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Solution Manual for Calculus for Engineers, 4th Edition

Struggling with textbook problems? Solution Manual for Calculus for Engineers, 4th Edition offers a clear breakdown of every exercise for easy understanding.

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    SOLUTIONS MANUAL Calculus for Engineers Fourth Edition Donald Trim University of Manitoba

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    Downloaded from StudyXY.com ® + StudyXY Sd Ye. o> \ | iF ’ pr E \ 3 S Stu dy Anything This ContentHas been Posted On StudyXY.com as supplementary learning material. StudyXY does not endrose any university, college or publisher. Allmaterials posted are under the liability of the contributors. wv 8) www.studyxy.com

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    CONTENTS This manual contains detailed solutions to all exercises in the text. We would appreciate being made aware of any errors. Chapter 1 Calculus Preparation 1 Chapter 2 Limits and Continuity 72 Chapter 3 Differentiation 99 Chapter 4 Applications of Differentiation 170 Chapter 5 Antiderivatives and the Indefinite Integral 300 Chapter 6 The Definite Integral 332 Chapter 7 Applications of the Definite Integral 360 Chapter 8 Techniques of Integration 443 Chapter 9 Parametric Equations and Polar Coordinates 506 Chapter 10 Infinite Sequences and Series 553 Chapter 11 Vectors and Three-dimensional Analytic Geometry 644 Chapter 12 Differential Calculus of Multivariable Functions 722 Chapter 13 Multiple Integrals 816 Chapter 14 Vector Calculus 935 Chapter 15 Differential Equations 1012 i

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    EXERCISES 1.2 1 CHAPTER 1 EXERCISES 1.2 1. The solution is z = 2/3. 2. The solution is 2 = 5/14. 3. Since 2% + 2x — 3 = (z+ 3)(z — 1), solutions are z = —3 and z = 1. 4. Since 122% + 11x — 5 = (3x — 1)(4z + 5), solutions are z = 1/3 and & = —5/4. 5. Since the discriminant 52 — 4(2)(10) = —55 is negative, the equation has no real solutions. —104 TOO 4(4)(9) -10+ v3 1 6. Quadratic formula 1.5 gives z = 0x 10-440) = clove = EEN 7. Since 22 + 8z + 16 = (z + 4)?, the solution is # = —4 with multiplicity 2. 8. Since 422 — 36x + 81 = (2x — 9)? the solution is = = 9/2 with multiplicity 2. 5+ BH AD) 51 VIB 9. Quadratic formula 1.5 gives z = Sox yB AAC = RENE 10. Since the discriminant (—8)% — 4(4)(9) = —80 is negative, the equation has no real solutions. 11. Since 23 — 32% + 32 — 1 = (x — 1)%, the solution is # ~ 1 with multiplicity 3. 12. Possible rational solutions are +1, -£1/2, 1/4, +1/8. We find that # = —1/2 is a solution. We factor 22 + 1 from the cubic, 82% +122” + 62 +1 = (20 + 1)(da? + do + 1) = (22 + 1)(2z + 1)? = (20 + 1)°. The only solution is z = —1/2 with multiplicity 3. 13. Possible rational solutions are 2 — +1, £2, #5, +10. We find that z = 2 is a solution. We factor z — 2 from the cubic, 28 — 20% +52 — 10 = (z — 2)(2® +5) = 0. The other two solutions are complex. 14. Possible rational solutions are 2 = +1, +3, £9, but it should also be clear that no positive value of = can satisfy the equation. We find that = = —1 is a solution. We factor 2 +1 from the cubic, Pde? +122 4+9= (a+)? +3z +9) =0. Since the discriminant of the quadratic is negative, the other two solutions are complex. 15. Possible rational solutions are +1, £2, +4, +8, 116, 1:32, +64, but it should also be clear that no pos- itive value of 2 can satisfy the equation. We find that x = —4 is a solution. We factor = + 4 from the cubic, 27 4 1222 + 485 + 64 = (x + 4)(2% + 82 + 16) = (z+ A) (x + 4)* = (z+ 4)* The only solution is x = ~4 with multiplicity 3. 16. Possible rational solutions are +1, £2, £3, £4, £6, £9, £12, £18, 4:36. We find that z = ~3 is a solution. We factor z + 3 from the quartic, 2% T2® + 92% — 21a — 36 = (+ 3)(a® + 42° — 3x. — 12). Possible rational zeros of the cubic are +1, +2, +3, 44, +6, +12. We find that x = —4 is a zero. We factor x + 4 from the cubic, a + 7% + 90% — 20a — 36 = (2 + 3) (x + 4)(a? — 3). The solutions are & = —3, 4, £V/3. 17. Since 2? — 16 = (22 + 4)(z® — 4) = (a® + 4)(z + 2)(z — 2), the real solutions are z = +2. + StudyXxy

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    2 EXERCISES 1.2 18. Possible rational solutions are £1, +3, +5, £15, £1/2, £3/2, £5/2, £15/2. We find that z = —5 is a solution. We factor z + 5 from the quartic, 22 + 92% — 62% — 8x — 15 = (z + 5)(22® — a? —2 — 3). Possible rational zeros of the cubic ave 1, £3, £1/2, £3/2. We find that x = 3/2 is a zero. We factor 2a — 3 from the cubic, 2 + 92% — 62% — 8x — 15 = (zx + 5)(2x — 3)(@% +z +1). Since the quadratic has a negative discriminant, the only real solutions are z = —5 and © = 3/2. 19. Possible rational solutions are +1, 438, £0, £1/2, £3/2, £9/2, +1/3, 1/6. We find that = —1/2 is a solution. We factor 2z + 1 from the quartic, 6a? 4 2° + 532% + 92 — 9 = (2x + 1)(3z° — a? + 272 — 9). Possible rational zeros of the cubic are +1, +3, +9, 21/3. We find that = 1/3 is a zero. We factor 3x — 1 from the cubic, 62 + 2° + 532% + 9x — 9 = (22 + 1)(3z — 1)(2? +9). Since the quadratic has complex zeros, the only real solutions are 2 = —1/2 and & = 1/3. 20. No real numbers can satisfy this equation. 21. Possible rational solutions are £1, £2, +3, +4, +6, £8, £9, +12, +18, 24, +36, £72. We find that x = 24 is a solution. We factor x — 24 from the cubic, 2% — 232% — 21a ~ 72 = (2 — 24)(2® +2 + 3). Since the quadratic has a negative discriminant, the only real solution is x = 24. 22. Possible rational solutions are +1, +2, £3, £4, +6, £8, £9, +12, £16, +18, +24, 4:32, +36, +48, +64, +72, +96, +144, +192, £192, £288, £576. We find that 2 = —4 is a solution. We factor 2 +4 from the quartic, 2% dp? — 442? 4 962 + 576 = (x + 4)(2® — 82% — 122 + 144). Possible rational zeros of the cubic are £1, £2, +3, +4, £6, 18, £9, £12, +16, +18, +24, £36, £48, +72, +144. We find that z = —4 is a zero. We factor z + 4 from the cubic, 2% — 4a? — 442 + 963 + 576 = (z + 4)(z + 4)(2® — 122 + 36) = (+ 4)*(z — 6). Thus, = —4 and 2 = 6 are solutions, each of multiplicity 2. 23. Since 32% + 2° + 52% = 2*(3z® + x + 5), and the quadratic has a negative discriminant, the only real solution is 2 = 0 with multiplicity 2. 24. Possible rational solutions are £1, £2, +4, +5, +10, +20, £1/2, 45/2, +1/3, £2/3, +4/3, £5/3, +10/3, £20/3, £1/6, £5/6. We find that 2 = 5/6 is a zero. We factor 6 — 5 from the cubic, 62° + 2% + 192 — 20 = (62 — 5)(2? + 2 +4). Since the quadratic has a negative discriminant, z = 5/6 is the only real solution. 25. Possible rational zeros are +1, £3, 5, 1:9, £15, +45. We find that = —5 is a solution. We factor 2 +5 from the polynomial, 20+ 5a — 92 — 45 = (z+ 5)(a! — 9) = (2 +5)(2® + 3) (x? — 3). Solutions are 2 = ~5 and © = +3. + StudyXY

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    EXERCISES 1.2 3 26. Possible rational solutions are +1, +2, +3, £4, 15, 16, £8, 1:10, +12, +15, +20, £24, 1:30, 3:40, 160, +120. We find that 2 = 1 is a solution. We factor z — 1 from the polynomial, 25 — 152% + 852% — 22522 + 27dw — 120 = (x — D(z? — 14z® + 71a? — 154z + 120). We use the same set of rational possibilitics for the quartic. We find that # = 2 is a zero. When we factor x — 2 from the quartic, 2° — 152% + 852° — 2262° + 27d — 120 = (2 — 1)(z — 2)(=® — 120° + 47a — 60). Possible rational zeros of the cubic are +1, +2, +3, +4, £5, £6, £10, +12, +15, +20, £30, £60. We find that z = 3 is a zero. We factor 2 — 3 from the cubic, 25 150% 4 852° — 225% + 274 — 120 = (x — 1)(z — 2)(z — 3)(2® — 9a + 20) = (z—1){z — 2)(z — 3)(z — 4)(z — 5). Solutions are therefore = 1, 2, 3, 4, 5. 27. Possible rational solutions are £1, +2, +4, +1/2, +1/4, but it should also be clear that no positive value of = can satisfy the equation. We find that z = —1/2 is a solution. We factor 2z'+ 1 from the quartic, dat 4 40% + 172% +162 +4 = (2x + 1)(2° + 2% + 82 + 4). Possible rational zeros of the cubic are £1, £2, +4, +1/2, and once again we reject the positive values. We find that © = —1/2 is a zero. We factor 22 +1 from the cubic, 4a + de® + 172% + 162 + 4 = (20 + 1)(2z + 1)(2? +4). Thus, & = —1/2 is the only real solution and it has multiplicity 2. 28. Possible rational solutions are £1, £2, 4, 1/5, £2/5, 34/5, £1/25, +2/25, +4/25. We find that » = 2/5 is a solution. We factor 52 — 2 from the quartic, 250 — 1200° + 1092 — 362 + 4 = (52 — 2)(bz® — 22% + 132 — 2). Possible rational zeros of the cubic are +1, +2, +1/5, £2/5. We find that © = 2/5 is a zero. We factor Sz — 2 from the cubic, 257 — 1202° + 1092° — 362 + 4 = (5x — 2)(52 — 2)(z* — dz +1). Thus, 2 = 2/5 is a real solution with multiplicity 2 and the quadratic formula gives the remaining two solutions po dEVI6 4 SARL —24 v3 29. Possible rational solutions are £1, £2, £4, +5, £8, £10, £20, #25, +40, £50, £100, £200. We find that x = 1 is a zero. We factor 2 — 1 from the polynomial, 2° + 9a? + 472% + 1252% + 18x — 200 = (z — 1)(=* + 102° + 572% + 1820 + 200). We use the same rational numbers for the quartic, but reject the positive values. We find that z = 2 is a zero. We factor z + 2 from the quartic, 25 + 9a + 470° + 12527 + 18 — 200 = (& — 1)(z + 2)(z® + 82% + 41x + 100). For zeros of the cubic we use —1, —2, —4, —5, —10, —20, —25, —-50, —100. We find that x = —~4 is a zero. When we factor it out, a® 4 92 + 472% + 1252° + 18% — 200 = (x — 1)(z + 2)(x + 4)(z* + 4x + 25). Since the quadratic has a negative discriminant, the real solutions are © = —4, -2, 1. + StudyXY

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    4 EXERCISES 1.3 80. Possible rational solutions are £1, £2, 3, +4, +6, +8, +9, +12, +16, £18, £24, +27, -:36, +48, +54, +72, +81, +108, +144, +162, £216, £324, +432, 1648, £1296. We find that z = 3 is a solution. When we factor z — 3 from the polynomial, 28 4 162% — 812% — 1296 = (x — 3)(2° + 327 + 250% + 7527 + 14x + 432). When we note that no positive value can satisfy the fifth degree polynomial, possible rational zeros arc —1, 2, —3, —4, —6, —8, —9, —12, —16, —18, —24, 27, —36, —48, —54, —72, —108, —144, —216, —432. We find that z = —3 is a zero. When we factor + 3 from the polynomial, 28+ 162 — 81a? — 1206 = (x — 8)(x + 3) (a + 252% + 144). Since there can be no real zeros of the quartic, the real solutions are x = £3. 31. 2(z +5) (xz —1) 32. 2x — (3+ v05)/4)fx ~ (3 - V65)/4] 33. (z+ 2)(z— 3)z 10) 34. 24(x + 5/8) (x ~ 1/4) (x — 1/2) 35. (z41)(z—1)° 36. 16(z — 1/2)%(z + 1/2)? 37. One polynomial is (x+41/3)(z—4/5)(z— 3)(z—4)*. This polynomial could be multiplied by any constant. 38. According to the rational root theorem, possible rational zeros must be divisors of ap divided by divisors of 1. This means that possible rational zeros are divisors of ao. EXERCISES 1.3 1. With formula 1.10, the distance is v/22 + 12 = /5. 2. With formula 1.10, the distance is 1/62 + (=3)Z = 3/5. 3. With formula 1.10, the distance is \/{—2)2 + (—6)% = 2V/T0. 4. With formula 1.10, the distance is \/{—7)2 + (3)? = v/58. 5. With formula 1.11, the midpoint is (259) — (=3)- 6. With formula 1.11, the midpoint is (5H 155 = (1-3) 7. With formula 1.11, the midpoint is (229) = (-2,-5). N oo (3-421 11 8. With formula 1.11, the midpoint is (3 2) = (-3 5) 9. With slope —2/4 = —1/2, equation 1.13 gives 10. The line is horizontal. 1 y-2= —le-1) = a+2=35. Tts equation is y = —6. eo 4 % (1.2) % 0 G6) (5-6) + StudyXxy

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    EXERCISES 1.3 5 11. With formula 1.13, the equation is 12. with formula 1.13, the equation is y+3=3(x+2) ==> y=238z13. yty=—3=1) = at+2+2=0 y id x (32) x (2:3) 13. The equation of the y-axis is @ = 0. 14. The equation of the z-axis is y = 0. 15. With m = (3 — 0)/(4 + 2) = 1/2, equation 1.13 16. Withm = (4+ 2)/(0 + 1) = 6, equation 1.13 gives y= (1/2)(z +2) => 2y =z +2. gives y —4=06(x— 0) => y=06x +4. y hy 3) 4 €l x 1-2) x 17. With slope (0+ 3)/(1 — 0) = 3, equation 1.13 18. The midpoint of the line segment, is 3-7 4+8 gives y —0=3(x —1) = y=3z—3. (55-2 With m = (6 — 0)/(—2 — 0) = —3, equation 1.13 gives y — 0 = —3(z — 0) = y = —3z. od Y £78) fe) 1 £3 p H 19. Since slopes ~1 and 1 are negative reciprocals, the lines are perpendicular. 20. Since slopes of both lines are —1/3, they are parallel. 21. Since slopes of both lines are 1/3, they are parallel. 22. Since slopes —2/3 and 3/2 are negative reciprocals, the lines are perpendicular. 23. Since slopes are 3 and —1/2, the lines are neither parallel nor perpendicular. 24. Since slopes arc 1 and —2/3, the lines are neither parallel nor perpendicular. 25. The lines are perpendicular. 26. Since slopes are —1 and 3, the lines are neither parallel nor perpendicular. 27. When we subtract the equations, y + 2y = 0 +3 => y = 1. The point of intersection is (=1,1). 28. The point of intersection is (1,2). 29. When we subtract 3 times the second equation from the first, 4y + 18y = 6 —9 —> y = —3/22. The point of intersection is (24/11, —3/22). 30. When we substitute y = 2a + 6 into the second equation, © = (22+ 6) +4 => x = —10. The point of intersection is (10,14). | vv 1 StudyXY

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    6 EXERCISES 1.3 31. When we substitute z = 2 — 2/3 into the second equation, 2(2 —2y/3) — y/4 = 15 => y = —132/19. ‘The point of intersection is (126/19, —132/19). 32. If we multiply the first equation by 5 and add the result to the socond equation, we obtain 73x = 37 = 2 = 37/73. The point of intersection is (37/73,153/146). CC IB+4-1 6 -6-3 8 33. Formula 1.16 Ee — aN 34. Formula 1.16 ——— = = ‘ormula gives Sixt 7 ‘ormula gives Nie 75 5-1- 34-1 4 35. Formula 1.16 gives bold =0. 86. Formula 1.16 gives Bed ~%5" 2V2. 44 3 67 37. The distance is 7. 38. Formula 1.16 gives mo - Ta 39. Since the point of intersection of the given lines is (3,1), and the slope of z + 2y = 15 is —1/2, the required equation is y — 1 = —(1/2)(z — 3) = x +2y=5. 40. Since the slope of z —y = 4 is 1, and the point of intersection of 2 + 3y = 3 and z —y = 4 is (3,-1), the required equation is y+ 1 = —1(z — 3) = z+y=2. 41. Since the slope of the line through (1,2) and (3,0) is 2/4, the required equation is y — 6 = (1/2)(z — 5) = 2y=x+T. 42, Since the slope of the line through (—3,4) and (1,2) is (4 + 2)/(~3 — 1) = —3/2, the equation of the required line is y + 2 = (2/3)(z + 3) =» 22 = 3y. 43. Since [|OA| = [|OB||, coordinates of A and B44. When we equate slopes of line segments AC and 5 are (a, 0) and (0, a), respectively. Since the AB, 2 = = = b= = area of AOAB is 8, it follows that Combine this with the fact that the area (1/2)a? = 8 => a = 4. The equation of AQAB = (1/2)ab = 30, and we obtain a = 6 of line ABisy=—(z—4) = z-+y=4 and b= 10. The equation of the line is y= —(5/3)(z — 6) = Sz +3y = 30. woof conf BG.5) ol Aw % Aap) x 45. Since the slope of line segment AB is 2, 46. If the equation is y = ¢, then A = (¢,¢) and and its length is 3, it follows that B= (4 —¢,c). Since triangle ABD has area 9, —bfa = 2 and Va? +b? = 3. These can it follows that 9 = le —)(4 — 2c), be solved for a = —3/v/5 and b= 6/5. solutions of which are 5, —1. The The equation of AB is y = 22 + 6/V5. required equation is therefore y = —1. 24 od BOY ANF r=4 =x > D(2,2) TN [oe po) % 1] a” 47. (a) The conversion equation is Tir = 5(Tr — 32)/9. (b) The conversion equation is Tr = 97¢/5 + 32. (¢) They are one and the same line if we plot Tp along the horizontal axis and Tc along the vertical axis (left figure below). If we plot Tr along the vertical axis and Te: along the horizontal axis in part + StudyXY

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    i EXERCISES 1.3 7 (b), we obtain the right figure below. 100 % nal * 32, 32 22 7, wo (d) If we set Tr = Tt: in the equation from part (a), we obtain Tg = 5(Tc — 32)/9 => 4T¢ = —160 = Te = —40. 48. (a) The conversion equation is Tx = 5(Tr — 32)/9 + 273.16. (1b) The conversion equation is Tp = 9(Tj ~ 273.16)/5 + 32. (c) They are one and the same line if we plot Tr along the horizontal axis and Tc along the vertical axis (left figure below). If we plot Tw along the vertical axis and Tx along the horizontal axis in part {b), we obtain the right figure below. Te TF 255.34 55.38 T 459.69 T 459.69) (d) If we set Tp = Tx in the equation from part (a), we obtain Tx = 5(Tx —32)/9 + 278.16 == 4Tx = 2208.44 ==> Tx = 574.61. 49. (a) If the temperature is changed to T, the change I in the length of the bar is aLo(T — Tp), and Po — therefore its length is Lo1-oT) L = Lo + aLo(T — To) = Lo[l + o(T — To). {b) The ends will be in contact when the rails have length L = 10.003 m. This occurs when 10.003 = 10[L + 1.17 x 1075(T ~ 20)] == T' = 45.6. E T 50. Coordinates of E, the midpoint of side AC, are (2,3). Since the slope of BE is —1/2, the equation of 4 D ES median BE is y —4 = —(1/2)2 — 2 +2y=8. BOA Similarly, equations for medians AD and CF are % 7x —y=6 and y = x + 2, respectively. Line segments rF AD and CF intersect in the point (4/3, 10/3), { and this point also satisfies x + 2y — 8. Thus, the ALD three medians intersect at the point (4/3,10/3). x 51. Coordinates of P, Q, R, and § are Y Clad Pl(@1 + 2)/2, (11 +92)/2), Qa +23)/2, (v2 + y3)/2), ) R((zs + 4) /2, (ya + y4)/2), and S((1 +24)/2 (51 +y4)/2). ® BGs.) Slopes of the line segments PS, RQ, PQ, and RS Deer are, respectively, ow fo @r—w)/2 (w-w)/2 (uw) (@-y)/2 § ® (wz —@4)/2" (za —x0)/2" (za —11)/2" (23 —21)/2 AG) Since PS and RQ are parallel, as arc PQ and RS, PQRS is a parallelogram. + StudyXxy

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    8 EXERCISES 1.3 52. The midpoint of the line segment is (1, ~1), and its slope is —3/2. The equation of the perpendicular bisector is y + 1 = (2/3}(z — 1) = 22 = 3y +5. 53. If coordinates of the point are (x,y), then (z—1)2 + (y—2)? = (z+1)*+(y—4)% and (z—1)* + (y—2)* = (2 +3)? + (y — 1)%. When cxpanded and simplified, these reduce to —y = —3 and 82 + 2y = 5, the solution of which is (—11/10,19/10). 54. Suppose two nonvertical, parallel lines have slopes m) and my. Their equations can be expressed in the form y = myz+b; and y = maz+by. To find their point of intersection we set miz+by = max +b. Since the lines do not intersect, there must be no solution of this equation. This happens only if my = ms; that is, the lines have the same slope. Conversely, if two nonvertical lines have the same slope, their equations must be of the form y = mz + by and y = ma + by, where by # ba. When we attempt to find their point of intersection by setting ma: + by = ma + by, we find that by = bz; a contradiction. In other words, the lines do not intersect. 55. Because triangles PTQ and PSR are similar, ratios of corresponding sides are equal, i. by 1PQl _ IPT) _ 22 —21 Qari) IPR] IPS] 2 —a RG) If we subtract 1 from each side of this equation, [25] x2 — LAIR 2 SU Rt BL . IPR] pr. 1 Py) Em Tle) IPQ Pal _ @a—2 ra _ wy ey ET =F 2.2 x IPR] Ta Om a-m Thus, rox — ro@y = 712 — ML —> & = hme, A similar proof gives the corresponding formula 2 for the y-coordinate of R. * 56. No. Definition 1.1 defines parallelism only for different lines. 57. We choose a coordinate system with the origin at one vertex of the triangle, and the positive z-axis along \y Bo one side. The coordinates of the vertices are then 0(0,0), A(a,0), and B(b,c). Using equation 1.11, coordinates of the midpoints of the sides are | FP a+b c be a (55) e(3s) #Go) The sum of the squares of the lengths of the medians is ol ® Aled x b\? ce)? b\? cy? a? Pl? 2 ir? = | (2H c _b _c _a 2 IOPIP + AI + | BR) VG) (eg) + (5) | + |(b-5) +e 3 = 5 +64 ab). Three-quarters of the sum of the squares of the lengths of the sides is 3 3 3 7 (fOAI + ABI? + [0B]? = i {a® + [b= a)? +c? + (B® +A} = 5(@® +62 + c* — ab). 58. If we choose the coordinate system in the figure, equations of AC and AB are y = —v/3(z — h/V/3) hom — V3z+y—h=0and y= V3(z+h/V3) —> Bx —y + h = 0. If P(d,e) is any point interior to the triangle, we can use formula 1.16 to » 5 > find the sum of the distances from P to the three sides IPEI+ 12D] + [PE] oT Gases + StudyXY

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    EXERCISES 1.4A 9 [V3d+e—h| |V3d—c+h| 1 1 —et ee po =e + | V3 — hl + z|vV3d—e+h|. e+ AT oT et plVidte il + 513 e+ hl Because P(d, e) is below and to the left of line AC, it follows that 3d + e—h <0. Because P is below and to the right of AB, 3d — e+ h > 0. Hence, 1 1 [|PF| + |PDI| + |IPE| = e+ (=V3d —eth)+ 5(v3d —e+h)=h EXERCISES 1.4A 1. This is the parabola y = 222 shifted 2. Factored in the form y = —(z — 3)(z — 1), 1 unit downward. the 2-intercepts of the parabola are 1 and 3. Its maximum is at © = 2. 4 od 7 % -1 1 x JS) 3. Factored in the form y = (x — 1)%, 4. This is the parabola 2 = 4y%/3 shifted the parabola has its minimum at (1,0). 1/3 unit to the left. 24 id 17 BR} 1 -12 * 1 . x 5. Factored in the form x = (2+), 6. Factored in the form y = —(x + 1)(z — 4)/2, the parabola opens to the right with 2-intercepts of the parabola are —1 and 4. y-intercepts 0 and —2. Its maximum is at z = 3/2. by id Ey x a 3 Nx H 3 7. The parabola z = 1 — 4? opens left with 8. Factored in the form x = —(2y — 5)(y + 1), x-intercept 1 and y-intercepis +1. y-intercepts of the parabola are —1 and 5/2. Its maximum in the z-direction occurs for y = 3/4. y y 37] 374] Tx 3! 5 x + Studyxy

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    10 EXERCISES 1.4A 9. Expressed in the form y = 4(x? + bz/4) + 10 10. This parabola opens to the right. = 4x + 5/8)* +10 — 25/16 = 4(z + 5/8)% -- 135/16, the parabola opens upward with minimum at (—5/8, 135/16). rd Ly x (-5/8,135/16) to xz 11. Expressed in the form y = —{x? — 6z + 9) 12. This parabola opens to the left = —(z — 3)?, the parabola opens downward and touches the y-axis at y = —4. from the point (3,0). y 4 3 x -16 x 4 9 13. (a) The y-intercept is —5, and z-intercepts are 2A D =1+ 6. (b) The a-intercept is 4, and from x = 4(y — 1)2, the y-intercept is 1. 14. The equation must be of the form y = az? +1. Since (2,3) is on the parabola, 3 = da+1 = a=1/2. 15. The equation must be of the form = = 2+ay?. Since (0,4) is on the parabola, 0 = 2+16a => a = —1/8. 16. Since the parabola crosses the z-axis at © = ~1 and z = 3, it must be of the form y = alz +1) (2 ~ 3). Since (0, —1) is on the parabola, —1 = a(1)(-3) = a= 1/3. 17. Since the parabola touches the z-axis at = 1, it must be of the form y = a(x — 1)%. Since (0,2) is on the parabola, 2 = a(—1)? => a = 2. 18. Wesctz +1=1-2 — 0=2"+2 19. Weset —20=1+a? = 0=2 12s +1 = a(z +1) = x = 0,—1. Points of =(z +1) = x = —1. The only point intersection are (—1,0) and (0,1). of intersection is (1,2). y 4 2 1] 2| 2 4 i % E; El i K 4 . 2 20. We set 22 — 22 — 6 =5+ 2/5 => 21. Wesety(y —1)=y—1/2 => 2° —dy+1=0. 522% — 9 + 55 = 0. With equation 1.5, With equation 1.5, y = (4 + 16 — 8)/4 2 = (94/81 = 4(5)(55))/10. Since = (2+ v/2)/2. Points of intersection are 81 — 4(5)(55) < 0, there are no points (1 V2)/2,(2 £ V2)/2). of intersection. + StudyXy

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    EXERCISES 1.4A 11 y y S| 3 ra) B i 3 A 10} 20} B i 3 * Bl 22. Weset —y* +1=y*+2y~3 23. Weset 62? —2—=2%+z+1=52—z—3=0. = 0=22 +2 —4=20y+2)(y—1) Equation 1.5 gives = = (1 + +/T+ 60)/10 ==3 7 = 1, —2. Points of intersection are = (1 + y/B1)/10. Points of intersection are (0,1) and (—3,-2). ((1 2 61) /10, (43 & 3/81) /25). Ly y I call BN ‘ Sell E) E] T * 24. The range of the shell is R = (v?/9.81) sin 26. Since 0 < 6 < 7/2, range is a maximum when sin 20 = 1 = 20 =1/2 => = 7/4 radians. i 25. With coordinates as shown, the equation of the parabola takes the form y = aa? + 10. Since the point (100,50) is on the parabola, 50 = (10000) + 10 => a = 1/250. When z = 70, we obtain y = (70)2/250 + 10 = 148/56 m for the oo length of the supporting rod. 100-10 * * 26. We set ba = (z— 2)" +4 —> (z 2)" 5c +4 = 0 => 2! — 82° +2422 — 372+ 20 = 0. Possible rational solutions are +1, £2, +4, +5, +10, £20. One solution is @ = 1, so that at 828 1 240% — 87x + 20 = (2 — 1)(2® — 7a? + 172 — 20) = 0. We find that 22 = 4 is a zero of the cubic, so that 21 — 82% + 242” — 3Tw +20 = (2 — 1)(x — 4)(2? — Bx +5). Since #2 — 8x + 5 = 0 has no real solutions, the only points of intersection are (1,1) and (4,4). 27. With the coordinate system shown, the pd equation of the parabola takes the form (0) y = (25/4 — 2%). Since the point (3/2,4) is on the parabola, 4 = (25/4 — 9/4) ==> c= 1. The arch is therefore 25/4 units high. Ei ERE "28. If the parabola is to pass through these points, then 2=a+b+ec 10 =9a—3b+c, 4=9a+3b+e The solution of these equations is a = 1/2, b= —1, c= 5/2. 29. (a) Since resistances at temperatures 0°, 100°, and 700° are 10.000, 13.946, and 24.172, 10.000 = Ro(1), 13.946 = Ro(1 + 100e + 10000b), 24.172 = Ro(1 + 700a + 490 0006). The solution of these equations is Rg = 10.000, + StudyXy

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    12 EXERCISES 1.4B a ~0.0042662, and b= —3.2024 x 1075. " (b) The parabola is plotted to the right. (c) The resistance is 20 ohms when 2 20 = 10(1 -- aT" + bT2) =» BT? + al ~1=0. N — faz Solutions of this equation are T' = a 5 Since only the positive solution is acceptable, T = (—a + Va? + 46)/(2b) ~ 304. Woe 7 80. If P(a,b) is the point at which the rope meets the parabola, the equation of the line PQ is y—4 = [(b—4)/(a — 8)](z — 3). The z-coordinates y of points of intersection of this line with A 06H the parabola are given by the equation 2 1= =) (z—3)+4 a—3 » =, (bd b—4 E ! * 2 (272 27%) 5= Bn » (9)=+3 (5) 5-0, Since the line is to meet the parabola in only one point, the discriminant of this quadratic must be zero b—4\? b—4 y 2 Jimi 228) gl = Z 4)? —12(b— 4)(a — 3) + 20(a — 3) = 0. (=) als (3) 3] 0 == (b—4)?—12(b—4)(a— 3) +20(a ) Since P(a,b) is on the parabola, b= a® — 1, and when we substitute this into the above equation, 0 = (a2 ~ 5)? — 12(a® — 5)(a — 3) + 20(a — 3) = o* — 120° + 46a” — 60a + 25. Possible rational solutions of this equation are £1, £5, £25. We find that a = 1 is a solution. When it is factored from the quartic, 0 =a 1243 + 4642 -- 60a + 25 == (a — 1)(a® — 11a® + 35a — 25). Once again a = 1 is a zero of the cubic, so that 0 = at — 120° + 460° — 60a + 25 = (a — 1)(a - 1)(a® — 10a +25) = (a — 1)*(a — 5)*. Clearly, ¢ = 5 is inadmissible, and the required point is (1,0). EXERCISES 1.4B 1. The circle is centred at the origin 2. The centre of the circle is (—5,2) with radius 5v/2. and its radius is vB. JY. y ah : x S—— % 3. When we complete the square on the a-terms, 4. When we complete the square on the y-terms, (z+ 1)? + y2 = 16. The centre of the circle a? + (y — 2)? = 3. The centre of the circle is (—1,0) and its radius is 4. is (0,2) and its radius is v3. + StudyXy

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    EXERCISES 1.4B 13 od y J % p —% 5. When we complete squares on 2- and y-terms, 6. When we complete the square on the z-terms, (z —1)% + (y — 1)? = 1. The centre of the circle (2 +3/2)% +y* = 59/4. The centre of the circle is (1,1) and its radius is 1. is (—3/2,0) and its radius is v/59/2. 21 4 IN NE Tx = 7. When we complete squares on z- and y-terms, 8. When we complete squares on @- and y-terms, (x +2/3)% + (y — 1/3) = 23/9. The centre of (z+ 2)% + (y — 1)® = 10. The centre of the the circle is (—2/3,1/3) and its radius circle is (—2,1) and its radius is vI0. is V23/3. y 2d 9. When we complete squares on «- and y-terms, (z — 1)* + (y — 2)? = 0. The only point satisfying this equation is (1,2). 10. When we complete squares on @- and y-terms, (2 3) + (y+ 3/2)? = —35/4. No point can satisfy this equation. 11. With centre (0,0) and radius 2, the equation of the circle is 2? fy? =4. 12. Since the centre is (1,0) and its radius is 1, the equation of the circle is (z — 1)% + yr =1 13. Since the centre is (3,4) and its radius is 2, the equation of the circle is (z — H+ (y— 4)? =4 14. The figure indicates that the centre of the circle is (3/2, —3/2). Its radius is then 1/(3/2)% + (3/22 = 3/2. The equation of the circle is therefore (z — 3/2)? + (y + 3/2) = 9/2. 15. If we take the equation of the circle in form 1.22, and substitute each of the points (3, 0), (2,7), and (5,6), 9+04+3f+e=0, 3f+e=-9, 4449127 +Tgte=0, => 2 + 7g +e= 53, 25436 —5f + 6g +e=0, —5f + 6g+e=—61. The solution of these equations is f = 2, g = —6, and e = —15. The equation of the circle is x2 + y? + 22 6y—15=0. + StudyXxy

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