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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 determine 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 discrim |
inant 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 cosine 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 Param |
eterization of Transformations Given the parent relation represent the relation, 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 solving 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 Hyper |
bolas 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 leads 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; vert |
ices (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 Algebraically 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 reduced row echelon form matrices • Solve applications by using mat |
rices • 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 3x 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 ainbnj 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 units 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 |
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