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Kunci Jawaban dan Pembahasan Soal || Buku Calculus 9th Purcell Chapter 0 - 0.3 Number 29 – 56

Pembahasan Soal Buku Calculus 9th Purcell Chapter 0 Section 0.3

◀ CHAPTER 0 PRELIMINARIES ▶

◀ SECTION 0.3 The Rectangular Coordinate System ▶


Problem Set 0.3, Number 29 – 56.
    In Problems 29-34, find an equation for each line. Then write your answer in the form Ax + By + C = 0.
  1. Through (2, 2) with slope –1
    2. Answer:
    y – 2
    =
    –1(x – 2)
    y – 2
    =
    –x + 2
    x + y – 4
    =
    0

    Calculus 9th Purcell Chapter 0 - 0.3

  2. Through (3, 4) with slope –1
    3. Answer:
    y – 4
    =
    –1(x – 3)
    y – 4
    =
    –x + 3
    x + y – 7
    =
    0

  3. With y-intercept 3 and slope 2
    4. Answer:
    y
    =
    2x + 3
    2x – y + 3
    =
    0

  4. With y-intercept 5 and slope 0
    5. Answer:
    y
    =
    0x + 5
    0x + y – 5
    =
    0

  5. Through (2, 3) and (4, 8)
    6. Answer:
    m
    =
    8 – 3
    =
    5
    ;
    4 – 2
    2
    y – 3
    =
    5
    (x – 2)
    2
    2y – 6
    =
    5x – 10
    5x – 2y – 4
    =
    0

  6. Through (4, 1) and (8, 2)
    7. Answer:
    m
    =
    2 – 1
    =
    1
    ;
    8 – 4
    4
    y – 1
    =
    1
    (x – 4)
    4
    4y – 6
    =
    x – 4
    x – 4y + 4
    =
    0

    In Program 35-38, find the slope and y-intercept of each line.
  7. 3y = –2x + 1
    8. Answer:
    3y
    =
    –2x + 6
    y
    =
    –2/3 · x + 1/3
    slope
    =
    –2/3 ;
    y-intercept
    =
    1/3

  8. –4y = 5x – 6
    9. Answer:
    –4y
    =
    5x – 6
    y
    =
    –5/4 · x + 3/2
    slope
    =
    –5/4 ;
    y-intercept
    =
    3/2

  9. 6 – 2y = 10x – 2
    10. Answer:
    6 – 2y
    =
    10x – 2
    –2y
    =
    10x – 8
    y
    =
    –5x + 4
    slope
    =
    –5 ;
    y-intercept
    =
    4

  10. 4x + 5y = –20
    11. Answer:
    4x + 5y
    =
    –20
    5y
    =
    –4x – 20
    y
    =
    –4/5 · x – 4
    slope
    =
    –4/5 ;
    y-intercept
    =
    –4

  11. Write an equation for the line through (3, –3) that is
    1. parallel to the line y = 2x + 5;
    2. perpendicular to the line y = 2x + 5;
    3. parallel to the line 2x + 3y = 6;
    4. perpendicular to the line 2x + 3y = 6;
    5. parallel to the line through (–1, 2) and (3, –1);
    6. parallel to the line x = 8;
    7. perpendicular to the line x = 8.
    Answer:
    1.  
      m = 2;
       
       
       
      y + 3
      =
      2(x – 3)
       
      y
      =
      2x – 9
    2.  
      m = –1/2;
       
       
       
      y + 3
      =
      –1/2 (x – 3)
       
      y
      =
      –1/2 · x – 3/2
    3.  
      2x + 3y
      =
      6
       
      3y
      =
      –2x + 6
       
      y
      =
      –2/3 · x + 2 ;
       
      m = –2/3;
       
       
       
      y + 3
      =
      –2/3 (x – 3)
       
      y
      =
      –2/3 · x – 1
    4.  
      m = 3/2;
       
       
       
      y + 3
      =
      3/2 (x – 3)
       
      y
      =
      3/2 · x – 15/2
    5.  
      m = –1 – 2/3 + 1 = –3/4;
       
      y + 3
      =
      –3/4 (x – 3)
       
      y
      =
      –3/4 · x – 3/4
    6.  
      x
      =
      3
    7.  
      y
      =
      –3

  12. Find the calue of c for which the line 3x + cy = 5
    1. passes through the point (3, 1);
    2. is parallel to the y-axis;
    3. is parallel to the line 2x + y = –1;
    4. has equal x- and x-intercepts;
    5. is perpendicular to the line y – 2 = 3(x + 3).
    Answer:
    1. 3x + cy
      =
      5
      3(3) + c(1)
      =
      5
      c
      =
      –4
    2. c
      =
      –4
    3.  
      2x + 3y
      =
      –1
       
      y
      =
      –2x – 1
       
      m = –2 ;
       
       
       
      3x + cy
      =
      5
       
      cy
      =
      –3x + 5
       
      y
      =
      –3/c · x + 5/c
      –2 = –3/c
       
       
      c = 3/2
       
       
    4.           c must be the same as the coefficient of x, so c = 3.
    5.  
      y – 2
      =
      3(x + 3) ;
      perpendicular slope
      =
      –1/3 ;
       
      –1/3
      =
      –3/c
       
      c
      =
      9

  13. Write the equation for the line through (–2, –1) that is perpendicular to the line y + 3 = –2/3 · (x – 5).
    14. Answer:
     
    m = 3/2 ;
     
     
     
    y + 1
    =
    3/2 · (x + 2)
     
    y
    =
    3/2 · x + 2

  14. Find the value of k such that the line kx – 3y = 10
    1. is parallel to the line y = 2x + 4;
    2. is perpendicular tp the line y = 2x + 4;
    3. is perpendicular tp the line 2x + 3y = 6
    Answer:
    1.  
      m = 2 ;
       
       
       
      kx – 3y
      =
      10
       
      – 3y
      =
      – 3x + 10
       
      y
      =
      k/3 · x – 10/3
       
      k/3
      =
      2
       
      k
      =
      6
    2.  
      m = – 1/2 ;
       
       
       
      k/3
      =
      – 1/2
       
      k
      =
      – 3/2
    3. 2x + 3y
      =
      6
      3y
      =
      –2x + 6
      y
      =
      – 2/3 · x + 2 ;
      m
      =
      3/2 ;
      k/3
      =
      3/2 ;
      k
      =
      9/2

  15. Does (3, 9) lie above or below the line y = 3x – 1?
    16. Answer:
    y = 3(3) – 1 = 8; (3, 9) is above the line.

  16. Show that the equation of the line with x-intercept a ≠ 0 and y-intercept b ≠ 0 can be written as
    17. x/a + y/b = 1
    Answer:
    (a, 0), (0, b); m = b – 0/0 – a = –b/a
    y = –b/a · x + b; bx/a + y = b; x/a + x/b = 1

    In Problems 45-48, find the coordinates of the point of intersection. Then write an equation for the line through that point perpendicular to the line given first.
  17. 2x + 3y = 4
    18. –3x + y = 5
    Answer:
    2x + 3y
    =
    4
    –3x + y
    =
    5
     
     
     
    2x + 3y
    =
    4
    9x – 3y
    =
    –15
    11x
    =
    –11
    x
    =
    –1
    –3(–1) + y
    =
    5
    y
    =
    2
    Point of intersection: (–1, 2)
    3y
    =
    –2x + 4
    y
    =
    –2/3 · x + 4/3
    m = 3/2
     
    y – 2
    =
    3/2 (x + 1)
    y
    =
    3/2 · x + 7/2

  18. 4x – 5y = 8
    19. 2x + y = –10
    Answer:
    4x – 5y
    =
    8
    2x + y
    =
    –10
     
     
     
    4x – 5y
    =
    8
    –4x – 2y
    =
    20
    –7y
    =
    28
    y
    =
    –4
    4x – 5(–4)
    =
    8
    4x
    =
    –12
    x
    =
    –3
    Point of intersection: (–3, –4);
    4x – 5y
    =
    8
    –5y
    =
    –4x + 8
    y
    =
    – 4/5 · x – 8/5
    m = – 5/4
     
    y + 4
    =
    – 5/4 (x + 3)
    y
    =
    – 5/4 · x – 31/4

  19. 3x – 4y = 5
    20. 2x + 3y = 9
    Answer:
    3x – 4y
    =
    5
    2x + 3y
    =
    9
     
     
     
    9x – 12y
    =
    15
    8x + 12y
    =
    36
    17x
    =
    51
    x
    =
    3
    3(3) – 4y
    =
    5
    –4y
    =
    –4
    y
    =
    1
    Point of intersection: (3, 1);
    3x – 4y = 5; –4y = –3x + 5
    y
    =
    3/4 · x – 5/4
    m = – 4/3
     
    y – 1
    =
    – 4/3 (x – 3)
    y
    =
    – 4/3 · x + 5

  20. 5x – 2y = 5
    21. 2x + 3y = 6
    Answer:
    5x – 2y
    =
    5
     
    2x + 3y
    =
    6
     
     
     
     
     
    15x – 6y
    =
    15
     
    4x + 6y
    =
    12
     
    19x
    =
    27
     
    x
    =
    27/19
     
    3(27/19) + 3y
    =
    6
     
    3y
    =
    60/19
     
    y
    =
    20/19
     
    Point of intersection: (27/19, 20/19);
     
    5x – 2y
    =
    5
     
    –2y
    =
    –5x + 5
     
    y
    =
    5/2 · x – 5/2
     
    m = – 2/5
     
     
    y – 20/19
    =
    – 2/5 (x – 27/19)
    y
    =
    – 2/5 · x + 54/95 + 20/19
    y
    =
    – 2/5 · x + 154/95

  21. The points (2, 3), (6, 3), (6, –1), and (2, –1) are corners of a square. Find the equations of the inscribed and circumscribed circles.
    22. Answer:
    center : (2 + 6/2, –1 + 3/2) = (4, 1)
    modpoint = (2 + 6/2, 3 + 3/2) = (4, 3)
    inscribed circle:
    radius = √[(4 – 4)2 + (1 – 3)2] = √4 = 2
    (x – 4)2 + (y – 1)2 = 4
    circumscribed circle:
    radius = √[(4 – 2)2 + (1 – 3)2] = √8
    (x – 4)2 + (y – 1)2 = 8

  22. A belt fits tightly around the two circles, with equations (x – 1)2 + (y + 2)2 = 16 and (x + 9)2 + (y – 10)2 = 16. How long is this belt?
    23. Answer:
    The radius of each circle is √16 = 4. The centers are (1, –2) and (–9, 10). The length of the belt is the sum of half the circumference of the first circle, half the circumference of the second circle, and twice the distance between their centers.
    L
    =
    1/2 · 2π(4) + 1/2 · 2π(4) + 2√[(1 + 9)2 + (–2 – 10)2]
    L
    =
    2π + 2√[100 + 144}
    L
    ≈
    56.37

  23. Show that the midpoint of the hypotenuse of any right triangle is equidistant from the three vertices.
    24. Answer:
    Put the vertex of the right angle at the origin with the other vertices at (a, 0) and (0, b). The midpoint of the hypotenuse is (a/2, b/2). The distances from the vertices are
    √[(a – a/2)2 + (0 – b/2)2] = √[a2/4 + b2/4] = 1/2 √[a2 + b2],
    √[(0 – a/2)2 + (b – b/2)2] = √[a2/4 + b2/4] = 1/2 √[a2 + b2], and
    √[(0 – a/2)2 + (0 – b/2)2] = √[a2/4 + b2/4] = 1/2 √[a2 + b2].
    which are all the same.

  24. Find the equation of the circle circumscribed about the right triangle whose vertices are (0, 0), (8, 0), and (0, 6).
    25. Answer:
    From Problem 51, the midpoint of the hypotenuse, (4, 3,), is equidistant from the vertices. This is the center of the circle. The radius is √[16 + 9] = 5. The equation of the circle is
    (x – 4)2 + (y – 3)2 = 25.

  25. Show that the two circles x2 + y2 – 4x – 2y – 11 = 0 and x2 + y2 + 20x – 12y + 72 = 0 do not intersect. Hint: Find the distance between their centers.
    26. Answer:
    x2 + y2 – 4x – 2y – 11
    =
    0
    (x2 – 4x + 4) + (y2 – 2y + 1)
    =
    11 + 4 + 1
    (x – 2)2 + (y – 1)2
    =
    16
    x2 + y2 + 20x – 12y + 72
    =
    0
    (x2 + 20x + 100) + (y2 – 12y + 36)
    =
    –72 + 100 + 36
    (x + 10)2 + (y – 6)2
    =
    64
    center of first circle: (2, 1)
     
     
    center of second circle: (–10, 6)
     
     
    d = √[(2 + 10)2 + (1 – 6)2]
    =
    √[144 + 25]
    d
    =
    √169 = 13
    However, the radii only sum to 4 + 8 = 12, so the circles must not intersect if the istance between their centers is 13.

  26. What relationship between a, b, and c must hold if x2 + ax + y2 + by + c = 0 is the equation of a circle?
    27. Answer:
    x2 + ax + y2 + by + c = 0
    (x2 + ax + a2/4) + (y2 + by + b2/4) = –c + a2/4 + b2/4
    (x + a/2)2 + (y + b/2)2 = a2 + b2 – 4c/4
    a2 + b2 – 4c/4 > 0 ⇒ a2 + b2 > 4c

  27. The ceiling of an attic makes an angle of 30° with the floor. A pipe of radius 2 inches is placed along the edge of the attic in such a way that one side of the pipe touches the ceiling and another side touches the floor (see Figure 17). What is the distance d from the edge of the attic to where the pipe touches the floor?
    28. Calculus 9th Purcell Chapter 0 - 0.3 Figure 17 and 18
    Answer:
    Label the points C, P, Q, and R as shown in the figure below. Let d = |OP|, h = |OR|, and a = |PR|. Triangles ΔOPR and ΔCQR are similar because each contains a right angle and they share angle ∠QRC. For an angle of 30°, d/h = √3/2 and a/h = 1/2 ⇒ h =2a. Using a property of similar triangles, |QC| / |RC| = √3/2,
    2/a – 2 = √3/2 → a = 2 + 4/√3
    By the Pythagorean Theorem, we have
    d = √[h2 – a2] = √3 · a = 2√3 + 4 ≈ 7.464
    Calculus 9th Purcell Chapter 0 - 0.3 Number 55

  28. A circle of radius R is placed in the first quadrant as shown in Figure 18. What is the radius r of the largest circle that can be placed between the original circle and the origin?
    29. Answer:
    The equations of the two circles are
    (x – R)2 + (y – R)2 = R2
    (x – r)2 + (y – r)2 = r2
    Let (a, a) denote the point where the two circles touch. This point must satisfy
    (a – R)2 + (a – R)2 = R2
    (a – R)2 = R2/2
    a = (1 ± √2/2)R
    Since a < R, a = (1 – √2/2)R.
    At the same time, the point where the two circles touch must satisfy
    (x – r)2 + (y – r)2 = r2
    a = (1 ± √2/2)r
    Since a > r, a = (1 + √2/2)r.
    Equating the two expressions for a yields
    (1 – √2/2)R = (1 + √2/2)r
    r
    =
    1 –
    √2/2
    · R
    2
    1 +
    √2/2
    2

    r
    =
     
     
    ⌈
    1 –
    √2/2
    ⌉2
     
     
    · R
     
    ⌊
    2
    ⌋
     
     
    ⌈
    1 +
    √2/2
    ⌉
    ⌈
    1 –
    √2/2
    ⌉
    ⌊
    2
    ⌋
    ⌊
    2
    ⌋

    r
    =
    1 - √2 + ½
    · R
    1 - ½

    r = (3 – 2√2)R ≈ 0.1716R
    Calculus 9th Purcell Chapter 0 - 0.3 Number 56

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