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Chapter 10: Vectors and Parametric Equations Lesson 10.1.1 10-1. a. Yes, assuming they made the correct moves. b. No, everyone was not in the same location. 10-2. a. Yes, the two vectors represent the same instruction. b. No, the starting points are not the same. c. d. e. Steps 1 and 5 are the same. Steps 4 and 7 are the same. 2, 0 + 0, 1 = 2, 1 Magnitude: 2 2 + 12 = 5 tan ! = 12 Angle: tan "1 tan ! = tan "1 ( 12 ) ! = 26.6! 10-3. a. Other equivalent vectors are s and r, and l and k. b. x = 6, !1 Any vector with this instruction will be equivalent, and everyone on the team should have equivalent vectors. 10-4. a. See diagram at right. b. See diagram at right. c. r = 7, 2 p = 5, !2 b = r + p = 7, 2 + 5, !2 = 13, 0 d. r + m = 7, 2 + 0, !3 = 7, !1 To add vectors in component form, add the horizontal components together to get the horizontal component of the resultant. The same applies for the vertical components. 10-5. v = 2, !3 2, !3 + u1, u 2 = !3, !1 w = !3, !1 u1, u 2 = !3, !1 ! 2, !3 u1, u 2 = !3 ! 2, !1 ! (!3) u1, u 2 = !5, 2 CPM Educational Program © 2012 Chapter 10: Page 1 Pre-Calculus with Trigonometry Review and Preview 10.1.1 10-6. a. c. Due East is 90º. 10º east of due south = 180º – 10º = 170º 10-7. a. 90º + 30º = 120º c. 270º + 75º = 345º 10-8. ( 2x ! 1x ) 5 ( 1x ) + 10(2x)3 ( ! 1x ) = (2x)5 + 5(2x)4 ! = 32x 5 ! 80 x 4 x + 80 x 3 x2 ! = 32x 5 ! 80x 3 + 80x ! 2 2 5 sin !1 sin x = sin !1 Southwest is 180º + 45 = 225º b. 180º + 67º = 247º ( 1x ) + 10(2x)2 ! 3 ( 1x ) + ( ! 1x ) 4 + 5(2x) ! 5 40 x 2 + 10 x ! 1 x3 x4 x5 40 + 10 ! 1 x x3 x5 10-9. 7 sin x ! 9 = 2 sin x ! 7 a. 5 sin x = 2 sin x = b. b. 4 sin 2 ! = 3 sin 2 ! = 3 4 sin ! = ± ( 25 ) != x = 0.412 x = " ! 0.412 = 2.73 " 3 3 4 =± 3 2 , 23" , 43" , 53" 10-10. a. See diagram at right. b. See diagram at right. c. See diagram at right. d. The quadrilateral is a parallelogram. e. Opposite sides have equal slope, thus are parallel, so the quadrilateral is a parallelogram. 10-11. f (x) = 2 x 2 !16 x 2 ! x!6 = 2(x 2 !8) (x! 3)(x+2) Vertical asymptote at x = 3 and x = !2 Horizontal asymptote at y = 2 . No holes. 10-12. CPM Educational Program © 2012 Chapter 10: Page 2 Pre-Calculus with Trigonometry The answer is c. x!y x"y = (x+ y)2 x 2 # y2 = (x+ y)(x+ y) (x+ y)(x# y) = x+ y x# y , for x $ y 10-13. 22 2 + 22 2 = c 2 Bearing of 135º or standard angle of –45°. 22 miles east 2 ! 22 2 = c 2 22 miles south 22 2 = c 10-14. This is a 30 - 60 - 90 right triangle. Thus the horizontal leg is 5 3 and the vertical leg is 5. Horizontal vector: 5 3, 0 Vertical vector: 0, 5 Resultant vector: 5 3, 5 Lesson 10.1.2 10-15. a. Answers vary. b. 3, 5 ! 7, 2 = 3 ! 7, 5 ! 2 = !4, 3 (!4)2 + 32 = 25 = 5 c. 10-16. a. 6i ! 2 j b. !1, 3 c. 2i 10-17. a. See diagram at right. b. This is a 45 - 45 - 90 right triangle. Therefore the horizontal and vertical components are equal. x = 50 = 25 2 2 c. d. 50 x 45º x i, j form: !25 2i ! 25 2 j The horizontal component. 25 2 ≈ 35.355 lbs 10-18. a. b. c. 4 2 + 32 = 25 = 5 The resultant vector is 1 unit long. 3 i+ 4 j 5 5 CPM Educational Program © 2012 Chapter 10: Page 3 Pre-Calculus with Trigonometry 10-19. a + a = 2a a. 0a is equivalent to 0. !1a is equivalent to !a b. c. They are in the opposite direction. b = 3, 2 e. 1 2 b= 3 ,1 2 3b = 3 3, 2 = 9, 6 10-20. a. Force, weight, wind; vector quantities must have both magnitude and direction. b. 35 mph c. 35 mph (no direction mentioned) d. The weight of a dictionary has direction, straight down. Review and Preview 10.1.2 10-21. a. a = 7, 3 b. r = 3, 6 b = 5, !5 s = 8, !5 !b = !5, 5 r ! s = 3, 6 ! 8, !5 = !5, 11 a + ! b = 7, 3 + !5, 5 = 2, 8 a ! b = 7, 3 ! 5, !5 = 7 ! 5, 3 ! (!5) = 2, 8 b 10-22. a. See diagram at right. b. a = 2, 3 a b = 1, !1 c. d. c a + b = 2, 3 + 1, !1 = 3, 2 Use a vector equivalent to b which begins at the end point of a. a + b is then the vector from the initial point of a to the end point of b. 3i + 2 j b = 1, !1 b + c = 1, !1 + !3, 2 = !2, 1 c = !3, 2 b + c = !2i + j c = !3, 2 c ! a = !3, 2 ! 2, 3 = !3 ! 2, 2 ! 3 = !5, !1 a = 2, 3 c ! a = !5i ! j CPM Educational Program © 2012 Chapter 10: Page 4 Pre-Calculus with Trigonometry 10-23. a. p = 8, !4 b. u = !3, 5 ,!z = !3, !5 ,!u + z = !3, 5 + !3, !5 = !6, 0 v= 1 2 8, !4 = 1 2 " 8, 12 " !4 = 4, !2 c. 1 2 d. u = !3, 5 ,!v = 8, !4 ,!u ! v = !3, 5 ! 8, !4 = !3 ! 8, 5 ! (!4) = !11, 9 10-24. a. An example is 5, 0 . There are many other answers. b. An example is 2 3, 4 = 6, 8 . There are many other answers. 10-25. a. cos 25! = sin 25! = x 20 x = 20 cos 25! = 20 ! 0.9063 = 18.126 y 20 y = 20 sin 25! = 20 ! 0.4226 = 8.452 b. Component form: 18.126, 8.452 18.126, 8.452 + 0, !5 = 18.126, 3.452 c. d. = 5.517 seconds 5.517 seconds ! 3.452ft/sec = 19.045 feet 100ft 18.126ft/sec 10-26. a. See diagram at right. 5 i + 12 j b. 13 13 c. Divide the coefficients of i and j by the magnitude of the vector. 10-27. Slope of v = ! 23 . ! slope = 1 12/13 5/13 3i + 2 j or ! 3i ! 2 j 2 3 Horizontal component = 3 Vertical component = 2 10-28. a. b. 3x 2 #5 x+2 = lim 3x 2 #5 x+2 = # 3 5 x!" 7#5 x 2 x!" #5 x 2 + 7 2 4(x+1)(x"1) lim 4 x3 " 4 = lim x(x+1)(x"1) = lim 4x = x!1 x " x x!1 x!1 lim 4 10-29. f (x) = 5 x ! x 3 f "(x) = 5 x # ln 5 ! 3x 2 f "(1) = 5 # ln 5 ! 3 = 5.047 CPM Educational Program © 2012 Chapter 10: Page 5 Pre-Calculus with Trigonometry 10-30. Vertical shift: 178+186 = 182 2 186!178 = 8 = 4 Amplitude: 2 2 2! ! Period: 11 = 5.5 Horizontal shift: 9 : 20 AM = 9 13 = 9.3333 ( 5.5! (x " 9.333) ) + 182 ! 184 = 4 cos ( 5.5 (x " 9.333) ) + 182 ! 2 = 4 cos ( 5.5 (x " 9.333) ) ! cos"1 ( 12 ) = cos"1 cos ( 5.5 (x " 9.333) ) y = 4 cos 1.047 = ! 5.5 (x " 9.333) 1.83333 = x " 9.333 x = 11.1667 x = 11.1667 " 3.6667 = 7.5 Earliest: 7:30 AM Latest: 11:10 AM Lesson 10.1.3 10-31. a. Draw a line due south from B, parallel to A. Call this point D. !BAD = 90! " 75! = 15! !ABD = 180! " 90! " 15! = 75! !B = 180! " 75! = 105! b. 105! + 32! = 137! c. c 2 = 675 2 + 140 2 ! 2(675)(140) cos(137! ) c 2 = 475225 + 189000 " 0.7314 c 2 = 613450.8496 c = 783.23 m CPM Educational Program © 2012 Chapter 10: Page 6 Pre-Calculus with Trigonometry d. 783.23 sin 137! = 140 sin ! 95.4798 = 783.23 sin ! 0.1219 = sin ! ! = 7! "BAC = 7! 75! # 7! = 68! 10-32. 68! a. 783.23 mph b. c. d. Bearing would be 68.0°. 783.23 ! 2 = 1566.46 miles Virtually the same as the last one. 10-33. a. Standard angle = 90! ! 75! = 15! cos 15! = x= b. x 783.23 675 cos 15! sin 15! = = 652 y 675 Component form = 652, 174.703 y = 675 sin 15! = 174.703 Standard angle = 90! ! 32! = 58! y x cos 58! = 140 sin 58! = 140 x = 140 cos 58! = 74.189 y = 140 sin 58! = 118.727 Component form = 74.189, 118.727 c. 652, 174.703 + 74.189, 118.727 = 726.187, 293.43 d. 726.187 2 + 293.432 = 783.23 e. 22°, bearing = 90! ! 22! = 68! 10-34. a. Standard angle = 90! ! 40! = 50! cos 50! = x 20 x = 20 cos 50! = 12.856 b. sin 50! = y 20 Component form = 12.856, 15.321 y = 20 sin 50! = 15.321 Standard angle = 90! + 20! = 110! y x cos 110! = 10 sin 110! = 10 x = 10 cos 110! = !3.420 y = 10 sin 110! = 9.397 Component form = !3.420, 9.397 c. 7, 0 d. 12.856, 15.321 + !3.420, 9.397 + 7, 0 = 16.436, 24.718 e. 16.436 2 + 24.718 2 = 881.1216 = 29.684 mph CPM Educational Program © 2012 Chapter 10: Page 7 Pre-Calculus with Trigonometry 16.436 29.684 = cos ! Bearing of 90! ! 56.38! = 33.62! 0.5537 = cos ! cos"1 0.5537 = cos"1 cos ! ! = 56.38! CPM Educational Program © 2012 Chapter 10: Page 8 Pre-Calculus with Trigonometry 10-35. a. 0, ! 450 b. c. d. v cos(35! ), v sin(35! ) u cos(165°) + v cos(35°) = 0 , u sin(165°) + v sin(35°) = 450 u cos(165°) + v cos(35°) = 0 !0.9659 u + 0.8192 v = 0 0.8192 v = 0.9659 u v = 1.1791 u u sin(165°) + 1.1791 u ! sin(35°) = 450 0.2588 u + 0.6763 u = 450 0.9351 u = 450 u = 481.2 pounds v = 1.1791 u v = 1.1791! 481.2 = 567.4 pounds Review and Preview 10.1.3 210 mph 64 mph 10-36. a/b. See diagram at right. c. c 2 = 210 2 + 64 2 ! 2(210)(64) cos 45! c 2 = 48196 ! 19007.03 64 sin = 29188.97 c = 170.848 mph c2 45º 64 sin ! = 170.848! 45! = 170.848 sin ! 45.2548 170.848 sin 45 = sin ! ! = 15.4 ! The bearing will be 270! ! 15.4 ! = 254.6! 10-37. 170.848 !1.5 = 256.272!miles a. c 2 = 256.272 2 + 315 2 " 2(315)(256.272) cos(15.4 ! ) c 2 = 164900.338 " 155654.51 = 9245.8238 c = 96 96 miles SE of Houston Solution continues on next page. → CPM Educational Program © 2012 Chapter 10: Page 9 Pre-Calculus with Trigonometry b. Distance from due south to Houston: tan 15.4 ! = y 315 y = 315 tan 15.4 ! y = 86.7655!miles Distance from New Orleans to due south: c 2 = 315 2 + 86.7655 2 , c = 326.73 miles 326.73 , t = 1.912 Time to travel from New Orleans to due south of Houston: t = 170.848 hours Speed traveling due south of Houston to Houston: b 2 = 210 2 ! 45.2538 2 , b = 205.07 , t = 0.543 Time to travel from due south of Houston to Houston: t = 86.7655 205.07 Total time: t = 1.912 + 0.543, t = 2.455, t ! 2 hours, 27 minutes The entire trip takes about 2 hours, 27 minutes. wind Houston 210 mph b sin ! = c. sin "1 sin ! = 45.2538 = 0.2155 210 sin "1 (0.2155) 45.25 mph 64 mph 45.25 ! = 12.4 ! mph Bearing 270! + 12.4 ! = 282.4 ! x 2 = 210 2 ! 45.2538 2 x = 205.0661 Speed = 205.0661 ! 45.2538 = 159.8123!mph 210 mph θ Houston 315 = 1.971 t = 159.8123 1.971 hours or 1 hour, 58.3 minutes. 10-38. a. Magnitude = !3 = 3 2 cos " , (!3)2 + 32 = 18 = 3 2 , cos " = ! "= b. Magnitude = 2 5 2 + 5 3 = 25 + 75 = 10 , 1 2 3# 4 3 = 3 2 sin ! sin ! = != 1 2 3" 4 5 = 10 cos ! , 5 3 = 10 sin ! cos ! = != 1 2 " 3 sin ! = != 3 2 " 3 10-39. a. x = 5 cos !6 = 5 " 3 2 = 2.5 3 b. y = 5 sin !6 = 5 " 12 = 2.5 y = 10 sin 54! = "10 # 2 = "5 2 2 = "5 2 2 2 "5 2, "5 2 2.5 3, 2.5 CPM Educational Program © 2012 x = 10 cos 54! = "10 # Chapter 10: Page 10 Pre-Calculus with Trigonometry c. x = 15 cos 23! = "15 # 12 = "7.5 y = 15 sin 23! = 15 # 3 2 = 7.5 3 "7.5, 7.5 3 CPM Educational Program © 2012 Chapter 10: Page 11 Pre-Calculus with Trigonometry 10-40. a. !5, 7 + 3, ! 3 = !5 + 3, 7 + (!3) = !2, 4 b. (!7i ! 2 j) ! (3i ! j) = !7, !2 ! 3, !1 = !10, !1 = !10i ! j !4, 4 + 4, ! 4 = !4 + 4, 4 + (!4) = 0, 0 c. 10-41. 15 sin 20° = 5.13 km/hr 10-42. a. b. 3 2 3x(2 x# 3)2 3x(4 x 2 #12 x+9) = lim = lim 12 x # 363 x +27 x 3 3 #3x +50 #3x +50 x!" 50# 3x x!" x!" 2 5(x"2)(x+2) 5(x+2) lim 5 x "20 = lim (x"2)(x+ 3) = lim (x+ 3) = 20 =4 5 x!2 x 2 + x"6 x!2 x!2 lim = 12 #3 = #4 10-43. h 20 = 1.8 20!d 36 = h(20 ! d), 36 (20!d ) =h 10-44. f (2+h)" f (2) 2+h"2 h!0 f (2 + h) = 2 x+h lim f (2) = 2 2 = 4 10-45. a. 2 ln 3 ! ln 5 = b. 1 2 ln 32 ln 5 = ln 2 2+h " 4 h h!0 = lim 4#2 h " 4 h h!0 = lim $ 2.773 ( 95 ) ( ln 9 + 4 ln 3 ) = 12 (2 ln 3 + 4 ln 3) = 12 (6 ln 3) = 3 ln 3 = ln 33 = ln 27 CPM Educational Program © 2012 Chapter 10: Page 12 Pre-Calculus with Trigonometry Lesson 10.1.4 10-46. v+u=w a. b. c. u=w!v u = w1, w 2 ! v1, v 2 = w1 ! v1, w 2 ! v 2 Law of cosines. v = (v1 )2 + (v 2 )2 d. 2 v e. = (v1 )2 + (v 2 )2 + (w1 )2 + (w 2 )2 ! 2 v w cos " = (v1 )2 + (v 2 )2 (w1 )2 ! 2w1v1 + (v1 )2 + (w 2 )2 ! 2w 2 v 2 + (v 2 )2 w = (w1 )2 + (w 2 )2 w 2 = (w1 )2 + (w 2 = (v1 )2 + (v 2 )2 + (w1 )2 + (w 2 )2 ! 2 v w cos " )2 u = (w1 ! v1 )2 + (w 2 ! v 2 )2 u 2 = (w1 ! v1 )2 + (w 2 ! v 2 (w1 ! v1 )2 + (w 2 ! v 2 )2 )2 !2w1v1 ! 2w 2 v 2 = !2 v w cos " w1v1 + w 2 v 2 = v w cos " 10-47. a. 3, 2 ! 4, " 2 = 3 ! 4 + 2 ! "2 = 12 " 4 = 8 b. !2, 4 " 6, ! 5 = !2 " 6 + 4 " !5 = !12 ! 20 = !32 c. 2 3, ! 5 " 5 3, 6 = 2 3 " 5 3 + !5 " 6 = 10 " 3 ! 30 = 0 10-48. a. v = 32 + 2 2 = 13 b. w = 6 2 + (!5)2 = 61 w = 4 2 + (!2)2 = 20 = 2 5 cos " = 8 13#2 5 = !32 20 # 61 cos " = !0.9162 4 13# 5 cos " = cos " = 0.4961 " = 60.3! c. v = (!2)2 + 4 2 = 20 " = 156.4 ! v = (2 3)2 + (!5)2 = 12 + 25 = 37 w = (5 3)2 + 6 2 = 75 + 36 = 111 cos " = 0 37 # 111 cos " = 0 " = 90! 10-49. They are perpendicular. 10-50. The dot product is zero. CPM Educational Program © 2012 Chapter 10: Page 13 Pre-Calculus with Trigonometry 10-51. a. cos 30! = sin 30! = x 30 y 30 F ! d = 10, 0 ! 25.981, "15 b. = 10 ! 25.981 + ("15) ! 0 = 259.81 foot pounds y = 30 sin 30! y = 15 x = 30 cos 30! x = 25.981 F = 25.981, 15 c. F ! d = 9.397, 3.420 ! 25.981, "15 d. x cos 20! = 10 = 9.397 ! 25.981 + 3.420 ! ("15) = 192.843 foot pounds x = 10 cos 20! x = 9.397 y sin 20! = 10 y = 10 sin 20! y = 3.420 F = 9.397, 3.420 10-52. v = (!2)2 + 4 2 = 20 a. m= y!6= y= t=2 b. 2, 4 + 2 3, 2 = 6! 4 = 2 5!2 3 2 (x ! 5) 3 2 (x ! 5) + 6 3 2, 4 + 6, 4 = 8, 8 8= 2 3 (8 ! 5) + 6 8=2+6 8=8 The point (8, 8) is on the line since 8 = 2 3 (8 ! 5) + 6 . 4, 3 ! !2, 5 = 4 ! (!2), 3 ! 5 = 6, !2 c. !2, 5 + t 6, !2 = !2 + 6t, 5 ! 2t 10-53. a. Velocity: t = 4!!! "2# sin Speed: b. ( #2 $ 4 ) , 2# cos ( #2 $ 4 ) = "2# sin(2# ), 2# cos(2# ) = 0, 2# 0 2 + (2! )2 = 2! ft/sec Velocity: t = 0.5!!! "2# sin = "2# $ 2 2 , 2# $ ( #2 $ 0.5 ) , 2# cos( #2 $ 0.5) 2 2 = "2# sin ( #4 ) , 2# cos ( #4 ) = " 2# , 2# (!" 2 )2 + (" 2 )2 = 2" 2 + 2" 2 = 4" 2 = 2" ft/sec c. 4! 2 sin 2 ( !2 t ) + 4! 2 cos2 ( !2 t ) = CPM Educational Program © 2012 ( ( t ) + cos ( t ) ) = 4! 2 sin 2 ! 2 Chapter 10: Page 14 2 ! 2 4! 2 = 2! Pre-Calculus with Trigonometry Review and Preview 10.1.4 10-54. !4, 5 " 2, 7 = !8 + 35 = 27 a. b. 2, !3 " !4, !2 = !8 + 6 = !2 v = (!4)2 + 5 2 = 41 v = 2 2 + (!3)2 = 13 w = 2 2 + 7 2 = 53 w = (!4)2 + (!2)2 = 20 !2 13 " 20 cos # = !0.1240 27 41 " 53 cos # = 0.5792 cos # = cos # = # = 54.6! # = 97.1! 10-55. cos 30! = sin 30! = x 50 y 50 F = 43.3, 25 x = 50 cos 30! = 43.3 y = 50 sin 30! = 25 W = 10, 0 ! 43.3, 25 = 10 ! 43.3 + 0 ! 25 = 433 ft-lbs of work 10-56. v!w = 0 3, 6a ! "16, 2a = 0 "16 ! 3 + 12a 2 = 0 12a 2 = 48 a2 = 4 a = ±2 10-57. 40 sin 90 = y sin 50 ! y= 200 sin 90 = 30.64 sin " ! " = sin #1 40 sin 50 sin 90 40 2 ! 30.64 2 = 25.713 = 30.64 ( 30.64200sin 90 ) = 8.8 bearing = 90 # " = 90 # 8.8 = 81.2! 200 2 ! 30.64 2 = 197.639 197.639 + 25.713 = 223.352 mph 10-58. 3 = !4 Slope: m = !1! 4!2 2 Vector Equation: (2 + 2t)i + (3 ! 4t)j or (4 + 2t)i + (!1 ! 4t)j . Other answers possible. CPM Educational Program © 2012 Chapter 10: Page 15 Pre-Calculus with Trigonometry 10-59. a. Look at vectors (v) and (vi). !4 /5 3/5 ( ! 45 ) + ( 53 ) 2 2 =! 4 3 and =1 ! 12 =! 9 4 3 and (!12)2 + (9)2 = 15 Thus, vectors (v) and (vi) have the same direction but not the same magnitude. b. Look at vectors (i) and (iii) !3 !3 = 1 and (!3)2 + (!3)2 = 18 0 3 2 c. = 0 and (0)2 + (3 2 )2 = 18 These vectors have the same magnitude but not the same direction. The magnitude of vector (v) is 1 (from part a), so this is a unit vector. 10-60. 3 2 2 +h 3 " x 3 f (x+h)" f (x) (x+h)3 " x 3 = lim = lim x + 3x h+ 3xh h h h h!0 h!0 h!0 2 2 3 +h = lim 3x 2 + 3xh + h 2 = lim 3x h+ 3xh h h!0 h!0 lim = 3x 2 at x = "2 = 3("2)2 = 12 10-61. a. 2 x 3 "8 x x!2 x 2 + x"6 b. 3 lim 22x "8 x x!"# x + x"6 lim 2 x(x"2)(x+2) x!2 (x"2)(x+ 3) = lim = 3 1 x3 lim 22x "8 x $ 3 x!"# x + x"6 1 x Use a table to figure out which: 3 lim 22x "8 x x!"# x + x"6 2 x(x+2) x!2 (x+ 3) = lim = "# = = 2(2)(2+2) (2+ 3) 2" 8 2 lim 1 1 x 6 x!"# + " x x2 x 3 x y = = 16 5 2 "0 = "# –10 –22.86 –100 –202.06 –1000 –2002.00 10-62. 150 ! sin 20 = 51.3 pounds CPM Educational Program © 2012 Chapter 10: Page 16 Pre-Calculus with Trigonometry Lesson 10.2.1 10-63. b. The car headed due south for about one minute, then due east for about 20 seconds, then southeast at an angle of 45º for about 20 seconds. c. McFreeze made a right turn at the point (600,600) at about 53 seconds. 10-64. The nickels will hit the floor at the same time. 10-65. a. x = 22t 0 11 22 33 Time (sec.) 0 0.5 1 1.5 b. c. d. e. f. y = !16t 2 + 256 256 252 240 220 See graph at right above. Half an upside down parabola. The shell hit the ground when y = 0 and this happened when t = 4 . y would be the same, but x = 10t or x = 40t . See graph at right below. As the speed of the wind increases, the horizontal distance the shell travels increases. 10-67. x = 2t ! t = y = t2 = ( 2x ) 2 x 2 = x2 4 4y = x 2 Review and Preview 10.2.1 10-68. c. Any point with z-coordinate equal to 0 lies in the xy-plane. e. The last point was below the paper. 10-69. When t = 4, x = 0, y is the horizontal displacement. x(4) = 22(4) = 88 ft . 10-70. See graph at right. CPM Educational Program © 2012 Chapter 10: Page 17 Pre-Calculus with Trigonometry ! 0 10-71. a. x = cos ! b. y = sin ! c. See table at right. 1 ! 3 ! 2 2! 3 y -1 1 x -1 d. e. f. x = cos ! , y = sin ! is a unit circle with radius 1, so x = 5 cos ! , y = 5 sin ! is a circle with radius 5 centered at the origin. The center of the circle has x -coordinate = 7 and y-coordinate = 9, so add these values to the x and y equations: x = 5 cos ! + 7, y = 5 sin ! + 9 . x cos 0 = 1 cos y sin 0 = 0 ( !3 ) = 12 sin ( !3 ) = 0 ! 3 2 1 ! 1 2 3 2 –1 0 4! 3 3! 2 5! 3 ! 1 2 ! 0 3 2 -1 1 2 ! 3 2 2! 1 0 7! 3 5! 2 8! 3 1 2 3 2 0 1 ! 1 2 3! 3 2 –1 0 10-72. x = 1 + 3 cos ! , y = 2 + 3 sin ! is a circle with radius 3 centered at (1,2). 10-73. 1250 ! 900 = 350 feet in 30 seconds, or a. b. c. 350 = 35 feet 30 3 second 200 = 20 feet 10 second 950 ! 750 = 200 feet in 10 seconds, or The distance between (1200,600) and (1450,350) is (1200 ! 1450)2 + (600 ! 350)2 = 353.55 feet in 20 seconds, or 353.55 20 feet = 17.68 second . 10-74. v ! u = v u cos " Solving 4a + 15 = 5 a 2 + 25 ! cos 60 for a you get: 4a + 15 = 5 a 2 + 25 ! 12 4a + 15 = 4 2 + 32 a 2 + 5 2 cos 60 (4a + 15)2 = 25(a 2 + 25) ! 14 4a + 15 = 25 a 2 + 5 2 cos 60 16a 2 + 120a + 225 = 25 4 a 2 + 156.25 9.75a 2 " 120a + 68.75 = 0 "120± 120 2 " 4(9.75)(68.75) 2!9.75 "120±108.253 19.5 10-75. a. There is not enough information for a specific time. All we know is the average rate at that time. CPM Educational Program © 2012 Chapter 10: Page 18 =a =a "0.602, "11.705 = a Pre-Calculus with Trigonometry b. 2!60+20+ 3!65 6 = 55.8 mph CPM Educational Program © 2012 Chapter 10: Page 19 Pre-Calculus with Trigonometry Lesson 10.2.2 10-76. a. The circumference is 2! " 3 feet . In one second, the bug traveled 3 feet, which makes up 1 radian. b. Similarly, in t seconds, the bug traveled 3t feet which makes up t radians. c. The equations of this circle are x = 3 cos ! , y = 3 sin ! assuming the center of the table is the origin. Since t radians = t seconds, the location of the bug after t seconds is x = 3 cos t, y = 3 sin t . d. y = 3 sin t . 10-77. a. A sample table: t x(t ) = t cos t 2! 2! 8! ! 43" 3 10! 3 b. ! 5" 3 3! "3! See graph at right. y(t) = t sin t 0 4 3! 3 5 3" ! 3 0 10-78. c. A circle centered at origin with radius 3. 10-79. 10-80. 10-81. b. CPM Educational Program © 2012 Chapter 10: Page 20 Pre-Calculus with Trigonometry 10-82. a. x = t 2 ! (x)1/2 = t ( y = t 4 ! y = (x)1/2 c. b. )4 = x 2 y = x2 The x-values are never negative, so the left half of the graph is missing. 10-83. a. x = t 3 ! (x)1/3 = t !10 " t " 10 # ! 10 " x1/3 " 10 b. (!10)3 " x " (10)3 !1000 " x " 1000 y = t 6 ! y = ((x)1/3 )6 = x 6/3 = x 2 y = x2 10.2.2 Review and Preview 10-84. a. Knowing sin 2 ! + cos2 ! = 1 , let 2! = " . Then, sin 2 2! + cos2 2! = 1 . b. Let ! 2 " 2 = # then sin 2 (! 2 " 2) + cos2 (! 2 " 2) = 1 . 10-85. Two concentric circles. 10-86. a. y = t 2 ! t = y1/2 , x = b. c. 1 t 2 +1 ! x= 1 (y1/2 )2 +1 = ( 1 y+1 ) ( ) 2 1/2 x= ! +1= ! = " 1 , ! t = " 1 , y = = " 1x ! 1 $ = # % Here, x and y would have negative values, as well as positive values. 1 2 t +1 t2 1 x t2 1 x 1 x 1/2 t2 1 x !1 10-87. a. x = sin 2 t ! t = sin "1 (x1/2 ) , y = sin t ! y = sin(sin "1 (x1/2 )) = x1/2 b. x = t 8 ! t = x1/8 , y = t 4 ! y = (x1/8 )4 = x1/2 10-88. a. x = tan t ! t = tan "1 x , y = tan 2 t ! y = tan 2 (tan "1 x) = x 2 b. x = log t ! t = 10 x , y = 1 + t 2 ! y = 1 + (10 x )2 = 1 + 10 2 x = 1 + 100 x 10-89. We know cos x = t = 1t ,!sin y = t = 1t . Drawing a diagram to fit this situation yields: y 1 x CPM Educational Program © 2012 Chapter 10: Page 21 t Pre-Calculus with Trigonometry Therefore x + y = 90º = !2 !!"!!y = !2 # x 10-90. "6, 3 # 2, 4 ' ! = cos"1 $& = cos"1 "12+12 = cos"1 (0) = 90! a. 45 20 % "6, 3 2, 4 )( b. 3i + 4 j = 3, 4 and ! 2 j = 0, !2 ( ! = cos"1 $& % 3, 4 # 0, "2 3, 4 0, "2 ' "1 )( = cos ) ( 5#2"8 ) = cos"1 ( " 108 ) = cos"1 ( " 45 ) = 143.13! 10-91. Let x be the miles to the cousin’s home. Then we know: 1 hour x miles + 1 hour x miles = 10 hours 15 miles 10 miles + 1 hour x miles = 10 hours ( 151 hour miles 10 miles ) x miles = ( 10 hours 1 hour + 1 hour 15 miles 10 miles ) = 10 hours ! 6 miles = 60 miles hour Answer: (c) 10-92. Let the length of the rectangle be L. Then the width is 0.78L. The diagonal creates a right triangle where L2 + (0.78L)2 = 30 2 . Solving for L : 1.6084L2 = 900 L2 = 559.562 L = 23.655 cm W = 0.78 ! L = 0.78 ! 23.655 = 18.451 cm Lesson 10.2.3 10-93. a. The vertical displacement is 36 ft ! 12 = 18 ft . After t seconds, the vertical displacement is 18t ft . b. The horizontal displacement is 36 ft ! c. d. After t seconds, the vertical displacement is 18t ! 16t 2 ft . When t = 1 second, the vertical displacement is 18 ! 16 = 2 feet. y(t) = !16t 2 + 18t + 3 e. y(t) = !16t 2 + 18t + 3 = 0 when t = f. Since time cannot be negative, t = 1.2724 seconds. 31.177t feet = 31.177(1.272) = 39.66 feet CPM Educational Program © 2012 3 2 t = 18 3t ft = 31.177t ft . !18± 18 2 ! 4(!16)(3) 2(!16) Chapter 10: Page 22 = !18± 516 !32 = !0.1473, 1.2724 Pre-Calculus with Trigonometry 10-94. a. Initial position: (0, 0) . Initial velocity: 102. Angle: ! = 38! : x(t) = 102 cos(38! )t y(t) = !16t 2 + 102 sin(38! )t b. c. The ball reaches the tree when: 90 !!!!!t = x(t) = 102 cos(38! )t = 30 yards = 90!feet cos(38! )t = 102 90 102cos(38! ) = 1.12 seconds The height of the ball at this time is: y(1.12) = !16(1.12)2 + 102 sin(38)(1.12) = 50.3 feet The ball will clear the tree by 5.3 feet. x(3.925) = 102(3.925) cos 38º = 315.5!feet 0 = !16t 2 + 102t sin 38º 16t 2 = 102t sin 38º 16t = 102 sin 38º 38º = 3.925!sec t = 102 sin 16 Distance to the pin = 100 yards + 60 feet = 360 feet 330 – 315.5 = 14.5 feet 10-95. a. b. c. d. e. f. g. t 0 0.25 0.5 0.75 1 x 0 –15 0 15 0 y 0 15 30 15 0 See graph at top right. y(t) = !15 cos(2" t) + 15 See graph at right middle. x(t) = !15 sin(2" t) The circumference of the wheel is 2! r = 2! "15 = 30! inches, so the center of the wheel has moved 30! inches. After t seconds, the center of the wheel has moved 30! t inches. x(t) = !15 sin(2" t) + 30" t See graph at bottom right. 10-96. a. See graph at right. x = t , y = t 3 ! 3 = x3 ! 3 b. c. See graph at right. d. x = t 3 ! 3 " t = (x + 3)1/3 y = t ! y = (x + 3)1/3 e. They are inverse functions. f. They are inverse functions. CPM Educational Program © 2012 Chapter 10: Page 23 Pre-Calculus with Trigonometry 10-97. a. g(x) = 2 x has inverse parametric equations x(t) = 2t y(t) = t y The inverse function is x = 2 log 2 x = log 2 2 y = y log 2 2 = y, g !1 (x) = log 2 x b. f (x) = 2x x+2 has inverse parametric equations x(t) = 2t t +2 y(t) = t x= The inverse function is: 2y y+2 x(y + 2) = 2y 2x = 2y ! xy = (2 ! x)y y = f !1 (x) = 2x 2! x 10.2.3 Review and Preview 10-98. a. A circle with radius 3 centered at (0, 0). b. t x y 0 3 0 5 0.85 –2.88 10 –2.517 –1.632 c. z 0 10 20 It forms a spiral like a staircase or a stripe on a barber pole. The spiral would be steeper. 10-99. a. We know cos2 t + sin 2 t = 1 so, x 2 + y 2 = 1 . b. We know cos2 ! + sin 2 ! = 1 so let ! = t 3 and we have x 2 + y 2 = 1 . c. y = t 2 ! x = t 4 " 2t 2 = y 2 " 2y x = y 2 ! 2y 10-100. a. x(t) = t , y(t) = cos(t 2 + 2t ) b. x(t) = cos(t 2 + 2t ) , y(t) = t 10-101. a. x(t) = t 2 , y(t) = t is an example. b. This is not possible. 10-102. x(t) = 3 + 2 cos t y(t) = 6 + 2 sin t CPM Educational Program © 2012 Chapter 10: Page 24 Pre-Calculus with Trigonometry 10-103. Due east is represented by 45! , x(t) = (200 cos(45! ) + 40)t y(t) = 200 sin(45! )t 10-104. x(t) = t has an inverse x(t) = t 3 + 1 y(t) = t y(t) = t 3 + 1 10-105. a. Initial velocity: 120, Angle: 40, initial position: (0,7): x(t) = (120 cos 40)t y(t) = !16t 2 + (120 sin 40)t + 7 b. The ball hits the ground when y(t) = 0 or when: y(t) = !16t 2 + (120 sin 40)t + 7 = 0 t= = = !120 sin(40)± (120 sin(40))2 ! 4(!16)(7) 2(!16) !77.135± (77.135)2 + 448 !32 !77.135± 79.986 = !0.089, 4.91 !32 t = 4.91 seconds At this time, the horizontal displacement is x(4.91) = (120 cos 40) ! 4.91 = 451.353 feet 10-106. If ! = 35! , x(t) = (120 cos 35)t and y(t) = !16t 2 + (120 sin 35)t + 7 = 0 when t = 4.4 seconds where x(4.4) = (120 cos 35) ! 4.4 = 432.12 . So, the player cannot throw the ball as far if ! = 35! . 10-107. Find t in terms of ! from the equation 0 = !16t 2 + (120 sin " )t + 7 : t= !120 sin " ! (120 sin " )2 ! 4(!16)(7) !32 = 15 sin " + 225 sin 2 " + 7 4 " Substitute this value into x = (120 cos ! )t = (120 cos ! ) $ 15 sin ! + # Graphing this, we get: 225 sin 2 ! + 7 4 % '& Where ! " 44.46! and x " 456.9 feet . If the ball is caught 7 feet above the ground, then the best angle is 45! and the ball goes 450 feet. CPM Educational Program © 2012 Chapter 10: Page 25 Pre-Calculus with Trigonometry Chapter 10 Closure Merge Problem 10-108. a. b. c. ( 15! t ) ; y(t) = 50 sin ( 15! t ) x(t) = 30 cos ( !2 t ) ; y(t) = "30 sin ( !2 t ) ! ! x(t) = 50 cos ( 15 t ) + 30 cos ( !2 t ) ; y(t) = 50 sin ( 15 t ) " 30 sin ( !2 t ) x(t) = 50 cos d. x(t) = !4 sin(" t); y(t) = 4 cos(" t) e. x(t) = 50 cos f. When t = 3 , < 31.736, 21.180 > , therefore speed = 38.154 ft/sec. ( 15! t ) + 30 cos ( !2 t ) " 4 sin(! t) ; y(t) = 50 sin ( 15! t ) " 30 sin ( !2 t ) + 4 cos(! t) Closure Problems 10-109. a. Look at vectors v and vi. !4 /5 3/5 ( ! 45 ) + ( 53 ) 2 2 =! 4 3 and =1 ! 12 =! 9 4 3 and (!12)2 + (9)2 = 15 Thus, vectors v and vi have the same direction but not the same magnitude. b. Look at vectors i and iii !3 !3 = 1 and (!3)2 + (!3)2 = 18 0 3 2 c. = 0 and (0)2 + (3 2 )2 = 18 These vectors have the same magnitude but not the same direction. The magnitude of vector v is 1 (from part a), so this is a unit vector. 10-110. The magnitude of 5i + 12 j is 5i + 12 j = 25 + 144 = 13 . The unit vector orthogonal to 5i + 12 j is: 12 13 5 j 5 j i ! 13 i + 13 or ! 12 . 13 10-111. !" !" !" !" !" A = 2, !3 and B = !4, 1 and C = A + 2B = 2, !3 + 2 !4, 1 = !6, !1 !" C = 2, !3 + 2 !4, 1 = 2, !3 + !8, 2 = !6, !1 = 6 2 + 12 = 37 CPM Educational Program © 2012 Chapter 10: Page 26 Pre-Calculus with Trigonometry 10-112. a. Channel: 0,!15 Boat: 40 sin 90 = ! x = 40 sin 60 = 34.64 x sin 60 40 2 " 34.64 2 = 20 34.64, 20 2 2 2 2 2 Wind: x + x = 30 ! 2x = 30 b. c. d. e. x 2 = 450 ! x = ±21.21 Because the wind is blowing northwest: !21.21, 21.21 0, !15 + !21.21, 21.21 + 34.64, 20 = !21.21 + 34.64, !15 + 21.21 + 20 = 13.43, 26.21 13.432 + 26.212 = 29.45 , 29.45 sin 90 = 26.21 sin ! " ! = sin #1 26.21 = 62.9! ( 29.45 ) 20 = 1.489 hours 13.43x = 20 ! x = 13.43 , 1 hour 29.4 minutes 26.21 mph !1.489 hours = 39.03 miles 10-113. a. 1, 4 cos " , cos!1 2, 3 ! 1, 4 = 2, 3 2, 3 " 1, 4 2, 3 1, 4 # = cos!1 b. !1, 2 " 6, 1 = !1, 2 6, 1 cos # , cos!1 =# 14 = 19.654 ! 13 17 !1, 2 " 6,1 =# !1, 2 6,1 # = cos!1 !4 5 37 or 0.343 radians = 107.103! or 1.859 radians 10-114. 2, 1 + b ! 4, 1 " b = 0 8 + (1 + b)(1 " b) = 0 8 + 1 " b2 = 0 b2 = 9 b = ±3 10-115. Slope: m = 2+1 7! 3 10-116. x = 2t ! t = a. y = t 2 ! 6t = b. = 3 4 so one option is 3 + 4t, !1 + 3t or 7 + 4t, 2 + 3t . x 2 ( 2x ) 2 ! 6 2x = x2 4 ! 3x x = t 3 + 1 ! t = (x " 1)1/3 y = t 6 ! 1 = (x ! 1)6/3 ! 1 = (x ! 1)2 ! 1 CPM Educational Program © 2012 Chapter 10: Page 27 Pre-Calculus with Trigonometry 10-117. v0 = 110 ft/sec (x0 , y0 ) = (0, 4) ! = 53! x(t) = (110 cos 53)t y(t) = "16t 2 + (110 sin 53)t + 4 The ball will travel 330 feet when: x(t) = (110 cos 53)t = 330 cos(53)t = 3 t= 3 cos(53) = 4.985 seconds Where the height of the ball is: y(4.985) = !16(4.985)2 + (110 sin 53)(4.985) + 4 = 44.334 feet Yes, Alex will hit a homerun. 2 The ball hits the ground when: y(t) = !16t + (110 sin 53)t + 4 = 0 t= = !110 sin(53)! (110 sin(53))2 ! 4(!16)(4) 2(!16) !87.8499!89.295 !32 = 5.536 seconds The ball will have traveled a distance of: x(5.536) = 110 cos(53) ! 5.536 = 366.467 feet y 10-118. x(t) = t 2 ! t ! 6 y(t) = t x 10-119. a. (15/ x)#20+(17/ x 3 ) 15 x 2 #20 x 3 +17 $ 1/ x 3 = lim . x!" 12 x 3 #60 x 2 + 75 x 1/ x 3 x!" 12#(60/ x)+(75/ x 2 ) lim # 60 = 0 , lim 752 = 0 , lim 15 = 0 , and lim 173 = x!" x x!" x x!" x x!" x 3 (15/ x)#20+(17/ x ) 15 x 2 #20 x 3 +17 x!" 3x(2 x#5)2 lim Since we know we know: lim = lim x!" 12#(60/ x)+(75/ x 2 ) 15/ x x!" 12#(60/ x)+(75/ x 2 ) lim 15/ x x!" 12 lim b. x 3 "1 x!1 x 2 "1 lim = lim x!1 # lim 20 12 x!" (x"1)(x 2 + x+1) (x"1)(x+1) CPM Educational Program © 2012 + 0 = 20 17/ x 3 + lim 2 x!" 12#(60/ x)+(75/ x ) x!" 12#(60/ x)+(75/ x 2 ) 3 20 + 0 = # 5 lim 17/12x = 0 # 12 3 x!" # lim (x"1)(x 2 + x+1) x!1 (x"1)(x+1) = lim (x 2 + x+1) x!1 (x+1) = lim Chapter 10: Page 28 = (1+1+1) (1+1) = = 3 2 Pre-Calculus with Trigonometry 10-120. f (x+h)" f (x) = #x h!0 2(x+h)2 " 3(x+h)"2 x 2 + 3x lim lim h!0 h = 2(x+h)2 " 3(x+h)"2 x 2 + 3x = h h!0 2 2 2 lim 2 x + 4 xh+2h "h3x" 3h"2 x + 3x h!0 lim 4 xh+2h 2 " 3h h h!0 lim = = lim 4x + 2h " 3 = 4x + 2(0) " 3 = 4x " 3 h!0 at x = 2 :!!4(2) " 3 = 5 CPM Educational Program © 2012 Chapter 10: Page 29 Pre-Calculus with Trigonometry