. v 50u u 50 b. v 50 0 0 100,000 the average cost can never be below Answers to Selected Exercises 1083 c. 100 Section 4.5.A, page 306 15. 4 17. i 19. i 21. i Section 4.6, page 313 1. 3. 5. 1 1 f 3 0.39.4521 0 0 ; 2 2 2 02 0.4521 . 0.5 i; 0.3; f 2 1 0.4521 2 0.5 0.5i; f 2 0 2 1 d 1.5207 0.25 1.5i ; 1.2 0.5i; 0.01 0.7i; f 2 0 2 1.6899 0.514i d 1.7663. The seventh iteration is more than 2 units from the origin. 9. The thirteenth iteration is more than 2 units from the origin. 11. The eighth iteration is more than 2 units from the origin. 13. i is the Mandelbrot set. 15. 1 is not in the Mandelbrot set. 17. The cycle approaches approximately The number 0.2 0.6i is in the 0.275 0.387i. Mandelbrot set. 1. g(x) is not a factor of f(x). 3. g(x) is not a factor of f(x). 5. g(x) is not a factor of f(x). 7. x 0 (multiplicity 54); x 4 5 (multiplicity 1) x p (multiplicity 14); (multiplicity 15); (multiplicity 13) 9. 11. 13. x 0 x p 1 x 1 2i 15. x 3 or 17. 19. 21 15 f x 1 2 A i b 21 1 2i; or x 1 2i 215 i or 3 215 x 1 a 3 3 313 2 x 3 2 3 2 x 3 2a 1 13i i b a 1 3 i; 215 3 x 1 2i 2 215 3 x 1 3 313 2 x 3 2 3 or 2 313 2 1 13i; i ba i; i or x 2 1 or i or x 1 or x 1 13i i; 2A 1 BA or x i x 1 21 15 or x 25 21 12i or x 15 or BA BA 2 x i 21 12i; x 22i x 1 13i B 23. Many correct answers, including 21 25. Many correct answers, including 21 1 2 f 27 2x 21 x 4 1 2 x 3 3 2 1 21 2 x 12i B BA 313 2 i b 50 0 35,000 d. If the object is close, a small change in u leads to a large change in v. However, when u is large, a small change in u leads to nearly no change in v, so that u may change substantially while the object stays in focus. 3. 2 10i 12 13 a 2 10 11i b 7. 11. 2i 25. 31. 1 3 i 10 17 11 17 i 37. 6i 27. 33. 12 41 7 10 15 41 11 10 i i 41. 45. 49. 4i 215 322 A 41 i i B Section 4.5, page 300 1. 8 2i 43. 11i 2i 1 2 1 13i 21 20i 2 29 i 4 41 i 5 29 5 41 113 170 214i 41 170 i A i B 2 3 2 522 A 12 ± 13 , y 3 2 17 2 ± 2 5. 9. 13. 23. 29. 35. 39. 47. 51. 53. 57. 61. 65. 67. 69. 25 2210 i 55. 59. 63. i B x 2, y 2 x 1 3 x 1 4 ± ± 114 3 131 4 i i 1 x 2, 1 13i, 1 13i x 1, 1, i, i 71. z a bi, 73. If with a, b real numbers, then 2 1 z a bi a bi and hence, 2bi. If z z 2bi 0. z z, Conversely, if then b 0. which implies that is Hence, a bi z z 2 1 b 0 real, then z z. Therefore, 0 z z 2bi, z a is real. 75. 1 z a a2 b2R Q b a2 b2R i Q 1084 Answers to Selected Exercises 2 7. 9. 11. x 44 7 1 4 ± 13 3 2 1 2, x f x 2 x 2 AA A 3 213 BB A 1 13. x A 3 213 BB 15. a. There are no rational zeros b. An irrational zero lies between 3 2. and 0. A third is and 1 Another lies between between 1 and 2. 17. 3 19. d 21. When x4 4x3 15 synthetically, the last row, 1 alternating signs. Therefore, for the real zeros. 1 is divided by 5 20, has is a lower bound x 1 5 5 1 23. rational zeros: and 4; irrational zero: 1.328 is a zero of multiplicity 2; 4 is a zero of is a zero of multiplicity 1; 3 is a 3 multiplicity 1; zero of multiplicity 1 3 2 i 25. z a bi c di 2 z w. and 27. Answers may vary. 29. i, iv, and v are false. 29. 31. 33. 35. 37. 39. 41. 43 x2 2x 5 2 1 1 21 21 21 x2 4x 5 x2 2x 2 x2 4x 5 x 2 x 3 x2 2x 5 x 4 1 x4 3x3 3x2 6x 6 2x3 2x2 2x 2 x2 6x 10 2 x2 2x 2 21 2 1 1 2 2 45. Many correct answers, including 1 i 47. Many correct answers, including x2 49. x f 2 1 3, 1 2 1, 2i, 2i 57. a. Since 53. x3 5x2 13 i, 1 2 2 2 1 7 2i 13 2 i x 3 6i 1 2 i, i, 1, 2 51. Hence 2 1 2 ac bd zw 2 ac bd ad bc 2 w c di, z w ad bc i. z w w c di, a c 1 b. Since zw and ac bd i, i, 55. b d 1 2 b d i. 2 a bi 2 i, i, 2 Since 1 2 z w ad bc i, 2 i. Since 2 a bi Hence 1 1 1 1 1 z a bi c di 21 zw z w. 2 (definition of f(z)) 59. a. 1 f z 1 2 2 1 2 az3 bz2 cz d az3 bz2 cz d a z3 b z2 c z d az3 bz2 cz d az3 bz2 cz d f z 1 b. Since f Hence 0, z we have 2 1 is a zero of f(x). z 2 (Exercise 57(a)) (Exercise 57(b)) r r for r real) ( (Exercise 57(b)) (definition of where gi1 g31 gk 1 61. If f(z) is a polynomial with real coefficients, then g11 degree g11 p degree g21 is a polynomial with real coefficients f(z) can be factored as each and degree 1 or 2. The rules of polynomial multiplication show that the degree of f(z) is the degree sum: gk1 z g31 z . degree 2, then this last sum is an even number. But f(z) has odd degree, so this can’t occur. Therefore, at least one of the is a first-degree polynomial and hence must have a real zero. This zero is also a zero of f(z). degree gi1 z g21 2 If all of the have gi 1 z z z 2 2 2 2 2 Chapter 4 Review, page 317 1. (a), (c), (e), (f) 5. 0 17 10 7 16 14 2 1 5 other factor: x5 3x4 2x3 5x2 7x 2 0 4 4 b. 31. Use 10 x 20 and 2 x 2 overall graph and for behavior around the origin. 10,000 y 500 and 10 y 5 for the 33. Use 0.2 x 2 and 20 y 40; 1 x 0.2, 20 y 20 35. Use 2 x 3 and 1, 0 , 5 y 5. a local maximum at (0, 3), There is an x-intercept at 1 2 and a local minimum at . a , 49 4 27b 3 5 y 5. x 0.618, 1.618 and There are and a local 37. Use 3 x 3 x-intercepts at minimum at x 0.909. 39. a. 120,000 0 0 25 y 0.04x4 29,552.18; 0.19x3 62.06x2 $79,223 in 2007 and $127,317 in 2015 3.1 y 3.1. 1615.35x There is a and x 2 and a horizontal 41. Use 4.7 x 4.7 vertical asymptote at asymptote at y 1. Answers to Selected Exercises 1085 43. Use 4.7 x 4.7 3.1 y 3.1. vertical asymptote at a hole at the x-axis is the horizontal asymptote. x 1, and There is a , and x 1 Chapter 5 Section 5.1, page 334 5. 0.09 11. 81 17. 23. 29. 23 12 125 A 1623 B 3 4 35. 22 825 41. 4 a4 b 47. 9 2 x 53. 12 5 9 2 a 2 34b4 59. 3 16 a 65. 1 3. 2 9. 0.125 15. 21. 27. 1 64 6 2 0.4 1 13 21 33. 7 39. 1 45. 1 4x 2y 2 2 1. 12 7. 0.2 13. 16 19. 1114 25. 8 31. 225 37. 1525 43. d5 21c 49. 42 5 d c 10 3 55. ax 61. 27 10 4t 51. 57. 63 BA 1 2 a 5 2 49b a2 b2 1 1 3 2 1 1 5y 2 5 x 69. 73. 77. 81 11 6 x y 1 313 3 4 3 3 5 9 x A x BA 1 2 x x BA A 1 1x h 1 1x 1 2x h 2 1 1x2 1 1 x h 1 2 67. 71. 75. 79. 83. 85. 87. x y 312 4 21x BA 89. a. The square (or any even power) of a real number is never negative. Graphically these equations lie strictly above or on the x-axis. 23 8 2, whereas 2 2 8 26 b. 91. 2n 1m c m2n c ; 2m cd 2m c 2m d ; 93. When n is an odd positive integer, if f xn Therefore, an 6 bn. x 1 function and thus is one-to-one. Therefore, f integer. The inverse is has an inverse if n is an odd positive 2n xn x. x x g 1 2 2 1 2 95. about 19° F 97. 49 mph 1 2 m c B d 2m c 2m d a 6 b, is an increasing 45. vertical asymptotes: x ± 23; parabolic asymptote: y x2 6x 5 47. Use 4.7 x 4.7 and 10 y 10 horizontal asymptote . vertical y 0; x 1; asymptote x-intercepts at 15 x 10, x 2, x 3; 0.5 y 0.5 for hidden behavior 49. Use 30 x 30 hidden behavior use 3 ± 131i 2 x and 1000 y 1000. For 5 y 5. 7 x 7, 53. x 3 ± 131i 10 51. 55. 57. 59. 2 A 3 1 23i or x 2 A 3 x 2 or i, i, 2, 1 or i or i or 1 23i 61. Many correct answers, including x4 2x3 2x2 x f 1 2 63. a fixed orbit of one point: 65. 67. 69. 1 1 1 21 x 1 x 1 21 x2 1 21 x 2 x 2 21 x2 1 x 3 ; 2 x2 1 1 ; 2 x i ; 2 1 21 . , 3 a 5 x 1 21 x 1 x i 4 5b x 2 21 x 2 21 x i 1 21 21 . x 3 2 x i 21 x i 2 21 x i . 2 21 Chapter 4 can do calculus, page 325 b. 1. a. 1, 4 1 2. a. (1, 0) 2 1 3. a. approximately b. 1 2 and (2, 4) 1, 4 2, 0 or (1, 0) 3.785 cm 6.980 cm 2 or c. (3, 20) c. 1 3, 16 2 b. 8.560 cm 1.365 cm V 126.49 cm3 approximately 6.324 cm and height is approximately 3.163 cm. when side length is 4. a. approximately 4.427 inches by 4.427 inches. b. The largest volume occurs when x 10 3 . 5. a. r 4.09977 b. r 1.996; 37.566 in2 6. a. Approximately 206 units are produced. b. The minimum value of about 577 dollars per unit occurs when about 269 units are produced. 7. r 1.769, h 58 pr 2 5.8996 8. The maximum area of about 220.18 square feet occurs when x is about 9.31 feet. 9. 4 sq. units 10. 5 x2 1.5. The point that is closest to (0, 1) has the exact value of 7 2 , 3 2 b aB are okay. , but approximations 1086 Answers to Selected Exercises 99. 1.5 19. g x 1 2 2x5 10 −3
a. c. x x 3 −1.. d2 −1 14 101. a. g is the graph of f moved 3 units left b. h is the graph of f moved 2 units down c. k is the graph of f moved 3 units left, then 2 units down. Section 5.2, page 343 21. 23. 25. 27. 29. 1 2 x : B 20 x 2 : C; h 1 and and and and 2 x : D : A; k 1 0 y 1 10 y 10 0 y 1 0 y 10 2 2 xh 2 h 1. Shift the graph of h vertically 5 units downward. 3. Stretch the graph of h vertically by a factor of 3. 31. Neither 35. When x is large, 5. Shift the graph of h horizontally 2 units to the left, then vertically 5 units downward. 37. 4 33. Odd x 0, e so 1 e1 e ex e 39. 1 x ex 0 ex. 3 e3 e 1 2 8.84 7. Shift the graph of h vertically 4 units upward. 9. Compress the graph of h vertically by a factor of 1 4 . 41. xh 51 2 2 5x 2 h 43. 1 exh e ex e x 1 2 45. The x-axis is a horizontal asymptote; local 11. Reflect the graph of h across the y-axis, then shift maximum at (1.443,0.531). horizontally 2 units to the right. 13. Reflect the graph of h across the y-axis, stretch horizontally by a factor of 1 0.15 6 2 3 , then stretch vertically by a factor of 4. x 15. f x 1 2 5 2b a 10 0 10 0 −5 17. x g 1 2 3 x 2 −5 5 5 47. No asymptotes; local minimum at (3, 0.0078). 49. No asymptotes; no extrema. 51. a. About 520 in 15 days; about 1559 in 25 days b. in 29.3 days 53. a. 1980: 74.06; 2000: 76.34 b. 1930 55. a. 100,000 now; 83,527 in 2 months; 58,275 in 6 months b. No. The graph continues to decrease toward zero. 57. a. The current population is 10, and in 5 years it will be about 149. b. After about 9.55 years. 59. a. Not entirely x b. The graph of f81 appears to coincide with the graph of g(x) on most calculator screens; when 2.4 x 2.4, the maximum error is at most 0.01. 2 c. Not at the right side of the viewing window; x f121 2 Section 5.3, page 353 1. Annually: $1469.33; quarterly: $1485.95; monthly: $1489.85; weekly: $1491.37 3. $585.83 5. $610.40 7. $639.76 Answers to Selected Exercises 1087 9. $563.75 11. $582.02 49. y 13. About $3325.29 15. About $3359.59 17. About $6351.16 19. About $568.59 21. Fund C 23. $385.18 25. About $1,162,003.14 27. $4000 29. About 5.00% 31. About 5.92% 33. a. About 9 years; about 9 years; about 9 years b. Doubling time is not dependent on the amount invested, but on the rate at which it is invested. 35. About 9.9 years 37. a. About 12.6% 39. a. b. 12.6%; about 12.7%; about 12.7% 3x1 c. No; yes f b. 3 18 6 3x or 41. a. g x 43. a. 2 x 1 E 1 b. $7966 2 100.4 5550 1.014 2 1 1.0368 1 x x 2 b. 115.38 million c. In the sixth year 45. About 256; about 654 b. 11.036 b. $0.86; $0.74 c. 16.242 47. a. 6.705 49. a. 51. a. x 2 1 1 f x 0.97 c. About 75 years 20 t 140 t f 2 0.5 A 2 1 B b. About 11.892 mg; about 3.299 mg c. About 325 days 53. About 5566 years old Section 5.4, page 361 3. 9. 1. 4 7. 13. 17. 21. 102.8751 750 ezw x2 2y log 3 0.4771 ln 5.5527 12 7 5. 103 1000 4.6052 0.01 e 2.5 e1.0986 3 15. 11. log 0.01 2 ln 25.79 3.25 ln w 2 r 19. 23. 25. 243 27. 15 29. 33. x y 35. x2 37. 1 41. They are exactly the same. 1 2 1, q 31. 931 2 39. q, 0 1 2 f(x) = log(x − 31 −2 51. y h(x) = −2 log 1 −2 x x 53. 55. 57. 0 x 9.4 asymptote at 10 x 10 0 x 20 and 6 y 6 and x 1 ) (vertical and 3 y 3 59. 0.5493 63. a. ln 1 3 h h 2 ln 3 6 y 3 61. 0.2386 b. h 2.2 65. a. About: 17.67, 11.90, 9.01, 6.12, 4.19, 3.22, 2.25 b. The rule of thumb is that the number of years it takes for your money to double at interest rate r% is 72 divided by r. 67. a. 77 69. a. 9.9 days n 30 error of 0.00001 when 71. b. 66; 59 b. About 6986 0.7 x 0.7. gives an approximation with a maximum 43. Stretch the graph of g away from the x-axis by a Section 5.5, page 369 factor of 2. domain: all positive reals; range: all reals 45. Shift the graph of g horizontally 4 units to the right. domain: all reals 7 4; 47. Shift the graph of g horizontally 3 units to the left, then shift it vertically 4 units downward. domain: all reals 7 3; range: all reals range: all reals 1. 103 3. About 3.63 5. About 0.9030 0.2219 9. About log 13. e 1 2 3 ln 1 u 2v 1 2 x 3 1 2 19. 25. log 2 20xy 1 u 1 6 v 2 3 17. 23. 0.1461 7. About x2y3 7 x ln ln 1 11. 15. 2 2 1 2u 5v 21. 27. a. For all x 7 0 1088 Answers to Selected Exercises logarithms on page 364, b. According to the fourth property of natural eln x x for every x 7 0. 29. False; the right side is not defined when x 6 0, but the left side is. 31. True by the Power Law 33. False; the graph of the left side differs from the graph of the right side. 35. Answers may vary: log 3 log 2 1.585 and 3 2b log a b e 0.1761 thus log 3 log 2 log 3 2b a 39. A 3, B 2 41. 2 37. 43. Approximately 2.54 45. 20 decibels 4 69. Horizontal shift of units to the right, then 3 compress horizontally by a factor of 1 3 . Domain: all real numbers Range: all real numbers 7 4 3 71. Compress the graph vertically by a factor of 1 3 , then a horizontal translation of 1 unit to the right, then a vertical translation of 7 units upward. Domain: all real numbers Range: all real numbers 7 1 47. Approximately 66 decibels 49. 100 times 73. True 75. True 77. False 79. 397398 51. a. 1.2553 b. 3.9518 c. log x ln x ln 10 Section 5.5.A, page 376 log 0.01 2 log r 7k 5. 1. 9. log3a 1 9b 2 3. 7. log 23 10 1 3 log7 5,764,801 8 11. 104 10,000 13. 102.8751 750 15. 53 125 19. 25. 29. 10zw x2 2y 1 2 x 21. 243 27. 6 0 log4x f x 1 2 Not defined 17. 2 1 2 4 23. 2x2 y2 1 0 2 0.5 4 1 31. x log6x h x 1 2 1 36 2 1 6 1 1 0 216 3 33. 35. 37. x 0 2 log7x f x 1 2 Not defined x 3 log21 h x 2 1 b 3 2.75 6 x 3 2 39. b 20 27 49 1 7 2 1 3 1 1 6 4 29 15 x2y3 z6 41. 5 43. 3 45. 4 47. log 49. log x2 3x 1 2 55. ln 1 x 1 2 x 2 b 2 a 51. log21 5c 2 53. log4a 1 49c2b 57. log21 x 2 59. ln 1 e2 2e 1 2 61. 3.3219 63. 0.8271 65. 1.1115 67. 1.6199 81. 85. 87. 89. logbu logau logab 83. log10u 2 log100u logbx 1 2 b3 2v logbA g f x 2 1 is false. x 2 1 logbv 3 logb2v logbb3 ; hence B only when x b32v. x 0.123, so the statement y 6 4 2 (−0.3679, 0.3195) −6 −4 −2 −2 −4 h(x) = x log x2 Hole at (0, 0) x 2 4 6 (0.3679, −0.3195) Section 5.6, page 386 1. x 4 3. x 1 9 5. x 1 2 or 3 7. x 2 or 1 2 9. x ln 5 ln 3 1.465 2.7095 11. 13. 15. x ln 3 ln 1.5 x ln 3 5 ln 5 ln 5 2 ln 3 x ln 2 ln 3 3 ln 2 ln 3 1.825 0.1276 17. x 1 ln 5 2 2 0.805 19. x 1 ln 3.5 1.4 2 0.895 5 2.1b 2 ln x a ln 3 x ln 2 0.693 21. 25. 1.579 23. x 0 or 1 or x ln 3 1.099 Answers to Selected Exercises 1089 510 20 20 20 0.792 b. 500 x ln 3 ln 4 41. x 3 47. x 5 −50 0 15. a. 2 27. 29. 31. 35. 43. x ln 3 1.099 x ln 2 ln 4 x ln A 1 2 t 2t2 1 or then x 5 x 9 ln u ln v, 37. 5 137 2 x 33. If 49. x ± 210001 u v so x 6 B eln u eln v, 39. x 9 1 x 45. 51. e 1 2 e 1 e 1 A 53. Approximately 3689 years old 55. Approximately 950.35 years ago 57. Approximately 444,000,000 years 59. Approximately 10.413 years 61. Approximately 9.853 days 63. Approximately 6.99% 65. a. Approximately 22.5 years b. Approximately 22.1 years 67. $3197.05 71. a. About 1.3601% k 21.459 73. a. 69. 79.36 years b. In the year 2027 b. t 0.182 75. a. There are 20 bacteria at the beginning and 2500 three hours later. b. ln 2 ln 5 0.43 77. a. At the outbreak: 200 people; after 3 weeks: about 2718 people b. In about 6.09 weeks k 0.229, c 83.3 b. 12.43 weeks 79. a. Section 5.7, page 396 1. Cubic, exponential, logistic 3. Exponential, quadratic, cubic 5. Exponential, logarithmic, quadratic, cubic 7. Quadratic, cubic 9. Quadratic, cubic 11. Ratios: 5.07, 5.06, 5.06, 5.08, 5.05; exponential is appropriate 13. a. For large values of x the term close to zero so the quantity is slightly larger than 1, which means 1 0.0216x 56.33e 1 56.33e is 0.0216x 0 0 b. 10 0 0 c. 1330 0 0 17. 19. ln x, ln y 5 1 26 appears the most linear. Power model ln x, ln y and 5 1 Power or logarithmic model ln x, y 5 1 26 26 are both nearly linear. 21. a. 105,000 2 442.1 1 56.33e close to) 442.1. 0.0216x is always less than (but very 0 92,000 8 1090 Answers to Selected Exercises b. 105,000 25. a. 1800 f(x) g(x) 8 0 92,000 c. 0 0 22 b. y 7.05x2 78.34x 398.73 d. The logarithmic model predicts continued but slowing growth while the logistic predicts a cap of about 102,520. Therefore, the logarithmic model seems the better one for the long haul. 23. a. 80 0 0 b. 5 0 0 c. Exponential. y 152.22 0.97x 1 175 0 0 110 110 110 2 c. y 6413.2 1 107.2e d. 2325.01, 2419.97 e. The quadratic model will give an ever 0.1815x increasing number of kids, and the rate of increase will continue to increase. Before too terribly long the number of kids home schooled by the quadratic model will exceed the number of kids in the world. The logistic model, on the other hand, gives us a maximum that can never be exceeded. y 17.5945 13.4239 ln x 27. a. b. 77.4 years c. 2012 29. a. 85 0 0 b. y 10.48 1.16x 1 2 13 Answers to Selected Exercises 1091 c.-d. 25. x f 1 2 56,000 1.065 1 x 2 Worldwide shipments (thousands) Predicted number shipments (thousands) 14.7 15.1 16.7 18.1 21.3 23.7 27 32.4 38.9 47.9 60.2 70.9 84.3 12.2 14.1 16.4 19 22 25.5 29.6 34.4 39.9 46.2 53.6 62.2 72.2 Worldwide shipments ratio (current to previous) 1.03 1.11 1.08 1.18 1.11 1.14 1.2 1.20 1.23 1.26 1.18 1.19 Year 1985 1986 1987 1988 1989 1990 1991 1992 1993 1994 1995 1996 1997 e. An exponential model may not be appropriate. Chapter 5 Review, page 403 1. c2 3. 10 3 b a 42 5 5. 1 2 v u 1 2 7. c2d4 2 9. 2 22x 2h 1 22x 1 11. Reflection across the x-axis, stretch vertically by a factor of 2 13. Reflection across the y-axis, stretch horizontally by a factor of 2 15. Vertical translation of 4 units upward 17. 3 x 3 and 0 y 2 19. a. 62,000 63,000 64,000 S 60,000 1000 33,708 35,730 37,874 t 1 b. c. Compunote is the best choice d. Calcuplay will be paying more this time, but 1.06 S 30,000 2 1 1 2 t1 your total earnings will be more from Compunote 21. a. About $1341.68 b. $541.68 23. a. About $2357.90 b. After about 32.65 years 1092 Answers to Selected Exercises 29. 27. About 3.75 grams ln 756 6.628 log 756 2.8785 et rs 37. 33. r2 1 1 2 31. u v ln e7.118 1234 39. Undefined 35. 41. Reflection across the y-axis, horizontal translation of 4 units to the right; Domain: all real numbers 6 4; Range: a
ll real numbers 43. Vertical stretch by a factor of 3, vertical translation of 5 units downward; Domain: all positive real numbers; Range: all real numbers 45. 3 47. 3 4 49. 2 ln x 51. ln 9y x2 b a 53. (c) 55. The domain consists of those values of x for which is positive; x x 1 dw uv x 3 ± 257 4 2 uc d x e1 x 101 q, 0 ´ 2 1 1, q 2 1 59. 2 65. x 1 2 69. x 2 73. About 1.64 mg 61. (c) 57. 63. 67. 71. 75. Approximately 12 years 77. $452.89 81. a. 11° F b. 30 79. 7.6 5 0 50 c. The points (x, ln (y)) are approximately linear. d. e. y 22.42 10.27° F 0.967x 2 1 Chapter 5 can do calculus, page 411 1. y x 1; 5 −1 5 −5 25 −1 31 −5 4. y e2 x 2 1 2 e2 10 −5 5. y x 3 −5 −1 10 −1 6. y e x 1 2 1 e 2 10 −1 −5 5 5 5 5 5 7. y e2 x 2 2 1 e2 2 10 −1 −5 8 10 −1 10 −1 −5 9. y x 1 ln 3 1 2 −5 10. y 3 ln 3 1 21 3 x 1 2 10 −5 −1 11. y 9 ln 3 1 21 9 x 2 2 10 −5 −1 5 5 5 5 5 Answers to Selected Exercises 1093 41. The area of the triangle is a h. The altitude h 1 2 forms a right triangle with side b as the hypotenuse and side h opposite sin u b S h b sin u. h so u, Thus, the area of the 5 43. triangle is 1 2 A 4320.123 a h 1 2 a 1 Section 6.2, page 429 b sin u ab sin u. 2 1 2 A 33.246 45. 1. 7. 13. 17. 19. 22 3 12. y 1 9 a ln 3 b1 x 2 1 9 2 5 −1 −5 Chapter 6 Section 6.1, page 419 1. 5. 9. 11. 13. 15. 19. 23. 27. 47.26° 23°9¿36– sin u 3. 15.4125° 4°12¿27– ˛, tan u 7. , cos u 3 2 11 3 sin u ˛, sec u ˛, tan u 211 211 3 ˛, csc u A cot u 3 22 7 ˛, cos u 2 27 27 ˛, sec u 2 ˛, csc u m ˛, tan u h d d ˛, csc u m A cot u 2 23 m ˛, cos u d h ˛, sec u m sin u h cot u d h 11 2 A 23 2 7 3 A sin 32° 0.5299 sec 47° 1.4663 u 45° 17. 21. 25. tan 6° 0.1051 u 30° u 60° 1 2 29. 3 8 31. 16 7 33. False; sin 50° 0.7660 2 sin 25° 0.8452 35. True; cos 28° 1 37. False; 2 2 0.7796 1 sin 28° 2 1 2 0.7796 39. tan 75° 3.7321 tan 30° tan 45° 1.5774 u 1° 0.1° 0.01° 0.0001° sin 0° 0; of 0°, cos u 0.9998 0.999998 0.99999998 0.9999999998 sin u 0.0175 0.00175 0.000175 0.0000175 the right triangle definitions do not apply. cos 0° 1 Since no triangle has an angle 1094 Answers to Selected Exercises c 36 h 2522 2 c 423 3 3. c 36 5. c 8.4 9. h 300 11. h 5023 15. a 1023 3 a 10 cos 50° 6.4, A 40°, c 10 sin 50° 7.7 b 6 C 76°, sin 14° 21. C 25°, a 5 tan 65° 10.7, 23. C 18°, c 3.5 cos 72° 1.1 a 3.5 sin 72° 3.3, 24.8, c 6 24.1 tan 14° b 5 cos 65° 11.8 25. About 48.59° 27. About 48.19° 29. 31. 33. 35. A 33.7°, C 56.3° A 44.4°, C 45.6° A 48.2°, C 41.8° A 60.8°, C 29.2° 37. a. b. 23.18 feet. 6.21 feet. 39. 460.2 ft 41. 8598.3 ft 43. No 45. Approximately 263.44 feet 47. 351.1 m 53. a. 56.7 ft 55. 173.2 mi 49. 10.1 ft b. 9.7 ft 51. 1.6 mi 57. 52.5 mph 59. 449.1 ft Section 6.3, page 441 1. 40°, 2p 9 radians 5. 10°, p 18 radians 9. 288°, 8p 5 radians 3. 20°, p 9 radians 7. 240°, 4p 3 radians 11. 36° 13. 18° 15. 135° 19. 75° 21. 972° 23. p 30 17. 4° 25. p 15 31. 39. 5p 4 3p 5 33. 31p 6 41. 7 2p 19. a. sin 10p 3 10p 3 tan 0.8660, cos 10p 3 0.5, 1.7321 b. Since the sine and cosine are both negative, the 5p 12 5p 3 9p 4 , 17p 4 3p 4 29. 37. 3p 4 , 15p 4 , 25p 6 , 7p 4 , 13p 6 23p 6 27. 35. 43. 45. 47. 55. 11p 6 , 4p 3 17 4 63. 8.75 69. 7p 77. 3 radians 49. 57. 7p 6 50 9 51. 41p 6 53. 8p cm 59. 2000 61. 5 65. 3490.66 mi 67. 942.48 mi 73. 42.5p 75. 2pk 4p 71. 171.9° 1 2 79. a. 400p radians per min b. 800p in. per min or 200p 3 ft per min 81. a. 5p radians per sec b. 6.69 mph 83. 15.92 ft 85. approximately 8.6 miles Section 6.4, page 452 sin t 7 1. , cos t 2 3. sin t 5. sin t 253 6 261 10 2103 , cos t , cos t 253 5 261 23 2103 , tan t 7 2 , tan t 6 5 , tan t 10 23 7. 9. sin t 1 25 sin t 4 5 , cos t 2 25 , tan t 4 3 , cos t 3 5 , tan t 1 2 21. a. terminal side is in the third quadrant. sin 9.5p 1, undefined cos 9.5p 0, tan 9.5p is b. Since the sine is 1 terminal side is on the negative y-axis. 0.2752, and the cosine is 0, the 17 cos 23. a. sin 1 tan 1 17 2 17 2 0.9614, 3.4939 1 2 b. Since the sine is positive and the cosine is negative, the terminal side is in the second quadrant. 17π 6 x 25. y reference angle = π 6 y 27. 1.75π x reference angle = π 4 11. sin 13p 6 1 2 ˛; cos 13p 6 23 2 13. sin 16p 0; cos 16p 1 29. y 15. a. sin 7p 5 7p 5 tan 0.9511, cos 7p 5 0.3090, 3.0777 b. Since the sine and cosine are both negative, the terminal side is in the third quadrant. 17. a. sin a 14p b 9 14p 9 b tan a 0.9848, cos 14p 9 b a 0.1736, 5.6713 b. Since the sine and cosine are both positive, the terminal side is in the first quadrant. – π 7 x reference angle = π 7 Answers to Selected Exercises 1095 31. sin 7p 3 b a 23 2 , cos 7p 3 b a 1 2 , tan 7p 3 b a 23 33. sin 11p a 4 b 22 2 , cos 11p a 4 b 22 2 , tan 11p a 4 b 1 35. sin 3p 2 b a 1, cos 3p 2 b a 0, tan 3p 2 b a sin u cos u is undefined. 37. sin 23p 6 b a 1 2 , cos 23p 6 b a 23 2 , tan 23p 6 b a 1 23 23 3 39. sin 19p 3 b a 23 2 , cos 19p 3 b a 1 2 , tan 19p 3 b a 23 41. sin 15p 4 b a 22 2 , cos 15p 4 b a 22 2 , tan 15p 4 b a 1 43. sin 5p 6 b a 1 2 , cos 5p 6 b a 23 2 , tan 5p 6 b a 1 23 23 3 45. 47. 49. 55. 57. 59. is undefined and tan u sin u 1, cos u 0, sin u 0, cos u 1, and tan u 0 22 2 1 23 A 51. 22 4 , cos t 3 234 B , tan t 5 3 sin t 5 234 53. 23 2 sin t 1 25 sin t 3 , cos t 2 25 cos t 1 , 210 210 , tan t 1 2 tan t 3 , 9. 11. 13. 15. 19. cos t 0.9457, sec t 1.0574, sec t 3.7646, cot t 0.2755, sin t 0.1601, cot t 6.1668, sin2 t cos2 t 1 4 tan t 0.3438, csc t 3.0760 cos t 0.2656, csc t 1.0372 cos t 0.9871, sec t 1.0131 cot t 2.9089, sin t 0.9641, tan t 0.1622, 17. cos t 21. cos t 2 23. cos t sin t cos t 25. 0 27. even 2sin t 0 33. sin t 23 2 39. 45. 51. 3 5 221 5 32 22 2 29. even 35. sin t 23 2 41. 47. 3 4 2 5 53. 32 22 2 31. odd 37. 43. 49. 3 5 3 4 221 5 55. possible 61. csc t csc cot t cot 1 1 57. not possible sec t sec ; t ± 2p t ± p 2 2 59. not possible t ± 2p ; 2 1 Chapter 6 Review, page 464 1. 41.115° 3. (d) 9. 13. 4 7 C 34°, b 13.3, c 7.4 11. 5. 4 265 7. 265 7 C 50°, a 6.4, c 7.7 15. 225.9 ft 17. The boat has moved about 95.3 feet. 19. 255° 21. p 5 23. ˛ 3p 4 25. 16p 3 27. 2 revolutions per minute 29. 37. 3 5 23 3 31. 0 39. 2 33. 41. ˛ 1 2 23 2 43. quadrants 2 and 3 45. 9 4 35. 23 61. r cos t, r sin t 1 2 63. Domain: all real numbers with q, 1 65. Domain: all real numbers with Range: 1, q ´ 2 1 1 2 Range: all real numbers Section 6.5, page 460 u a multiple of p u a multiple of p 47. sin t cos t cos t sin t tan2 t 49. e 51. ˛ 3 5 53. 1 55. b 1. cos t 3. 1 5. 1 7. csc2 t 1096 Answers to Selected Exercises Chapter 6 can do calculus, page 471 1. sin t cos t 2 t f 1 p 12 2 cos t 2 1 1 sin t 4. t f 1 p 12 p 144 max of 0.5 when x 0.7854 max of 0.5 when x 0.78538 p 144 sin t 2 cos t 2. 2 t f 1 p 12 5. a. b. 200 sin t cos t p radians; 4 width 1022 height 522 meters max of 1.1375 when x 0.2618 max of 1.1387 when x 0.28362 meters and p 144 3. 3 sin t sin p 2 a t b p 12 p 144 max of 2.2321 when x 0.5236 max of 2.236 when x 0.45815 max of 3.1566 when x 1.309 max of 3.1622 when x 1.2435 6. approximately 13.23 feet from the statue 7. a. road cost 10,000˛a b. approximately $117,321 10 1 tan tb 20,000˛a 1 sin tb Chapter 7 Section 7.1, page 483 1. 3. 5. 13π 2 −π ≤ x ≤ −2 ≤ y ≤ 2 , −π ≤ x ≤ 3π, −4 ≤ y ≤ 4 −π ≤ x ≤ 4π, −2 ≤ y ≤ 2 Answers to Selected Exercises 1097 7. 11. 15. 17. p 2 3p 2 3p 2 ˛, 3p 3p 2 , 5p 6 t 6 3p 2 4 p 2 7p 4 and 9. 1 13. 1 3p 2π 45. 47. 19. all values on the interval p, 2p 3 4 except 3p 2 21. t p 4 2np or t 3p 4 2np, where n is any −2π 2np or t 4p 3 2np, where n is any 2np or t 5p 3 2np, where n is any 49. d integer t 2p 3 integer t 4p 3 integer t p 6 integer t p 6 integer t 3p 4 integer t p 3 23. 25. 27. 29. 31. 33. 2np 2np or t 11p 6 2np, where n is any or t 5p 6 2np, where n is any 2np or t 5p 4 2np, where n is any np, where n is any integer 35. Reflect the graph of f across the horizontal axis. 1 domain: all real numbers; range: 1 g t 1 2 7 −7 1 −1 2π 2π 53. f 55. a. odd; b. even; c. odd; d. even; e. odd; 51. e t sin 2 1 t cos 1 t tan 2 1 t sin 2 1 t tan 2 1 2 sin t cos t tan t sin t tan t 57. 1.4 61. −2 59. 11 40 2 −40 37. Shift the graph of f vertically 5 units upward. domain: all real numbers except odd multiples of p 2 ; range: all real numbers a. 0 c. 15.4 yards e. When t 0.25, b. 0 d. 3.6 yards d is undefined. The beam is parallel to the wall at this time. 39. Stretch the graph of f away from the horizontal axis by a factor of 3. domain: all real numbers; range: 3 g 3 t 1 2 41. Stretch the graph of f away from the horizontal axis by a factor of 3, then shift the resulting graph vertically 2 units upward. domain: all real 1 g numbers; range: 43. Shift the graph of f vertically 3 units upward. 4 domain: all real numbers; range 1098 Answers to Selected Exercises Section 7.2, page 490 1. The graph of s t 3 sec t 2 is the graph of stretched vertically by a factor of 3 and 1 2 t sec t g shifted down 2 units. 2 1 3. The graph of csc t f t 2 1 t m 4 csc shifted 4 units up. t 1 2 1 2 is the graph of 5. The graph of p 1 2 t 2 1 sec t 1 is the graph of sec t g t 2 1 compressed vertically by a factor of 1 2 and shifted up 1 unit. 7. The graph of t sec t g shifted 8 units down. 2 1 t q sec 8 reflected across the vertical axis, and is the graph of t 1 2 2 1 9. amplitude: 0.3; period: 6p 11. amplitude: 1 2 ˛; period: 2p 3 9. The graph of v t p csc t is the graph of stretched vertically by a factor of p. 13. amplitude: 5; period: 20p 17 13. g t 2 1 1 4 sec t 5 17. g t 2 1 cot t 2 1 23. A 31. A 25. A 33 21. B 29 11. csc t 3 sec 15. g t 2 1 csc 19. D 27. B 35. − π 2 −2π 37. π 2 2π −1 4 −4 y sec t 39. Look at the graph of y t, draw in the line sec t intersect the graph of p 2 t p 2 and Q , . on page 488. If you it will pass through and obviously will not when But it will intersect each part of the graph that lies above the horizontal axis, to the right of t p 2 ; it will also intersect those parts that lie below the horizontal axis, to the left of p 2 The first coordinate of each of these infinitely . many intersection points will be a solution of sec t t. Section 7.3, page 498 1. amplitude: 1; period: 2p 3. amplitude: 1; period: 2p 3 5. amplitude: 4; p
eriod: 2p 7. amplitude: none; period: p 2 15. amplitude: none; period: 4 17. a. 2 c. t 0 or 2 b. d. t 1 2 t 1 or 3 2 19. g is the graph of f horizontally compressed by a factor of 1 5 ˛; amplitude: 1; period: 2p 5 ˛. 21. g is the graph of f horizontally compressed by a factor of 1 8 ˛; amplitude: 1; period: p 4 ˛. 23. g is the graph of f reflected across the y-axis; amplitude: none; period: p. 25. g is the graph of f horizontally compressed by a factor of 5 8 ˛; amplitude: 1; period: 5p 4 ˛. 27. g is the graph of f vertically stretched by a factor of 3; amplitude: 3; period: 2p. 29. g is the graph of f vertically compressed by a factor of 1 3 ˛; amplitude: none; period: p. 31. g is the graph of f vertically stretched by a factor 1 2 ˛; of 5 and horizontally compressed by a factor of amplitude: 5; period: p. 33. g is the graph of f reflected across the x-axis, vertically stretched by a factor of 2, and horizontally stretched by a factor of 5; amplitude: none; period: 5p. 35. g is the graph of f vertically compressed by a factor of factor of 2 5 1 8 ˛; and horizontally compressed by a amplitude: 2 5 ˛; period: p 4 ˛. 37. g is the graph of f vertically compressed by a and horizontally compressed by a 1 3 1 p ˛; factor of factor of 39. 5 0 −5 amplitude: none; period: 1 4π Answers to Selected Exercises 1099 11. amplitude: 7; period: 2p 7 ; phase shift: 1 49 ; vertical shift: 0 13. amplitude: 3; period: p; phase shift: 5p 8 ; vertical shift: 0 15. amplitude: 97; period: p 7 ; phase shift: 5 14 ; vertical shift: 0 17. amplitude: 1; period: 1; phase shift: 0; vertical shift: 7 19. amplitude: 3; period: 6; phase shift: 3 p ; vertical or f 3 sin t 2 1 8t 8p 5 b a a t p 5 b t 2p 3 b 3 t 4p 2 3 a 2 a 2 3 2 3 sin sin 9 b t 1.5 2 0.8pt 1.2p 1 0.5 sin 0.8p 0.5 sin 2 or 2 or 0.6 0.6 2 t 0 2 1 or f t 2 1 6 sin 6 5 t b a 1 shift: 5 21. f t 2 1 3 sin 8 23. 25. 27. 29 sin 5 2 5 2 sin sin a 0 or 10p 9 1 10p 2 t 0.2 9 t 2p 9 b 10t p 2 b 31. a. f b. g t t 2 2 1 1 12 sin 12 cos 10t a 2t p 2 b a 8t p a 2 b 35. a. f t 2 1 1 2 sin 8t b. g t 2 1 1 2 cos 37. a. b. 39. a. f f f b sin 4 cos 2 sin 2 cos 3p 41. 12 −12 π 6 1 − π 6 43. 4 0 −4 45. d 47. b 49. f 51. 55. 59 sin t 2 sin pt 3 2 2 5 sin 5t 1 53. 57.8 sin 4t 3 2 sin 4t 2 61. local maximum of 1 at t 0 and t 2p 3 ; local minimum of 1 at t p 3 . 63. there is a local maximum of 23 2 at t p; local Section 7.4, page 508 1. amplitude: 1; period: 2p; phase shift: 1; vertical shift: 0 3. amplitude: 5; period: p; phase shift: 0; vertical shift: 0 5. amplitude: 1; period: 2p; phase shift: p; vertical shift: 4 7. amplitude: 6; period: 2 3 ; phase shift: 1 3p ; vertical shift: 0 9. amplitude: 4; period: 2p 3 ; phase shift: p 18 ; vertical shift: 1 1100 Answers to Selected Exercises minimum of 1 at t 3p 2 . 65. 1 900,000 33. a. f t 2 1 sin 2t b. g t 2 1 cos 41. y −2π −π 3 1 −1 −3 43. y −2π −π 1 −1 45. y 3 2 1 −1 −2 −3 47. 3 0 −1 2.6180; local t 5p 6 5.7596 minimum at 49. Local maximum at t 11p 6 t p 6 51. Local maxima at 0.5236, 2.6180, t 5p 6 t p 2 1.5708, π x 2π t 3p 2 t 7p 6 4.7124; 3.6652, local minima at t 11p 6 5.7596 53. The graph of f the graph of intercepting the y-axis at 1. sin2 t cos2 t 1, t 1 x 2 2 f 1 a horizontal line is the same as 55. Not an identity 57. Possibly an identity 59. 61. A 3.8332, b 4, c 1.4572 A 5.3852, b 1, c 1.1903 63. All waves in the graph of g are of equal height, which is not the case with the graph of f. It cannot be constructed from a sine curve through translations, stretches, or contractions. x Section 7.4A, page 515 1. 3. 5. 7. 9. 11. 13. 15.2361 sin 5.3852 sin 5.1164 sin t 1.1072 2 4t 1.1903 3t 0.7442 f 1 0 t 2p and 5 y 5 10 t 10 and 10 y 10 0 t p 50 and 2 y 2 1 0 t 0.04 and 7 y 7 0 t 10 and 6 y 10 1 2 one period 2 2 one period one period 1 one period 2 2 1 17. To the left of the y-axis, the graph lies above the t-axis, which is a horizontal asymptote of the graph. To the right of the y-axis, the graph makes waves of amplitude 1, of shorter and shorter period. Window: 3 t 3.2 2 y 2 and 19. The graph is symmetric with respect to the y-axis and consists of waves along the t-axis, whose amplitude slowly increases as you move farther from the origin in either direction. Window: 30 t 30 and 6 y 6 21. There is a hole at point (0, 1). The graph is symmetric with respect to the y-axis and consists of waves along the t-axis whose amplitude rapidly decreases as you move farther from the origin in either direction. Window: 30 t 30 and 0.3 y 1 2π 23. The function is periodic with period The graph lies on or below the t-axis because the logarithmic function is negative for numbers p. Answers to Selected Exercises 1101 π 2π h(t) = 3 sin 2t +( π 2 ) π 2 π 3π 2 2π x 19. 6 0 0 , cos t between 0 and 1 and is always between 0 and 1. The graph has vertical asymptotes when t ± p ± 7p 1 2 2 points and ln 0 is not defined). Window: 2p t 2p and periods) 3 y 1 cos t 0 ± 5p 2 ± 3p 2 at these (four , . . . , , Chapter 7 Review, page 517 1. (c) 3. 5. 7. 0 n2p, p 3 np, where n is an integer where n is an integer 9. The graph of g is the graph of f reflected across the horizontal axis and compressed horizontally 1 2 ˛. domain: all real numbers except −2π where n is an integer; range: all real by a factor of t p 4 numbers n p 2 , 11. 4 −2π 2π −2π −2π 21. 23. even 25. 27. 29. 400 33. 37. 2π 2π 2π −2 8 −8 4 −4 13. −2π −4 6 −1 15. A 17. C 2π −3π 31. C 35. 2 cos 5t 2 b a 6π 8 sin f t 2 1 2pt 28p 5 b a 2 −2 1102 Answers to Selected Exercises 39. Not an identity 41. Possibly an identity 43. 45. 2 t f 1 0 t p 50 10.5588 sin 4t 0.4580 1 and 5 y 5 2 one period 2 1 Chapter 7 can do calculus, page 521 1. 2 6x 18x2 54x3 162x4 . . . . ; 1 6 x 1 6 2. 3 6x 12x2 24x3 48x4 . . . . ; 6 16 x 5 16 3. 2 2x 2x2 2x3 2x4 . . . . ; 9 16 x 5 8 4. 3 6x 12x2 24x3 48x4 . . . . ; 1 4 x 11 32 5. cos x; q x q 6. 1 x ; 0 x 2 Chapter 8 7. ex; q x q 1. 3. 5. Section 8.1, page 528 x 0.5275 kp x 0.4959 2kp 1.8877 2kp or x 0.1671 2kp 4.5453 2kp or x 1.2161 2kp x 2.4620 2kp x 0.5166 2kp 13. a. The graph of 11. 9. 7. f 2 shows that 1 to 2p or 1.6868 kp 1.2538 2kp or 2.6457 2kp or 1.5708 2kp or 4.7124 2kp or or 1.8256 2kp or 2.8867 2kp or or or x 5.0671 2kp 3.8212 2kp 5.6766 2kp sin x sin x 1 on the interval from 0 only when x p 2 . Since sin x has period obtained by adding or subtracting integer all other solutions are 2p, from p 2 , that is, , 2p multiples of 2p 5p p 2 2 13p 2 7p 2 p 2 p 2 p 2 3 2 2p 2p 2 1 1 2 2 2p 1 2 , etc., and , p 2 3 9p , 2 2p 3p 2 11p 2 p 2 2p , etc. 2 1 , b. Similarly, the graph shows that sin x 1 only when x 3p 2 , so that all solutions are obtained by adding or subtracting integer multiples of 2 1 2 : , 2p 2p 2p 3 2 3p from 2 3p 2p 7p 2 2 15p 2 5p 2 3p 2 3p 2 3p 1 2 x 0.1193 u 82.83°, 262.83° u 210°, 270°, 330° u 60°, 120°, 240°, 300° u 120°, 240° 29. Sin x k cos x k k 7 1 or and k 6 1. or 3.0223 2 2p 1 2 , etc., and , 11p 2 2p p 2 9p 2 3p 2 2p , etc. 2 , 1 x 1.3734 or 4.5150 u 114.83°, 245.17° 3 , 3p 2 17. 21. a 65.38° 31. a 30° have no solutions when 5. 0 7. p 6 13. 2p 3 0.8584 15. 0.3576 23. 2.2168 1. Section 8.2, page 536 p 4 p 3 11. 3. 9. p 2 p 4 1.2728 cos u 1 2 19. 0.7168 21. ; tan u 23 p 2 p 6 29. 37. 5p 6 4 5 31. 39. p 3 4 5 33. 41. p 3 5 12 43. cos 45. tan sin 1 v sin 1 v 1 1 2 2 47 21 v2 1 v 21 v2 1 y π 2 15. 19. 23. 25. 27. 33. 17. 25. 27. 35. f(x) = cos −1 (x + 1) f(x) = sin x x 2π −2 −1 2 0 −2 Answers to Selected Exercises 1103 49. y 2π 3 π 3 x −2π −π π 2π −2π 3 51. a. u sin 1 40 x b a 53. y csc x b. 9.2° 5 −5 π 2 y csc x is one-to-one and has an − π 2 The graph of inverse. y csc 1 x − π 2 5 −5 55. a. Let 1 b. Let Then cos 2 and cos u w 1 w, u cos 1 v. u cos 1 v cos cos tan u w 1 w, u tan 1 v tan u tan Let tan 2 with 0 u p. 1 cos u and cos 1 cos u v, Then cos u v. p 2 1 with tan u and tan 1 1 v. tan u v, Then tan u v. u p . 2 tan and 2 1 w u. Let Then 1 w u. 1 u tan 1 2 4 xb a 57. a. tan 1 2 xb a b. x 9.13 feet Section 8.3, page 545 1. x p 3 2kp or 2p 3 2kp 3. x p 3 kp 1104 Answers to Selected Exercises 5. 9. 13. 15. 19. 23. 27. 31. 35. 37. 41. 43. 45. 47. 49. 51. 55. x ± 5p 6 2kp 7. x p 6 2kp or 7p 6 2kp 11. 17. 14.18° 27.57° x 0.4836 2kp or 3.6252 2kp x ± 2.1700 2kp x 0.4101 kp x p 6 x p 9 2kp or kp 3 kp 2p 3 21. 29. 25. x 0.2327 kp x ± 1.9577 2kp x ± p 2 4kp x ± 0.7381 2kp 3 x 3.4814, 5.9433 33. x 2.2143 2kp x 3p 4 x p 4 7p 4 p 2 , , 2.1588, 5.3004 x p 6 , , , 39. 5p 4 3p 2 x 0.8481, 1.7682, 2.2935, 4.9098 x 0.8213, 2.3203 x 0.3649, 1.2059, 3.5065, 4.3475 x 1.0591, 2.8679, 4.2007, 6.0095 x p 7p 4 4 x p 4 5p 4 3p 4 53 , 5p 6 , 3p 2 , 5p 4 kp 6 1.2682, 0.7446, 0.2210, 1.7918 5p 4 1 4 t tan 6 1. 3. 5 125 sin pt 5 b cos 20pt 216 sin2 a 1 6 sin pt 2 b a 7. h t 2 1 20pt 2 6 cos pt 2 b a 9. d t 2 1 10 sin pt 2 b a 11. a. 10 0 0 b. Roughly periodic; y 1.358 sin 0.4778x 0.569 1 20 7.636 2 57. not possible x p 6n 63. 2kp n , 59. 5p 6n 74.0° 61. No solution 16.0° or 2kp n π 2 Section 8.4, page 555 c. No, the unemployment total will only be predicted in the range 6.2781 to 8.9944. 2.376 y 2.9138 sin 0.400x 1.809 b. About 15.7, which is somewhat reasonable but 2 1 13. a. may not be the best model to use. c. 5 5. (graph of C-major chord: C E G ) 4 0 −4 0.03 −1 −1 25 y sin 1 262 2p x sin 1 2 330 2p x sin 1 2 392 2p x 2 Chapter 8 Review, page 564 The model is not a good fit in the second year. y 2.1663 sin 0.513x 1.051 1.71 d. 1 2 , where k is an integer. 1. 3. 5. 7. x kp x 0.8419 2kp 4.1784 2kp or x 0.5236 2kp tan 3t 3 5 , 3t tan 1 3 5b a kp, 2.2997 2kp or 5.2463 2kp 2.6180 2kp or or or 4.7124 2kp 25 1 tan 3 5b a 3 kp 3 t . In the first 2 seconds the solutions are 0.1801, 1.2273. 5 −1 −1 e. About 12.2 This model provides a much better fit. Section 8.4.A, page 562 1. 1 0.01 0 −1 y sin 1 294 2px 2 3. 1 0 −1 y sin 1 440 2px 2 9. p 3 19. π 2 0 − π 2 21. 25. 27. 2 515 x p 6 x 4p 9 11. p 3 13. 0 15. 2 17. 3p 4 4 23. 75° or 255° 2kp or 2kp x 11p 6 x 5p 9 2kp 3 or 2kp 3 x 3.78509 2kp 31. x tan 1 5 2b a kp 1.19029 kp 33. x p 4 kp or 37. x 0.8959 2kp 3p 4 or kp 35. x ± p 3 kp 2.2457 2kp Answers to Selected
Exercises 1105 0.01 29. x 2.49809 2kp or 39. 9.06° or 80.94° 41. a. 19 feet b. 3 ft below water c. 20 seconds d. Answers may vary: g 11 cos t 2 1 p 10 a t b 8 t e. Answers may vary: p 11 sin h 1 10 t 4.418 20k where k is any integer. and f. a 2 8 t p 2 b t 15.582 20k 43. Using approximate values, y 0.006 cos 2094.768x 8.379 1 2 seconds, 31. 33. Chapter 8 can do calculus, page 569 1. 1 2. 0 3. does not exist 4. 1 5. 1 6. 1 7. does not exist 8. does not exist 9. 1 10. 1 11. 0.75 12. 8 7 13. does not exist 16. does not exist 14. 3 17. 6 19. does not exist 20. 0.5 22. does not exist 23. 3 25. lim xS0 sin bx sin cx b c 15. does not exist 18. 4 21. 5 24. 0 Chapter 9 Section 9.1, page 580 1. Possibly an identity 3. Possibly an identity 5. b 7. e 9. tan x cos x sin x cos xb a cos x sin x 11. cos x sec x cos x 1 cos xb a 1 13. tan x csc x sin x cos xb a 1 sin xb a 1 cos x sec x 15. tan x sec x sin x cos x 1 cos x sin x 1 cos x 1 21 17. 1 cos x 19. Not an identity x 2 x 2 21. sin x cos x sin 1 cos 1 x 1 2 23. cot x cos 1 x sin 1 2 2 1 cos2x sin2x 2 tan x cos x sin x cot x 1106 Answers to Selected Exercises 27. 25. Not an identity sec2x csc2x tan2x cot2x cot x 1 sin2x cos x sin x 1 cos2x 2 2 2 3 3 2 tan x 1 29. 1 2 1 2 sin2x tan2x 1 tan2x 1 2 1 cot2x 2 1 sin x cos x cot x 1 1 1 tan x 2 2 4 2 2 4 1 sin2x sin2x cos2x 1 sec2x sin2x 1 2 sin2x 1 1 a tan2x cos2xb sin2x 1 cos2x 1 sin2x 1 2a 1 21 2 tan2x 1 1 cos2xb sin2x tan2x 2 1 tan2x 35. sec x csc x 1 cos x 1 sin x sin x cos x tan x sin2x sec2x 2 21 37. cos4x sin4x cos2x sin2x 39. Not an identity 1 41. sec x csc x sin x cos x cos2x sin2x cos2x sin2x 2 21 1 cos x 1 sin x tan x sin x cos x tan x tan x tan x 2 tan x 43. sec x csc x 1 tan x sin x cos x sin x cos x sin2x 1 csc x sin x 1 1 cos x 1 sin x cos x sin x sin x cos x sin x cos x sin x cos x cos x sin x sin x 1 2 45. 1 csc x sin x 1 sin x sin x sin x sin x 1 sin2x 1 sin x sin x a cos2x sec x tan x 1 cos xba sin x cos xb 47. Not an identity 49. Conjecture: cos x. Proof: 1 cos x sin2x 1 cos x cos x cos2x 1 cos x 1 sin2x 1 cos x 1 sin2x 1 1 cos x 2 cos x cos x 1 cos x 1 1 cos x 2 cos x 51. Conjecture: tan x: Proof: sec x csc x sin x cos x 1 sin x sec x cos x csc x cot x 2 sin x 21 sin x csc x 2 cot x 2 cos x sec x 1 cos x 53. 55. 57. 59. 1 sin x cot x 2 cos x sin x 1 sin x sin x cot x 2 cos x tan x cot x 1 sin x sec x 1 sin x sec x 1 cot x 1 sin2x 1 sin x sec x 1 2 1 cos x cos x 1 cos x sin x tan x 1 sin x 1 sin x cos2x 1 sin x 1 1 2 2 1 cos x 1 cos x 1 2 1 sin x cos x 1 sin x cos x 1 sin x 1 sin x 1 sin x 1 cos2x 2 sec x tan x 1 sin2 x 1 1 sin x cos x sin x cos x cos x cos3x 1 sin x 2 sin x sin x cos x cot x cot x cos x cos2x 1 sin x cos x cos x cos x cos x cos x sin x 1 sin x 1 sin x 1 sin x cos x cos x sin x cos x cos2x 1 sin x 1 sin x cot x cos x cos x cot x 41. 1 1 cos x 1 sin x 1 sin2x 2 2 cos x sin x cos x cos x sin x b cos x a cot x 1 , so tan x 61. csc x cot x 1 log10 1 cot x 2 log101 csc x cot x tan x 1 log10 a tan xb 1 log101 tan x 2 csc x cot x 2 csc x cot x 2 1 csc x cot x 1 1 2 csc2x cot2x csc x cot x log10 a 2 1 log101 1 csc x cot xb csc x cot x 2 ; 2 2 so, csc x cot x csc x cot x cot x cot y log10 1 log101 tan x tan y 1 tan y tan y tan x tan x tan y cos x sin y cos y sin x cos x sin y cos x sin y 2 tan x tan x cot x 1 1 2 63. 65. tan y cot y 2 cos2x sin2y 1 1 1 1 1 1 cos y sin x 1 sin2x 2 cos y sin x 21 cos x sin y 1 cos2y 2 1 cos x sin y 21 cos2y sin2x cos y sin x cos y sin x cos y sin x cos x sin y cos y sin x cos x sin y 21 21 21 2 2 2 2 2 cos y sin x cos x sin y Section 9.2, page 587 1. 16 12 4 7. 2 23 9. 2 23 19. 13. cos x sin 2 4 12 6 25. 31. 0.993 15. sin x 21. cos x 216 13 10 2.34 27. 33. 16 12 4 1/cos x 2 sin x sin y 11. 17. 23. 29. 0.393 f 1 35. x h 2 h f x 1 2 cos 1 cos x cos h sin x sin h cos x x h 2 h cos x 2 1 h cos x cos h cos x h a cos x cos h 1 h 44 2 125 the third quadrant. x y sin 1 ; b 37. sin x sin h h sin h sin x a h b 44 117 x y ; is in tan 1 x y 2 39. cos x y 1 56 65 2 ; tan 1 x y 33 56 2 x y ; is in the cos x. third quadrant. u v w sin cos u sin v cos w cos u cos v sin w sin u sin v sin w sin u cos v cos w 2 1 43. Since Hence, 1 1 Q p 2 x, x R y p sin y sin 2 sin2x sin2y sin2x cos2x 1 sin x cos p cos x sin p x p 2 1 sin x 0 21 2 21 cos p cos x sin p sin x p x 1 1 sin x cos x 1 2 x p sin x sin x cos p cos x sin p 2 1 0 21 2 cos x sin x sin x cos x cos x 21 45. sin 47. cos 49. sin 51. By Exercises 49 and 50, tan sin x cos x tan x x p 2 1 sin 1 cos 1 x p 2 x p 2 cos x cos y 1 cos 2 4 cos x cos y sin x sin y sin x sin y 2 3 1 2 4 53. 55. 57. 2 1 1 1 2 cos cos x y 2 sin x sin sin x sin y 1 2 1 x y 1 1 cos x cos y sin x sin y 1 2 cos x cos y 1 cos 2 cos2x cos2y sin2x sin2y x y cos 2 sin x cos y cos x sin x sin y cos y 2 1 1 1 sin x sin y 2 2 2 cos x cos y sin x sin y sin x cos y cot x tan y cos x cos y sin x sin y 2 3. 2 23 5. 2 23 59. Not an identity Answers to Selected Exercises 1107 61. sin sin 1 1 x y x y 2 2 sin x cos y cos x sin y sin x cos y cos x sin y 1 cos x cos y 1 cos x cos y sin x cos x sin x cos x sin y cos y sin y cos y tan x tan y tan x tan y 63. Not an identity 65. Not an identity 59. Section 9.2.A, page 592 1. 0.64 radians 3. 2.47 radians 7. 1.37 radians or 1.77 radians 11. 1.39 radians or 1.75 radians 5. 9. p 2 p 4 or 3p 4 45. cos x 47. sin 4y 49. 1 51. 53. sin 16x sin 2 1 cos4x sin4x cos2x sin2x cos 2x 8x 2 4 cos2x sin2x 1 3 2 sin 8x cos 8x cos2x sin2x 2 21 55. Not an identity 1 cos 2x sin 2x 57. 1 1 2 cos2x 1 2 sin x cos x 2 2 cos2x 2 sin x cos x 1 2x x cot x 2 sin x cos x cos x sin x sin 3x sin 2 1 2 sin x cos2x sin x 2 sin3x 1 sin2x 2 sin x sin x sin x 2 1 2 2 sin2x 1 2 sin2x 3 4 sin2x 2 cos x 2 sin x 2 sin3x 1 2 sin2x 1 2 1 1 2 sin 2x cos x cos 2x sin x sin x 61. Not an identity 63. csc2 x 2b a sin2 Section 9.3, page 600 22 12 2 1. 5. 2 23 9. 22 12 2 3. 7. 22 12 2 22 13 2 11. 22 1 65. 67. sin x sin 3x cos x cos 3x tan x sin 4x sin 6x cos 4x cos 6x cot x 1 1 1 cos x x 2 2R Q 2 cos 2x sin 1 2 cos 2x cos 1 x 2 x 2 2 1 cos x sin x cos x 2 sin 5x cos 1 2 sin 5x sin x 2 x 1 cos x sin x 2 2 sin a 2 sin x y cos 2 R x y 2 sin cot x y 2 a b 69. sin x sin y cos x cos y x y cos 2 Q x y 2 Q 1 cos x sin x sin R 71. a. and part (a), tan x 2 73. R vt cos a a R v 2v sin a g R 2v2 sin a cos a b cos a g R v2 sin 2a g 1 1 cos x sin x 21 1 cos x 2 1 1 cos x sin2x 1 cos x sin x 2 1 cos2x sin x 1 cos x 1 2 b. By the half-angle identity proved in the text sin x 1 cos x 2 1 1 cos x sin x sin x 1 cos x 13. 17. 21. 23. 25. 27. 29. 1 2 1 2 sin 10x 1 2 cos 20x 1 2 sin 2x 15. 1 2 cos 6x 1 2 cos 2x cos 14x 19. 2 sin 4x cos x 2 sin 2x cos 7x sin 2x 120 169 sin 2x 24 25 sin 2x 24 25 115 8 sin 2x , tan 2x 120 119 , tan 2x 24 7 , cos 2x 119 169 , cos 2x 7 25 , tan 2x 24 7 115 7 , cos 2x 7 25 , cos 2x 7 8 , tan 2x 31. sin 33. sin 35. sin 0.5477, cos x 2 0.8367, tan x 2 0.6547 1 110 , cos x 2 15 2 215 B , tan , cos B 3 110 x 2 1 x 3 2 15 2 215 , tan 29 415 25 2 x 2 x 2 x 2 x 2 37. sin 2x 0.96 39. cos 2x 0.28 41. sin x 2 0.3162 43. cos 3x 4 cos3x 3 cos x 1108 Answers to Selected Exercises 3. x 0, p, 2p, 7p 6 , 11p 6 7. x 0, 2p 11. x 3p 8 , 7p 8 , 11p 8 , 15p 8 15. cos x sec x cos x 1 1 cos2 x cos x sin x cos x cos x sin2 x cos x sin x tan x sin x 75. 2 sin2 1 a 2 a sin2 1 2 1 2 1 cos cos a 1 cos u 2 cos u cos a a 1 cos 2 1 2 Q u R ¢ 2 b Section 9.4, page 608 1. no solution 5. 9. 13. 15. 19. 23. 27. 31. 35 3p 4 x 3p 4 x 5p 12 x p 3 x p 4 3p 2 3p 2 3p 4 3p 4 , , 5p 4 5p 4 , , 7p 4 7p 4 , , 7p 4 7p 4 , 13p 12 , p, p , 3p 4 , 5p 4 37. a. f x 1 2 2 sin b. 2 c. x p 6 39. a. f x 1 2 x 7p 4 b a 22 sin 22 b. c. x f 2 1 x 3p 4 , p 12 , 5p 12 17. 21. 25. , , 13p 12 x p 3 x p 3 x p 3 17p 12 , , , 5p 3 5p 3 5p 3 , p 29. x p, 0, p 33. x 2p 5 , 6p 5 , 2p, 0, 4p 3 , 0, p, 2p , 7p 4 x p a 3 b 11. 13. 2 1 2 sin 2x 9. Not an identity sin x cos x sin2x 2 sin x cos x cos2x 2 sin x cos x sin2x cos2x 1 cos4 x sin4 x 1 tan4 x cos2 x sin2 x 1 tan2 x sec2 x cos2 x sin2 x 1 tan2 x 21 21 cos2 x sin2 x 1 sin2 x cos2 x b 1 cos2 x 1 1 1 cos2 x sin2 x 1 tan2 x 2 2 2 1 2 1 cos2 x sin2 x cos2 x sin2 x a cos4 x 2 cos4 x 17. (a) 19. 3 2 2 cos x y x y cos 1 1 cos x cos y sin x sin y cos2x cos2y sin2x sin2y 1 sin2y cos2x 1 cos2x cos2x sin2y sin2y cos2x sin2y cos2x sin2y 1 cos2x sin2y 4 3 1 2 2 cos x cos y sin x sin y 4 21. a. 3 5 23. 120 169 b. 117 44 c. 44 125 25. 142 212 10 29. 1.23 radians 2 2 sin2 x tan x 27. 31. 33. 1 1 1 1 2 sin2 x tan x 2 sin2 x cos x sin x cos x 1 cos 2x tan x 2 sin2 x sin x cos x 2 sin x cos x 2 cos x 2 cos3 x 2 cos x 2 cos x sin2 x sin 2x sin x 1 1 2 1 cos2 x 2 sin x cos x 2 2 1 35. 120 169 37. Yes. cos x 0 0 1 2 sin 2x 2 sin x cos x 2 p 12b p 6 b d 1 2 a a cos c p 6 1 cos 1 13 2 R B 2 2 13 4 R 2 22 13 2 ; cos p 12b a Chapter 9 Review, page 611 39. cos 1. 5. 7. 3. sin4 x cot t 1 3 sin4t cos4t sin2t sin t 1 cos t sin t 1 3 1 1 cos2t 1 sin2t 2 4 1 sin t 1 cos t sin2t cos2t 1 1 2 1 1 sin t 21 2 sin2t 1 1 cos t 1 cos t 2 2 1 cos t sin2t 1 1 cos t 2 sin2t cos2t 2 2 1 cos t sin t Answers to Selected Exercises 1109 19. 334.9 km 23. 84.9° 27. 231.9 ft 21. 63.7 ft 25. 8.4 km 29. 154.5 ft 31. 4.7 cm and 9.0 cm 33. 33.44° 35. 978.7 mi 37. AB 24.27, B 56.8°, AC 21.23, C 73.0° 39. 16.99 m Section 10.2, page 634 BC 19.5, A 50.2°, 1. 3. 5. 7. C 110°, b 2.5, c 6.3 B 14°, b 2.2, c 6.8 A 88°, a 17.3, c 12.8 C 41.5°, b 9.7, c 10.9 11. 32.5 9. 7.3 13. 82.3 17. No solution 55.2°, 124.8°, A1 A2 19. C1 C2 104.8°, 35.2°, 14.1; 8.4 c1 c2 B2 21. No solution 65.8°, 9.8°, 23. a1 10.3; 58.2°, B1 A1 A2 2.1 a2 C 72°, b 14.7, c 15.2 a 9.8, B 23.3°, C 81.7° A 18.6°, B 39.6°, C 121.9° c 13.9, A 60.1°, B 72.9° C 39.8°, A 77.7°, a 18.9 25. 27. 29. 31. 33. 15. 31.4 114.2°, 35. No solution 37. 6.5 39. About 7691 41. 135.5 m 43. 5.4° 45. 5 ft 47. 5.3° 49. 30.1 km 51. About 9642 ft then EBA 180° ABD. 53. a. Use the Law of Cosines in triangle ABD to find is ABD; Use the Law of Cosines in triangle ABC to find CAB; have two of the angles in triangle EAB and can easily find the third. Use these angles, side AB, and the Law of Sines to find AE. 180° CAB. You
now EAB (Why?) then is b. 94.24 ft 55. 13.36 m 57. 5.8 gal 59. 11.18 sq units 61. No such triangle exists because the sum of the lengths of any two sides of a triangle must be greater than the length of the third side, which is not the case here. cos p a 4 12 2 a 22 13 2 p 6 b cos p 4 cos 13 2 b 12 1 2b 2 a 16 12 4 x kp 43. 41. 0.96 sin p p 6 4 16 12 4 sin p 6 . So, or 22 13 12 16 2 45. x 2.0344 kp Chapter 9 can do calculus, page 615 1. a. 1; sin x is changing at a rate of 1 unit per increase in x when 12 2 ; b. x 0. sin x is increasing approximately 0.7071 per unit increase in x when x p 4 . c. 0; sin x is not changing per unit increase in x when x p 2 d. 13 2 ; sin x is increasing approximately 0.8660 per unit increase in x when x p 6 . 2. sin x 3. a. 0; cos x is not changing per unit increase in x x 0. when 12 2 ; b. c. cos x is decreasing approximately 0.7071 units per unit increase in x when 1; cos x is decreasing 1 unit per unit increase x p 4 in x when x p 2 . d. 1 2 ; cos x is decreasing unit per unit increase 1 2 in x when x p 6 . Chapter 10 Section 10.1, page 622 1. 3. 5. 7. 9. 11. 13. 15. 17. a 4.2, B 125.0°, C 35.0° c 13.9, A 22.5°, B 39.5° a 24.4, B 18.4°, C 21.6° c 21.5, A 33.5°, B 67.9° A 120°, B 21.8°, C 38.2° A 24.1°, B 30.8°, C 125.1° A 38.8°, B 34.5°, C 106.7° A 34.1°, B 50.5°, C 95.4° 54.2° at vertex at vertex (0, 0); 1, 4 1 2 48.4° at vertex 5, 2 1 2 ; 77.4° 1110 Answers to Selected Exercises real i i 1 2 1 real 1 21. 23. 25. 27. 29. 31. Section 10.3, page 642 1.–7. i (2i)(3, − 5 2 i) real 3 + 2i (1 + i)(1 − i) − 8 3 − i5 3 9. 13 11. 23 15. Many correct answers, including 13. 12 z 2i, w i 17. 19. real 1 i 1 5 i 1 cos 0.9273 i sin 0.9273 2 cos 5.1072 i sin 5.1072 2 cos 1.1071 i sin 1.1071 5 1 13 1 25 1 218.5 2 cos 2.1910 i sin 2.1910 1 2p 2p 3 b 3 i sin 3 323i 2 33. 6 cos a 35. 42 cos a 7p 6 i sin 7p 6 b 2123 21i 37. 3 2 a cos p 4 i sin p 4 b 322 a 4 b 322 a 4 b i 39. 222 cos a 7p 12 i sin 7p 12 b real 41. cos p 2 i sin p 2 43. 12 cos a 2p 3 i sin 2p 3 b 45. 222 cos a 19p 12 i sin 19p 12 b 47. The polar form of i is 1 cos 90° i sin 90° 1 . Hence, 2 by the Polar Multiplication Rule i sin zi r 1 u 90° cos u 90° . 1 1 1 2 You can think of z as lying on a circle with center at the origin and radius r. Then zi lies on the same circle (since it too is r units from the origin), but 90° direction). farther around the circle (in a counterclockwise 2 Answers to Selected Exercises 1111 y b d c b1 a x a 2 29. 24 2 a 1 2 or 24 2 i b 1 a 2 or i b lies on L since b d b 2 1 a c, b d lies on M since 2 49. a. d. f. 51. a. b. c. 2 and d c x c b b. a y d b a 1 a c sin u221 sin2u22 i sin u121 cos u1 cos u2 1 r21 r21 r11 r13 1 sin u1 cos u2 i 1 r13 cos 2 cos u2 cos2u2 cos u1 u1 1 1 a c c . 4 i sin u22 i sin u22 2 cos u2 r2 cos u2 sin u1 sin u22 cos u1 sin u22 4 i sin u22 u22 4 u1 1 Section 10.4, page 652 24323 2 1. i 3. 243 2 i 5. 64 7. 1 2 23 2 i 9. i 11. 1, 1, i, i i sin p 15b , 4 cos a 11p 15 i sin 11p 15 b , i sin 7p 5 b 13. 4 cos a 4 cos a 15. 3 cos a p 15 7p 5 p 48 3 cos a 49p 48 i sin 49p 48 b , 3 cos a 73p 48 i sin 73p 48 b 17. cos a p 5 i sin p 5 b , cos a 3p 5 i sin 3p 5 b , cos p i sin p 1 , 2 cos a 7p 5 i sin 7p 5 b , cos a 19. cos a cos a 9p 5 p 10 9p 10 i sin 9p 5 b i sin p 10b , cos a p 2 i sin p 2 b , i sin 9p 10 b , cos a 13p 10 i sin 13p 10 b , cos a 17p 10 i sin 17p 10 b 21. 24 2 cos a p 8 i sin p 8 b , 24 2 cos a 9p 8 i sin 9p 8 b 1 2 i or 23 2 1 2 i or 23 2 1 2 i or 23. x 23 2 1 2 23 2 i or i or i 23 2 23 2 a 24 2 23 2 23 2 1 2 i b or 24 2 1 2 0.2225 ± 0.9749i, i b a 31. 1, 0.6235 ± 0.7818i, 0.9010 ± 0.4339i ± 1, 35. 1, 0.7660 ± 0.6428i, ± i, 33. 0.7071 ± 0.7071i, 0.7071 ± 0.7071i 0.1736 ± 0.9848i, 37. 1 0.5 ± 0.8660i, 0.9397 ± 0.3420i x6 1 x5 x4 x3 x2 x 1 x 1 so 21 x5 x4 x3 x2 x 1 0 the solutions of the sixth roots of unity other than 1; namely, 23 2 i, 1 2 i, 1 2 23 2 23 2 1, 1 2 1 2 i, , 2 are 23 2 i. 39. 12 41. For each i, ui n vui2 Hence 1 vui and is a solution of the equation. If is an nth root of unity, so n vn 1 r n 1. u12 cos u i sin u vuj, ui2 vui vn 21 1 1 1 2 then multiplying both sides by 1 v shows that In other words, if ui Thus, the solution is not equal to vu1, p , vun uj, are all then uj. ui vuj. vui distinct. 7. 11. 13. 15. 17. 19. 21. 23. 25. 27. 29. 31. H 5 4, ; I 6, 10 I H u v 18, 10 H I u v H 1 322 I 23 4 u v h ; 13 5 , 2 5 i 9. h u v 8, H 3 ; I 3u 2v 3 422, 3u 2v u v 1 322 ; I 9 822, H 3 422, H 2 922 I u v ; , 13 i 9 4 h , 7 ; i , 24 i 2 h 3u 2v 17 2 u v 2130 v w 1022 2 0 w 2u 1 v 225 7 v 2 3 u3 u2 u4 v c, d 2 I H v r s rc, rd u1 v 1 r s 1v 1 6 1 2 3, 9 I H c, d H c, d 2H I c, d c, 9 I H 0 0, 2 r v sv d 2 1 I H1 I sc, sd I H v, 0v 0 c, d 0, 0 H H 2 2c2 d2 v b d b b d d 2 2a2 b2 u 0 I I 2 1 2 i sin p 48b , 3 cos a 25p 48 i sin 25p 48 b , Section 10.5, page 660 1. 325 3. 234 5. 6, 6 H I 25. 23 2 1 2 i or 23 2 27. x 3i or 323 2 3 2 i i or 1 2 or 323 2 i 3 2 i 33. a. 2 b. 2 1 1 1112 Answers to Selected Exercises c. The slope is which is the same as the slope d c , of the vector v. d. The slope of the line is which is the same as b a , the slope of the vector u. Section 10.6, page 667 1. 3. 5. 7. u v 3i, u v i 2j, 3u 2v i 5j u v i 4j, u v 7i 4j, 3u 2v 18i 12j u v 23 i 22 j, u v 23 i 22 j , 3u 2v 223 i 322 j i 2j 5 2 9. 7i 7j 523 2 , 5 2 i h 7.5175, 2.7362 H v 11. 9i 18j 13. v 15. 19. 21. 23. 25. 27. 29. I 17. 10, 1023 H I 1.9284, 2.2981 H v v v 10, u 60° v 241 6.4031, u 51.34° v 425, u 296.6° v 5213, u 213.7° i 8 7 31. j 2113 2113 1 210 i 3 210 j 33. Direction: 82.5°; magnitude: 9.52 lb 35. Direction: 37. u sin 1 a 18.4°; 894.8 1500 b 36.6° magnitude: 80.4 kg 39. Parallel to plane: 68.4 lb; perpendicular to plane: 187.9 lb 41. 1931.85 pounds 43. Ground speed: 401.1 mph; course: 154.3° 45. Ground speed: 448.7 mph; course: 174.2° 47. Air speed: 96.6 mph; direction: 326° 49. 0.8023 mph 4° 6°; 53. 1002 lb on lies on the straight line through (0, 0) and which has slope 2 a c b d 51. 1005 lb on u v a c through (a, b) and (c, d), which also has slope a c b d they actually have the same direction by considering the relative positions of (a, b), (c, d), points and and w are parallel. Verify that For instance, if a c, b d Similarly, w lies on the line u v u v So, . . . 1 2 upward to the right, then (a, b) lies to the right d 6 b, and above (c, d). Hence so that a c 7 0 which means that the and endpoint of w lies in the first quadrant, that is, w points upward to the right. b d 7 0, c 6 a and Section 10.6.A, page 679 1. 3. 5. u v 7, u u 25, v v 29 u v 6, u u 5, v v 9 u v 12, u u 13, v v 13 7. 6 9. 20 11. 13. 1.75065 radians 15. 2.1588 radians 17. radians 28 p 2 19. Orthogonal k 2 25. 21. Parallel k 22 27. 23. Neither I 29. 31. 33. 37. 39. ; h 12 17 projuv , 20 17 i ; projvu projuv 0, 0 I H compvu 22 213 v w 1 a, b H I u a, b 1H I H 2 c r, d s I H ac ar bd bs u v u w a, b I H ar bs ac bd projvu 6 5 , 2 5 i h 0, 0 H 35. I compvu 3 210 c, d I a r, s H c r 1 I2 b 2 d s 1 2 I a, b c, d I H H ac ar bd bs 2 r, 0 I p, H or 1 a, b I H then u and v are parallel, so 0a 0b 0 41. If v v ku u k . 0 0 k 7 0, 0 1 and ku u then and so Since u v u v then for some real number k. We know that u 0, cos u 1 k 7 0. If k v ku. k Therefore, 0 u ku ku2 ku u 2 u v cos u. u p, On the other hand, if k cos u 1 k 6 0, k and and 0 v ku. u ku, ku u so u ku2 ku u v cos u. In u v u v cos u. both cases we have shown C , ! I BC A vector Since so the angle at vertex B is a right angle. Since , 2 ! I H 0, BC 1 . and are perpendicular, , 3, 4 and ! 2 AC ! AB k 6 0. Then then the 2, 2 1, 2 ! 1 AB ! AB H and , ! 2 BC 5, 2 4, 0 I H 1 2 0 43. If 45. Many possible answers: One is v 47. 300 , I 1, 1 H lb w and 600 cos 60° 1 1, 1 H . I 2 51. 24 49. 13 u 1, 0 , I H 53. The force in the direction of the lawnmower’s motion is is 75 15 30 cos 60° 15 lb. 1125 ft-lb. Thus, the work done 2 55. 1368 ft-lb 1 Answers to Selected Exercises 1113 Chapter 10 Review, page 684 1. 3. 5. A 52.9°, B 41.6°, C 85.5° A 20.6°, b 21.8, C 29.4° A 35.5°, b 8.3, C 68.5° 7. Approximately 301 mi 9. 11. 13. 15. 17. 71.89° A 25°, a 2.9, b 5.6 A 52.03°, B 65.97°, b 86.9 B 81.8°, C 38.2°, c 2.5 B 98.2°, C 21.8°, c 1.5 a 41.6; C 75°, c 54.1 and 19. 147.4 21. 13.4 km 23. Joe is 217.9 m from the pole and Alice is 240 m from the pole. 25. a. 3617.65 ft b. 4018.71 ft c. 3642.19 ft 27. 10 31. 210 220 29. 37.95 33. The graph is a circle of radius 2 centered at the origin. 35. 2 cos a p 3 i sin p 3 b 39. 223 2i 37. 422 422i 41. 81 2 8123 2 i 43. 1, cos i sin p 3 4p i sin 3 p 3 4p 3 , cos , cos 2p 3 5p 3 i sin i sin 2p 3 5p 3 1, , cos cos cos p 8 9p 8 11, 1 H 210 I , i sin p 8 9p i sin 8 cos 5p 8 , cos i sin , 5p 8 i sin 13p 8 2229 49. 55. 522 2 , 522 2 h i 1 25 i 2 25 j 59. Ground speed: 321.87 mph; course: 126.18° 63. 3 26 projvu v 2i u2 v2 0 the same magnitude since u and v have 71. 1750 lb 45. 47. 53. 57. 61. 67. 69. 13p 8 51. 11i 8j −6 ; 0.7071 0.7071i 4. cos p 4 b i sin a a cos p i sin p p 4 b ; 2.7183 2 e 5. 1 6. 1 7. e a 8. ep i sin p 3 b cos p 3 cos 1 i sin 1 1 ; 12.5030 19.4722i 2 ; 1.3591 2.3541i Chapter 11 1. x2 49 Section 11.1, page 698 y2 4 y2 49 1 1 x2 9 5. 9. 2x2 y2 12 13. Ellipse −6 15. Ellipse 4 −4 4 −4 3. x2 36 y2 16 1 7. x2 6y2 18 11. x2 16 y2 9 1 6 6 21. 7p 23 65. 0.70 radians 17. 8p 19. 223p 23. approximately 1,507,964 sq. ft. x2 a2 y2 a2 25. If a b, then 1. Multiplying both sides a2 gives x2 y2 a2, by radius a with center at the origin. the equation of a circle of Chapter 10 can do calculus, page 689 1. 1 cos 2 3 1 1 ; 0.4161 0.9093i 2. i sin i sin 3. cos .9900 0.1411i 27. As b gets larger, the ellipse becomes more elongated horizontally. As b gets closer to 0, the graph becomes very close to being a vertical line. However, it will never be a vertical line, because b cannot have a value of 0. 1114 Answers to Selected Exercises 10 10 10 29. Approximately 226,335 mi and 251,401 mi 19. Hyperbola 31. of OC; b length c 2a2 b2, Pythagorean Theorem c length or a length c2 a2 b2 of OF; since a2 b2 c2; of CF. 5. 1. 1 Section 11.2, page 707 y2 36 y2 5 y2 8
y2 8 x2 9 x2 4 x2 1 x2 1 1 11. 9. 1 or 8x2 y2 8 1 or 8x2 y2 8 3. 7. x2 4 y2 16 y2 1 x2 9 1 by the −10 10 −10 Because of limited resolution, this calculator-generated graph does not show that the top and bottom halves of the graph are connected. 13. 10 21. 10 −10 −10 23. −10 10 −10 −10 10 −10 foci are at A y ± b a x; are 222, 0 and 222, 0 ; asymptotes B y 4 26 A and x B y 4 26 x 15. 10 −10 10 −10 foci are at A y ± b a x; are 220, 0 B y 2x and A and 220, 0 y 2x 17. 2 ; asymptotes B 25. The two branches of the hyperbola are very “flat” when b is large. With very large b and a small viewing window, the hyperbola may look like two horizontal lines, but it isn’t because its asymptotes y ± 2 b ˛x ± 2 b close to, but not equal to, 0 when b is large). are not horizontal (their slopes, are , 27. 8000 −3 3 −200 3800 −2 213 and 0, foci are at 0, a y ± a b x; 6 b y 2 3 x a and y 2 3 are 213 6 b −8000 ; asymptotes 29. y x 2 x 31. The distance between the vertices is 2a. One point on the hyperbola is the vertex at (a, 0). This point is the distance from the closest focus and a c a distance c a from the other focus. The Answers to Selected Exercises 1115 1 c a c a differences of these distances is 2 2a. Therefore, by the definition of a hyperbola, when the difference of the distances from any point on the hyperbola to the two foci is constant, this constant difference is 2a; this is the distance between the two vertices. OV a, OP b; a2 b2 c2 c2 a2 b2, each focus to the center O. since in the right triangle and in the equation of a hyperbola PV c, which is the distance from 33. Section 11.3, page 714 1. 7. y 3x2 y x2 4 11. Parabola 3. y2 20x 9. x2 8y 5. y2 8x −10 13. Parabola −10 4 −4 2 10 10 −10 15. Focus: 1 12b 0, a ; directrix: y 1 12 17. Focus: (0, 1); directrix: y 1 19. x2 1 2 y 21. y2 4x 1 1 2 2 x 2 6 y 2 16 13 x 3 36 x 3 4 2 1 5. 7. 9. 13. 15 16 y 2 2 1 17. Ellipse 23. 9 11 19. Parabola 21. Hyperbola −5 10 −1 25. 9 27. −7 −5 29. 7 −1 9 −1 7 5 23. The point closest to the focus is the vertex at (0, 0). 25. y2 8x 27. y2 2x −2 13 2 1 1. Section 11.4, page 726 y 3 16 y 4 36 x 2 4 x 7 25 1 1 33 1116 Answers to Selected Exercises 5910aans_1114-1147 9/21/05 2:03 PM Page 1117 31. 9 −18 18 5. 7. u 53.13°; x 3 5 u 36.87°; ; y 4 5 v; y 3 5 33. −15 10 −10 10 −10 35. Ellipse; 37. Hyperbola; 6 x 3 and 7 x 13 2 y 4 and 3 y 9 39. Parabola; 41. Ellipse; 1 x 8 1.5 x 1.5 and 3 y 3 1 y 1 and 43. Hyperbola; 15 x 15 and 10 y 10 45. Parabola; 19 x 2 and 1 y 13 47. Hyperbola; 15 x 15 and 15 y 15 49. Ellipse; 6 x 6 and 4 y 4 51. Parabola; x 5 49 53. 2 1 2 9 x 4 y 3 16 2 2 1 1 and 2 y 10 x 5 16 1 2 2 or 55. The asymptotes of 1 are x2 a2 1 with slopes y2 a2 and 1, Since 1. these lines are perpendicular. 9, 1 a 2 ± 1 2 234 59. b y ± x, 1 1 2 1 1 b 0 2 57. 61. 63. 65. 67 105 200.45 1 2 2 1 1 1 2 y 55 2 1 y2 5,759,600 x2 1,210,000 The exact location cannot be determined from the given information. measurement in feet 49 y ± a a x or Section 11.4A, page 733 1. u2 2 v2 2 1 3. u2 4 v2 1 v u 3 5 u 4 5 u2 v 9. a. b. c. E cos u 2 1 2 u A cos2 u B cos u sin u C sin2 u 1 B cos2 u 2A cos u sin u 2C cos u sin u 1 C cos2 u B cos u sin u uv B sin2 u 1 2 D cos u E sin u v2 A sin2 u 1 2 2 v F 0 D sin u B¿ B cos2 u 2A cos u sin u 2C cos u sin u B sin2 u 2 B since Since sin 2u 2 sin u cos u and cos 2u cos2 u sin2 u, B¿ 1 cot 2u A C then B B¿ B cot 2u we have 2 B cos 2u B cos 2u 0. A C B cot 2u cos2 u sin2 u C A B¿ 1 1 2 2 and sin 2u B cos 2u sin 2u B cos 2u. C A sin u cos u is the coefficient of uv 1 2 d. If 11. a. From Exercise 9 (a) we have B¿ 2 4A¿C¿ 1 2 2 2 4 C cos2 u 2 1 2 1 1 2 B2 C A 1 C A cos u sin u 2cos2 u sin2 u 4B 2 A2 C2 B2 2 2 1 cos2 u sin2 u 2 cos2 u A cos2 u B cos u sin u 2 4 C cos2 u B cos u sin u A sin2 u C A cos u sin u B cos2 u 2A cos u sin u 2C cos u sin u 1 B sin2 u 2 C sin2 u 2 1 cos2 u sin2 u B 2 1 3 A cos2 u B cos u sin u C sin2 u 4 1 B cos u sin u A sin2 u 4 1 sin2 u 4 1 4AB B2 4BC 1 2 cos2 u sin2 u sin4 u 4 cos2 u sin2 u 2 cos2 u sin2 u cos4 u sin4 u 4AC 1 (everything else cancels) B2 4AC 2 1 B2 4AC 2 1 B2 4AC 6 0, B¿ 0, 4AC 2 cos2 u sin2 u 2 cos3 u sin u cos u sin3 u 1 cos3 u sin u cos u sin3 u 1 cos4 u 2 cos2 u sin2 u sin4 u cos2 u sin2 u 2 then also 4A¿C¿ 6 0 B¿ 2 1 and so cos4 u sin4 u 2 B2 4AC cos4 u 2 4A¿C¿ 6 0. A¿C¿ 7 0. By Since Exercise 10, the graph is an ellipse. The other two cases are proved in the same way. 522 2 23 2 , 1 2 b 22 2 b , 15. a 1 1 b. If Section 11.5, page 743 13. a 1. P T 2, a 4, a 5, 2p , 1 1 2 5p 6 b , 1, a others 3. 5. 7. 13. a , U , Q p 4 b 3p 2 b 5, 2p , 1 2 1, 7p 6 b a 3, 2p 3 b 6, p 3 b a 5, 3p 2 1, a , , R 5, p , S 1 7p 6 b , 7, a 2 5p 3 b , V 7, 0 1 2 or 6, a 5, p , 1 11p 6 b , and others 2 1, 13p 6 b a , , and 6, p 6 b , 323 3 a 2 225, 1.1071 2 b A B 9. 23 2 a , 1 2 b a 231.25, 2.6779 11. B 15. A Answers to Selected Exercises 1117 25. θ = π 2 −1 −0.5 0.5 1 θ = 0 −0.5 −1 −1.5 −2 θ = 3π 2 27. 29. θ = π 2 1 0.5 −0.5 −1 θ = 0 θ = π −2 −1.5 −1 −0.5 θ = π 2 1 0.5 −1 −0.5 0.5 1 θ = 0 −0.5 −1 θ = 0 5 10 31. θ = π 2 θ = 0 0.5 0.5 −0.5 −0.5 θ = 3π 2 17. π 2 θ = 4 −4 2 −2 −2 −4 19 11 − π 3 −1 θ = 0 21. θ = π 2 3 2 1 −1 −2 −3 −2 −1 1 radian 1 2 θ = 0 23. θ = − 3π 2 −10 −5 7.5 5 2.5 −2.5 −5 −7.5 −10 θ = − π 2 1118 Answers to Selected Exercises 33. −2 −1 θ = π 2 0.5 −0.5 θ = 3π 2 θ = 0 1 2 35. θ = π 2 3 2 1 −1 −2 − = 3π 2 37. θ = π 2 1.2 1 0.8 0.6 0.4 0.2 −0.2 0.2 0.2 0.4 0.6 0.8 1 1.2 θ = 0 θ = 3π 2 θ = π 2 0.75 39. −1 −0.75 θ = 3π 2 41. θ = π 2 2 1.5 1 0.5 −0.5 −1 −1.5 −2 0.2 0.4 0.6 0.8 1 θ = 0 θ = 3π 2 43 45. θ = π 2 0.8 0.6 0.4 0.2 −0.2 −0.2 0.2 0.4 θ = 0 θ = 3π 2 47. a. 3 −4 4 θ = 0 1 −3 Answers to Selected Exercises 1119 b. 3 19. a. 4 6 6 4 b. 215 4 , 210 4 , 22 4 c. The smaller the eccentricity, the closer the shape is to circular. 21. 10 5 (4, π) −10 −5 5 10 −5 −10 23. 10 5 (−2, 0) −10 −5 −5 −10 ( , π2 ) 3 5 10 25. (10, )π 2 10 5 −5 −5 −10 −10 5 ( 10 7 , 10 )3π 2 −4 c. 3 4 3 −3 3 3 49. r a sin u b cos u 1 r2 ar sin u br cos u 1 x2 y2 ay bx 1 x2 bx y2 ay 0 1 a2 b2 x2 bx b2 a 4 b 4 y a a 2b y2 ay a2 4 b a2 b2 4 x b 2b a circle with center b 2 , a 2b a and radius 2a2 b2 2 . 51. Using the Law of Cosines in the following d2 r2 s2 2rs cos so . diagram, d 2r2 s2 2rs cos r, θ) d r θ − β s (s, β) Section 11.6, page 752 1. (d) 3. (c) 5. (a) 7. Hyperbola, e 4 3 11. Ellipse, e 2 3 9. Parabola, e 1 13. 0.1 15. 25 17. 5 4 1120 Answers to Selected Exercises 10 5 (3, π) −5 −5 −10 −10 27. 29. 31. (15, 0) 5 10 15 10 5 −5 −5 −10 15 10 5 ( 3 2 , )π 2 5 10 )3π 2 (−10, )π (2, 2 −10 −5 −5 5 10 33. r 6 1 cos u 37. r 3 1 2 cos u 41. r 3 1 sin u 45. r 2 1 2 cos u Section 11.7, page 763 35. r 39. r 16 5 3 sin u 8 1 4 cos u 43. 47. r 2 2 cos u r 3 107 1 cos u 2 1 and 4, 7 2, 5 25. Both give a straight line segment between Q P 1 equations in (a) move from P to Q, and the parametric equations in (b) move from Q to P. t x a c a The parametric 27. Solving t . 2 x a substituting in ˛, 2 ˛ 1 for t gives 1 2 then gives 1 or x a . 2 and This is a linear equation and therefore gives a straight line. You can check by substitution that (a, b) and (c, d) lie on this straight line; in fact, these points correspond to respectively. x 6 18t, y 12 22t t 1, 0 t 1 t 0 and 1 2 29. 31. Local minimum at 6, 2 33. Local maximum at (4, 5) 1 2 35. Solve the first equation for t x 140 cos 31° ˛. Substitute into the second to get y 140 sin 31° a which is the equation of a parabola. x 140 cos 31°b 16 2a 1 x 140 cos 31°b 2 , 37. a. 120 0 0 250 88 cos 48° 88 sin 48° x 1 y 1 0 t 4.5 t 2 t 16t2 4 2 b. Yes v 8024 50 39. 41. a. 3 105.29 ft/sec 1. 3. 5. 7. 9. 11. 13. 15. 21. 5 x 6 and 2 y 2 3 x 4 and 2 y 3 0 x 14 and 15 y 0 2 x 20 and 11 y 11 12 x 12 and 12 y 12 2 x 20 and 20 y 4 25 x 22 and 25 y 26 y 2x 5 y 2x 7 17. 19. y ln x 0 0 350 t 2 t 16t2 2 110 cos 28° 110 sin 28° x 1 y 1 0 t 3.5 b. About 3.2 sec c. 41.67 ft x2 y2 9 23. 16x2 9y2 144 Answers to Selected Exercises 1121 80° 60° 40° 20° 43. a. 200 0 0 b. c. 40° 200 45° 40° 0 0 An angle of distance. c. 6 0 0 12 The particles do not collide; they are closest when 8 t 1.13. d. 0 −1 2 350 350 45° seems to result in the longest d is smallest when t 1.1322. Section 11.7A, page 769 1. 3. 5. x 9 5 cos t, y 12 5 sin t x 2 cos t 2, y 2 sin t 3 x 210 cos t, y 6 sin t 0 t 2p 1 0 t 2p 2 2 1 0 t 2p 2 1 −10 7 −7 10 7. x 1 2 cos t, y 1 2 sin t 1 0 t 2p 2 −3 2 −2 3 45. a. Since Then 3 cos . , y CT PQ 3 3 sin u u t 3p 2 u, TCQ 3p t 3p 2 2 x OT CQ 3t 3 cos u 3t t 3p And . 2 b t 3p 2 b cos t cos 1 3p 2 sin t. sin t sin 3p 2 Therefore, . b. a cos a 3 3 sin t 3p a 2 b cos t sin t 0 1 t 3p a 2 b t 3p 2 b 21 3t 3 cos t sin t . Sin 3 a cos t sin sin t 2 1 2 1 2 3p 2 21 3t 3 sin t sin t cos 3p 2 cos t. 21 1 3 3 sin 0 1 2 t 3p a 2 b cos t 1 2 21 3 3 cos t Therefore, 1 cos t 3 1 . 2 47. a. 6 0 0 12 The particles do not collide. b. t 1.1 1122 Answers to Selected Exercises 9. x 210 cos t , y 6 tan t 0 t 2p 2 1 17. x 4 cos t 1, y 28 sin t 4 0 t 2p 1 2 −10 10 −10 10 −7 11. x 1 cos t , y 1 2 tan t 1 0 t 2p 2 19. x t any real number t 2 9 −1 7 9 −1 5 6 −5 −6 4 −4 9 −1 −5 13. x t2 4 , y t 1 t any real number 2 −5 10 −10 15 15. x 2 cos t 1, y 3 sin t 5 0 t 2p 2 1 21. x 2 t 2 1 2 2, y t 1 any real number t 2 7 −3 −2 13 23. x 4 tan t 1, y 5 cos t 3 1 0 t 2p 2 9 −18 18 10 −15 Answers to Selected Exercises 1123 1; 0, ± 25 y2 x2 1 4 points A Shown is the graph on the window 6 y 6. This is a hyperbola with foci at the 0, ± 2 1 9 x 9, and vertices at the points B 2 . 25. x 1 cos t 3, y 2 tan t 2 0 t 2p 1 2 9. −10 10 −10 10 Chapter 11 Review, page 772 1. This is an ellipse with foci at the points and vertices at the points A graph on the window 0, ± 225 9 x 9, B . 0, ± 2 2 1 Shown is the 6 y 6. 11. y2 25x 13. 15. y2 16x 17. Focus: 5 14b , 0, a directrix: y2 16x y 5 14 19. Ellipse, foci: (1, 6), (1, 0), vertices: (1, 7), 1, 1 1 2 9 −3 9 −9 3. This is the same graph as in Exercise 2; the equations are equivalent. 5. This is an ellipse with foci at the y2 16 1; 0, ± 2 x2 12 points 1 Shown is the graph on the window 6 y 6. and vertices at the points 2 0, ± 4 9
x 9, . 1 2 21. This is an ellipse with center at 0, 5 and 1 the points 1 3 25, 5 1 2 and 1 2 6, 5 3 25, 5 1 2 . 2 3, 5 2 , vertices at and foci at the points 1 2 3 4 5 6 −3 −4 −5 −6 −7 2 7. This is a hyperbola with foci at the points and vertices at the points graph on the window ± 3, 0 9 x 9, . 2 1 Shown is the 6 y 6. ± 5, 0 1 1124 Answers to Selected Exercises 23. This is a hyperbola with center at (0, 0), vertices at 6, 0 , foci at the points and asymptotes 252, 0 2 1 the points (6, 0) and 252, 0 A y ± 2 and 1 B 3 ˛x. 2 37. Hyperbola 39. Ellipse 41. 43. 45. 47. 9 x 9 and 6 y 6 15 x 10 and 10 y 20 6 x 6 and 4 y 4 23 x 1 2 2 23 u 1 2 2 u y v v 10 49. 45° 51. θ = π 2 −10 10 −10 25. This is a parabola with vertex at the point (3, 3), y 23 25 8 b 8 focus at the point and directrix 3, a , . 20 15 10 5 1 2 3 4 5 6 27. This is a parabola with vertex at the point 5 4 focus at the point and directrix , 1 1 x 3 4 b a , 1, 1 (2, 3π/4) (−3, −2π/3) θ = π θ = 0 53. θ = π 2 2π 3 θ = 0 , 2 . 5520 −10 1 −1 −2 −3 θ = 0 10 20 15 10 5 −5 −10 −15 −20 θ = − π 2 4, 6 2 29. Center: y2 4 1 31. 1 x 3 16 2 2 1 33. 35. 1 y 1 2b a y 1 2 1 x 3 2b 2 2 1 Answers to Selected Exercises 1125 57. θ = π 2 69. Ellipse −2 − ( 12, π 2 ) 59. 61. −1 −2 −3 −4 θ = π 2 −1 −0.5 0.2 −4 −4 −2 θ = 0 2 4 θ = π Pole θ = 0 3π 2 ) 4,( 3π 2 θ = 71. r 2 1 cos u 73. r 24 5 cos u 75. 77. 79. 35 x 32 and 2 y 16 15 x 15 and 10 y 10 y 2x2 2 y 2 x −1 1 1, 0 as t goes from 0 1 Then point retraces its path, moving from 2 Point moves from (1, 0) to p. to 1, 0 1 x 3 2y or y 1 to (1, 0) as t goes from 2 2 ˛x 3 x 5 cos t 3, y 5 sin t 5, 0 t 2p 2 p to 2p. 83. numbers 74 and 75 11 −1 9 −9 63. 2 ˛, 323 3 2 b a 67. Hyperbola 65. Eccentricity 2 3 B 0.8165 81. 85. θ = π Pole (−2, π) (4, 0) θ = 0 1126 Answers to Selected Exercises 87. x cos t 2, y 2 sin t 3, 0 t 2p 95. x 212 tan t 2, y 2 sec t 3, 0 t 2p −9 6 −6 9 −9 9 3 −9 Note that the diagonal lines shown here are not part of the graph but are just about where the asymptotes should be. t 4 x 32 2 5, y t, 10 t 10 1 2 97. 1.5 −25 5 −5 5 89. x 1 2 cos t, y 1 3 sin t, 0 t 2p −1.5 1 −1 91. x sec t, y 1 36 tan t, 0 t 2p 0.1 −2 2 −1.5 93. x 7 cos t 2, y 8 sin t 5, 0 t 2p 15 −5 15 −15 Chapter 11 can do calculus, page 777 1. approximately 30 2. approximately 20 3. approximately 8 4. approximately 4 5. approximately 10 6. approximately 5 Chapter 12 Section 12.1, page 788 1. Yes 3. Yes 5. No 7. 11. 15. 5 x 11 5 ˛, y 7 2 ˛, s 5 2 x 28, y 22 r 5 9. 13. 17. 7 x 2 7 ˛, y 11 2 ˛, y x 2, y 1 x 3c c 2d 2 19. Inconsistent 21. x b, 23. x b, y 3b 4 2 y 3b 2 4 25. Inconsistent ˛, where b is any real number ˛, where b is any real number 27. x 6, y 2 29. x 0.185, y 0.624 31. 33. x 3, y 1, z 4 x 66 5 ˛, y 18 5 Answers to Selected Exercises 1127 35. c 3, d 1 2 y 960x 2000; y 114x 14,000 37. a. Electric: solar: b. Electric: c. Costs same in fourteenth year; electric; solar $14,570 $6800; solar: 39. a. b. c. y 7.50x 5000 y 8.20x 130,000 ≈ (7143, 58564) 0 0 15,000 3. 50, 2, −3) y y Costs equal at approximately 7143 cases. d. The company should buy from the supplier any number of cases less than 7143 and produce their own beyond that quantity. 2 4 2 (3, 0, 0) 4 x 41. 140 adults, 60 children 43. $19,500 at 2% and $15,500 at 4% 45. 3 4 lb cashews and 2 1 4 lb peanuts 47. 60cc of 20% and 40cc of 45% 49. 80 bowls; 120 plates Section 12.1.A, page 794 1. z (1, 4, 5) 4 2 y 2 2 4 4 x 1128 Answers to Selected Exercises 7. z 4 2 (2, −3, −1) 2 4 x 9. y 2 4 x 6 intercepts: y 2 z 6 x 3 intercepts: y 9 4 z 3 2 19. intercepts: no y-intercept x 2 z 3 21. Each equation can be represented by a plane. Two planes are parallel, are coincident, or they intersect in a line. The system of equations either has no solution, an infinite number of solutions which lie on the plane, or an infinite number of solutions, all of which lie on a straight line. Section 12.2, page 801 x 3 intercepts: y 3 z 3 1 11. 13. 15. 17. x 4 intercepts: y 2 z 8 intercepts: no y-intercept x 4 z 12 3. 9. 5. 11. y 5, z 2, £ 2x 3y 1 4x 7y 2 4x 2y 4z 2w 1 4x 4y 4z 2w 3 4x 2y 5z 2w 2 x 3 2 x 2 t, any real number x 1, y 1, y 3 x 1 2 ˛, 4 ˛, x 14, y 6, 19. No solution 2t, z t, 2t, 21. 17. 15. 13. x t, w 0 z 4, w t, where t is where t is any real 25. No solutions x t 2, for any real number t y t 1, y 0, z 0 number x 1, y 2 z t, x 0, x 1, x 3, x 3 4 ˛, 23. 27. 29. 31. 33. 35. y 1, z 3, y 1, z 2, w 2 w 5. y 10 3 ˛, z 5 2 Answers to Selected Exercises 1129 37. 10 quarters; 28 dimes; 14 nickels 39. $3000 from her friend; $6000 from the bank; $1000 from the insurance company 41. $15,000 in the mutual fund; $30,000 in bonds; $25,000 in food franchise 43. Three possible solutions: 18 bedroom, 13 living room, 0 whole house 16 bedroom, 8 living room, 2 whole house 14 bedroom, 3 living room, 4 whole house 45. Tom: 8 hours; George: 24 hours; Mario: 12 hours. 47. 2000 chairs; 1600 chests; 2500 tables 49. 20 model A; 15 model B; 10 model C 1. 2 A B Section 12.3, page 811 11 4 7 A C 4 3 10 ≥ 23. 25. 27. 29. 31. To ≥ £ ≥ £ ≥ £ ≥ From: 0 1 0 1 1 1 £ 3.75 6.5 14.75 The total cost to bake and decorate each giant ≥ £ cookie is $3.75; sheet cake: $6.50; 3-tiered cake: $14.75. ; a11 $91,000 18200 5080 represents the total 91000 25400 amount of tuition the college got from lecture; a22 ¢ the college got from lab; are not meaningful in the context of the problem. represents the total amount of tuition and $5080 a12 a21 ≤ 33. a Ma Mil Min SL C Ma Mil Min SL . • A2 . • A A2 A3 12 11 11 11 11 A3 , 8 8 8 8 8 µ 11 7 10 10 7 11 10 7 7 10 • 11 10 7 7 10 8 4 8 8 4 11 7 10 10 7 • This matrix represents the total number of flights that are direct, have one layover, or have two layovers between each pair of cities. µ 5. 2C 10 14 2 8 4 2 £ 12 2 4; 3 3; 3 2; £ 7. AB defined, 9. AB defined, 11. AB defined, 13. 3 2 0 8 11 10 ¢ 17 17 3 33 19 5 19. § AB BA not defined ≥ BA defined, 2 2 BA not defined 1 3 2 1 6 5 15. £ ≥ ¥ ; BA 4 24 9 2 21. AB 1130 ¢ £ 8 2 2 ≤ 3 21 24 8 6 15 ; BA ¢ 19 10 ≤ 0 9 2 0 8 0 0 Answers to Selected Exercises ≥ £ ≥ 5910aans_1114-1147 9/21/05 2:03 PM Page 1131 Section 12.4, page 819 35. a. ; CI2 3 2 4 1 ; I2C 3 2 4 1 CI3 ¢ ; ≤¢ 2u 0 ≥ • 4u . I2 1 0 3. I3 ¢ 1 0 0 0 1 0 1 ≤ 0 I3C £ 5. 1 ≤ 1 1 9. 2x 1 £ 4x y 0 11. x 2y 2z 1 y 2z 0 3x 2y 2z 0 u 2v 2w 0 v 2w 1 3u 2v 2w 0 r 2s 2t 0 s 2t 0 3r 2s 2t 1 13. 2 3 2 1 1 2 ≥ 17. No inverse £ 15. No inverse 3 1 19. 2 4 1 1 10 8 5 1 2 6 4 3 7 The A matrix and X matrix are the same; no . The first system has an infinite number of ≤¢ ¢ ≤ ¢ x t, solutions of the form ¢ ≤ ≤ ≤¢ t 3 y 1 2 2 ≤ ¢ the ; second system has no solution. If a coefficient matrix does not have an inverse, then any system with those coefficients will have an infinite number of solutions or no solution. b 0, y x2 1 c 1; 37. 39. 41. 43. 45. a 1, a 0.5, x, y 2 0, 5 2 2, 1 2 4, 7 8, y 2 1 5, 1 1 2, 0 1 1, 3 2 1 2, 5 2 1 10, 4 1 a 1, b 1.5, y 0.5x2 1.5x 2. c 2; y ax 3 bx 2 cx d 5 d 1 8a 4b 2c d 7 64a 16b 4c d 3 512a 64b 8c d y ax 4 bx 3 cx 2 dx e 1 625a 125b 25c 5d e 0 16a 8b 4c 2d e 3 a b c d e 5 16a 8b 4c 2d e 4 10,000a 1000b 100c 10d e c 1 ; y e x 4e x 1 47. A 0, F, C 0, D 0, E 0, and F t, 2 b 4, B 1 12 where t is any real number. The equation is t 12 which reduces to xy t 0, xy 12; hyperbola l 10,128.2, 49. h 224.4, b 2339.7 Section 12.5, page 824 ≥ 1. x 3, 21. 23. 25. 27. x 3, y 1 x 1, y 0, z 3 x 8, y 16, z 5 x 0.5, y 2.1, z 6.7, w 2.8 £ 29. no solution 31. 33. x 2 t, y 3.5 2.5t, where t is any real number. x 1149 161 y 426 161 , , z 1124 161 , w 579 161 z 1 2t, 3. 5. 7. 9. 11. w t, x 1, y 1 or x , y 3 2 y 9 or y 3 2 , or or x 221, B x 7, x 0, x 2, x 4, or or x 3, x 6, or x 4, x 4 or x 221, y 2 Answers to Selected Exercises 1131 x 1.3163, or y 2.4814 or x 0.3634, y 0.9578 15. 13. x 1.6237, y 1.0826 or x 1.9493, x 1.4184, or x 0.9519, 19. No solutions 17. y 8.1891 x 2.8073, y 0.4412 y 0.5986 y 0.8145 5. 10 x 10; 10 y 10 21. x 13 2105 x 13 2105 8 8 y y ˛, ˛, 3 2105 8 3 2105 8 or 7. 10 x 10; 10 y 10 x 3.1434, or y 7.7230 or x 2.8120, or x 0.9324, or y 1.4873 x 0.0480, or 25. 23. y 2.2596 y 19.3201 x 2.1407, x 4.8093, y 7.7374 or y 11.7195 x 3.8371, y 7.7796 x 1.4873, y 0.0480 y 1.4873 x 0.0480, or x 1.4873, y 0.0480 r 5 29. center (0, 0); 31. center (7.5, 12.5); r 12.75 27. 33. (440.2, 38205.5) and (1893.1, 81794.5). 37. 35. Two possible boxes: one is 2 by 2 by 4 m and the other is approximately 3.123 by 3.123 by 1.640 m. 4 and 12 15 and 12 8 15 inches 43. 12 ft by 17 ft y 6x 9 39. 1.6 and 2.6 47. 45. 41. Section 12.5.A, page 832 10 x 10; 1. 10 y 10 9. 10 x 10; 10 y 20 11. 10 x 10; 10 y 10 3. 10 x 10; 10 y 10 13. 0 x 10; 0 y 10 corner points: 0, 0 , 1 2 1 0, 3 2 1132 Answers to Selected Exercises 15. 0 x 10; 0 y 10 Chapter 12 can do calculus, page 841 corner points: 0, 0 , 1 2 1 0, 4 2, 0 , 1 2 , 2 a 20 7 , 18 7 b 17. At (0, 6), the objective function has a value of 30. 19. At (0, 0), the objective function has a value of 0. 21. 24 roast beef sandwiches for a profit of $72 23. 3000 peach trees and 27,000 almond trees Chapter 12 Review Exercises, page 835 1. 5. 9. x 5, x 35, x t 1, 19 11. 37 and y 7 y 70, 3. x 0, 7. y 2 x 2, y 4, z 6 z 22 y 2t, for any real number t z t 13. (c) 15. 100 17. 2x 6y 16 2x 3y 7 19. 2x 3z 2 4x 3y 7z 1 8x 9y 10z 3 21. x 26 9 , y 11 9 ; consistent 23. no solution; inconsistent t 37 11 y 3 11 x 1 11 25. , t 10 11 z t, , for any real number t; consistent 2 3 4 1 9 ¢ 4 7 3 ≤ 29. Not defined 33 46 85 , , , £ z 21 34 x 4, y 3, z 2 ≤ ¢ x 1 y 14 85 85 y 5x 2 2x 1 x 3, y 9 or x 1, y 1 x 1 27, y 1 27 x 1 27, y 1 27 x 1.692, y 3.136 or or x 1.812, y 2.717 47. maximum is 150 at (5, 0); minimum is 20 at (0, 2) 49. minimum of 97.5 pounds at (85, 12.5) 27. 31. 35. 37. 39. 41. 43. 45 3x 1 x 2 x 1 2. 2 2 5. 1 25 x 2 5 2x 1 3 5 x 4 2x 2 1 3 25 x 3 3x 1 x 2 4 5 ˛x 1 x 2 2x 2 3 5 3 x 1 x 3 3. 4. 6. 7. Chapter 13 Section 13.1, page 851 1. The population is the entire student body; the sample is 50 students from each grade level. 3. The population is the total number of American families; the sample is 50 families in each of 10
counties in each of 5 states. 5. In #1 the data is qualitative. In #2 the data is qualitative. In #3 the data is quantitative and discrete. In #4 the data is quantitative and continuous. 7. 200 cartons. 9. 2500 families. 11. Exercise Aerobics Kick boxing Tai chi Stationary bike 13. Relative frequency 40% 16% 16% 28% 50% 40% 30% 20% 10% Aerobics Kick boxing Tai chi Stationary bike Answers to Selected Exercises 1133 Frequency 6 8 5 4 1 1 Relative frequency 24% 32% 20% 16% 4% 4% 15. Color red blue purple green yellow orange 17. 40% 30% 20% 10% red blue purple green yellow orange 19. symmetric 21. skewed left 23. uniform 25 27 10 0 0 Key: 2 Key: 5 0 0 29. The number 90 is an outlier since it is quite a distance from the rest of the data. 31. The shape of the fall semester data is basically symmetric; the summer semester data is skewed right. It may be that in the summer enrolled students live closer to campus. 33. 25 0 0 5 10 15 20 25 30 35 1134 Answers to Selected Exercises 35. Sample answer: the histogram is not as symmetric as the stem plot. The histogram more accurately shows the distribution of the data due to the smaller class interval of 5. Section 13.2, page 862 1. approximately 43.429 3. approximately 7.583 5. 42 9. mean: 12.2 median: 12 mode: 13 7. 5.15 11. 53 13. Grains 15. The median is larger than the mean. 17. The mean and the median are the same. 19. approximately 2.828 23. 21. approximately 7.071 s 5.249; s 5.447 s 13.176; s 13.518 25. 27. 19 31. 7 29. 44 33. 21 35. five-number summary: 8, 17, 18.5, 24, 27 37. five-number summary: 50, 62.5, 73.5, 83.5, 94 50 94 39. mean: 43.167 median: 42.5 41. standard deviation: 20.3 range: 80 interquartile range: 35 43. The sample standard deviation is a good measure of dispersion because the data set is relatively small. 3 represents 23 6 represents 56 8 27 5910aans_1114-1147 9/21/05 2:33 PM Page 1135 45. 90.4 47. median mean skewed right 49. The median more accurately describes the “typical” salary, since the outlier of 105,000 has a large effect on the mean. 51. Answers may vary. Sample: data 26,28,30,32,34 20,25,30,35,40 mean 30 30 standard deviation 3.162 7.906 53. The mean will increase by the value of k. 55. The mean will be multiplied by the constant k. 57. All data values must be the same. Section 13.3, page 872 1. {A, B, C, D} 5. Outcome Probability 7. 1 216 3. 0.8 black 1 2 red 1 3 white 1 6 Section 13.4, page 882 1. Outcome Probability red 0.50 blue 0.31 green 0.15 yellow 0.04 3. 0.0016 5. Outcome Probability 7. 0.0001 nun 0.1 gimel 0.45 hay 0.24 shin 0.21 Outcome 9. Answers may vary. Sample: 6 2 5 1 50 50 Probability 3 4 50 5 6 50 4 6 50 7 7 50 8 6 50 9 10 11 12 5 1 50 50 5 50 4 50 15. 0.0016 19. 2048 23. 362,880 27. 12,870 11. Answers may vary. Sample: 6.89 13. approximately 0.0027 17. 0.56; approximately 0.176 21. 3,268,760 25. approximately 0.04 29. approximately 3.108 10 4 31. 48,228,180; 1.03669 1051; 365! 365 n 1 ! 2 33. probability 1 365 Pn 365n n 3: approximately 0.008 n 20: approximately 0.411 n 35: approximately 0.814 35. 366 people 9. approximately 0.05 or 5% Section 13.4A, page 889 11. {0, 1, 2} 13. approximately 39% 15. Number of blue Probability 0 25 64 1 15 32 30 64 2 9 64 17. 12.52 19. 17.2 21. 0.41 23. 0.026 25. 0.5 0.4 0.3 0.2 0.1 1 2 3 4 5 27. 1 to 5 31. approximately 0.55 29. 1 2 33. The median appears to occur at 15 inches of rain. 35. 5 10 or 50% 3. 0.015 0.111 0.311 0.384 0.179 . 0.35 5.316 0.422 0.211 0.047 0.004 7. approximately 0.738 . approximately 0.0796 11. approximately 0.000023 13. expected value: 50; standard deviation: 5 15. approximately 0.160; approximately 0.185 Section 13.5, page 896 1. 20 3. approximately 0.68 Answers to Selected Exercises 1135 5910aans_1114-1147 9/21/05 2:34 PM Page 1136 5. 0.1 5. quantitative −40 60 −0.01 7. 0.1 0 −0.01 1000 9. the mean is 70. The standard deviation is 5.228. 11. 0.5 13. 0.68 15. 0.8385 17. 19. 560 S 0.6 z-value 450 S 0.5 z-value 640 S 1.4 z-value 530 S 0.3 z-value $0.95 S 4.29 z-value $1.00 S 3.57 z-value $1.35 S 1.43 z-value 23. 0.48 25. 0.16 21. 0.19 27. 0.815 31. 63.25; Q3 Q1 fall between 63.25 and 76.75. 29. 0.48 76.75; Fifty percent of the scores Chapter 13 Review, page 900 1. qualitative 3. 25 20 15 10 5 sparrow purple finch chickadee cardinal bluejay 1136 Answers to Selected Exercises 7. 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 Key 18 0 2 represents 18.2 9. The distribution is skewed left. 11. 31.2 13. 4.72 15. Q1 27.7; Q3 33 17. 18.2 39 19. Outcome Probability 1 0.1 21. 0.7 2 0.2 3 0.4 23. 0 to 3 4 0.2 5 0.1 25. Number of red Probability 0 0.512 1 0.384 2 0.096 3 0.008 27. 1.0 0.5 1 2 3 4 29. 0.4; 0.53; 0.07 31. 0.5 33. 0.86 35. approximately 0.0048 37. 3.08; 1.3 39. 0.1 40 −0.01 110 for h 4; area 1.856 41. 0.68 43. 99.85% 45. 0.55 Chapter 13 can do calculus, page 907 In Exercises 1–6, estimates may vary. 1. 1 3 0.14 2. 2 3. 4.5 4. 5. 0.51 5.67 7. Your answer should be close to 0.7301. A sample is shown of the program using the normal curve 1 1812p with equation a 100, x 138 648 y 1 e 6. ; 2 2 b 150, and h 0.025. h 5; area 2.02 3.7 x 5.7, 1 y 5 for (Scales The area estimates get larger for larger values of h. h 5 The value for is close to 2, which is the exact area under the curve from 0 to 1. 2 9. A sample is shown of the program using the equation of the standard normal curve with a 3, h 0.5. close to the expected area of 0.997, or 99.7% of the area under the curve. The estimate is very b 3, and (Scales 84 X 192, 0.01 Y 0.05 ) 8. Samples of the program are shown for h 3; area 1.74 (Scales 3 x 3, 0.5 y 1 2 Chapter 14 Section 14.1, page 916 1. x f(x) 2.9 2.99 2.999 0.25641 0.25063 0.25006 Answers to Selected Exercises 1137 x 3.001 3.01 3.1 lim xS3 3. f(x) 0.25 x f(x) 0.24994 0.24938 0.2439 0.1 0.01 0.001 0.35809 0.354 0.3536 x 0.001 0.01 0.1 0.35351 0.35311 0.34924 f(x) 0.3535 x f(x) 7.1 7.01 7.001 0.1662 0.1666 0.1667 lim xS0 5. x 6.999 6.99 6.9 0.1667 f(x) 0.1667 0.1667 0.1671 x f(x) 0.9 0.99 0.999 0.1149 0.1115 0.1111 lim xS7 7. x 1.001 1.01 1.1 0.1111 0.1107 0.1075 f(x) 0.1111 lim xS1 0.1 0.01 0.001 0.04996 0.005 0.0005 29. 31. lim xS3 f x 1 2 2; lim xS0 f x 1 2 does not exist; lim xS2 f x 1 2 0 lim xS3 f x 1 2 2; lim xS0 f x 1 2 1 ; lim xS2 f x 1 2 1 33. a. y 5 4 3 2 1 −2 −1 1 2 x b. c. lim xS2 lim xS1 f f 2 x 1 x 2 1 does not exist. 3 d. lim xS2 f x 1 2 does not exist. 35. No matter how close x gets to c, there are still an infinite number of both rational and irrational numbers between x and c, so t(x) will take the values 0 and 1 an infinite number of times, but never get close to a single number for all values of x that are very close to c. Section 14.2, page 923 1. 5 5. 0 13. 214 17. 1 2 3. Limit does not exist. 7. 3 5 9. 47 11. 7 15. 1 25 19. 0 23. The limit does not exist. Values of x less than are mapped by the function to more than 3 are mapped by the function to 1. 1 1 21. 222 3 and values of x 0.001 0.01 0.0005 0.005 0.1 0.05 25. 12 27. 1 222 29. As the angle t in standard position gets closer to 13. 1.25 21. 0.2887 1; lim xS0 15 1.5 23. 0 17. 0.333 25. 1 1; f x 1 2 lim xS2 f x 2 1 does not , the x-coordinate of the point at which the p 2 terminal side of t crosses the unit circle gets closer to 0. Since this x-coordinate is the cosine of the angle t, we have cos t 0. lim tS p 2 x f(x) x f(x) 0 9. lim xS0 11. 1.5 2 19. 27. f x 1 2 lim xS3 exist 1138 Answers to Selected Exercises 31. Using the function , when x is When x is close to 2 but slightly less 2. x 2 close to 2 but slightly more than 2, we have 3. x 3 x than 2 we have In this case, the difference is either 5 or 4 depending on whether x is more than 2 or less than 2 and so the limit doesn’t exist. 1 x 2, 3 4 33. 35. Many correct answers, including f 0 x 0x , , and c 0. In this case, lim xS0 f x 1 2 does not exist by Example 10; a similar argument shows that f g 2 by Exercise 22. does not exist. But g 1 0 lim xS0 x x lim xS0 1 21 2 Section 14.2A, page 927 1. 0 3. 0 5. 6 7. a. 0 b. 1 c. 1 d. 1 9. a. 1 b. 0 d. Limit does not exist. c. Limit does not exist. 11. a. 5 b. 0 c. 4 d. 2 13. 3 19. 2.5 15. 1 8 21. 2 17. Limit does not exist. 23. lim xS2 x 3 4 2 and lim xS2 3 x 4 1 Section 14.3, page 935 The symbol means “implies.” 1 1. Given e 7 0, let Then 0 6 . d e 3 x 3 0 3x 2 0 d e 6x 20 Then 6 e 0 6 0 x 5 0 Then 0 6 6 e 1 15 3x 9 f x 0 7 2 1 3. Given x 5 0 5. Given , 6 e 1 let f 0 1 0 e 7 0, let 0 x 2 0 6x 12 f x 0 15 . Given 0 e 7 0, 0 0 0 0 0 let be any positive number. Then d for every number x (including those satisfying let d e. x 6 Then . Then let 2x 5 2 0 1 22 . Given x 4 0 x f 1 2 11. Given e 7 0 2x , 2 1 13. Given 0 6 x2 0 0 15. Let x 0 0 6 e 1 e 7 0. such that if let 6 d 1 x2 0 d 2e Then 6 2e Then there exist positive numbers 6 d1, x c then L 0 6 x f 0 d1, d2 6 e 2 0 x c 0 6 d2, and if 0 6 Choose d then g x 0 0 0 to be the smaller of 2 1 d1, d2. 0 Then if 2 1 0 M 0 6 e 2 . we have both and 0 0 f x x c 6 d Then . rewritten as 0 Now using the triangle inequality we have x f 0 1 lim xSc 2 L M 2 0 L M. 0 6 e. Therefore. 2 0 x 1 1 1 22 1 2 2 1 1 e. This can be Section 14.4, page 946 x 3, x 6 3. Continuous at 1. x 0 x 2 and x 3; discontinuous at 5. Continuous at x 2 x 0 and x 3; discontinuous at 7. f lim 1 xS3 lim 5 1 xS3 9 5 x 2 x 2 1 3 2 lim xS3 7 x2 5 1 lim xS3 2 2 7 14 f 1 2 x 2 7 1 x2 2 2 lim xS3 1 5 2 lim xS3 lim xS3 1 x 2 x 2 1 22 7 3 1 2 x2 9 9. lim xS2 lim xS2 1 f 2 1 x lim xS2 1 lim xS2 1 x2 x 6 2 22 9 x2 x 6 x2 9 lim xS2 1 x2 6x 9 21 2 x2 6x 9 2 5 2 1 4 25 2 21 2 22 6 2 9 21 22 2 6 1 1 f 20 2 1 2 Answers to Selected Exercises 1139 f has a removable discontinuity at Chapter 14 Review, page 961 11. lim xS36 f x 1 2 lim xS36 x 6 lim xS36 1 36 6 30 30 2 B 2 2x x 6 x2x x 6 lim xS36 A lim xS36 1 x lim xS36 lim xS36 1 6 1 25 x 3. x 1. 1; , but f 0 f 2 2 216 900 0 36 1 2 1 2 13. f is not defined at 15. f is not defined at 17. lim xS0 f x 1 2 36136 36 6 36 6 21 2 hence lim xS0 f x 2 1 f 0 1 2 19. Continuous 23. Continuous 21. Continuous 25. Continuous at every real numbe
r except 27. Continuous at every real number except x 3 x 0 and . 29. 31. 33. 1 2 2x f has a removable discontinuity at g x 2 1 x 4. 35. If and 1 for all 1 x x 0, then x g 1 x 6 0. 2 1 Thus for all for all g lim xS0 x 2 2 does not exist. Hence the definition of continuity cannot be satisfied, no matter what g(0) is. 1 9. No vertical asymptotes; 0 x xSq f lim 1 2 xSq f lim x 1 2 q; 11. No vertical asymptotes; xSq f lim x 1 2 1 xSq f lim x 1 2 2; 13. Vertical asymptote x 10; q x 1 2 xSq f lim y 3 4 15. xSq f lim x 1 2 q; 17. y 2 19. No horizontal asymptote 21. 1 2 23. q 25. 5 27. 2 29. 3 22 31. 22 3 33. 23 35. 1 37. 1 39. 0 41. 0 1140 Answers to Selected Exercises 43. With a parachute: 20 ft/sec Without a parachute: 177.78 ft/sec 45. 1 47. 1 49. The first part of the informal definition is included in the second part, which says “the values of f(x) can be made arbitrarily close to L by taking large enough values of x.” This means that whenever you specify how close f(x) should be to L, we can tell you how large x must be to guarantee this. In other words, you specify how close you want f(x) to be to L by giving a positive number and we tell you how large x must be to guarantee that f(x) is within of L, that is, to guarantee that f 0 0 number k such that x 7 k. e, positive number e (depending on ) with this property: 6 e. 0 This can be reworded as follows: For every We do this by giving a positive there is a positive number k e 6 e. whenever x 7 k, then L L L 6 e If . 2 9. 1 2 3. 2 5. 5 7. 2 11. 4 13. 1 15. Given e 7 0, let Then 2x 6 0 7 f x 17. Continuous at 2x ; discontinuous at x 2 x lim xS2 2 1 x2 x 6 x2 9 2 lim 1 xS2 lim 1 xS2 2 22 2 6 22 9 4 5 f 2 4 5 2 b. f is not defined at 1 x 3 and hence is discontinuous there. 21. Vertical asymptotes at x 1 and graph moves upward as x approaches 1 the left and downward as x approaches the right. The graph moves downward as x approaches 2 from the left and upward as x approaches 2 from the right. x 2. 1 The from from 23. 1 2 25. 10 3 27. y 1 2 Chapter 14 can do calculus, page 967 1. a. upper estimate: 259 ft. lower estimate: 157 ft. Section 14.5, page 957 1. 1 3. 0 5. 0 7. 0 19. a. lim xS2 f x2 x 6 x2 9 b. 110 0 −5 77. Many possible examples, including but 6 3 22 4 2 26 23 64 8 8, 79. False for all nonzero a; for instance, 3 32 9 9, but 3 3 2 1 2 1 21 2 5 Section A.2, page 976 c. less than 5 ft: less than 1 ft: ¢t 6 0.04902 sec. ¢t 6 0.0098 sec. 2. Lower estimate: 21 Upper estimate: 25 For the lower estimate, count all of the complete squares beneath the curve. For the upper estimate, count all of the complete squares below the curve and estimate the number of partial squares below the curve. 3. a. 1.798 b. The degree of accuracy can be ¢t. increased by lowering the 5. 1.0 4. between 15 and 17 ft. 6. between 0.67 and 0.72 ft. Algebra Review Section A.1, page 972 1. 36 7. 125 64 13. 19. 25. 31. 37. 43. 81 16 x10 9x4y2 ab3 3xy x7 49. a12b8 55. a7c b6 3. 73 9. 1 3 15. 21. 27. 33. 39. 45. 51. 57. 211 216 0.03y9 21a6 1y3 8x 212 ce9 1 c10d6 1 c3d6 2 5. 5 11. 112 17. 23. 29. 35. 41. 47. 53. 59. 129 8 24x7 384w6 3 a8x 12 2 b2c2d6 1 108x 1 a 1 a 2 61. Negative 63. Negative 67. 3s 69. a6tb4t 65. Negative brsst c2rt 71. 73. Many possible examples, including 32 42 9 16 25, but 2 75. Many possible examples, including 1 3 4 2 72 49 32 23 9 8 72; 3 2 23 65 7776 but 1 2 2a2b 5u3 u 4 5xy x 15. 15y3 5y 12a2b 18ab2 6a3b2 2x2 2x 12 6x2 x 35 x2 16 y2 22y 121 16x6 8x3y4 y8 2y3 9y2 7y 3 x3 6x2 11x 6 61. 1 9 61y y 3ax2 3b 2a 1 63. 5 x 2b 2 ab ac bc x abc xmn 2xn 3xm 6 2 2 2 32; 2; 2 1 3 4 1 2 3y 6 2 2 3 then 2 x2 2xy y2 7 2 1 7y 21 2 2 2 23; 7 3 2 21 49xy then 7x 1 2 2 correct 1. 5. 9. 11. 17. 19. 23. 27. 31. 35. 39. 43. 47. 49. 53. 55. 65. 69. 73. 75. 77. 81. 3. 25. 29. 37. 33. 45. 41. 21. 8x x3 4x2 2x 3 7. 4z 12z2w 6z3w2 zw3 8 3x3 15x 8 13. 12a2x2 6a3xy 6a2xy 12z4 30z3 x2 x 2 y2 7y 12 3y3 9y2 4y 12 16a2 25b2 25x2 10bx b2 9x4 12x2y4 4y8 15w3 2w2 9w 18 24x3 4x2 4x 51. 3x3 5x2y 26xy2 8y3 6 59. 6 x ab x2 a2 b2 2 1 a b c 3 x 25 13x2 4x 13 abx2 x3 1 34rt 2x4n 5x3n 8x2n 18xn 5 y 4, 4 2 correct statement: x 2, correct statement: x 2, 3 then 1 y 2 3 1 y 3, x y 2 1 y 3, 67. 79. 57. 71. 1 2 83. Example: if 85. Example: if 87. Example: if 7 2 3; 89. Example: if statement: 91. Example: if correct statement: then y 2, y y y 3y x 4, then 42 5 4 6; x2 5x 6 4 3 1 correct statement: 21 4 2 2 x 3 1 x 2 2 21 93. If x is the chosen number, then adding 1 and squaring the result gives 1 from the original number x and squaring the result gives Subtracting the second of these Subtracting 1 x 1 2. 2 x 1 2. 1 2 Answers to Selected Exercises 1141 squares from the first yields: 1 x2 2x 1 x2 2x 1 1 1 the original number x now gives 2 2 2 4x. 4x x x 1 2 1 2 2 x 1 Dividing by 4. So the answer is always 4, no matter what number x is chosen. 95. Many correct answers Section A.3, page 981 21 21 9x 2 7 2z x 3 21 y 9 x 5 21 3x 1 21 1 x 8 9x 4u 3 x 5 x 2 2 x 5 2x y 21 x3 23 21 2u 3 21 x2 5x 25 2 3. 7. 11. 15. 19. 23. 27. 31. 35. 41. 45. 2 B x y 2 1 1 A 21 3y 5 2 15 x BA x y 3y 5 21 1 15 x x2 y2 21 z 1 z 3 21 x 3 2 x 9 21 2z 3 21 1 x 1 2 21 1 2x 5y x 2 2 z 4 2 5x 21 21 4 2x x2 x2 x 1 2 2 x2 2x 4 53. 2 21 4x2 2xy y2 x3 23 2 x 2 21 1 y2 2 21 z2 z 1 x2 y 21 a2 b 21 x 4 21 y2 5 z 1 21 x2 3y a 2b x2 8 21 x2 1 x 2 x2 2x 4 21 3 y 3 y 2 21 2 1 21 9 y2 21 z2 z 1 x z 21 1 21 59. 2 x y 65. If 1 2 x c x 18 A x d cd 1. and hence that 21 and c d 0 c d then that or equivalently, that number with this property, factor in this way. 2 d2 1. x 18 x 4 1 2 x cd, BA x2 But 1 cd B 1 c d 2 c d 0 implies d d2, d Since there is no real cannot possibly x2 1 2 1 Section A.4, page 985 3. 9. 195 8 1 x 15. 21. ce 3cd de x 3 x 4 2 2 1 5. x 2 x 1 11. 17. 23. 29 35 b2 c2 bc 2x 4 3x 4 x 1 9 7 a b a2 ab b2 121 42 1 x 1 2 x 1 x2 xy y2 x y x3 y3 1. 5. 9. 13. 17. 21. 25. 29. 33. 37. 39. 43. 47. 49. 51. 55. 57. 61. 63. 1. 7. 13. 19. 25. 1142 Answers to Selected Exercises 27. 6x5 38x4 84x3 71x2 14x 1 3 3 4x x 1 x 2 1 2 1 2 29. 2 31. 37. 43. 49. 55. 2 5y2 y 5 2 3 1 35 24 x2y2 x 2y x y 21 1 3y 3 y 2 2 3c 1 2 39. u 1 u 45. u2 v w 33. 3y x2 35. 41. 47. u v 21 1 2u v 21 1 x 3 2x 12x x 3 4u 3v 2 2u 3v 2 53. 59. y x xy xy x y 1 1 2 ; cd 51. 57 then 1 b a ab then 1 1 a b b 9, ; correct statement: 61. Example: if a 1, b 2, correct statement: a 4, 2 1 4 9 63. Example: if 1 14 19b 2 1 1a 1bb a a 1 a 21ab b 65. Example: if statement: 67. Example: if 24 29 A u 1, v 2, u2 v2 vu y 9, 1 24 29 u v u v x 4, B then 1 2 2 1 1; correct then 4 9; correct 2 statement: 1x 1y A B 1 1x 1y 1 Section A.5, page 991 1. 13; 1 a 2 , 1 b 3. 217; 3 2 a , 3 b , 3 2b 22 23 a 2 a b 2 b , (6, 36) 5. 26 216 1.05. 22 9. a. 40 M 30 M 20 M 10 M (0, 12) 2 0 (1992) 1 2 3 4 5 6 7 b. 40 M 30 M 20 M 10 M (3, 24) 0 (1992) 1 2 3 4 5 6 7 c. In 1995, about 24 million personal computers were sold. We must assume that sales increased steadily. 2 x 3 2 1 1 x2 y2 2 y 4 2 4 2 35. x2 y2 4x 2y 0 3, 4 and 2 1 k 7 d. 39. Assume 2, 1 37. 2 1 The other two vertices of one 1 possible square are those of another square are , d c k d those of a third square are 41. (0, 0), (6, 0) 43. 45. 3, 5 111 A x 6 , A B 3, 5 111 B 47. M has coordinates r 2b formula. Hence the distance from M to (0, 0) is 2 by the midpoint and the distance from M to (0, r) is the same: B a b s2 4 r2 as is the distance from M to 1 s 2b a 2 s 2b r2 4 s2 4 2 2 s 2b r2 4 s2 49. Place one vertex of the rectangle at the origin, 11. 13. 15. 17. 4 1 −2 −4 1169.25 2, 2 has length 2 18. and Since 1 12 this is a right triangle. 110; 1 other sides have lengths 2 2 1 2 145 2, to 3, 2 15 has length 145. 2 and Since this is a right triangle. 19. Center 21. Center 23. Center 1 1 1 radius radius 2 , 4, 3 3, 2 , 2 12.5, 5 2110 217 , radius 2 25. Hypotenuse from (1, 1) to other sides have lengths 110 12 2, B 2, 3 A 27. Hypotenuse from 18 2 2 B A B A A 150; 15 29. 31. 33 150 A 2 8 2 8 2 16 1 1 2 the be the coordinates 0, b with one side on the positive x-axis and another on the positive y-axis. Let a, 0 2 of the vertex on the x-axis and coordinates of the vertex on the y-axis. Then the fourth vertex has coordinates picture!). One diagonal has endpoints and 2 2a2 b2. b 0 2 diagonal has endpoints (0, 0) and has the same length: 2 2 so that its length is , 2 The other a, b a, 0 2 1 0 a 2 (draw a 0, b and hence a 2a2 b2. 0 a 2 1 51. The circle k, 0 to (0, 0)). So the radius family consists of every circle that is tangent to the y-axis and has center on the x-axis. 1 x k 2 y2 k2 1 (the distance from has center and k 53. The points are on opposite sides of the origin because one first coordinate is positive and one is negative. They are equidistant from the origin because the midpoint on the line segment joining them is , 0 . 2 1 Answers to Selected Exercises 1143 35 8 5 u 33. 4032 35. 160 1 Advanced Topics Section B.1, page 1000 1. 720 3. 220 5. 0 7. 64 9. 3,921,225 x5 5x4y 10x3y2 10x2y3 5xy4 y5 a5 5a4b 10a3b2 10a2b3 5ab4 b5 32x5 80x4y2 80x3y4 40x2y6 10xy8 y10 x3 6x21x 15x2 20x1x 15x 61x 1 1 10c 45c2 120c3 210c6 120c7 12 4x 8 6x x 8i 45c8 10c9 4 4 x4 10x3y2 210c4 252c5 23. 56 29. 27. c10 35c3d4 11. 13. 15. 17. 19. 21. 25. 31. 37. a. b. 9 1b a n 1b a 9; 9! 1!8! n n 1b a 39. 2n 1 1 1 2 n 1n 9 8b a 9! 8!1! 9 n! n 1 ! 2 1n1 1 1! 1 n 1b 2b a 1n2 12 n 3b a 1n3 13 p 11 1n1 1n n 0b a n 1b a n nb n n 1b a 6 cos2 u sin2 u 41. 43. a. n n 1b a n 2b n 3b p a a a cos4 u 4i cos3 u sin u 4i cos u sin3 u x h f f 2 1 x5 5 1b 1 2 1 5 x4h 2b a xh4 h5 x5 sin4 u x a a 5 4b 5 2b x h 2 h 1 a f b. 5 3b a f x 1 2 x h 2 x3h2 5 x5 5 3b a x2h3 x4h 5 1b a 5 4b a 5 1b a x4 5 2b a x3h 5 3b a x2h2 5 4b a xh3 h4 c. When h is very close to 0, so are the last four f f x terms in part b, so 1 2 1 x h 2 h x4 5x4. 5 1b a x h 2 h f 1 45. f x 1 2 1 x h 12 2 b a x10h 12 3 b a x9h2 12 4 b a 12 x12 2 h 12 1 b a x8h3 p x11 1144 Answers to Selected Exercises n r 3 1 21 47. a. x11 x2h9 a 12 11 b a xh10 h11 12 1 b when h is very close to 0 21 r 1 r 1 2 21 n r n r 12 10 b a 12x11, n r 1 n r 1 b. Since , n r ! 2 n 1b n! n r n! 1 r 1 1
3 1 a r rb n. For example, rows 2 and 3 of Pascal’s triangle 1b are 1 1 2 1 3 3 1 that is, 2 0b a 2 1b a 2 2b a a 3 1b a 3 2b a 3 3 0b 3b a The circled 3 is the sum of the two closest entries in the row above: 2 , 1b 2 3 0b a 1b a r 0. Similarly, in the general case, and n 2 verify that the two closest entries in the row But this just says that 1 2. which is part c with a n 1 r 1b a are n rb a and n r 1b a and use above part c. 1. Step 1: For n 1 1 21 1, to 2k Add n k: the statement is which is true. Step 2: Assume that the statement is true for that is, 1 2 22 23 p 2k1 2k 1. both sides, and rearrange terms: 1 2 22 23 p 2k1 2k 2k 1 2k 1 2 22 23 p 2k1 21 1 2 1 2k 1 1 2 22 23 p 2k1 21 1 2k1 1 But this last line says that the statement is true for n k 1. Therefore, by the Principle of Mathematical Induction the statement is true for every positive integer n. k1 k1 2 2 2 x3h2 x2h3 xh4 h5 Section B.2, page 1009 Note: Hereafter, in these answers, step 1 will be omitted if it is trivial (as in Exercise 1), and only the essential parts of step 2 will be given. n k: 3. Assume that the statement is true for 2 2 2 2k 1 k2. 1 to both sides: 2 1 3 5 p 1 k 1 Add 1 1 3 5 p 2k 1 1 k2 2k 1 k2 2 k 1 The first and last parts of this equation say that the statement is true for n k 1. 1 2. Assume that the statement is true for k 1 12 22 32 p k2 k k 1 Add to both sides: 2 12 22 32 p k2 k 1 1 1 2 n k: 2k 1 21 2k 1 2 21 6 k 1 k k 1 1 21 2 2k 1 6 2k 1 6 2k2 7k 6 k 1 2 2 2k 3 2 21 2 3 21 21 The first and last parts of this equation say that the statement is true for n k 1. 7. Assume that the statement is true for n k 21 k 1 1 k 2 . 2 to 1 k 1 1 4 2 Adding k 1 2 3 1 1 both sides yields 21 21 21 21 k2 2k 21 2 k 1 k 1 1 2 1 The first and last parts of this equation show that the statement is true for n k 1. n k: 9. Assume the statement is true for k 2 1 7 Adding 1 to both sides, we have: k 1 k 1, . 1 Therefore, the statement is true for n k 1. or equivalently. 11. Assume the statement is true for and hence that 3k1 3 3k. Multiplying both sides by 3 yields: or equivalently, 3k 3 know that 2 3k 3k 3 3k, Therefore, 3 3k 3k 3. with the fact that 3k1 3k 3, Therefore, the statement is true for 3k1 3 3k, or equivalently, n k: 3k 3k. 3 3k 3 3k, we k 1, Now since 2 3k 3. or equivalently, we see that 3k1 3 k 1 2 1 n k 1. . Combining this last inequality 13. Assume the statement is true for n k: Adding 3 to both sides yields: k 1 3k 7 k 1. 3k 3 7 k 1 3, 3. 7 3 2 1 2 certainly greater than k 1 k 1 that statement is true for k 1 7 3 1 2 1 1 3 or equivalently, k 1 1 1, 2 Therefore, the Since k 1 1 1. 2 n k 1. 2 is we conclude then 3 is a n k ; 15. Assume the statement is true for 22k1 1; that is, 22k1 1 3M for Now 22k1 3M 1. 1 k1 2 1 3M 1 4M 1 factor of some integer M. Thus, 22k21 22 4 1. 3 1 1 this equation we see that k1 1 1. 22 3 2 1 n k 1. Therefore, the statement is true for 2 From the first and last terms of 2 22 22k1 3 1 222k1 4M 3 Hence, 3 is a factor of 12M 4 4M 1 k1 22 1 1 2 2 . 1 2 1 1 17. Assume the statement is true for n k: 64 is a 32k2 8k 9 32k2 9 32 2k2 k1 2 Then 32k2 8k 9. factor of 8k 64N for some integer N so that k1 32k22 2 32 64N. Now 2 1 1 32 32k2 8k 9 64N Consequently, 9 . 2 1 2 8 k1 9 32 k 1 32 2 8k 17 k1 2 1 32 9 72K 81 9 64N 8k 17 64k 64 9 64N 64 8k 9 64N k 1 9N 8k 17 8k 8 9 From the first and last parts of this equation we 2 8 9. see that 64 is a factor of Therefore, the statement is true for k1 32 2 2 1 19. Assuming that the statement is true for c 2d 1 d p c k 1 2c c d c 2 Adding 3 c kd 2 . 4 3 1 2 to both sides, we . n k: d k 1 have c 1 c d 1 2 c 2d kd 2 d 2 4 c kd 2c 2c 2ck k 1 4 1 2 c kd d 2 2 k 1 d 2c 2kd 2 2 2 Answers to Selected Exercises 1145 2 1 k 1 2ck 2c kd 1 2 2c kd 2 2c kd 2 2c k 1 k 1 1 2 1 21 k 1 2 2kd 2c kd 21 . 21. a. Therefore, the statement is true for x y 21 x y 21 x y 21 xn yn x2yn3 x y ; 2 x2 xy y2 2 x3 x2y xy2 y3 xn1 xyn2 yn1 x2 y2 1 x3 y3 1 x4 y4 1 b. Conjecture: x y 21 ; 1 2 xn2y . 2 n 2, 3, 4, xn3y2 p Proof: The statement is true for by part a. Assume that the statement is true for n k: xk yk x y xk1 xk2y p xy k2 y k1 yxk yxk 0 to write 21 1 Now use the fact that xk1 yk1 as follows: xk1 yk1 xk1 yxk yxk yk1 xk1 yxk x y x y 2 xk y 1 xk y 1 1 xk yk 2 xk1 xk2 y x y 21 xk3y2 p xyk2 yk1 yxk yk1 xk2y2 xk3y3 p xyk1 yk xk1y xk 21 1 2 1 x y xk xk1y xk2y2 2 3 xk3y3 p xyk1 yk The first and last parts of this equation show that the conjecture is true for Therefore, by mathematical induction, the n 2. conjecture is true for every integer n k 1 12 1, 25. True: Proof: Since n 1. 23. False; counterexample: 1 1 the statement 2 1 Assume the statement is true for 2 7 k2 1. 2 k 1 1 k2 2k 2 2 7 is true for n k: k 1 2 1 1 k2 2k 1 1 The first and last terms of this inequality say that the statement is true for statement is true for every positive integer n. k 1 3 1 1 2 1 k2 2k 2 Therefore, by induction the Then 1 7 k2 1 2 k 1 n k 1. 2 2 1. k 1 4 2 2 2 1 27. False; counterexample: n 2 the statement is true for n k 2 5 4 7 5, 29. Since n 5. Assume the statement is true for k 5 (with shows that k 1 2 2 1 k 1 that 2 n k 1. for 4 7 k 1. Therefore, by the Extended : 2 2k 4 2 7 k 2, Adding 2 to both sides or equivalently, k 2 7 k 1 , we see So the statement is true 4 7 k 2. 2k 4 7 k. Since 2 1 1146 Answers to Selected Exercises 31. Since 33. Since n 2. n 5. the statement is true for k 2 Principle of Mathematical Induction, the statement is true for all 22 7 2, Assume that n k: for k2 1 7 k 1. inequality show that the statement is true for n k 1. is true for all 34 81 and that the statement is true k2 2k 1 7 Therefore, by induction, the statement 24 10 4 16 40 56, The first and last terms of this k2 7 k. n 2. k 1 Then 2 2 1 we So the statement is true n 4. and 34 7 24 10 4. Assume that see that for statement is true for Multiplying both sides by 3 yields: 2k 10k 3 3k 7 3 , 2 1 3k1 7 3 2k 30k. or equivalently, But n k: k 4 and that the 3k 7 2k 10k. 3 2k 30k 7 2 2k 30k 2k1 30k. Now we shall show 3k1 7 2k1 30k. Since . 20k 7 80 7 10. k 4, Therefore, 30k 7 10 that 20k 20 4, both sides of equivalently, k 1 2 1 so that 20k 7 10 30k 7 10 yields: k 1 . 1 3k1 7 2k1 30k 7 2k1 10 2 we have Adding 30k 7 10k 10, Consequently, k 1 . 1 2 to 10k or The first and last terms of this inequality show that the statement is true for the statement is true for all n k 1. n 4 Therefore, 35. a. 3 (that is, n 3; for 22 1 for 2 15 (that is, n 2; 24 1 b. Conjecture: The smallest possible number of by induction. 23 1 7 (that is, n 4. for 2 2 2n 1. n 1 or and that we Assume it is true for k 1 n k rings to move. In order to move the moves for n rings is Proof: This conjecture is easily seen to be true for n 2. have bottom ring from the first peg to another peg (say, the second one), it is first necessary to move the top k rings off the first peg and leave the second peg vacant at the end (the second peg will have to be used during this moving process). If this is to be done according to the rules, we will end up with the top k rings on the third peg in the same order they were on the first peg. According to the induction assumption, the least possible number of moves needed to do this is one move to transfer the bottom ring [the k 1 st] from the first to the second peg. 1 Finally, the top k rings now on the third peg must be moved to the second peg. Once again by the induction hypothesis, the least number of moves for doing this is smallest total number of moves needed to k 1 transfer all rings from the first to the second peg is 1 1 2k 2k 1 2k 1 2 2 2k 1 2k 1 2 1 2k1 1. It now takes 2k 1. 2k 1. Hence, Therefore, the 2 1 2 the conjecture is true for by induction it is true for all positive integers n. Therefore, n k 1. 37. De Moivre’s Theorem: For any complex number and any positive integer n, . Proof: The theorem is Assume that the zk that is, 3 2 1 cos cos u i sin u nu z r 2 1 i sin zn rn nu 2 4 1 n 1. obviously true when n k, theorem is true for rk ku ku . 1 zk1 z zk r cos u i sin u i sin cos rk 2 4 1 2 3 Then cos 2 4 1 3 3 1 According to the multiplication rule for complex numbers in polar form (multiply the moduli and add the arguments) we have: cos 3 cos u ku k 1 zk1 r rk rk1 u ku 2 4 k 1 This statement says the theorem is true for n k 1. true for every positive integer n. i sin 1 i sin Therefore, by induction, the theorem is u 2 . 4 6 ku 1 2 i sin ku 1 2 4 2 . Answers to Selected Exercises 1147 Index angles (continued) applications (continued) direction, 662–664 double-angle identities, 593–595, 602–603, 611 of elevation and depression, 425–429 with the horizontal, 620–621 identity, 620 inclination, 589–592 intersecting lines, 590–592 inverse trigonometric functions as, 530 negative angle identities, 459–460, 464 parts of, 413, 433–434 radian measure, 436–439, 444–446 reference, 448–451 rotation, 730–731, 771 solving triangles, 421, 422 special, 418–419, 437, 462 standard position, 434 between vectors, 671–673, 683 angle-side-angle (ASA) information, 631–632 Angle Theorem, 672 angular speed, 439–440 applications box construction, 103–104, 323 break-even point, 824 composition of functions, 195–196 compound interest, 345–349, 382, 402 distance, 101–102 exponential equations, 345–352, 379–384 food webs, 810–811 gravity, 677–678 guidelines, 97 height and elevation, 425–429 interest, 100–101 ladder safety, 426 linear programming, 831–832 LORAN, 724–725 lottery probabilities, 870, 881–882, 886–887 matrices, 795–801 mixtures, 104 multiple choice exams, 887–888 optimization, 322–324, 468–471 parabolas, 711–714 population growth, 340–342, 349–350, 382–384, 389–392, 395 radiocarbon dating, 352, 381–382 response times, 895–896 rotating wheels, 550–551 sequences, 13–19, 26–29 solutions in context, 98–100 sound waves, 558–562 spring motion, 551–553 trigonometry, 421–429 vector, 661–667 width of walkway, 102–103 approaching infinity, 201, 303–305, 914, 948–957 arccosine function, 532–534, 538, 539, 541 Archimedian spiral, 742 arc length, 434–435, 437–439, 776–777 arc
sine function, 529–534, 539 arctangent function, 534–536, 540 area circular puddle, 195 under a curve, 904–906 maximum, 138–139, 169–170, 468–471 quarter-circle, 905–906 Riemann sums, 964–967 triangle, 632–633, 682 z-values, 895–896 A absolute value complex numbers, 638–639 definitions, 107–108, 134 deviations, 858 distance and, 108, 128 equations, 109–111 extraneous solutions, 110–111 functions, 156, 173–174 inequalities, 127–132 properties, 108–109 square roots, 109 acoustics, 696 addition identities, 582–585, 593–600, 604, 610 matrix, 804–805 vector, 657–659, 662, 683 adjacency matrix, 809–811 adjacent sides, 415 algebraic expressions, 973–977 amplitude, 494–497, 502–505, 516, 563 amplitude modulation (AM), 472, 499 analytic geometry, definition, 691 angle-angle-side (AAS) information, 626 Angle of Inclination Theorem, 589–590 angles arc length, 434–435, 438–439, 776–777 argument, 639 central, 434 coterminal, 434, 436, 450 degree measure, 437–438 1148 Index arguments, 639 arithmetic progression. see arithmetic sequences arithmetic sequences definition, 21, 66 explicit forms, 22–24, 34, 66 finding terms, 24 graphs, 22, 33–34 lines and, 34 partial sums, 26–29 recursive forms, 22, 24, 66 summation notation, 25–26 arithmetic series, 25–29 asymptotes horizontal, 284–285, 288, 951–953 hyperbolas, 701–706, 720 oblique, 285–286 parabolic, 286 slant, 285–286, 289 vertical, 280–281, 288–289, 479, 486, 950 augmented matrices, 795–798 auxiliary rectangles, 703 average rates of change, 214–220, 226 averages, 853–854 axes (singular: axis) definition, 5 ellipses, 692, 694–695 hyperbolas, 701–702 imaginary, 638 parabolas, 709, 711 polar, 734 rotation of, 728–732 three-dimensional, 790–791 axis of symmetry, 163, 711 B bacteria growth, 350, 382–383 bar graphs, 845 beats, sound, 561 Bernoulli experiments, 884–889 Big-Little Concept, 281, 954 bimodal data, 855 binomial distributions, 887–888, 898 binomial expansion, 997–1000 binomial experiments, 842, 884–888 Binomial Theorem, 994–1000 boiling points, 335 bounds, 254–256 bounds test, 256 box plots, 861–862 box volume, 99–100, 323 break-even point, 824 C Calculator Explorations, 8, 16, 27, 201, 299, 409, 411, 457, 857, 861, 862, 877, 911, 994, 995, 1025 calculators absolute value, 110, 638 area under the normal curve, 895 complex numbers, 297, 299, 302–303, 638 composite functions, 193 conic sections, 721 continuity, 945 discontinuity, 115, 910, 937 dot mode, 157 ellipses, 695 factorials, 520 function graphers, 34–35 function notation, 143 geometric sequences, 60 graphical root finder, 84–85, 122 graphic solutions, 84 greatest integer function, 147 histograms, 849–850 holes in graphs, 115, 910, 937 horizontal shifts, 175–176 hyperbolas, 703–704 inequality symbols, 156 infinite geometric series, 76–78 instantaneous rates of change, 237 intersection finders, 525 inverse sine function, 531 iterations of functions, 200 limits of functions, 913 linear regression, 47–51 logarithms, 357, 375–376, 393 matrix operations, 299, 798–801, 808–809 numerical derivatives, 237 orbits, 302–303 parabolas, 712 parametric mode, 159, 756, 760–761 periodic graphs, 493 piecewise-functions, 157 points of inflection, 267 polar form, 641 polynomial regression, 274–275 probabilities, 875–876, 887 radian mode, 445, 480 radical equations, 113 rational exponents, 330 calculators (continued) rational functions, 283–285 rational zeros, 252 reflections, 177 sequences and sums, 16, 26–27 shading on graphs, 827–828 sine functions, 526 sound wave data collection, 559–561 statistics, 857 stretches and compressions, 177 sum and difference functions, 191 systems of equations, 780 tables of values, 910 trigonometric ratios, 416–417, 421–423, 426, 454, 473–479 tuning programs, 559 vertical shifts, 174 viewing windows, 512 calculus, 76, 138. see also Can Do Calculus Can Do Calculus approximating functions with infinite series, 520–521 arc length of a polar graph, 776–777 area under a curve, 904–906 Euler’s formula, 688–689 infinite geometric series, 76–79 instantaneous rates of change, 234–237, 614–615 limits of trigonometric functions, 566–568 maximum area of a triangle, 138–139 optimization applications, 322–324, 468–471 partial fractions, 838–841 Riemann sums, 964–967 tangents to exponential functions, 408–411 carbon-14 dating, 352, 381–382 cardioid graphs, 742 Cartesian coordinate systems, 5, 739 Cassegrain telescopes, 705 center, statistical, 857 chances, 866. see also probability change in x or y, 31 change-of-base formula, 374–375, 402 chord frequency, 561–562 chords, 18 circle equation, 989 circle graphs, 742 circles, parameterization of, 766–767 Index 1149 circles, unit, 445–446, 448–449, 474, conic sections (continued) cosines (continued) 475, 478, 650–651 circumference, ellipses, 699 closed intervals, 118 coefficient of determination, 47 coefficients, 239 cofunction identities, 585–587, 611 combinations, in counting, 880–882, 899 common differences, 22–24, 34 common logarithmic functions, 356–357, 364 common ratios, 58–61, 66, 77 complements, event, 866–867 completing the square, 90–92, 169 complex numbers absolute values, 638–639 arithmetic of, 295–296 definition, 294 equal, 294 Euler’s formula, 688–689 factorization, 308, 310–313 imaginary powers of, 688–689 Mandelbrot set, 304–306 nth roots, 645–651 orbits, 301–304 polar form, 639–640 polar multiplication and division, 640–642 polynomial coefficients, 307 powers of, 644–645, 688–689 properties, 293–294 quotients of, 296 real and imaginary parts, 300 roots of unity, 648–651 square roots, 297 zeros, 308–310, 317 complex number system, 293–294 complex plane, 301, 637–642, 650 components, vector, 655, 676–677 composite functions, 193–195, 211–212 compound inequalities, 118–120 compound interest, 345–349, 360–361, 382, 402 compressions, 177–180 concavity, 154 conic sections definitions, 691, 747 degenerate, 691 discriminants, 723–724 eccentricity, 745–748 ellipses, 692–698, 716–722, 745–747, 767, 771–772 horizontal and vertical shifts, 716–717 hyperbolas, 700–706, 721–725, 745–750, 771–772 1150 Index identifying, 717–719, 722–723 nonstandard equations, 721–722 parabolas, 163, 709–714, 719, 756–760, 771–772 parameterizations, 766–769 polar equations, 745–752, 772 rotations, 722–724, 728–732 standard equations, 720 conjugate pairs, 296 conjugates, complex, 296, 309 conjugate solutions, 299 conjugate zeros, 309–313 Conjugate Zero Theorem, 309 consistent systems, 781 constant functions, 152, 173, 192, 953 constant polynomials, 240 constants, 239 constraints, 829 continuity analytic description, 937–939 calculators and, 937 composite functions, 944 definition, 939 at endpoints, 941–942 informal definition, 936–937 on intervals, 940–942 from the left and right, 941 at a point, 938–940 polynomial equations, 261–262 properties of continuous functions, 942–943 removable discontinuities, 947 of special functions, 940 continuous compounding, 347–349 convergence, 77, 200, 203, 520–521 coordinate planes, 5, 790 coordinates, 5 coordinate systems comparison of, 792 conversions, 737–738 polar, 734–743, 776–777 rectangular, 5, 736–739 three-dimensional, 790–793 corner points, 829, 831–832 correlation coefficients, 47, 52, 66 cosecants, 416–419, 444–446, 485–487, 490 cosines addition and subtraction identities, 610 amplitude, 493–498, 516 basic equations, 538–541 coterminal angles, 451 damping, 512–514 definition, 416, 444–445 domain and range, 447, 477–478, 483 double-angle identities, 593–595, 602–603, 611 exact values, 448–451, 536 graphs of, 475–478, 497–498 half-angle identities, 596–597, 611 inverse function, 532–534, 539–541, 563 law of, 617–622, 682 oscillating behavior, 568 periodicity, 456–458, 493–497, 516 phase shifts, 501–505, 516, 549 power-reducing identities, 595–596 product-to-sum identities, 599 property summary, 483 restricted, 532–533 roots of unity, 648–651, 682 special angles, 418–419, 462 sum-to-product identities, 599–600 transformations, 481–482, 503–505 trigonometric identities, 454–460, 463 unit circle, 445–446 cotangents, 416–419, 444–446, 489–490 coterminal angles, 434, 436–437, 450–451 counting numbers, 3 counting techniques, 879–882, 899 Critical Thinking, 20, 21, 64, 132, 172, 214, 250, 259, 273, 300, 344, 363, 388, 420–421, 433, 453, 492, 529, 538, 547, 624, 637, 643–644, 652, 700, 708, 727, 733, 744, 765–766, 794, 820, 852, 864, 884, 935, 947, 959 crystal lattices, 802 cube root of one, 299 cubic functions, 173, 240 cubic models, 396 cubic regression, 274–276 curve fitting, 818. see also models; regression cycles, 492, 559 cycloids, 761–763 cylinder surface area, 149, 324 D damping, 512–514 data. see also statistics comparing, 893–894 definition, 843 displays of, 844–850 data (continued) division (continued) end behavior, 262–264, 284–287, distribution shapes, 846–847 outliers, 847, 862 qualitative, 843, 845–846 quantitative, 843, 846–850 standardized, 893 types of, 843 variability, 857 z-values, 894–896 decomposition, partial fraction, 838–841 definite integrals, 967 degenerate conic sections, 691 degree measure, 94, 436–437, 462, 528 degree of a polynomial, 240–242, 260–261, 263, 313 DeMoivre’s Theorem, 644–645, 682 denominators, partial fractions, polynomial, 240–245 remainders and factors, 243–245 synthetic, 241–242 domains convention, 145–146 exponential and logarithmic functions, 359–360, 375–376 functions, 142, 145 inverse functions, 533 rational functions, 279 of relations, 6–7 restricting, 210–211, 529–530, 532, 534–535 sequences, 14–15 sum, difference, product, quotient functions, 192 trigonometric functions, 447, 477, 480, 483, 486–488, 490 838–841 dot products density functions, 871–872, 891, angles between vectors, 898 depression, angles of, 425–429 derivatives, numerical, 237 deviations, data, 857–859 difference functions, 191, 943 difference of cubes, 298 difference quotients, 143–144, 671–673 gravity, 677–678 projections and components, 674–677 properties, 670–671 double-angle identities, 593–595, 602–603, 611 219–220, 234–235,
408, 584, 922–923 dreidels, 882 dynamical systems, 199 differential calculus, 76 directed networks, 809–811 direction angles, 662–664 directrix, 709–711, 720 discriminants, 93–94, 172, 723–724 distance absolute value and, 108, 128 applications, 101–102, 113–114 average rates of change, 214–220 between two moving objects, 619 formula, 987–988 from velocities, 964–967 distance difference, 700 distributions binomial, 887–888, 898 definition, 846 mean, median, and mode, 856–857 normal, 889–896 probability, 865–866 shapes, 846–847, 857 divergence, 77 division algorithm, 243 checking, 242–243 polar, of complex numbers, 640–642 E e, 341, 347–349. see also exponential functions; logarithmic functions eccentricity, 745–748 effective rate of interest, 354 elementary row operations, 795–796 elevation, angles of, 425–429 eliminating the parameter, 757–759 elimination method, 783–786, 797–798, 821 ellipses applications, 696–698 characteristics, 694 circumference, 699 definitions, 692, 745–747 eccentricity, 745–747 equations, 692–694, 696, 720–722, 771–772 graphing, 695 parameterization, 767 polar equations, 749 translations, 716–718 empirical rule, 892–893, 899 289, 316, 954–955 endpoints, 118 Engelsohn’s equations, 685 equations. see also quadratic equations; systems of equations; trigonometric equations absolute value, 109–111 applications of, 97–104 basic, 524–528, 538–542 conditional, 523 conic sections, 720 degrees, 94 ellipses, 692–694, 696, 720–722, 770–771 Engelsohn’s, 685 exponential, 379–384 fractional, 114–115 functions, 144–145 graphical solutions, 81–86, 524–528 hyperbolas, 701–702, 704, 720, 771–772 linear, 33–37, 39 logarithmic, 379, 384–386 matrix, 814, 817–818 normal curve, 891, 899 number relations, 97–98 parabolas, 709–710, 712, 720 parametric, 157–159, 755–757, 767–769 polar, 745–752, 772 polynomial, 94–95, 260–262 radical, 111–113 rotation, 728–730 second-degree, 722–724 solutions in context, 98–100 tangent lines, 236–237 translated conics, 718 equivalent inequalities, 119 equivalent statements, 81, 134 equivalent systems, 795 equivalent vectors, 653–655 Euler’s formula, 688–689 even degree, 261, 263, 313 even functions, 188, 482–483, 489–490 events, definition, 865. see also probability eventually fixed points, 202–203 eventually periodic points, 202 expected values, 869–870 experiments binomial, 842, 884–888 definition, 864–865 probability estimates from, 874–877 Index 1151 exponential decay, 350–352, 402 exponential equations, 379–384 exponential functions. see also logarithmic functions applications, 345–352 bases, 336–337, 371, 380, 402 bases other than e, 410–411 common logarithms and, 356–357 compound interest, 345–347, 382 graphs, 59, 336–342, 375 growth and decay, 339–342, 349–352, 402 horizontal stretches, 338–339 natural, 341 natural logarithms and, 358–359 Power Law, 366–367 Product Law, 365 solving, 379–381 tangent lines to, 408–410 translations, 338 trigonometric functions, 454 vertical stretches, 339 exponential growth, 349–350, 402 exponential models, 389–390, 392–394 exponents complex numbers, 295–296 fractional, 294 irrational, 333 laws of, 330–331, 402, 969–973 rational, 329–331, 333, 402 extraneous solutions, 110–111 extrema, 153, 170, 266, 468–469, 607–608, 829–830 F factorials, 520–521, 880 factoring common factors, 978 complex numbers, 308–313 cubic factoring patterns, 979 partial fractions, 839–841 polynomials, 243–245, 253–254, 310–313 quadratic equations, 88–89, 978 trigonometric equations, 542–544 Fibonacci sequence, 20–21 filtering, 350–351 finance, exponential change, 339–340 finance, interest applications, 100–101, 339–340, 345–349, 354, 360–361, 382 finite differences, 43–44, 47–48 first octant, 790 five-number summary, 861–862 fixed points, 202–203 flagpole height, 427 focal axis, 701 foci (singular: focus) ellipses, 692, 694, 696–697, 720 hyperbolas, 700–702, 720 parabolas, 709, 711, 720 food webs, 810–811 force, gravitational, 677–678 force, resultant, 664–667 fractional expressions, 114–115 fractions, partial, 838–841. see also rational functions free-fall, 958–959 frequency, definition, 844 frequency, wave, 558–562 frequency tables, 844–845 functions. see also logarithmic functions; polynomial functions; trigonometric functions absolute-value, 156, 173 composite, 193–195, 211–212, 944 concavity and inflection points, 154, 266 constant, 152, 173, 192, 953 continuity, 261–262, 936–945 cube root, 173 cubic, 173, 240 defined by graphs, 150–152 definition, 142–143 density, 871–872, 891, 900 difference quotients, 143–144, 219–220, 234–235, 408–410, 584, 922–923 domain, 142, 145 equations, 144 evaluating, 10, 143 even and odd, 188–189, 482–483, 489–490 exponential, 336–342 fractional, 114–115 graphs of, 8–9 greatest integer, 147, 157, 173 horizontal line test, 208–209, Factor Theorem, 245–246, 226 functions (continued) as infinite series, 520–521 input and output, 141–142, 204 instantaneous rates of change, 234–237 inverse, 204–212, 529–536, 539–540, 563 iterations of, 199–200 limits of, 566–568, 909–915 linear, 34–36 local maxima and minima, 153, 266, 468–471, 607–608, 829–831 notation, 9–10, 143 numerical representation, 7–8 objective, 829–831 odd, 188–189, 482–483, 489–490 one-to-one, 208–211 orbits, 200–203, 301–303 parametric, 157–159 parent, 172–173 piecewise-defined, 146–147, 155 polynomial, 145, 165–168, 188–189, 239–248 probability density, 871–872, 891, 898 product and quotient, 192 profit, 146 quartic, 240 range, 142 rates of change, 214–220 rational, 278–289, 954–957 reciprocal, 173 sinusoidal, 510–511, 521, 547–555 square root, 173 step, 157 sums and differences, 191 symmetry, 186–189 transformations, 172–181 trigonometric ratios, 443–444 vertical line test, 151–152, 225 zeros, 240, 245–248, 250–257, 265, 308–313, 316–317 Fundamental Counting Principle, 879–882, 899 Fundamental Theorem of Algebra, 307–313 Fundamental Theorem of Linear Programming, 829–831 G Gateway Arch (St. Louis, MO), 342 Gauss-Jordan elimination, 797–798, 252–253 feasible regions, 829–831 Ferris wheels, 522, 555 1152 Index identity, 173, 918 increasing and decreasing, 152–154 817 geometric sequences applications, 62–63 geometric sequences (continued) graphs (continued) definition, 58, 66 examples of, 58 explicit form, 59–61, 66 infinite, 76–79 partial sums, 61–63 recursive form, 59, 66 geometric series, 76–79, 520–521 graphical root finder, 84–85 graphical testing, 572–573 graphing calculators. see calculators Graphing Explorations, 83, 122, 156, 157, 158, 174, 175, 176, 177, 181, 185, 186, 187, 207, 260, 261, 262, 266, 274, 337, 342, 365, 366, 446, 475, 482, 486, 487, 494, 496, 510, 513, 514, 562, 572, 581, 582, 651, 741, 751, 756, 761, 784, 891, 910, 913, 950, 1023, 1024 graphs absolute-value functions, 156 amplitude, 493–498, 516 arithmetic sequences, 22 bar, 845 binomial distributions, 888 box plots, 861–862 concavity, 154 continuity, 261–262 cosecant function, 486–487 cosine function, 475–478, 493–494, 501 cotangent function, 488–489 damped and compressed, 512–514 defining functions, 150–152 definition, 30 ellipses, 695–696 end behavior, 262–264, 289 of equations, 30, 173 equation solutions, 81–86, 524–528 exponential functions, 336–341 finding slopes, 32 function values from, 8–9 geometric sequences, 59 greatest integer functions, 157 histograms, 849–850 holes, 282–283, 289 horizontal asymptotes, 284, 951–953 horizontal shifts, 175–176 horizontal stretches, 338–339 hyperbolas, 702–704 identifying, 505 identities, 506–507 increasing and decreasing functions, 152–154 inequalities, 121–123 inflection points, 154, 266 inverse cosine function, 533 inverse relations, 205–207 inverse sine function, 530 inverse tangent function, 535 limits of functions, 566–568, 950–953 of lines, 34–37 local maxima and minima, 153–154, 266 logarithmic functions, 359–361, 375–376 maximum area of a triangle, 139 multiplicity, 265, 283 nth roots, 328 one-to-one, 208–209 open circles, 150 oscillating behavior, 514–515, 568 parabolas, 286, 711–712 parametric, 157–159, 206–207, 755–757 parent functions, 172–173 periodicity, 487, 489, 493–497 phase shifts, 501–505 piecewise-defined functions, 155 plane curves, 754–755 polar, 739–743 polynomial functions, 260–268 quadratic functions, 163–167 rational functions, 279–289 reflections, 176–177, 206, 226 roots of unity, 651, 681 rotated conics, 731–732 scatter plots, 5 secant function, 487–488 sequences, 14–15 sine function, 473–475, 494–495 sinusoidal, 511–512 slant asymptotes, 285–286 stem plots, 847–848 stretches and compressions, 177–181 symmetry, 184–189, 482–483 systems of equations, 780–788, 822–824 tangent function, 478–481 in three dimensions, 790–794 translations, 338, 716–717 in two dimensions, 792 vertical asymptotes, 282–283, 288–289, 950 vertical shifts, 174 vertical stretches, 339 viewing windows, 512 gravitational acceleration, 293 greatest integer functions, 147, 157, 173 H half-angle identities, 596–598, 604, 611 half-life, 351–352 half-open intervals, 118 Halley’s Comet, 700, 754 headings, 432 height, from trigonometry, 425–429 Heron’s formula, 633, 682 Hertz, 559 histograms, 849–850 holes, 282–283, 289, 910, 937–939 horizontal asymptotes, 284, 288, 951–953 horizontal lines, 37, 152, 792 horizontal line test, 209–210, 226, 530 horizontal shifts, 175–176 horizontal stretches and compressions, 178–180, 338–339 Hubble Space Telescope, 705 Huygens, Christiaan, 761 hyperbolas applications, 705–706, 724–725 characteristics, 702 definitions, 700–701, 747 eccentricity, 745–748 equations, 701–702, 704, 720, 771–772 graphing, 702–704, 721–724 parametric equations, 768 polar equations, 749 hypotenuses, 415 I identities. see trigonometric identities identity functions, 173, 918 identity matrices, 815 imaginary axis, 638 imaginary numbers. see complex numbers inconsistent systems, 781, 784–785 independent events, 867–868 index of refraction, 545 indirect measurement, 427–429 inequalities absolute value, 127–131 algebraic methods, 128 applications, 123–124 Index 1153
inequalities (continued) compound, 118–120 equivalent, 119 interval notation, 118–119 linear, 119–120, 827 multiplying by negative numbers, 119 nonlinear, 121–122 quadratic and factorable, 122–123, 129–130 solving, 119–123 systems of, 826–832 test-point method, 826 infinite geometric series, 76–79, 520–521 infinite sequences, 13 infinite series, 520–521 infinity concept, 119, 201, 948 horizontal asymptotes, 951–953 limits approaching, 201, 303–305, 914, 948–953 negative, 948, 951–953, 956 properties of limits, 953–957 vertical asymptotes and, 950 inflection points, 154, 266, 317 initial point (vectors), 653 initial sides, of rays, 433 input, 7–9 instantaneous rates of change, 234–237 integers, 3–4 integral calculus, 76 integrals, definite, 967 intensity, 562 interest applications, 100–101, 339–340, 345–349, 354, 360–361, 382, 404 Intermediate Value Theorem, 944–945 interquartile range, 860–861 intersection method, 86, 127, 134, 524–525 interval notation, 118–119 interval of convergence, 520–521 inverse functions composition of, 532 cosine, 532–534, 538–539, 541, 563 definition, 210 horizontal line test, 530 restricting the domain, 210–211 sine, 529–532, 539, 563 tangent, 534–536, 540, 563 inverse matrices, 815–817 inverse relations algebraic representations, 207–208, 534 1154 Index inverse relations (continued) composite, 211–212 definition, 205 graphs, 205–207 horizontal line test, 209–210, 226 one-to-one functions, 208–211 restricting the domain, 210–211, 529, 532, 534 irrational exponents, 333 irrational numbers, 4, 341, 688 irreducible polynomials, 253, 315 iterations, 199–200 L ladder safety, 426 latitude, 442 law of cosines, 617–622, 682 law of sines AAS information, 626 ambiguous case, 627–628 area of a triangle, 632–633 ASA information, 631–632 definition, 625, 682 SSA information, 628–633 supplementary angle identity, 628 laws of exponents, 330–331, 402 laws of logarithms, 373–374, 402 least squares regression lines, 47–52 lemniscate graphs, 743 length, maximum, 469 light, reflection, 705–706, 712–714 limaçon graphs, 743 limits of functions approaching infinity, 201, 303–305, 914, 948–957 approaching two values, 914–915 of constants, 918, 953 definition, 909–913, 929–935 difference quotients, 922–923 function values, 912–913 identity function, 918 nonexistence of, 913–915 notation, 910–911, 925, 931, 951 one-sided, 924–927, 935 oscillating functions, 915 polynomial functions, 919–920 properties of, 919, 933–934, 953–957 proving, 931–934 rational functions, 920–923, 954–957 trigonometric functions, 566–568 two-sided, 926–927 limits of sequences, 76–77 Limit Theorem, 922, 954 linear combinations of vectors, 662 linear depreciation, 35–36 linear equations, 33–37, 39 linear functions, 34–36, 240 linear inequalities, 119–120, 827 linear models corresponding function, 396 finite differences, 43–44 least squares regression lines, 47–52 modeling terminology, 44–46 prediction from, 51–52 residuals, 44–46, 49–51, 66 linear programming, 829–832 linear regression, 47–52 linear speed, 439–440 linear systems, 779, 781–782, 796–797 lines. see also slope angles between intersecting, 590–592 arithmetic sequences and, 34 least squares regression, 47–52 parallel and perpendicular, 38–39, 66 parameterizations of, 755 point-slope form, 36–37, 39, 66, 792 secant, 218–219, 408, 409 slope-intercept form, 33–36, 39, 66, 792–793 standard form, 39, 66 tangent, 235–237, 408–411 vertical and horizontal, 37–38, 66 local maxima, 153, 266 local minima, 153, 266 logarithmic equations, 379, 384–386 logarithmic functions change-of-base formula, 374–375, 402 common, 356–357, 364 graphs, 359–361, 375–376 laws of, 373–374, 402 natural, 358–359, 364 other bases, 370–376 Power Law, 367 Product Law, 365 properties, 363–364, 372–373 Quotient Law, 366 solving, 384–386 transformations, 359–360, 375–376 logarithmic models, 389, 394 logistic models, 342, 389, 391–392, 395, 401 LORAN, 724–725 lottery probabilities, 870, 881–882, 886–887 lower bounds, 255–256 M Mach numbers, 529 magnitude, vector, 653, 655 Mandelbrot set, 304–306 mathematical induction, 1002–1010 mathematical models, definition, 43. see also models mathematical patterns, 13–19 matrices (singular: matrix) addition and subtraction, 804–805 adjacency, 809–810 applications, 800–801, 808–811 augmented, 795–798 dimensions, 804 directed networks, 809–811 elementary row operations, 795–796 equivalent, 795–796 Gauss-Jordan elimination, 797–798, 817 identity, 815 inverse, 815–817 matrix equations, 814, 817–818 multiplication, 805–809 notation, 299, 795, 797, 804 reduced row echelon form, 797–801 square systems, 814–818 maxima, 153, 170, 266, 468, 607–608, 829–830 mean, 853–854, 856–857, 869, 898 measures of center, 853–857 measures of spread, 857–862 median, 854–856, 861–862 midpoint formula, 998 minima, 153, 266, 829–830 mixtures, 104, 787–788 mode, 855–856, 898 modeling terminology, 44–46 models cubic, 396 definition, 43 exponential, 389–390, 392–394, 396 linear, 43–52, 396 logarithmic, 389, 394–395, 396 logistic, 341–342, 389, 391–392, 396, 402 polynomial, 273–276 models (continued) power, 389, 392–394, 396 quadratic, 396 quartic, 274–276, 392 simple harmonic motion, 549–553 terminology, 44–46 modulus. see absolute value motion parameterization, 759–763 pendulum, 335 planetary, 393–394 projectile, 546, 602, 761–762 multimodal data, 855 multiple choice exams, 887–888 multiplication matrix, 805–809 by negative numbers, 119 polar, of complex numbers, 640–642 scalar, 655–656, 659, 661, 683, 805–806 multiplicity, 265, 283, 308–309, 316 music, 558–562 mutually exclusive events, 866–868 N natural logarithmic functions, 358–359, 364 natural numbers, 3–4 navigation systems, 724–725 negative angle identities, 459–460, 463, 574 negative correlation, 52 negative infinity, 948, 951–953, 956 negative numbers, 119, 297, 330 n factorial, 520–521, 880 no correlation, 52 nonlinear systems, 779, 821–824 nonnegative integers, 3 nonrepeated linear factor denominators, 838–839 nonrepeated quadratic factor denominators, 838–839 nonsingular matrices, 815 normal curve area under, 906 definition, 889–890 empirical rule, 892–893, 899 equation of, 891, 900 properties, 890–892 quartiles, 897 normal curve (continued) standard, 890, 893–896 z-values, 894–896, 900 normal distributions, 889–896 notation angles, 413–415 complex numbers, 294, 296, 301 ellipses, 694 functions, 9–10, 143, 191–192 interval, 118–119 inverse functions, 210, 532, 534 iterated functions, 199–200 limits of functions, 910–911, 925, 929–930, 951 matrix, 299, 795, 797, 804 sequences, 14, 16–17 summation, 25–26, 61 triangles, 617 trigonometric functions, 483, 523, 581 vectors, 653, 655, 662 nth roots, 327–329, 645–651 nuclear wastes, 340 number e, 341, 347–349. see also exponential functions; logarithmic functions number lines, 4, 107–108, 128 number relations, 6–7, 97–98 numerical derivatives, 237 O objective functions, 829–830 oblique asymptotes, 285–286 oblique triangles, 617–622, 625–633 octants, 790–791 odd degree, 260–261, 313 odd functions, 188–189, 482, 489–490 one-stage paths, 809 one-to-one functions, 208–211 open intervals, 118 opposite sides, 415 optimization applications, 322–324, 468–469 orbits, 200–203, 301–304, 697–698, 705, 754 ordered pairs, 5, 118 order importance, in counting, 879–880 orientation, 757 origin, 5, 734 origin symmetry, 186–189 orthogonal vectors, 673, 683 Index 1155 oscillating behavior, 514–515, 568, polar coordinates Power Law of Logarithms, 367, 915 outliers, 847, 862 output, 7, 9 P parabolas applications, 712–714 asymptotes, 286–287 characteristics, 711 curve fitting, 818 definitions, 163, 709, 747 equations, 712, 720, 770–771 graphs, 711–712 parameterization, 756–760, 769 polar equations, 749 translations, 719 parallel lines, 38–39, 66 parallel planes, 793 parallel vectors, 671 parameterization, 755, 757–763, 766–769 parameters, definition, 157, 785 parametric graphing, 157–159, 206–207, 755–757 parent functions, 172–173 parentheses, 9 partial fractions, 838–841 partial sums, 26–29 pendulum motion, 335 perfect square trinomials, 91 periodicity identities, 456–458, 460, 463, 574 periodic orbits, 202 periodic points, 202 periods determining, 495–497, 501–502 pendulum, 335 trigonometric functions, 483, 487, 489, 493–495, 516, 563 permutations, 880–881, 899 perpendicular lines, 38–39, 66 phase shifts, 501–503, 516, 549, 563 pi, 4 piecewise-defined functions, 146–147 pie charts, 845–846 plane curves, 754–755 planetary motion, 393–394, 697–698 plutonium, 340 point-slope form, 36–37, 39, 66, 792 points of ellipsis, 13 points of inflection, 154, 266, 317 1156 Index arc lengths, 776–777 graphs, 739–743 polar coordinate system, 734– 736 rectangular coordinates and, 736–738 polar equations, 745–752, 772 polar form of complex numbers, 639–640 373–374 power models, 389, 392–394, 396 Power Principle, 112–113 power-reducing identities, 595–596 powers of i, 295–296 principal, 100 probability area under a curve, 904–906 binomial experiments, 885–888, polar multiplication and division, 900 640–642 polynomial functions. see also polynomials complete graphs, 267–268 continuity, 261–262 definition, 240 end behavior, 262–264, 267, 316 graphs, 260–268 intercepts, 264–265, 267 limits of, 919–920 local extrema, 266–268 multiplicity, 265 points of inflection, 266, 317 polynomial inequalities, 127 polynomial models, 273–276 polynomials. see also polynomial functions bounds, 254–256 complex coefficients, 307–310 complex zeros, 309–313, 317 conjugate solutions, 299 constant, 240 definition, 239–240 degree of, 240, 242, 248, 260– 261, 263, 308 division, 240–245 equations, 94–95, 240 factoring, 246–247, 253–254, 310–313, 317 complements, 866–867 counting techniques, 879–882, 899 definitions, 864–865 density functions, 871–872, 891, 900 distributions, 865–866 expected values, 869–870 experimental estimates, 874– 877 independent events, 867–868 mutually exclusive events, 866–868 random variables, 869–870 simulations, 875–876, 905–906 theoretic
al estimates, 877–878 probability density functions, 871–872, 891 product functions, 192, 943 Product Law of Exponents, 365 Product Law of Logarithms, 365, 373–374 product-to-sum identities, 599 profit functions, 146, 170, 196, 216–217 projectile motion, 546, 602, 761–762 projections, 674–677, 681 proofs, identities, 573–579 Pythagorean identities, 456, 460, Factor Theorem, 245–246, 463, 574–577 252–253 rational zero test, 251–253 regression, 273–276 remainders, 244–245 solutions, 246–247 x-intercepts, 246, 316 zeros, 240, 245–248, 250–257, 265, 308–313, 316–317 population growth, 340–342, 349–350, 382–384, 389–392, 395 populations, statistical, 843, 858–859, 899 positive correlation, 52 positive integers, 3 Power Law of Exponents, 366–367 Pythagorean Theorem, 421, 1012–1015 Q quadrants, 5 quadratic equations algebraic solutions, 88–95 applications, 169–170 changing forms, 167–169 completing the square, 90–92 complex solutions, 298 definition, 88 discriminant, 93–94 quadratic equations (continued) rates of change factoring, 89 graphs of, 173 irreducible, 253 number of solutions, 93–94 parabolas, 163 polynomial form, 94–95, 164–167, 169, 225, 240 regression, 274–276 summary of forms, 169 taking the square root of both sides, 90 transformation form, 164–165, 168–169, 225 vertex, 163–166, 169 x-intercept form, 164, 166–169, 225 quadratic formula, 92–94, 134, 165, 544–545 quadratic inequalities, 122, 129– 130 quadratic models, 396 quadratic regression, 274–276 quartic functions, 240 quartic regression, 274–276 quartiles, 860–861, 897 quotient functions, 192, 943 quotient identities, 455, 460, 463, 574 Quotient Law of Exponents, 365 Quotient Law of Logarithms, 366, 373–374 R radian/degree conversion, 436–437, 463 average, 214–219, 226 difference quotient, 219–220, 234 instantaneous, 234–237, 614–615 logistic models, 391 slope of tangent lines, 235–237 rational exponents, 329–331, 402 rational functions complete graphs, 287–289 definition, 278 domains, 279 end behavior, 284–285, 287, 289 holes, 282–283, 289 horizontal asymptotes, 284 intercepts, 279–280, 288–289, 317 limits, 920–923, 954–957 maximum of, 322 parabolic asymptotes, 286 partial fractions, 838–841 slant asymptotes, 285–286, 289 trigonometric identities, 578–579 vertical asymptotes, 281–282, 288–289, 317, 950 rational inequalities, 127–128 rationalizing denominators and numerators, 332–333 rational numbers, 4, 78–79 rational zeros, 250–254, 316 rays, rotation, 433 real axis, 638 real numbers, 3–4 real solutions, 89, 94 real zeros, 245, 248, 250–257 reciprocal identities, 455, 460, 463, 574 radian measure, 435–438, 444–445 radicals, 111–113, 327–329, 332– rectangles, 98–99, 703 rectangular box volume, 99–100, 333. see also roots radioactive decay, 340, 351–352, 402 radiocarbon dating, 352, 381–382 radio signals, 472, 500, 724–725 radio telescopes, 713–714 Ramanujan, 699 random samples, definition, 843 random variables, 869–870 ranges definition, 142 exponential and logarithmic functions, 359, 375 relations, 6–7 statistical, 859–860 trigonometric functions, 446, 476, 479, 482, 486–489 323 rectangular coordinate systems, 5, 736–738 recursively defined sequences, 15, 66 reduced row echelon form, 797–801 reference angles, 449–451 reflection light, 705–706, 712–714 radio signals, 712–714 sound, 696–697 reflections, 177, 206, 226, 481, 487, 501, 696–697, 705–706, 712–714 refraction, 545 regression. see also models cubic, 274–276 regression (continued) exponential, 390 least squares, 47–52 linear, 47–52 polynomial, 273–276 power, 394 quadratic, 274–276 quartic, 274–276, 392 sinusoidal functions, 553–554 relations, 6–7, 97–98 relative frequency, 844–845 remainders, 244–245 Remainder Theorem, 244 removable discontinuities, 947 repeating decimals, 78–79 replacement, in counting, 879–880 residuals, 44–46, 48, 50–51, 66 response times, 895–896 restricted domains, 210–211, 529, 532, 534 resultant force, 664–667 Richter magnitudes, 368–369 Riemann sums, 964–967 right triangles, 415–418, 421–426 roots absolute value, 109 of complex numbers, 646–648 cube root of one, 299 extraneous, 110–111 graphical root finder, 84–85, 122 limits at infinity, 955–957 nth, 327–329, 645–651 square, 90, 109, 297, 955–956 of unity, 648–651, 682 roots of unity, 648–651, 682 rose graphs, 742 rotation angles, 730–731, 771 rotations, 722–723, 728–732, 771 rounding, 426, 525 rule of a relation, 7 rule of the function, 7 S samples, definition, 843 sample space, definition, 864–865 sample standard deviation, 858, 899 scalar multiplication, 655–656, 659, 661, 683, 805–806 scatter plots, 5–7, 48, 50–51 Schwarz inequality, 673, 683 secant lines, 218–219, 408 secants, 408, 416–418, 444–445, 486–487, 489 Index 1157 second-degree equations, 722–724. sines (continued) see also quadratic equations self-similar under magnification, 305 sequences applications, 17–19, 28–29 arithmetic, 21–29, 34, 66 definition, 13, 66 explicit forms, 34, 66 geometric, 58–63, 76–79 graphs, 14–16 limits, 76–77 notation, 14, 16–17 partial sums, 26–29 recursive forms, 15–16, 22, 66 summation notation, 25 series arithmetic, 25–29 infinite geometric, 76–79, 520–521 shading on graphs, 827–828 side-angle-side (SAS) information, 618–621 side-side-angle (SSA) information, 627–631 side-side-side (SSS) information, 619, 633 sides of angles, 413, 422, 424 simple harmonic motion. see also sinusoidal functions bouncing springs, 551–553 characteristics, 547 definition, 549 examples of, 522, 550–553 rotating wheel, 550–551 simple interest, 100 sines addition and subtraction identities, 582–583, 604, 610 amplitude, 497–498, 516 basic equations, 539–540, 542 calculators, 423 coterminal angles, 451 damping, 512–513 definition, 416, 444 domain and range, 446, 476, 482 double-angle identities, 593–594, 602–603, 611 exact values, 448–450 finding values, 449–450 graphical transformations, 481–482, 503–504 graph of, 473–475, 497–498, 510–511 half-angle identities, 596–597, 604, 611 instantaneous rates of change, 614–615 inverse function, 529–532, 539, 563 1158 Index law of, 625–633, 682 oscillating behavior, 514–515 periodicity, 456–457, 493–494, 496–497, 516 phase shifts, 501–503, 516, 549 polar graphs, 740–741 power-reducing identities, 595–596 product-to-sum identities, 599 restricted, 529, 531 roots of unity, 648–651, 682 sinusoidal graphs, 510–511 special angles, 418, 462 summary of properties, 483 sum-to-product identities, 599–600 trigonometric identities, 454–460, 463 unit circle, 445 1 SIN key, 423 sinusoidal functions amplitude, 547–549 constructing, 548–549 examples, 522 graphs, 510–511, 548 modeling, 553–555 rotating wheels, 550–551 sound waves, 560–561 spring motion, 551–553 skewed distributions, 846 skid mark length, 335 slant asymptotes, 285–286, 289 slope correlation and, 52, 66 definition, 31–32, 66 from a graph, 32 horizontal and vertical lines, 37–38 parallel and perpendicular lines, 38–39 point-slope form, 36–37, 39, 66, 792 properties, 33 secant lines, 408, 410 slope-intercept form, 33–36, 39, 66, 792–793 tangent lines, 235–237, 408–409 Snell’s Law of Refraction, 545 solutions, definition, 81 solving a triangle, 421 sound, speed of, 705 sound waves, 558–562, 696–697 special angles, 418–419, 437, 462 speed angular, 439–440 average, 215–216, 219–220 instantaneous, 234–236 speed (continued) linear, 439–440 skid mark length, 335 sound, 705 spring motion, 551–553 square roots, 90, 109, 173, 297, 955–956 square systems, 814–818 standard deviation, 857–859, 888, 894, 899 standard form, of a line, 39, 66 standard normal curve, 890, 893–896 standard position, of angles, 434 standard viewing window, 84 statistics. see also data box plots, 861–862 data displays, 844–850 five-number summary, 861–862 interquartile range, 860–861 mean, 853–854, 856–857, 869, 899 median, 854–856, 861 mode, 855–856, 898 range, 859–860 standard deviation, 857–859, 888, 894, 899 variance, 858 z-values, 894–896, 900 stem plots, 847–848 step functions, 157 substitution method, 782, 821–822 subtraction identities, 582–585, 593–600 matrix, 804–805 vector, 658–659 sum functions, 191, 943 summation notation, 25–26, 61 sum of the convergent series, 77 sum-to-product identities, 599–600 surface area of a cylinder, 149, 324 symmetric distributions, 846 symmetry, 184–189, 482–483 synthetic division, 241–242 systems of equations. see also matrices algebraic solutions, 782–786 applications, 786–787 augmented matrices and, 795–798 definition, 779 elimination method, 783–786, 797–798, 821 equivalent, 795 graphs of, 780–781, 783–785, 822–824 inconsistent and consistent, 781, 784–785, 799 systems of equations (continued) linear, 779, 781–782, 796–797 nonlinear, 779, 821–824 number of solutions, 781–782, 821 square, 814–818 substitution method, 782, 821–822 systems of inequalities, 826, 828–832 T tangent lines, 235–237, 408–411 tangents addition and subtraction identities, 584–585, 610 basic equations, 540 coterminal angles, 451 definition, 416, 444 domain and range, 447, 483 exact values, 448–450, 536 graphical transformations, 481–482 graph of, 478–479 half-angle identities, 596, 598, 611 inverse function, 534–536, 540, 563 periodicity, 456–457, 495, 516 restricted, 534 special angles, 418, 462 summary of properties, 483 trigonometric identities, 454–460, 463 two intersecting lines, 590–593 unit circle, 446 technology tips. see calculators telescopes, 705, 713–714 temperature, rate of change, 217–218 terminal points, of vectors, 653 terminal sides, of rays, 433 terminal velocity, 908, 959 terms, of a sequence, 14 test-point method for inequalities, 826–827 theorems Angle of Inclination Theorem, 589–590 Angle Theorem, 672–673 Conjugate Zero Theorem, 309 DeMoivre’s Theorem, 644–645, 682 Factor Theorem, 245–246, 252–253 Fundamental Theorem of Algebra, 307–313 Fundamental Theorem of Linear Programming, 829–830 Intermediate Value Theorem, 944–945 theorems (continued) Limit Theorem, 922, 954 Pythagorean Theorem, 421 Remainder Theorem, 244 Triangle Sum Theorem, 421 three-dimensional c
oordinates, 790–791 TRACE feature, 176, 185 transformation form, 164–165, 168–169, 225 transformations amplitude, 497–498, 503–505, 516, 563 combined, 180, 503–505 conic sections, 716–725 horizontal shifts, 175–176 logarithmic functions, 359–360, 375–376 parameterization of, 757–759 parent functions, 172–173 phase shifts, 501–504, 516, 549, 563 reflections, 176–177, 206, 226, 481, 487, 501 rotations, 722–723, 728–732, 771 stretches and compressions, 177–181, 338–339, 481, 487–489, 497 transforming identities, 574–576 trigonometric functions, 481–482, 487–489, 501–507 vertical shifts, 174, 481–482, 501, 504, 516, 563 translations, 338 trapezoids, area of, 468–469 triangles angle of inclination, 589–592 angles, 413–414 area, 632–633, 682 Heron’s formula, 633, 682 hypotenuses, 415 maximum area, 138–139 oblique, 617–622, 625–633 right, 414–419, 421–426 similar, 415–417 solving, 421–426 special, 418–419, 437, 462 standard notation, 617 trigonometric ratios, 415–417 Triangle Sum Theorem, 421 trigonometric equations algebraic solutions, 538–545 basic, 524–526, 538–542 complex numbers, 644–648 conditional, 523 factoring, 542–544 fractional, 577–579 graphical solutions, 524–528 trigonometric equations (continued) inverse trigonometric functions, 529–536, 539–540, 563 quadratic formula and, 544–545 roots of unity, 648–651, 682 rotation of conic sections, 728–730 solution algorithm, 540–541 special values, 541 substitutions, 542 trigonometric form of complex numbers, 639–640 trigonometric functions. see also trigonometric identities for all angles, 460 angle notation, 413 applications, 425–426 coordinate plane, 443–444 cosecants, 416, 418, 444–445, 486–487, 490 coterminal angles, 451 damped and compressed graphs, 512–514 definitions, 462 domain, 447, 477, 480, 483, 490 even and odd, 482–483, 489–490 exact values, 448–449, 536 finding values, 450–451 graphs of, 472–482, 510–514 identities, 454–460, 463 instantaneous rates of change, 614–615 inverse cosine, 532–534 inverse sine, 529–532 inverse tangent, 534–536 limits of, 566–569 maximum and minimum, 607–608 optimization with, 468–469 oscillating behavior, 514–515, 568 phase shifts, 501–504, 516, 549 polar coordinates, 736–738 powers of, 454 ranges, 447, 477, 480, 483, 487–488, 490 ratios, 415–419 restricted, 529, 531–532, 534 secants, 408, 419, 444–445, 487–488, 490 signs, 447–448 special angles, 418–419, 462 summary of properties, 483, 490 trigonometric functions. see also transformations, 481–482, 487–489, 501–507, 446–449 unit circle, 446–449 trigonometric identities addition and subtraction, 581– 587, 593–600, 604, 610 Index 1159 trigonometric identities (continued) alternate solutions, 577–579 a sin x b cos x c, cofunction, 585–587, 611 definition, 454–455 double-angle, 593–595, 602–603, 605–608 611 Euler’s formula, 688–689 factoring and, 544 graphical testing, 572–573 graphs and, 506–507 half-angle, 596–598, 604, 611 negative angle, 457–459, 462 periodicity, 456–458, 460, 463 power-reducing, 595–596 product-to-sum, 599 proofs, 573–579 Pythagorean, 456, 460, 463, 575–577 quadratic formula and, 544–545 quotient, 455, 460, 463 reciprocal, 455, 460, 463 summary of, 460, 463, 574 sum-to-product, 599–600 using, 602–608 trimodal data, 855 trivial solutions, 799 tuning forks, 558–562 two-stage paths, 810 typing, 388 U uniform distributions, 846 unit circles, 445–446, 449, 473, 475, 478, 650 unit vectors, 661–664 upper bounds, 255–256 V variability, 857–860 variance, 858 vectors angles between, 671–673, 681 arithmetic, 655–659, 662, 681 components, 674–677, 681 components and magnitudes, 655 direction angles, 662–664 dot products, 670–678 equivalent, 654–655 gravity, 677–678 linear combinations, 662 magnitude, 653, 655 1160 Index vectors (continued) x-coordinate transformations, notation, 653, 655, 662 orthogonal, 673, 681 parallel, 671 projections, 674–677, 681 properties, 659 resultant force, 664–667 Schwarz inequality, 673, 683 unit, 661–664 velocity, 663 work calculation, 677–678 zero, 658 velocity average, 235 free-fall, 958–959 instantaneous, 234–236 terminal, 908, 959 total distance from, 964–965 vectors, 663 vertical asymptotes, 281–282, 288–289, 950 vertical lines, 37–38, 66, 792 vertical line test, 151–152, 186, 225 vertical shifts, 174, 481–482, 501, 504, 516 vertical stretches and compressions, 177–180, 339, 481, 487, 501 vertices (singular: vertex) directed matrices, 809–810 ellipse, 692, 720 hyperbolas, 701, 720 parabolas, 709, 720 quadratic functions, 163–166, 169, 225 triangle, 415 Very Large Array (Socorro, New Mexico), 713–714 volume, 216, 323 W waves, 493–498, 558–562 whole numbers, 3–4 wind chill, 335 work, 677–678 X x-axis symmetry, 185–186, 188 x-coordinates, 5 180–181 x-intercept form of quadratic functions, 164, 166–169, 225 x-intercept method, 84–86, 94, 112, 127–128, 134, 525–528 x-intercepts ellipses, 694 hyperbolas, 702 parabolas, 163 polynomial functions, 264–265 quadratic functions, 163–165, 169, 225 rational functions, 279–280, 288–289 x-variables, 6 Y y-axis symmetry, 184–185, 188 y-coordinates, 5, 166 y-coordinate transformations, 178–179 y-intercepts ellipses, 694 hyperbolas, 702 parabolas, 163 polynomial functions, 264 quadratic functions, 163–166, 169 rational functions, 279, 288–289 in three dimensions, 792 y-variables, 6 Z z-axis, 790 Zero Product Property, 89 zeros bounds, 254–256 complex, 310–313, 317 complex polynomials, 307–313 conjugate, 309–313 Factor Theorem and, 252–253 multiplicity, 265, 283, 308–309 orbits, 302–304 polynomials, 240, 245–248, 250–257, 265, 308–313, 316–317 rational, 250–254, 316 of unity, 298–299 zero vectors, 658 z-values, 894–896, 900 Exponents crcs crs 2 crs cr c s s crs r crdr r cr dr 2 cr 1 cd 1 c db a Algebra Multiplication & Factoring Patterns Difference of Squares: Perfect Squares: Difference of Cubes: u2 v2 u v u v 1 21 2 2 u2 2uv v2 2 u2 2uv v2 2 2 u v u v 1 1 u3 v3 u v u v 1 1 21 21 u2 uv v2 u2 uv v2 2 2 d 0 2 1 Sum of Cubes: u3 v3 c r 1 cr c 0 2 1 Perfect Cubes u3 3u2v 3uv2 v3 3 u3 3u2v 3uv2 v3 The Quadratic Formula If a 0, then the solutions of ax2 bx c 0 are x b ± 2b2 4ac 2a . Equations and Graphs The solutions of the equation are the x-intercepts of the graph of Natural Logarithms Logarithms to Base b Special Notation 7 0 and any u: For v, w ln v u means eu v ln v ln w vw ln 2 1 1 v wb a v k ln ln 2 ln v ln w k ln v 1 2 w 7 0 For v, and any u: log bv u means bu v log b 1 log bv log bw vw 2 log bv log bw v log b a wb v k log b 1 2 k log b v 2 1 ln v means log e v log v means log 10 v Change of Base Formula log bv ln v ln b The Pythagorean Theorem c2 a2 b2 c b a Geometry Area of a Triangle A 1 2 bh Circle Diameter 2r h b Circumference 2pr r Area pr2 Distance Formula Length of segment PQ d 2 x1 1 x22 2 y1 1 2 y22 Slope x2 Slope of line m y2 x2 PQ, x1 y1 x1 Midpoint Formula Midpoint M of segment PQ x1 y1 x2 2 , y2 2 b M a Q (x2, y2) M P (x1, y1) The equation of the straight line through The equation of line with slope m and y-intercept b is x1, y12 1 with slope m is y mx b. y y1 m x x12 1 . Rectangular and Parametric Equations for Conic Sections h 1 x r cos t h y r sin t k Circle Center (h, k), radius r2 1 Ellipse Center (h, k) y k b2 1 2 2 2 x h a2 k k h 2 1 Parabola Vertex (h, k) x h 2 4p 2p 2 x a cos t h y b sin t k 0 t 2p 2 1 x t y 1 2 t h 4p 2 k (t any real) Parabola Vertex (h, k) y k 1 2 2 4p 1 x h 2 1 Hyperbola Center (h, k) y k b2 1 2 2 x h a2 2 1 1 2 Hyperbola Center (h, k) x h b2 1 2 2 y k a2 4p 2 h x 1 y t (t any real) x a h cos t y b tan t k 0 t 2p 2 1 x b tan t h y a k cos t 0 t 2p 2 1 Trigonometry If t is a real number and P is the point where the terminal side of an angle of t radians in standard position meets the unit circle, then cos t x-coordinate of P sin t y-coordinate of P tan t sin t cos t csc t 1 sin t sec t 1 cos t cot t cos t sin t Trigonometric Ratios in the Coordinate Plane For any real number t and point (x, y) on the terminal side of an angle of t radians in standard position: (x, y) r y t sin t y r csc t r y x y 0 1 2 cos t x r sec t r x x 0 1 2 tan t y x cot If a 0 and b 7 0, then each of f t 2 1 a sin 1 Periodic Graphs bt c and g t 2 a sin bt c has 1 2 phase shift c b • 2 1 2p b • amplitude a 0 0 • period Right Triangle Trigonometry Special Values opposite sin u opposite hypotenuse tan u opposite adjacent cos u adjacent hypotenuse adjacent Special Right Triangles 45° 2 1 45° 1 2 30° Law of Cosines a2 b2 c2 2bc cos A b2 a2 c2 2ac cos B c2 a2 b2 2ab cos C 3 A 60° 1 c U Degrees Radians sin U cos U tan U 0° 30° 45° 60° 90 12 2 13 2 1 1 13 2 12 2 1 2 0 0 13 3 1 13 undefined B b a C Law of Sines c a sin A b sin B sin C Area 1 2 ab sin C Heron’s Formula: Area 1s s a 1 21 s b 21 s c where Area Formulas for Triangles Reciprocal Identities Pythagorean Identities Negative Angle Identities Trigonometric Identities sin x 1 csc x csc x 1 sin x cos x 1 sec x sec x 1 cos x cot x 1 tan x tan x 1 cot x sin2x cos2x 1 tan2x 1 sec2x 1 cot2x csc2x sin 1 x 2 sin x cos tan x x 1 1 2 2 cos x tan x Periodicity Identities Cofunction Identities x ± 2p x ± 2p sin csc 1 1 x ± p tan 1 sin x csc x tan x 2 2 2 x ± 2p x ± 2p cos sec 1 1 x ± p cot 1 cos x sec x cot x 2 2 2 sin x cos tan x cot sec x csc cos x sin cot x tan csc x sec Quotient Identities Addition and Subtraction Identities tan x sin x cos x cot x cos x sin x sin sin cos cos sin x cos y cos x sin y sin x cos y cos x sin y cos x cos y sin x sin y cos x cos y sin x sin y tan tan 1 1 x y x y 2 2 tan x tan y 1 tan x tan y tan x tan y 1 tan x tan y Double Angle Identities Half-Angle Identities sin 2x 2 sin x cos x cos 2x 1 2 sin2x sin x 2 ± B cos 2x cos2x sin2x cos 2x 2 cos2x 1 tan 2x 2 tan x 1 tan2x cos x 2 ± B 1 cos x 2 1 cos x 2 tan tan x 2 x 2 1 cos x sin x sin x 1 cos x Product-to-Sum Identities Sum-to-Product Identities sin x cos y 1 2 1 sin 1 x y sin 1 2 x y 22 sin x sin y 1 2 1 cos cos x cos y 1 2 1 cos x y x y 2 2 1 1 cos cos x y x y 1 1 22 22 cos x sin y 1 2 1 sin 1 x y sin 1 2 x y 22 sin x sin y 2 sin a sin x sin y 2 cos cos x cos y
2 cos cos x cos y 2 sin cos sin a a cos a sin
s as π 6 , and so on. If you and your friends carry , π , 3π 4 , π 2 , π 3 backpacks with books in them to school, the numbers of books in the backpacks are discrete data and the weights of the backpacks are continuous data. Example 1.5 Data Sample of Quantitative Discrete Data The data are the number of books students carry in their backpacks. You sample five students. Two students carry three books, one student carries four books, one student carries two books, and one student carries one book. The numbers of books (three, four, two, and one) are the quantitative discrete data. 1.5 The data are the number of machines in a gym. You sample five gyms. One gym has 12 machines, one gym has 15 machines, one gym has ten machines, one gym has 22 machines, and the other gym has 20 machines. What type of data is this? Example 1.6 Data Sample of Quantitative Continuous Data The data are the weights of backpacks with books in them. You sample the same five students. The weights (in pounds) of their backpacks are 6.2, 7, 6.8, 9.1, 4.3. Notice that backpacks carrying three books can have different weights. Weights are quantitative continuous data because weights are measured. 1.6 The data are the areas of lawns in square feet. You sample five houses. The areas of the lawns are 144 sq. feet, 160 sq. feet, 190 sq. feet, 180 sq. feet, and 210 sq. feet. What type of data is this? Example 1.7 You go to the supermarket and purchase three cans of soup (19 ounces) tomato bisque, 14.1 ounces lentil, and 19 ounces Italian wedding), two packages of nuts (walnuts and peanuts), four different kinds of vegetable (broccoli, cauliflower, spinach, and carrots), and two desserts (16 ounces Cherry Garcia ice cream and two pounds (32 ounces chocolate chip cookies). Name data sets that are quantitative discrete, quantitative continuous, and qualitative. Solution 1.7 One Possible Solution: • The three cans of soup, two packages of nuts, four kinds of vegetables and two desserts are quantitative discrete data because you count them. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 1 \| SAMPLING AND DATA 15 • The weights of the soups (19 ounces, 14.1 ounces, 19 ounces) are quantitative continuous data because you measure weights as precisely as possible. • Types of soups, nuts, vegetables and desserts are qualitative data because they are categorical. Try to identify additional data sets in this example. Example 1.8 The data are the colors of backpacks. Again, you sample the same five students. One student has a red backpack, two students have black backpacks, one student has a green backpack, and one student has a gray backpack. The colors red, black, black, green, and gray are qualitative data. 1.8 The data are the colors of houses. You sample five houses. The colors of the houses are white, yellow, white, red, and white. What type of data is this? NOTE You may collect data as numbers and report it categorically. For example, the quiz scores for each student are recorded throughout the term. At the end of the term, the quiz scores are reported as A, B, C, D, or F. Example 1.9 Work collaboratively to determine the correct data type (quantitative or qualitative). Indicate whether quantitative data are continuous or discrete. Hint: Data that are discrete often start with the words "the number of." a. b. the number of pairs of shoes you own the type of car you drive c. where you go on vacation d. e. f. g. the distance it is from your home to the nearest grocery store the number of classes you take per school year. the tuition for your classes the type of calculator you use h. movie ratings i. political party preferences j. weights of sumo wrestlers k. amount of money (in dollars) won playing poker l. number of correct answers on a quiz m. peoples’ attitudes toward the government n. IQ scores (This may cause some discussion.) Solution 1.9 Items a, e, f, k, and l are quantitative discrete; items d, j, and n are quantitative continuous; items b, c, g, h, i, and m are qualitative. 16 CHAPTER 1 \| SAMPLING AND DATA 1.9 Determine the correct data type (quantitative or qualitative) for the number of cars in a parking lot. Indicate whether quantitative data are continuous or discrete. Example 1.10 A statistics professor collects information about the classification of her students as freshmen, sophomores, juniors, or seniors. The data she collects are summarized in the pie chart Figure 1.2. What type of data does this graph show? Figure 1.3 Solution 1.10 This pie chart shows the students in each year, which is qualitative data. 1.10 The registrar at State University keeps records of the number of credit hours students complete each semester. The data he collects are summarized in the histogram. The class boundaries are 10 to less than 13, 13 to less than 16, 16 to less than 19, 19 to less than 22, and 22 to less than 25. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 1 \| SAMPLING AND DATA 17 Figure 1.4 What type of data does this graph show? Qualitative Data Discussion Below are tables comparing the number of part-time and full-time students at De Anza College and Foothill College enrolled for the spring 2010 quarter. The tables display counts (frequencies) and percentages or proportions (relative frequencies). The percent columns make comparing the same categories in the colleges easier. Displaying percentages along with the numbers is often helpful, but it is particularly important when comparing sets of data that do not have the same totals, such as the total enrollments for both colleges in this example. Notice how much larger the percentage for part-time students at Foothill College is compared to De Anza College. De Anza College Foothill College Number Percent Number Percent Full-time 9,200 40.9% Full-time 4,059 28.6% Part-time 13,296 59.1% Part-time 10,124 71.4% Total 22,496 100% Total 14,183 100% Table 1.2 Fall Term 2007 (Census day) Tables are a good way of organizing and displaying data. But graphs can be even more helpful in understanding the data. There are no strict rules concerning which graphs to use. Two graphs that are used to display qualitative data are pie charts and bar graphs. In a pie chart, categories of data are represented by wedges in a circle and are proportional in size to the percent of individuals in each category. In a bar graph, the length of the bar for each category is proportional to the number or percent of individuals in each category. Bars may be vertical or horizontal. A Pareto chart consists of bars that are sorted into order by category size (largest to smallest). Look at Figure 1.5 and Figure 1.6 and determine which graph (pie or bar) you think displays the comparisons better. It is a good idea to look at a variety of graphs to see which is the most helpful in displaying the data. We might make different choices of what we think is the “best” graph depending on the data and the context. Our choice also depends on what we are using the data for. 18 CHAPTER 1 \| SAMPLING AND DATA Figure 1.5 (a) (b) Figure 1.6 Percentages That Add to More (or Less) Than 100% Sometimes percentages add up to be more than 100% (or less than 100%). In the graph, the percentages add to more than 100% because students can be in more than one category. A bar graph is appropriate to compare the relative size of the categories. A pie chart cannot be used. It also could not be used if the percentages added to less than 100%. Characteristic/Category Full-Time Students Percent 40.9% Students who intend to transfer to a 4-year educational institution 48.6% Students under age 25 TOTAL Table 1.3 De Anza College Spring 2010 61.0% 150.5% This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 1 \| SAMPLING AND DATA 19 Figure 1.7 Omitting Categories/Missing Data The table displays Ethnicity of Students but is missing the "Other/Unknown" category. This category contains people who did not feel they fit into any of the ethnicity categories or declined to respond. Notice that the frequencies do not add up to the total number of students. In this situation, create a bar graph and not a pie chart. Frequency Percent Asian Black Filipino Hispanic 8,794 1,412 1,298 4,180 Native American 146 Pacific Islander 236 5,978 White TOTAL 36.1% 5.8% 5.3% 17.1% 0.6% 1.0% 24.5% 22,044 out of 24,382 90.4% out of 100% Table 1.4 Ethnicity of Students at De Anza College Fall Term 2007 (Census Day) Figure 1.8 20 CHAPTER 1 \| SAMPLING AND DATA The following graph is the same as the previous graph but the “Other/Unknown” percent (9.6%) has been included. The “Other/Unknown” category is large compared to some of the other categories (Native American, 0.6%, Pacific Islander 1.0%). This is important to know when we think about what the data are telling us. This particular bar graph in Figure 1.9 can be difficult to understand visually. The graph in Figure 1.10 is a Pareto chart. The Pareto chart has the bars sorted from largest to smallest and is easier to read and interpret. Figure 1.9 Bar Graph with Other/Unknown Category Figure 1.10 Pareto Chart With Bars Sorted by Size Pie Charts: No Missing Data The following pie charts have the “Other/Unknown” category included (since the percentages must add to 100%). The chart in Figure 1.11b is organized by the size of each wedge, which makes it a more visually informative graph than the unsorted, alphabetical graph in Figure 1.11a. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 1 \| SAMPLING AND DATA 21 (a) (b) Figure 1.11 Sampling Gathering information about an entire population often costs too much or is virtually impossible. Instead, we use a sample of the population. A sample should have
the same characteristics as the population it is representing. Most statisticians use various methods of random sampling in an attempt to achieve this goal. This section will describe a few of the most common methods. There are several different methods of random sampling. In each form of random sampling, each member of a population initially has an equal chance of being selected for the sample. Each method has pros and cons. The easiest method to describe is called a simple random sample. Any group of n individuals is equally likely to be chosen by any other group of n individuals if the simple random sampling technique is used. In other words, each sample of the same size has an equal chance of being selected. For example, suppose Lisa wants to form a four-person study group (herself and three other people) from her pre-calculus class, which has 31 members not including Lisa. To choose a simple random sample of size three from the other members of her class, Lisa could put all 31 names in a hat, shake the hat, close her eyes, and pick out three names. A more technological way is for Lisa to first list the last names of the members of her class together with a two-digit number, as in Table 1.5: ID Name ID Name ID Name 00 01 02 Anselmo Bautista Bayani 11 12 13 King 21 Roquero Legeny 22 Roth Lundquist 23 Rowell 03 Cheng 14 Macierz 24 Salangsang 04 Cuarismo 15 Motogawa 25 Slade 05 Cuningham 16 Okimoto 06 Fontecha 07 Hong 17 18 Patel Price 26 27 28 Stratcher Tallai Tran 08 Hoobler 19 Quizon 29 Wai 09 10 Jiao Khan 20 Reyes 30 Wood Table 1.5 Class Roster Lisa can use a table of random numbers (found in many statistics books and mathematical handbooks), a calculator, or a computer to generate random numbers. For this example, suppose Lisa chooses to generate random numbers from a calculator. The numbers generated are as follows: 0.94360; 0.99832; 0.14669; 0.51470; 0.40581; 0.73381; 0.04399 22 CHAPTER 1 \| SAMPLING AND DATA Lisa reads two-digit groups until she has chosen three class members (that is, she reads 0.94360 as the groups 94, 43, 36, 60). Each random number may only contribute one class member. If she needed to, Lisa could have generated more random numbers. The random numbers 0.94360 and 0.99832 do not contain appropriate two digit numbers. However the third random number, 0.14669, contains 14 (the fourth random number also contains 14), the fifth random number contains 05, and the seventh random number contains 04. The two-digit number 14 corresponds to Macierz, 05 corresponds to Cuningham, and 04 corresponds to Cuarismo. Besides herself, Lisa’s group will consist of Marcierz, Cuningham, and Cuarismo. To generate random numbers: • Press MATH. • Arrow over to PRB. • Press 5:randInt(. Enter 0, 30). • Press ENTER for the first random number. • Press ENTER two more times for the other 2 random numbers. If there is a repeat press ENTER again. Note: randInt(0, 30, 3) will generate 3 random numbers. Figure 1.12 Besides simple random sampling, there are other forms of sampling that involve a chance process for getting the sample. Other well-known random sampling methods are the stratified sample, the cluster sample, and the systematic sample. To choose a stratified sample, divide the population into groups called strata and then take a proportionate number from each stratum. For example, you could stratify (group) your college population by department and then choose a proportionate simple random sample from each stratum (each department) to get a stratified random sample. To choose a simple random sample from each department, number each member of the first department, number each member of the second department, and do the same for the remaining departments. Then use simple random sampling to choose proportionate numbers from the first department and do the same for each of the remaining departments. Those numbers picked from the first department, picked from the second department, and so on represent the members who make up the stratified sample. To choose a cluster sample, divide the population into clusters (groups) and then randomly select some of the clusters. All the members from these clusters are in the cluster sample. For example, if you randomly sample four departments from your college population, the four departments make up the cluster sample. Divide your college faculty by department. The departments are the clusters. Number each department, and then choose four different numbers using simple random sampling. All members of the four departments with those numbers are the cluster sample. To choose a systematic sample, randomly select a starting point and take every nth piece of data from a listing of the population. For example, suppose you have to do a phone survey. Your phone book contains 20,000 residence listings. You must choose 400 names for the sample. Number the population 1–20,000 and then use a simple random sample to pick a number that represents the first name in the sample. Then choose every fiftieth name thereafter until you have a total of 400 names (you might have to go back to the beginning of your phone list). Systematic sampling is frequently chosen because it is a simple method. A type of sampling that is non-random is convenience sampling. Convenience sampling involves using results that are readily available. For example, a computer software store conducts a marketing study by interviewing potential customers This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 1 \| SAMPLING AND DATA 23 who happen to be in the store browsing through the available software. The results of convenience sampling may be very good in some cases and highly biased (favor certain outcomes) in others. Sampling data should be done very carefully. Collecting data carelessly can have devastating results. Surveys mailed to households and then returned may be very biased (they may favor a certain group). It is better for the person conducting the survey to select the sample respondents. True random sampling is done with replacement. That is, once a member is picked, that member goes back into the population and thus may be chosen more than once. However for practical reasons, in most populations, simple random sampling is done without replacement. Surveys are typically done without replacement. That is, a member of the population may be chosen only once. Most samples are taken from large populations and the sample tends to be small in comparison to the population. Since this is the case, sampling without replacement is approximately the same as sampling with replacement because the chance of picking the same individual more than once with replacement is very low. In a college population of 10,000 people, suppose you want to pick a sample of 1,000 randomly for a survey. For any particular sample of 1,000, if you are sampling with replacement, • • • the chance of picking the first person is 1,000 out of 10,000 (0.1000); the chance of picking a different second person for this sample is 999 out of 10,000 (0.0999); the chance of picking the same person again is 1 out of 10,000 (very low). If you are sampling without replacement, • • the chance of picking the first person for any particular sample is 1000 out of 10,000 (0.1000); the chance of picking a different second person is 999 out of 9,999 (0.0999); • you do not replace the first person before picking the next person. Compare the fractions 999/10,000 and 999/9,999. For accuracy, carry the decimal answers to four decimal places. To four decimal places, these numbers are equivalent (0.0999). Sampling without replacement instead of sampling with replacement becomes a mathematical issue only when the population is small. For example, if the population is 25 people, the sample is ten, and you are sampling with replacement for any particular sample, then the chance of picking the first person is ten out of 25, and the chance of picking a different second person is nine out of 25 (you replace the first person). If you sample without replacement, then the chance of picking the first person is ten out of 25, and then the chance of picking the second person (who is different) is nine out of 24 (you do not replace the first person). Compare the fractions 9/25 and 9/24. To four decimal places, 9/25 = 0.3600 and 9/24 = 0.3750. To four decimal places, these numbers are not equivalent. When you analyze data, it is important to be aware of sampling errors and nonsampling errors. The actual process of sampling causes sampling errors. For example, the sample may not be large enough. Factors not related to the sampling process cause nonsampling errors. A defective counting device can cause a nonsampling error. In reality, a sample will never be exactly representative of the population so there will always be some sampling error. As a rule, the larger the sample, the smaller the sampling error. In statistics, a sampling bias is created when a sample is collected from a population and some members of the population are not as likely to be chosen as others (remember, each member of the population should have an equally likely chance of being chosen). When a sampling bias happens, there can be incorrect conclusions drawn about the population that is being studied. Example 1.11 A study is done to determine the average tuition that San Jose State undergraduate students pay per semester. Each student in the following samples is asked how much tuition he or she paid for the Fall semester. What is the type of sampling in each case? a. A sample of 100 undergraduate San Jose State students is taken by organizing the students’ names by classification (freshman, sophomore, junior, or senior), and then selecting 25 students from each. b. A random number generator is used to select a student from the alphabetical listing of all undergraduate students in the Fall semester. Starting with that student, every
50th student is chosen until 75 students are included in the sample. c. A completely random method is used to select 75 students. Each undergraduate student in the fall semester has the same probability of being chosen at any stage of the sampling process. 24 CHAPTER 1 \| SAMPLING AND DATA d. The freshman, sophomore, junior, and senior years are numbered one, two, three, and four, respectively. A random number generator is used to pick two of those years. All students in those two years are in the sample. e. An administrative assistant is asked to stand in front of the library one Wednesday and to ask the first 100 undergraduate students he encounters what they paid for tuition the Fall semester. Those 100 students are the sample. Solution 1.11 a. stratified; b. systematic; c. simple random; d. cluster; e. convenience 1.11 You are going to use the random number generator to generate different types of samples from the data. This table displays six sets of quiz scores (each quiz counts 10 points) for an elementary statistics class. #3 10 9 8 10 9 9 10 9 8 10 #1 5 10 9 9 7 9 7 8 9 8 #2 7 5 10 10 8 9 7 8 7 8 Table 1.6 #4 #5 #6 9 8 6 9 5 10 9 10 Instructions: Use the Random Number Generator to pick samples. 1. Create a stratified sample by column. Pick three quiz scores randomly from each column. ◦ Number each row one through ten. ◦ On your calculator, press Math and arrow over to PRB. ◦ For column 1, Press 5:randInt( and enter 1,10). Press ENTER. Record the number. Press ENTER 2 more times (even the repeats). Record these numbers. Record the three quiz scores in column one that correspond to these three numbers. ◦ Repeat for columns two through six. ◦ These 18 quiz scores are a stratified sample. 2. Create a cluster sample by picking two of the columns. Use the column numbers: one through six. ◦ Press MATH and arrow over to PRB. ◦ Press 5:randInt( and enter 1,6). Press ENTER. Record the number. Press ENTER and record that number. ◦ The two numbers are for two of the columns. ◦ The quiz scores (20 of them) in these 2 columns are the cluster sample. 3. Create a simple random sample of 15 quiz scores. ◦ Use the numbering one through 60. ◦ Press MATH. Arrow over to PRB. Press 5:randInt( and enter 1, 60). This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 1 \| SAMPLING AND DATA 25 ◦ Press ENTER 15 times and record the numbers. ◦ Record the quiz scores that correspond to these numbers. ◦ These 15 quiz scores are the systematic sample. 4. Create a systematic sample of 12 quiz scores. ◦ Use the numbering one through 60. ◦ Press MATH. Arrow over to PRB. Press 5:randInt( and enter 1, 60). ◦ Press ENTER. Record the number and the first quiz score. From that number, count ten quiz scores and record that quiz score. Keep counting ten quiz scores and recording the quiz score until you have a sample of 12 quiz scores. You may wrap around (go back to the beginning). Example 1.12 Determine the type of sampling used (simple random, stratified, systematic, cluster, or convenience). a. A soccer coach selects six players from a group of boys aged eight to ten, seven players from a group of boys aged 11 to 12, and three players from a group of boys aged 13 to 14 to form a recreational soccer team. b. A pollster interviews all human resource personnel in five different high tech companies. c. A high school educational researcher interviews 50 high school female teachers and 50 high school male teachers. d. A medical researcher interviews every third cancer patient from a list of cancer patients at a local hospital. e. A high school counselor uses a computer to generate 50 random numbers and then picks students whose names correspond to the numbers. f. A student interviews classmates in his algebra class to determine how many pairs of jeans a student owns, on the average. Solution 1.12 a. stratified; b. cluster; c. stratified; d. systematic; e. simple random; f.convenience 1.12 Determine the type of sampling used (simple random, stratified, systematic, cluster, or convenience). A high school principal polls 50 freshmen, 50 sophomores, 50 juniors, and 50 seniors regarding policy changes for after school activities. If we were to examine two samples representing the same population, even if we used random sampling methods for the samples, they would not be exactly the same. Just as there is variation in data, there is variation in samples. As you become accustomed to sampling, the variability will begin to seem natural. Example 1.13 Suppose ABC College has 10,000 part-time students (the population). We are interested in the average amount of money a part-time student spends on books in the fall term. Asking all 10,000 students is an almost impossible task. Suppose we take two different samples. First, we use convenience sampling and survey ten students from a first term organic chemistry class. Many of these students are taking first term calculus in addition to the organic chemistry class. The amount of money they spend on books is as follows: $128; $87; $173; $116; $130; $204; $147; $189; $93; $153 26 CHAPTER 1 \| SAMPLING AND DATA The second sample is taken using a list of senior citizens who take P.E. classes and taking every fifth senior citizen on the list, for a total of ten senior citizens. They spend: $50; $40; $36; $15; $50; $100; $40; $53; $22; $22 It is unlikely that any student is in both samples. a. Do you think that either of these samples is representative of (or is characteristic of) the entire 10,000 part-time student population? Solution 1.13 a. No. The first sample probably consists of science-oriented students. Besides the chemistry course, some of them are also taking first-term calculus. Books for these classes tend to be expensive. Most of these students are, more than likely, paying more than the average part-time student for their books. The second sample is a group of senior citizens who are, more than likely, taking courses for health and interest. The amount of money they spend on books is probably much less than the average parttime student. Both samples are biased. Also, in both cases, not all students have a chance to be in either sample. b. Since these samples are not representative of the entire population, is it wise to use the results to describe the entire population? Solution 1.13 b. No. For these samples, each member of the population did not have an equally likely chance of being chosen. Now, suppose we take a third sample. We choose ten different part-time students from the disciplines of chemistry, math, English, psychology, sociology, history, nursing, physical education, art, and early childhood development. (We assume that these are the only disciplines in which part-time students at ABC College are enrolled and that an equal number of part-time students are enrolled in each of the disciplines.) Each student is chosen using simple random sampling. Using a calculator, random numbers are generated and a student from a particular discipline is selected if he or she has a corresponding number. The students spend the following amounts: $180; $50; $150; $85; $260; $75; $180; $200; $200; $150 c. Is the sample biased? Solution 1.13 c. The sample is unbiased, but a larger sample would be recommended to increase the likelihood that the sample will be close to representative of the population. However, for a biased sampling technique, even a large sample runs the risk of not being representative of the population. Students often ask if it is "good enough" to take a sample, instead of surveying the entire population. If the survey is done well, the answer is yes. 1.13 A local radio station has a fan base of 20,000 listeners. The station wants to know if its audience would prefer more music or more talk shows. Asking all 20,000 listeners is an almost impossible task. The station uses convenience sampling and surveys the first 200 people they meet at one of the station’s music concert events. 24 people said they’d prefer more talk shows, and 176 people said they’d prefer more music. Do you think that this sample is representative of (or is characteristic of) the entire 20,000 listener population? As a class, determine whether or not the following samples are representative. If they are not, discuss the reasons. 1. To find the average GPA of all students in a university, use all honor students at the university as the sample. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 1 \| SAMPLING AND DATA 27 2. To find out the most popular cereal among young people under the age of ten, stand outside a large supermarket for three hours and speak to every twentieth child under age ten who enters the supermarket. 3. To find the average annual income of all adults in the United States, sample U.S. congressmen. Create a cluster sample by considering each state as a stratum (group). By using simple random sampling, select states to be part of the cluster. Then survey every U.S. congressman in the cluster. 4. To determine the proportion of people taking public transportation to work, survey 20 people in New York City. Conduct the survey by sitting in Central Park on a bench and interviewing every person who sits next to you. 5. To determine the average cost of a two-day stay in a hospital in Massachusetts, survey 100 hospitals across the state using simple random sampling. Variation in Data Variation is present in any set of data. For example, 16-ounce cans of beverage may contain more or less than 16 ounces of liquid. In one study, eight 16 ounce cans were measured and produced the following amount (in ounces) of beverage: 15.8; 16.1; 15.2; 14.8; 15.8; 15.9; 16.0; 15.5 Measurements of the amount of beverage in a 16-ounce can may vary because different people make the measurements or because the exact amount, 16 ounces of liquid, was not put into the
cans. Manufacturers regularly run tests to determine if the amount of beverage in a 16-ounce can falls within the desired range. Be aware that as you take data, your data may vary somewhat from the data someone else is taking for the same purpose. This is completely natural. However, if two or more of you are taking the same data and get very different results, it is time for you and the others to reevaluate your data-taking methods and your accuracy. Variation in Samples It was mentioned previously that two or more samples from the same population, taken randomly, and having close to the same characteristics of the population will likely be different from each other. Suppose Doreen and Jung both decide to study the average amount of time students at their college sleep each night. Doreen and Jung each take samples of 500 students. Doreen uses systematic sampling and Jung uses cluster sampling. Doreen's sample will be different from Jung's sample. Even if Doreen and Jung used the same sampling method, in all likelihood their samples would be different. Neither would be wrong, however. Think about what contributes to making Doreen’s and Jung’s samples different. If Doreen and Jung took larger samples (i.e. the number of data values is increased), their sample results (the average amount of time a student sleeps) might be closer to the actual population average. But still, their samples would be, in all likelihood, different from each other. This variability in samples cannot be stressed enough. Size of a Sample The size of a sample (often called the number of observations) is important. The examples you have seen in this book so far have been small. Samples of only a few hundred observations, or even smaller, are sufficient for many purposes. In polling, samples that are from 1,200 to 1,500 observations are considered large enough and good enough if the survey is random and is well done. You will learn why when you study confidence intervals. Be aware that many large samples are biased. For example, call-in surveys are invariably biased, because people choose to respond or not. Divide into groups of two, three, or four. Your instructor will give each group one six-sided die. Try this experiment twice. Roll one fair die (six-sided) 20 times. Record the number of ones, twos, threes, fours, fives, and sixes you get in Table 1.7 and Table 1.8 (“frequency” is the number of times a particular face of the die occurs): 28 CHAPTER 1 \| SAMPLING AND DATA Face on Die Frequency 1 2 3 4 5 6 Table 1.7 First Experiment (20 rolls) Face on Die Frequency 1 2 3 4 5 6 Table 1.8 Second Experiment (20 rolls) Did the two experiments have the same results? Probably not. If you did the experiment a third time, do you expect the results to be identical to the first or second experiment? Why or why not? Which experiment had the correct results? They both did. The job of the statistician is to see through the variability and draw appropriate conclusions. Critical Evaluation We need to evaluate the statistical studies we read about critically and analyze them before accepting the results of the studies. Common problems to be aware of include • Problems with samples: A sample must be representative of the population. A sample that is not representative of the population is biased. Biased samples that are not representative of the population give results that are inaccurate and not valid. • Self-selected samples: Responses only by people who choose to respond, such as call-in surveys, are often unreliable. • Sample size issues: Samples that are too small may be unreliable. Larger samples are better, if possible. In some situations, having small samples is unavoidable and can still be used to draw conclusions. Examples: crash testing cars or medical testing for rare conditions • Undue influence: collecting data or asking questions in a way that influences the response • Non-response or refusal of subject to participate: The collected responses may no longer be representative of the population. Often, people with strong positive or negative opinions may answer surveys, which can affect the results. • Causality: A relationship between two variables does not mean that one causes the other to occur. They may be related (correlated) because of their relationship through a different variable. • Self-funded or self-interest studies: A study performed by a person or organization in order to support their claim. Is the study impartial? Read the study carefully to evaluate the work. Do not automatically assume that the study is good, but do not automatically assume the study is bad either. Evaluate it on its merits and the work done. • Misleading use of data: improperly displayed graphs, incomplete data, or lack of context This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 1 \| SAMPLING AND DATA 29 • Confounding: When the effects of multiple factors on a response cannot be separated. Confounding makes it difficult or impossible to draw valid conclusions about the effect of each factor. 1.3 \| Frequency, Frequency Tables, and Levels of Measurement Once you have a set of data, you will need to organize it so that you can analyze how frequently each datum occurs in the set. However, when calculating the frequency, you may need to round your answers so that they are as precise as possible. Answers and Rounding Off A simple way to round off answers is to carry your final answer one more decimal place than was present in the original data. Round off only the final answer. Do not round off any intermediate results, if possible. If it becomes necessary to round off intermediate results, carry them to at least twice as many decimal places as the final answer. For example, the average of the three quiz scores four, six, and nine is 6.3, rounded off to the nearest tenth, because the data are whole numbers. Most answers will be rounded off in this manner. It is not necessary to reduce most fractions in this course. Especially in Probability Topics, the chapter on probability, it is more helpful to leave an answer as an unreduced fraction. Levels of Measurement The way a set of data is measured is called its level of measurement. Correct statistical procedures depend on a researcher being familiar with levels of measurement. Not every statistical operation can be used with every set of data. Data can be classified into four levels of measurement. They are (from lowest to highest level): • Nominal scale level • Ordinal scale level • Interval scale level • Ratio scale level Data that is measured using a nominal scale is qualitative. Categories, colors, names, labels and favorite foods along with yes or no responses are examples of nominal level data. Nominal scale data are not ordered. For example, trying to classify people according to their favorite food does not make any sense. Putting pizza first and sushi second is not meaningful. Smartphone companies are another example of nominal scale data. Some examples are Sony, Motorola, Nokia, Samsung and Apple. This is just a list and there is no agreed upon order. Some people may favor Apple but that is a matter of opinion. Nominal scale data cannot be used in calculations. Data that is measured using an ordinal scale is similar to nominal scale data but there is a big difference. The ordinal scale data can be ordered. An example of ordinal scale data is a list of the top five national parks in the United States. The top five national parks in the United States can be ranked from one to five but we cannot measure differences between the data. Another example of using the ordinal scale is a cruise survey where the responses to questions about the cruise are “excellent,” “good,” “satisfactory,” and “unsatisfactory.” These responses are ordered from the most desired response to the least desired. But the differences between two pieces of data cannot be measured. Like the nominal scale data, ordinal scale data cannot be used in calculations. Data that is measured using the interval scale is similar to ordinal level data because it has a definite ordering but there is a difference between data. The differences between interval scale data can be measured though the data does not have a starting point. Temperature scales like Celsius (C) and Fahrenheit (F) are measured by using the interval scale. In both temperature measurements, 40° is equal to 100° minus 60°. Differences make sense. But 0 degrees does not because, in both scales, 0 is not the absolute lowest temperature. Temperatures like -10° F and -15° C exist and are colder than 0. Interval level data can be used in calculations, but one type of comparison cannot be done. 80° C is not four times as hot as 20° C (nor is 80° F four times as hot as 20° F). There is no meaning to the ratio of 80 to 20 (or four to one). Data that is measured using the ratio scale takes care of the ratio problem and gives you the most information. Ratio scale data is like interval scale data, but it has a 0 point and ratios can be calculated. For example, four multiple choice statistics final exam scores are 80, 68, 20 and 92 (out of a possible 100 points). The exams are machine-graded. The data can be put in order from lowest to highest: 20, 68, 80, 92. The differences between the data have meaning. The score 92 is more than the score 68 by 24 points. Ratios can be calculated. The smallest score is 0. So 80 is four times 20. The score of 80 is four times better than the score of 20. 30 CHAPTER 1 \| SAMPLING AND DATA Frequency Twenty students were asked how many hours they worked per day. Their responses, in hours, are as follows: 5; 6; 3; 3; 2; 4; 7; 5; 2; 3; 5; 6; 5; 4; 4; 3; 5; 2; 5; 3. Table 1.9 lists the different data values in ascending order and their frequencies. DATA VALUE FREQUENCY Table 1.9 Frequency Table of Student Work Hours A frequency is the number of times a value of the data occurs. Accordin
g to Table 1.9, there are three students who work two hours, five students who work three hours, and so on. The sum of the values in the frequency column, 20, represents the total number of students included in the sample. A relative frequency is the ratio (fraction or proportion) of the number of times a value of the data occurs in the set of all outcomes to the total number of outcomes. To find the relative frequencies, divide each frequency by the total number of students in the sample–in this case, 20. Relative frequencies can be written as fractions, percents, or decimals. DATA VALUE FREQUENCY RELATIVE FREQUENCY or 0.15 or 0.25 or 0.15 or 0.30 or 0.10 or 0.05 3 20 5 20 3 20 6 20 2 20 1 20 Table 1.10 Frequency Table of Student Work Hours with Relative Frequencies The sum of the values in the relative frequency column of Table 1.10 is 20 20 , or 1. Cumulative relative frequency is the accumulation of the previous relative frequencies. To find the cumulative relative frequencies, add all the previous relative frequencies to the relative frequency for the current row, as shown in Table 1.11. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 DATA VALUE FREQUENCY RELATIVE FREQUENCY CUMULATIVE RELATIVE FREQUENCY CHAPTER 1 \| SAMPLING AND DATA 31 20 5 20 3 20 6 20 2 20 1 20 or 0.15 0.15 or 0.25 or 0.15 or 0.30 or 0.10 or 0.05 0.15 + 0.25 = 0.40 0.40 + 0.15 = 0.55 0.55 + 0.30 = 0.85 0.85 + 0.10 = 0.95 0.95 + 0.05 = 1.00 Table 1.11 Frequency Table of Student Work Hours with Relative and Cumulative Relative Frequencies The last entry of the cumulative relative frequency column is one, indicating that one hundred percent of the data has been accumulated. NOTE Because of rounding, the relative frequency column may not always sum to one, and the last entry in the cumulative relative frequency column may not be one. However, they each should be close to one. Table 1.12 represents the heights, in inches, of a sample of 100 male semiprofessional soccer players. HEIGHTS (INCHES) FREQUENCY RELATIVE FREQUENCY CUMULATIVE RELATIVE FREQUENCY 59.95–61.95 5 61.95–63.95 3 63.95–65.95 15 65.95–67.95 40 67.95–69.95 17 69.95–71.95 12 71.95–73.95 7 73.95–75.95 1 5 100 3 100 15 100 40 100 17 100 12 100 7 100 1 100 = 0.05 0.05 = 0.03 = 0.15 = 0.40 = 0.17 = 0.12 = 0.07 = 0.01 0.05 + 0.03 = 0.08 0.08 + 0.15 = 0.23 0.23 + 0.40 = 0.63 0.63 + 0.17 = 0.80 0.80 + 0.12 = 0.92 0.92 + 0.07 = 0.99 0.99 + 0.01 = 1.00 Total = 100 Total = 1.00 Table 1.12 Frequency Table of Soccer Player Height 32 CHAPTER 1 \| SAMPLING AND DATA The data in this table have been grouped into the following intervals: • 59.95 to 61.95 inches • 61.95 to 63.95 inches • 63.95 to 65.95 inches • 65.95 to 67.95 inches • 67.95 to 69.95 inches • 69.95 to 71.95 inches • 71.95 to 73.95 inches • 73.95 to 75.95 inches NOTE This example is used again in Section 2., where the method used to compute the intervals will be explained. In this sample, there are five players whose heights fall within the interval 59.95–61.95 inches, three players whose heights fall within the interval 61.95–63.95 inches, 15 players whose heights fall within the interval 63.95–65.95 inches, 40 players whose heights fall within the interval 65.95–67.95 inches, 17 players whose heights fall within the interval 67.95–69.95 inches, 12 players whose heights fall within the interval 69.95–71.95, seven players whose heights fall within the interval 71.95–73.95, and one player whose heights fall within the interval 73.95–75.95. All heights fall between the endpoints of an interval and not at the endpoints. Example 1.14 From Table 1.12, find the percentage of heights that are less than 65.95 inches. Solution 1.14 If you look at the first, second, and third rows, the heights are all less than 65.95 inches. There are 5 + 3 + 15 = 23 players whose heights are less than 65.95 inches. The percentage of heights less than 65.95 inches is then 23 100 or 23%. This percentage is the cumulative relative frequency entry in the third row. 1.14 Table 1.13 shows the amount, in inches, of annual rainfall in a sample of towns. Rainfall (Inches) Frequency Relative Frequency Cumulative Relative Frequency 2.95–4.97 4.97–6.99 6.99–9.01 9.01–11.03 11.03–13.05 Table 1.13 6 7 15 8 9 = 0.12 = 0.14 = 0.30 = 0.16 = 0.18 6 50 7 50 15 50 8 50 9 50 0.12 0.12 + 0.14 = 0.26 0.26 + 0.30 = 0.56 0.56 + 0.16 = 0.72 0.72 + 0.18 = 0.90 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 1 \| SAMPLING AND DATA 33 Rainfall (Inches) Frequency Relative Frequency Cumulative Relative Frequency 13.05–15.07 5 = 0.10 5 50 0.90 + 0.10 = 1.00 Total = 50 Total = 1.00 Table 1.13 From Table 1.13, find the percentage of rainfall that is less than 9.01 inches. Example 1.15 From Table 1.12, find the percentage of heights that fall between 61.95 and 65.95 inches. Solution 1.15 Add the relative frequencies in the second and third rows: 0.03 + 0.15 = 0.18 or 18%. 1.15 From Table 1.13, find the percentage of rainfall that is between 6.99 and 13.05 inches. Example 1.16 Use the heights of the 100 male semiprofessional soccer players in Table 1.12. Fill in the blanks and check your answers. a. The percentage of heights that are from 67.95 to 71.95 inches is: ____. b. The percentage of heights that are from 67.95 to 73.95 inches is: ____. c. The percentage of heights that are more than 65.95 inches is: ____. d. The number of players in the sample who are between 61.95 and 71.95 inches tall is: ____. e. What kind of data are the heights? f. Describe how you could gather this data (the heights) so that the data are characteristic of all male semiprofessional soccer players. Remember, you count frequencies. To find the relative frequency, divide the frequency by the total number of data values. To find the cumulative relative frequency, add all of the previous relative frequencies to the relative frequency for the current row. Solution 1.16 a. 29% b. 36% c. 77% d. 87 e. quantitative continuous f. get rosters from each team and choose a simple random sample from each 34 CHAPTER 1 \| SAMPLING AND DATA 1.16 From Table 1.13, find the number of towns that have rainfall between 2.95 and 9.01 inches. In your class, have someone conduct a survey of the number of siblings (brothers and sisters) each student has. Create a frequency table. Add to it a relative frequency column and a cumulative relative frequency column. Answer the following questions: 1. What percentage of the students in your class have no siblings? 2. What percentage of the students have from one to three siblings? 3. What percentage of the students have fewer than three siblings? Example 1.17 Nineteen people were asked how many miles, to the nearest mile, they commute to work each day. The data are as follows: 2; 5; 7; 3; 2; 10; 18; 15; 20; 7; 10; 18; 5; 12; 13; 12; 4; 5; 10. Table 1.14 was produced: DATA FREQUENCY RELATIVE FREQUENCY CUMULATIVE RELATIVE FREQUENCY 3 4 5 7 10 12 13 15 18 20 19 1 19 3 19 2 19 4 19 2 19 1 19 1 19 1 19 1 19 0.1579 0.2105 0.1579 0.2632 0.4737 0.7895 0.8421 0.8948 0.9474 1.0000 Table 1.14 Frequency of Commuting Distances a. Is the table correct? If it is not correct, what is wrong? This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 1 \| SAMPLING AND DATA 35 b. True or False: Three percent of the people surveyed commute three miles. If the statement is not correct, what should it be? If the table is incorrect, make the corrections. c. What fraction of the people surveyed commute five or seven miles? d. What fraction of the people surveyed commute 12 miles or more? Less than 12 miles? Between five and 13 miles (not including five and 13 miles)? Solution 1.17 a. No. The frequency column sums to 18, not 19. Not all cumulative relative frequencies are correct. b. False. The frequency for three miles should be one; for two miles (left out), two. The cumulative relative frequency column should read: 0.1052, 0.1579, 0.2105, 0.3684, 0.4737, 0.6316, 0.7368, 0.7895, 0.8421, 0.9474, 1.0000. c. d. 5 19 7 19 , 12 19 , 7 19 1.17 Table 1.13 represents the amount, in inches, of annual rainfall in a sample of towns. What fraction of towns surveyed get between 11.03 and 13.05 inches of rainfall each year? Example 1.18 Table 1.15 contains the total number of deaths worldwide as a result of earthquakes for the period from 2000 to 2012. Year Total Number of Deaths 2000 231 2001 21,357 2002 11,685 2003 33,819 2004 228,802 2005 88,003 2006 6,605 2007 712 2008 88,011 2009 1,790 2010 320,120 2011 21,953 2012 768 Total 823,356 Table 1.15 Answer the following questions. 36 CHAPTER 1 \| SAMPLING AND DATA a. What is the frequency of deaths measured from 2006 through 2009? b. What percentage of deaths occurred after 2009? c. What is the relative frequency of deaths that occurred in 2003 or earlier? d. What is the percentage of deaths that occurred in 2004? e. What kind of data are the numbers of deaths? f. The Richter scale is used to quantify the energy produced by an earthquake. Examples of Richter scale numbers are 2.3, 4.0, 6.1, and 7.0. What kind of data are these numbers? Solution 1.18 a. 97,118 (11.8%) b. 41.6% c. 67,092/823,356 or 0.081 or 8.1 % d. 27.8% e. Quantitative discrete f. Quantitative continuous 1.18 Table 1.16 contains the total number of fatal motor vehicle traffic crashes in the United States for the period from 1994 to 2011. Year Total Number of Crashes Year Total Number of Crashes 1994 36,254 1995 37,241 1996 37,494 1997 37,324 1998 37,107 1999 37,140 2000 37,526 2001 37,862 2002 38,491 2003 38,477 Table 1.16 2004 38,444 2005 39,252 2006 38,648 2007 37,435 2008 34,172 2009 30,862 2010 30,296 2011 29,757 Total 653,782 Answer the following questions. a. What is the frequency of deaths measured from 2000 through 2004? b. What percentage of deaths occurred after
2006? c. What is the relative frequency of deaths that occurred in 2000 or before? d. What is the percentage of deaths that occurred in 2011? e. What is the cumulative relative frequency for 2006? Explain what this number tells you about the data. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 1 \| SAMPLING AND DATA 37 1.4 \| Experimental Design and Ethics Does aspirin reduce the risk of heart attacks? Is one brand of fertilizer more effective at growing roses than another? Is fatigue as dangerous to a driver as the influence of alcohol? Questions like these are answered using randomized experiments. In this module, you will learn important aspects of experimental design. Proper study design ensures the production of reliable, accurate data. The purpose of an experiment is to investigate the relationship between two variables. When one variable causes change in another, we call the first variable the explanatory variable. The affected variable is called the response variable. In a randomized experiment, the researcher manipulates values of the explanatory variable and measures the resulting changes in the response variable. The different values of the explanatory variable are called treatments. An experimental unit is a single object or individual to be measured. You want to investigate the effectiveness of vitamin E in preventing disease. You recruit a group of subjects and ask them if they regularly take vitamin E. You notice that the subjects who take vitamin E exhibit better health on average than those who do not. Does this prove that vitamin E is effective in disease prevention? It does not. There are many differences between the two groups compared in addition to vitamin E consumption. People who take vitamin E regularly often take other steps to improve their health: exercise, diet, other vitamin supplements, choosing not to smoke. Any one of these factors could be influencing health. As described, this study does not prove that vitamin E is the key to disease prevention. Additional variables that can cloud a study are called lurking variables. In order to prove that the explanatory variable is causing a change in the response variable, it is necessary to isolate the explanatory variable. The researcher must design her experiment in such a way that there is only one difference between groups being compared: the planned treatments. This is accomplished by the random assignment of experimental units to treatment groups. When subjects are assigned treatments randomly, all of the potential lurking variables are spread equally among the groups. At this point the only difference between groups is the one imposed by the researcher. Different outcomes measured in the response variable, therefore, must be a direct result of the different treatments. In this way, an experiment can prove a cause-and-effect connection between the explanatory and response variables. The power of suggestion can have an important influence on the outcome of an experiment. Studies have shown that the expectation of the study participant can be as important as the actual medication. In one study of performance-enhancing drugs, researchers noted: Results showed that believing one had taken the substance resulted in [performance] times almost as fast as those associated with consuming the drug itself. In contrast, taking the drug without knowledge yielded no significant performance increment.[1] When participation in a study prompts a physical response from a participant, it is difficult to isolate the effects of the explanatory variable. To counter the power of suggestion, researchers set aside one treatment group as a control group. This group is given a placebo treatment–a treatment that cannot influence the response variable. The control group helps researchers balance the effects of being in an experiment with the effects of the active treatments. Of course, if you are participating in a study and you know that you are receiving a pill which contains no actual medication, then the power of suggestion is no longer a factor. Blinding in a randomized experiment preserves the power of suggestion. When a person involved in a research study is blinded, he does not know who is receiving the active treatment(s) and who is receiving the placebo treatment. A double-blind experiment is one in which both the subjects and the researchers involved with the subjects are blinded. Example 1.19 Researchers want to investigate whether taking aspirin regularly reduces the risk of heart attack. Four hundred men between the ages of 50 and 84 are recruited as participants. The men are divided randomly into two groups: one group will take aspirin, and the other group will take a placebo. Each man takes one pill each day for three years, but he does not know whether he is taking aspirin or the placebo. At the end of the study, researchers count the number of men in each group who have had heart attacks. Identify the following values for this study: population, sample, experimental units, explanatory variable, response variable, treatments. Solution 1.19 The population is men aged 50 to 84. The sample is the 400 men who participated. 1. McClung, M. Collins, D. “Because I know it will!”: placebo effects of an ergogenic aid on athletic performance. Journal of Sport & Exercise Psychology. 2007 Jun. 29(3):382-94. Web. April 30, 2013. 38 CHAPTER 1 \| SAMPLING AND DATA The experimental units are the individual men in the study. The explanatory variable is oral medication. The treatments are aspirin and a placebo. The response variable is whether a subject had a heart attack. Example 1.20 The Smell & Taste Treatment and Research Foundation conducted a study to investigate whether smell can affect learning. Subjects completed mazes multiple times while wearing masks. They completed the pencil and paper mazes three times wearing floral-scented masks, and three times with unscented masks. Participants were assigned at random to wear the floral mask during the first three trials or during the last three trials. For each trial, researchers recorded the time it took to complete the maze and the subject’s impression of the mask’s scent: positive, negative, or neutral. a. Describe the explanatory and response variables in this study. b. What are the treatments? c. d. Identify any lurking variables that could interfere with this study. Is it possible to use blinding in this study? Solution 1.20 a. The explanatory variable is scent, and the response variable is the time it takes to complete the maze. b. There are two treatments: a floral-scented mask and an unscented mask. c. All subjects experienced both treatments. The order of treatments was randomly assigned so there were no differences between the treatment groups. Random assignment eliminates the problem of lurking variables. d. Subjects will clearly know whether they can smell flowers or not, so subjects cannot be blinded in this study. Researchers timing the mazes can be blinded, though. The researcher who is observing a subject will not know which mask is being worn. Example 1.21 A researcher wants to study the effects of birth order on personality. Explain why this study could not be conducted as a randomized experiment. What is the main problem in a study that cannot be designed as a randomized experiment? Solution 1.21 The explanatory variable is birth order. You cannot randomly assign a person’s birth order. Random assignment eliminates the impact of lurking variables. When you cannot assign subjects to treatment groups at random, there will be differences between the groups other than the explanatory variable. 1.21 You are concerned about the effects of texting on driving performance. Design a study to test the response time of drivers while texting and while driving only. How many seconds does it take for a driver to respond when a leading car hits the brakes? a. Describe the explanatory and response variables in the study. b. What are the treatments? c. What should you consider when selecting participants? d. Your research partner wants to divide participants randomly into two groups: one to drive without distraction and one to text and drive simultaneously. Is this a good idea? Why or why not? e. Identify any lurking variables that could interfere with this study. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 1 \| SAMPLING AND DATA 39 f. How can blinding be used in this study? Ethics The widespread misuse and misrepresentation of statistical information often gives the field a bad name. Some say that “numbers don’t lie,” but the people who use numbers to support their claims often do. A recent investigation of famous social psychologist, Diederik Stapel, has led to the retraction of his articles from some of the world’s top journals including Journal of Experimental Social Psychology, Social Psychology, Basic and Applied Social Psychology, British Journal of Social Psychology, and the magazine Science. Diederik Stapel is a former professor at Tilburg University in the Netherlands. Over the past two years, an extensive investigation involving three universities where Stapel has worked concluded that the psychologist is guilty of fraud on a colossal scale. Falsified data taints over 55 papers he authored and 10 Ph.D. dissertations that he supervised. Stapel did not deny that his deceit was driven by ambition. But it was more complicated than that, he told me. He insisted that he loved social psychology but had been frustrated by the messiness of experimental data, which rarely led to clear conclusions. His lifelong obsession with elegance and order, he said, led him to concoct sexy results that journals found attractive. “It was a quest for aesthetics, for beauty—instead of the truth,” he said. He described his behavior as an addic
tion that drove him to carry out acts of increasingly daring fraud, like a junkie seeking a bigger and better high.[2] The committee investigating Stapel concluded that he is guilty of several practices including: • creating datasets, which largely confirmed the prior expectations, • altering data in existing datasets, • changing measuring instruments without reporting the change, and • misrepresenting the number of experimental subjects. Clearly, it is never acceptable to falsify data the way this researcher did. Sometimes, however, violations of ethics are not as easy to spot. Researchers have a responsibility to verify that proper methods are being followed. The report describing the investigation of Stapel’s fraud states that, “statistical flaws frequently revealed a lack of familiarity with elementary statistics.”[3] Many of Stapel’s co-authors should have spotted irregularities in his data. Unfortunately, they did not know very much about statistical analysis, and they simply trusted that he was collecting and reporting data properly. Many types of statistical fraud are difficult to spot. Some researchers simply stop collecting data once they have just enough to prove what they had hoped to prove. They don’t want to take the chance that a more extensive study would complicate their lives by producing data contradicting their hypothesis. Professional organizations, like the American Statistical Association, clearly define expectations for researchers. There are even laws in the federal code about the use of research data. When a statistical study uses human participants, as in medical studies, both ethics and the law dictate that researchers should be mindful of the safety of their research subjects. The U.S. Department of Health and Human Services oversees federal regulations of research studies with the aim of protecting participants. When a university or other research institution engages in research, it must ensure the safety of all human subjects. For this reason, research institutions establish oversight committees known as Institutional Review Boards (IRB). All planned studies must be approved in advance by the IRB. Key protections that are mandated by law include the following: • Risks to participants must be minimized and reasonable with respect to projected benefits. • Participants must give informed consent. This means that the risks of participation must be clearly explained to the subjects of the study. Subjects must consent in writing, and researchers are required to keep documentation of their consent. • Data collected from individuals must be guarded carefully to protect their privacy. These ideas may seem fundamental, but they can be very difficult to verify in practice. Is removing a participant’s name from the data record sufficient to protect privacy? Perhaps the person’s identity could be discovered from the data that remains. What happens if the study does not proceed as planned and risks arise that were not anticipated? When is informed 2. Yudhijit Bhattacharjee, “The Mind of a Con Man,” Magazine, New York Times, April 26, 2013. Available online at: http://www.nytimes.com/2013/04/28/magazine/diederik-stapels-audacious-academic-fraud.html?src=dayp&_r=2& (accessed May 1, 2013). 3. “Flawed Science: The Fraudulent Research Practices of Social Psychologist Diederik Stapel,” Tillburg University, November 28, 2012, http://www.tilburguniversity.edu/upload/064a10cdbce5-4385-b9ff-05b840caeae6_120695_Rapp_nov_2012_UK_web.pdf (accessed May 1, 2013). 40 CHAPTER 1 \| SAMPLING AND DATA consent really necessary? Suppose your doctor wants a blood sample to check your cholesterol level. Once the sample has been tested, you expect the lab to dispose of the remaining blood. At that point the blood becomes biological waste. Does a researcher have the right to take it for use in a study? It is important that students of statistics take time to consider the ethical questions that arise in statistical studies. is fraud in statistical studies? You might be surprised—and disappointed. There is a website How prevalent (www.retractionwatch.com) (http://www.retractionwatch.com) dedicated to cataloging retractions of study articles that have been proven fraudulent. A quick glance will show that the misuse of statistics is a bigger problem than most people realize. Vigilance against fraud requires knowledge. Learning the basic theory of statistics will empower you to analyze statistical studies critically. Example 1.22 Describe the unethical behavior in each example and describe how it could impact the reliability of the resulting data. Explain how the problem should be corrected. A researcher is collecting data in a community. a. She selects a block where she is comfortable walking because she knows many of the people living on the street. b. No one seems to be home at four houses on her route. She does not record the addresses and does not return at a later time to try to find residents at home. c. She skips four houses on her route because she is running late for an appointment. When she gets home, she fills in the forms by selecting random answers from other residents in the neighborhood. Solution 1.22 a. By selecting a convenient sample, the researcher is intentionally selecting a sample that could be biased. Claiming that this sample represents the community is misleading. The researcher needs to select areas in the community at random. b. c. Intentionally omitting relevant data will create bias in the sample. Suppose the researcher is gathering information about jobs and child care. By ignoring people who are not home, she may be missing data from working families that are relevant to her study. She needs to make every effort to interview all members of the target sample. It is never acceptable to fake data. Even though the responses she uses are “real” responses provided by other participants, the duplication is fraudulent and can create bias in the data. She needs to work diligently to interview everyone on her route. 1.22 Describe the unethical behavior, if any, in each example and describe how it could impact the reliability of the resulting data. Explain how the problem should be corrected. A study is commissioned to determine the favorite brand of fruit juice among teens in California. a. The survey is commissioned by the seller of a popular brand of apple juice. b. There are only two types of juice included in the study: apple juice and cranberry juice. c. Researchers allow participants to see the brand of juice as samples are poured for a taste test. d. Twenty-five percent of participants prefer Brand X, 33% prefer Brand Y and 42% have no preference between the two brands. Brand X references the study in a commercial saying “Most teens like Brand X as much as or more than Brand Y.” This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 1 \| SAMPLING AND DATA 41 1.1 Data Collection Experiment Class Time: Names: Student Learning Outcomes • The student will demonstrate the systematic sampling technique. • The student will construct relative frequency tables. • The student will interpret results and their differences from different data groupings. Movie Survey Ask five classmates from a different class how many movies they saw at the theater last month. Do not include rented movies. 1. Record the data. 2. In class, randomly pick one person. On the class list, mark that person’s name. Move down four names on the class list. Mark that person’s name. Continue doing this until you have marked 12 names. You may need to go back to the start of the list. For each marked name record the five data values. You now have a total of 60 data values. 3. For each name marked, record the data. ______ ______ ______ ______ ______ ______ ______ ______ ______ ______ ______ ______ ______ ______ ______ ______ ______ ______ ______ ______ ______ ______ ______ ______ ______ ______ ______ ______ ______ ______ ______ ______ ______ ______ ______ ______ ______ ______ ______ ______ ______ ______ ______ ______ ______ ______ ______ ______ ______ ______ ______ ______ ______ ______ ______ ______ Table 1.17 Order the Data Complete the two relative frequency tables below using your class data. Number of Movies Frequency Relative Frequency Cumulative Relative Frequency 0 1 2 3 4 5 6 42 CHAPTER 1 \| SAMPLING AND DATA Number of Movies Frequency Relative Frequency Cumulative Relative Frequency 7+ Table 1.18 Frequency of Number of Movies Viewed Number of Movies Frequency Relative Frequency Cumulative Relative Frequency 0–1 2–3 4–5 6–7+ Table 1.19 Frequency of Number of Movies Viewed 1. Using the tables, find the percent of data that is at most two. Which table did you use and why? 2. Using the tables, find the percent of data that is at most three. Which table did you use and why? 3. Using the tables, find the percent of data that is more than two. Which table did you use and why? 4. Using the tables, find the percent of data that is more than three. Which table did you use and why? Discussion Questions 1. Is one of the tables “more correct” than the other? Why or why not? 2. In general, how could you group the data differently? Are there any advantages to either way of grouping the data? 3. Why did you switch between tables, if you did, when answering the question above? This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 1 \| SAMPLING AND DATA 43 1.2 Sampling Experiment Class Time: Names: Student Learning Outcomes • The student will demonstrate the simple random, systematic, stratified, and cluster sampling techniques. • The student will explain the details of each procedure used. In this lab, you will be asked to pick several random samples of restaurants. In each case, describe your procedure briefly, including how you might have used the random number generator, and then li

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