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Kunci Jawaban dan Pembahasan Soal || Buku Calculus 9th Purcell Chapter 0 - 0.3 Number 57 – 77

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 57 – 77.
  1. Construct a geometric proof using Figure 15 that shows two lines are perpendicular if and only if their slopes are negative reciprocals of one another.
    2. 💬 SOLUTION:
    Refer to figure 15 in the text. Given ine l1 with slope m, draw ΔABC with vertical and horizontal sides m, 1.

    Chapter 0 - 0.3 Number 57 – 77

    Line l2 is obtained from l1 by rotating it around the point A by 90° counter-clockwise. Triangle ABC is rotated into triangle AED . We read off
    slope of l2 =
    1
    =
    –
    1
    – m
    m
    ■

  2. Show that the set of points that are twice as far from (3,4) as from (1,1) form a circle. Find its center and radius.
    3. 💬 SOLUTION:
    2√[(x – 1)2 + (y – 1)2]
    =
    √[(x – 3)2 + (y – 4)2]
    4(x2 – 2x + 1 + y2 – 2y + 1)
    =
    x2 – 6x + 9 + y2 – 8y + 16
    3x2 – 2x + 3y2
    =
    9 + 16 – 4 – 4;
    3x2 – 2x + 3y2
    =
    17;
    x2 – (2/3)x + y2
    =
    (17/3);
    [x2 – (2/3)x + (1/9)] + y2
    =
    (17/3) + (1/9)
    [x – (1/3)]2 + y2
    =
    (52/9)
    center: (1/3, 0);
     
    radius: (√52/3)
    ■

  3. The Pythagorean Theorem says that the areas A, B, and C of the squares in Figure 19 satisfy A + B = C. Show that semicircles and equilateral triangles satisfy the same relation and then guess what a very general theorem says.
    4. Chapter 0 - 0.3 Number 59
    💬 SOLUTION:
    Let a, b, and c be the lengths of the sides of the right triangle, with c the length of the hypotenuse. Then the Pythagorean Theorem says that a2 + b2 = c2.
    Thus,
    πa2
    +
    πb2
    =
    πc2
    or
    8
    8
    8
    1/2 π (a/2)2 + 1/2 π (b/2)2 = 1/2 π (c/2)2
    1/2 π (x/2)2 is the area of a semicircle with diameter x, so the circles on the legs of the triangle have total area equal to the area of the semicircle on the hypotenuse.
    From a2 + b2 = c2.
     
    √3
    a2
    +
    √3
    b2
    =
    √3
    c2
    4
    4
    4
    (√3/4) x2 is the area of an equilateral triangle with sides of length x, so the equilateral triangles on the legs of the right triangle have total area equal to the area of the equilateral triangle on the hypotenuse of the right triangle.
    ■

  4. Consider a circle C and a point P exterior to the circle. Let line segment PT be tangent to C at T, and let the line through P and the center of C intersect C at M and N. Show that (PM)(PN) = (PT)2.
    5. 💬 SOLUTION:
    See the figure below. The angle at T is a right angle, so the Pythagorean Theorem gives
    (PM + r)2 = (PT)2 + r2
    ⇔ (PM)2 + 2rPM + r2 = (PT)2 + r2
    ⇔ PM(PM + 2r) = (PT)2
    PM + 2r = PN so this gives (PM)(PN) = (PT)2
    Chapter 0 - 0.3 Number 60
    ■

    Semoga Bermanfaat 😁

  5. A belt fits around the three circles x2 + y2 = 4, (x – 8)2 + y2 = 4, and (x – 6)2 + (y – 8)2 = 4, as shown in Figure 20. Find the length of this belt.
    6. Chapter 0 - 0.3 Number 61
    💬 SOLUTION:
    The lengths A, B, and C are the same as the corresponding distances between the centers of the circles:
    A = √[(–2)2 + (8)2] = √68 ≈ 8.2
    B = √[(6)2 + (8)2] = √100 = 100
    C = √[(8)2 + (0)2] = √64 = 8
    Each circle has radius 2, so the part of the belt around the wheels is
    2(2π – a – π) + 2(2π – b – π) + 2(2π – c – π)
    = 2[3π – (a + b + c)] = 2(2π) = 4π
    Since a + b + c = π , the sum of the angles of a triangle.
    The length of the belt is
    ≈
    8.2 + 10 + 8 + 4π
     
    ≈
    38.8 units.
    Chapter 0 - 0.3 Number 61
    ■

  6. Study Problems 50 and 61. Consider a set of nonintersecting circles of radius r with centers at the vertices of a convex n-sided polygon having sides of lengths d1,d2,...,dn. How long is the belt that fits around these circles (in the manner of Figure 20)?
    7. 💬 SOLUTION:
    As in Problems 50 and 61, the curved portions of the belt have total length 2πr The lengths of the straight portions will be the same as the lengths of the sides. The belt will have length 2πr + d1 + d2 + … + dn.
    ■

    It can be shown that the distance d from the point (x1,y1) to the line Ax + By + C = 0 is
    Chapter 0 - 0.3 Number 62
    Use this result to find the distance from the given point to the given line.
  7. (–3, 2): 3x + 4y = 6
    8. 💬 SOLUTION:
    A = 3, B = 4, C = –6
    d =
    |3(–3) + 4(2) + (–6)|
    =
    7
    √ [(3)2 + (4)2]
    5
    ■

  8. (4, –1): 2x – 2y + 4 = 0
    9. 💬 SOLUTION:
    A = 2, B = –2, C = 4
     
    d =
    |2(4) – 2(–1) + (4)|
    =
    14
    =
    7√2
    √[(2)2 + (2)2]
    √8
    2
    ■

    Semoga Bermanfaat 😁

  9. (–2, –1): 5y = 12x + 1
    10. 💬 SOLUTION:
    A = 12, B = –5, C = 1
     
    d =
    |12(–2) – 5(–1) + 1|
    =
    18
    √ [(12)2 + (–5)2]
    13
    ■

  10. (3, –1): y = 2x – 5
    11. 💬 SOLUTION:
    A = 2, B = –1, C = –5
     
    d =
    |2(3) – 1(–1) – 5|
    =
    2
    =
    2√5
    √[(2)2 + (–1)2]
    √5
    5
    ■

    In Problems 67 and 68, find the (perpendicular) distance between the given parallel lines. Hint: First find a point on one of the lines.
  11. 2x + 4y = 7, 2x + 4y = 5
    12. 💬 SOLUTION:
    2x + 4(0)
    =
    5
     
    x
    =
    5/2
     
    d =
    |2(5/2) + 4(0) – 7|
    =
    2
    =
    √5
    √[(2)2 + (4)2]
    √20
    5
    ■

  12. 7x – 5y = 6, 7x – 5y = –1
    13. 💬 SOLUTION:
    7(0) – 5(y)
    =
    –1
     
    y
    =
    1/5
     
    d =
    |7(0) – 5(1/5) – 6|
    =
    7
    =
    7√74
    √[(7)2 + (–5)2]
    √74
    74
    ■

    Semoga Bermanfaat 😁

  13. Find the equation for the line that bisects the line segment from (–2, 3) to (1, –2) and is at right angles to this line segment.
    14. 💬 SOLUTION:
    m =
    –2 – 3
    =
    –
    5
    ;
     
    1 + 2
    3
     
    m =
    3
    ;
     
    5
     
    passes through
    ⌈
    –2 – 3
    ,
    3 – 2
    ⌉
    ⌊
    2
    2
    ⌋
    =
    ⌈
    –
    1
    ,
    1
    ⌉
     
    ⌊
    2
    2
    ⌋
     
    y – 1/2
    =
    3/5 (x + 1/2)
    y
    =
    3/5 x + 4/5
    ■

  14. The center of the circumscribed circle of a triangle lies on the perpendicular bisectors of the sides. Use this fact to find the center of the circle that circumscribes the triangle with vertices (0,4), (2, 0), and (4,6).
    15. 💬 SOLUTION:
    m =
    0 – 4
    =
    –2
    ;
    2 – 0
    m =
    1
    ;
    2
    passes through
    ⌈
    0 + 2
    ,
    4 + 0
    ⌉
    ⌊
    2
    2
    ⌋
    =
    (1, 2)
     
    y – 2
    =
    1/2 (x – 1)
    y
    =
    1/2 x + 3/2
    m =
    6 – 0
    =
    3
    ;
    4 – 2
    m =
    –
    1
    ;
    3
    passes through
    ⌈
    2 + 4
    ,
    0 + 6
    ⌉
    ⌊
    2
    2
    ⌋
    =
    (3, 3)
     
    y – 3
    =
    –1/3 (x – 3)
    y
    =
    –1/3 x + 4
    1/2 x + 3/2
    =
    –1/3 x + 4
    5/6 x
    =
    5/2
    x
    =
    3
    y
    =
    1/2 (3) + 3/2
     
    =
    3
    center
    =
    (3, 3)
    ■

  15. Find the radius of the circle that is inscribed in a triangle with sides of lengths 3, 4, and 5 (see Figure 21).
    16. Chapter 0 - 0.3 Number 71
    💬 SOLUTION:
    Let the origin be at the vertex as shown in the figure below. The center of the circle is then (4–r, r), so it has equation (x – (4–r))2 + (y – r) = r2. Along the side of length 5, the y-coordinate is always 3/4 times the x-coordinate. Thus, we need to find the value of r for which there is exactly one x-solution to (x – 4 + r)2 + (3/4 x – r)2 = r2.
    Solving for x in this equation gives x = 16/25 (16 – r ± √[24(–r2 + 7r – 6)]. There is exactly one solution when –r2 + 7r – 6 = 0, that is, when r = 1 or r = 6 . The root r = 6 is extraneous. Thus, the largest circle that can be inscribed in this triangle has radius r = 1.
    Chapter 0 - 0.3 Number 71 Solution
    ■

  16. Suppose that (a, b) is on the circle x2 + y2 = r2. Show that the line ax + by = r2 is tangent to the circle at (a, b).
    17. 💬 SOLUTION:
    The line tangent to the circle at (a, b) will be perpendicular to the line through (a, b) and the center of the circle, which is (0, 0). The line through (a, b) and (0, 0) has slope m = 0 – b/0 – a = b/a ; ax + by = r2 ⇒ y = –a/b x + r2/b so ax + by = r2 has slope –a/b and is perpendicular to the line through (a, b) and (0, 0) so it is tangent to the circle at (a, b).
    ■

    Semoga Bermanfaat 😁

  17. Find the equations of the two tangent lines to the circle x2 + y2 = 36 that go through (12, 0). Hint: See Problem 72.
    18. 💬 SOLUTION:
    12a + 0b = 36
    a = 3
    23 + b3 = 36
    b = ±3√(3)
    3x – 3√(3)y = 36
    x – √(3)y = 12
    3x + 3√(3)y = 36
    x + √(3)y = 12
    ■

  18. Express the perpendicular distance between the parallel lines y = mx + b and y = mx + B in terms of m, b, and B. Hint: The required distance is the same as that between y = mx and y = mx + B – b.
    19. 💬 SOLUTION:
    Use the formula given for problems 63–66, for (x, y) = (0, 0).
    A = m, B = –1, C = B – b ; (0, 0)
    d =
    |m(0) – 1(0) + B – b|
    =
    |B – b|
    √[(m)2 + (–1)2]
    √[m2 + 1]
    ■

  19. Show that the line through the midpoints of two sides of a triangle is parallel to the third side. Hint: You may assume that the triangle has vertices at (0, 0), (a, 0), and (b, c).
    20. 💬 SOLUTION:
    The midpoint of the side from (0, 0) to (a, 0) is (0+a/2, 0+0/2) = (a/2, 0)
    The midpoint of the side from (0, 0) to (b, c) is (0+b/2, 0+c/2) = (b/2, c/2)
    m1 = c–0/b–a = c/b–a
    m2 = c/2–0/b/2–a/2 = c/b–a
    m1 = m2
    ■

  20. Show that the line segments joining the midpoints of adjacent sides of any quadrilateral (four-sided polygon) form a parallelogram.
    21. 💬 SOLUTION:
    See the figure below. The midpoints of the sides are
    P(x1+x2/2 , y1+y2/2) ,
    Q(x2+x3/2 , y2+y3/2) ,
    R(x3+x4/2 , y3+y4/2) , and
    S(x1+x4/2 , y1+y4/2) .
    The slope of PS is
     
    1/2 [y1 + y4 – (y1 + y2)]
    =
    y4 – y2
    1/2 [x1 + x4 – (x1 + x2)]
    x4 – x2
    The slope of QR is
     
    1/2 [y3 + y4 – (y2 + y3)]
    =
    y4 – y2
    1/2 [x3 + x4 – (x2 + x3)]
    x4 – x2
    Thus PS and QR are parallel. The slopes of SR and PQ are both
     
    y3 – y2
    x3 – x2
    so PQRS is a parallelogram.
    Chapter 0 - 0.3 Number 76 Solution
    ■

  21. A wheel whose rim has equation x2 + (y – 6)2 = 25 is rotating rapidly in the counterclockwise direction. A speck of dirt on the rim came loose at the point (3,2) and flew toward the wall x = 11. About how high up on the wall did it hit? Hint: The speck of dirt flies off on a tangent so fast that the effects of gravity are negligible by the time it has hit the wall.
    22. 💬 SOLUTION:
    x2 + (y – 6)2 = 25; passes through (3, 2) tangent line: 3x – 4y = 1; The dirt hits the wall at y = 8.
    ■

Semoga Bermanfaat 😁

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