Conics and Polar Coordinates
10 Coordinates
Conics and Polar
CHAPTER
10.1 Concepts Review
3. 2 x = –4(3) y
1. e < 1; e = 1; e > 1
Focus at (0, –3) Directrix: y = 3
2 2. y = 4 px
3. (0, 1); y = –1
4. parallel to the axis
Problem Set 10.1
1. y 2 = 4(1) x 4. x 2 = –4(4) y
Focus at (1, 0)
Focus at (0, –4)
Directrix: x = –1
Directrix: y = 4
2. y 2 = –4(3) x 5. y 2 4 =⎜⎟ ⎛⎞ x
Focus at (–3, 0) Directrix: x = 3
⎛ Focus at 1 ,0 ⎞
Directrix: x = –
Section 10.1 Instructor’s Resource Manual
Focus at ⎜ – ,0 ⎟ Focus at ⎜ 0,
Directrix: x =
Directrix: y = –
7. 2 x 2 = 6 y 9. The parabola opens to the right, and p = 2.
10. The parabola opens to the left, and p = 3.
Focus at ⎜ 0,
y 2 = –12 x
Directrix: y = –
4 11. The parabola opens downward, and p = 2.
x 2 = –8 y
12. The parabola opens downward, and p = .
13. The parabola opens to the left, and p = 4.
y 2 = –16 x
14. The parabola opens downward, and p = .
x 2 = –14 y
Instructor’s Resource Manual
Section 10.1 591
15. The equation has the form y 2 = cx , so
18. The equation has the form x 2 = cy , so
(–1) 2 = 3. c (–3) 2 = 5c .
19. y 2 = 16 x
2 yy ′= 16
2 16. The equation has the form 16 y = cx , so y ′=
2 y (4) = –2 . c At (1, –4), y ′= –2.
c = –8 2
Tangent: y + 4 = –2(x – 1) or y = –2x – 2
y = –8 x 1 1 9
Normal: y += 4 ( – 1) x or y = x –
17. The equation has the form x 2 = cy , so
x 2 = – y At 25 ( 2 5, – 2 , ) y ′= – .
–25 ) or
Tangent: y += 2 –
Normal: y += 2 ( x –25 ) or y = x –7
Section 10.1 Instructor’s Resource Manual
21. x 2 = 2 y 25 21 5
y ′= x At (4, 8), y ′= 4 . Tangent: y – 8 = 4(x – 4) or y = 4x – 8
Normal: y –8 = – ( – 4) x or y = – x + 9
Tangent: y – 4 = 2(x – 4) or y = 2x – 4
2 yy ′= –9 1 1 Normal: y –4 = – ( – 4) x or y = – x + 6
y ′= –
At (–1, –3), y ′=
Tangent: y += 3 ( x + 1) or y = x –
Normal: y += 3 – ( x + or 1) y = – x –
23. y 2 = –15 x Tangent: y += 3 –2 ( x –32 ) or y = –2 x + 3
Normal: y += 3 ( x –32 ) or y = x –6
At ( –3, – 3 5 , ) y ′= .
Tangent: y + 35 =
Normal: y + 35 = –
( x + 3) or
Instructor’s Resource Manual
Section 10.1 593
26. y 2 = 20 x 29. The slope of the line is 3 .
2 y ⎛⎞=− ⎜⎟ 18; y =− 6
At ( 2, – 2 10 , ) y ′= – .
2 ( 6) − 2 =− xx 18 ; 2 =−
10 The equation of the tangent line is
Tangent: y + 2 10 = –
30. Place the x-axis along the axis of the parabola
Normal: y + 2 10 =
such that the equation y 2 = 4 px describes the
( – 2) x or
parabola. Let ⎜
be one of the
extremities and
be the other.
First solve for y 1 in terms of y 0 and p. Since the focal chord passes through the focus (p, 0), we
have the following relation.
y ′= when y = 2 5, so x = 4.
4 ⎛ 4 p 3 4 p 2 ⎞ The point is
4, 2 5 . Thus, the other extremity is ⎜
Implicitly differentiate 2 y = 4 px to get
2 yy ′= 4, p so y ′=
At ⎜ 0 ⎜ ,, y 0 ⎟ ⎟ y ′=
. The equation of the
tangent line is y – y 0 =
⎜ ⎟ . y When 0
⎟ , y ′= – 0 . The equation of 27
y′ = –
when x = 2 7, so y = –2.
the tangent line is y +
The point is ( 2 7, – 2 . ) y 0 2 p ⎜ y 2 ⎝ ⎟ 0 ⎠
Section 10.1 Instructor’s Resource Manual
When x = –p, y = 0 – p .
2 2 35. Let the y-axis be the axis of the parabola, so the
Earth’s coordinates are (0, p) and the equation of
the path is Thus, the two tangent lines intersect on the 2 x = 4 py , where the coordinates are
2 directrix at ⎜ –,– p 0 2 p
in millions of miles. When the line from Earth to
the asteroid makes an angle of 75 with the axis
of the parabola, the asteroid is at
31. From Problem 30, if the parabola is described by
( 40 sin 75 , p + 40 cos 75 ). (See figure.)
the equation 2 y = 4 px , the slopes of the tangent
2p
lines are
and – 0 . Since they are negative
y 0 2 p reciprocals, the tangent lines are perpendicular.
32. Place the x-axis along the axis of the parabola such that the equation 2 y = 4 px describes the
parabola. The endpoints of the chord are ⎜ 1, ⎟
(40 sin 75 ) °= 2 4( pp +
40 cos 75 ) ° ⎛ 2 1 ⎞ 1 1 2 2
and ⎜ 1, – ⎟ , so ⎛⎞= ⎜⎟ 4(1) p or p = . The
1600 cos 75 2 °+ 1600 sin 75 2 °
distance from the vertex to the focus is
33. Assume that the x- and y-axes are positioned such
− 20 20 cos 75 °≈
14.8 ( p > 0)
that the axis of the parabola is the y-axis with the
The closest point to Earth is (0, 0), so the asteroid
vertex at the origin and opening upward. Then
will come within 14.8 million miles of Earth.
the equation of the parabola is x 2 = 4 py and
36. Let the equation x 2 = 4 py describe the cables.
(0, p) is the focus. Let D be the distance from a
point on the parabola to the focus.
The cables are attached to the towers at
D 2 = ( x − 0) 2 + ( y − p ) 2 = x 2 + ⎜ − p ⎟ (400) = p 4 (400), 100 p =
The vertical struts are at x =±300.
x 4 x 2 x 2 (300) 2 = 4(100) , 225 yy =
+ p The struts must be 225 m long.
be the distance from R to the directrix.
Observe that the distance from the latus rectum to D′′ > so at x = 0, D is minimum. y = 0 0 the directrix is 2p so RG = 2– p RL . From the
Therefore, the vertex (0, 0) is the point closest to
definition of a parabola, RL = FR . Thus,
the focus.
FR + RG = RL + 2– p RL = 2. p
34. Let the y-axis be the axis of the parabola, so Earth’s coordinates are (0, p) and the equation of
the path is x 2 = 4 py , where the coordinates are in millions of miles. When the line from Earth to the asteroid makes an angle of 90 with the axis of the parabola, the asteroid is at (40, p).
(40) 2 = pp 4 ( ), 20 p = The closest point to Earth is (0, 0), so the asteroid
will come within 20 million miles of Earth.
Instructor’s Resource Manual
Section 10.1 595
38. Let the coordinates of P be ( x 0 , y 0 ). 40. Let C denote the center of the circle and r the
radius. Observe that the distance from a point P
2 yy ′= 4, p so y ′=
. Thus the slope of the
to the circle is PC –. r Let
l be the line and
PL the distance from the point to the line. Thus,
normal line at P is – 0 .
be the line parallel to l, r The equation of the normal line is
PC – r = PL . Let ' l
units away and on the side opposite from the y 0 circle. Then PL ′ , the distance from P to ' l , is
(– x x 0 ). When y = 0,
PL + r ; so PL = PL ′ –. r Therefore,
x = 2 p + x 0 , so B is at (2 p + x 0 , 0).
A is at
PC – r = PL ′ – r or PC = PL′ . The set of ( x 0 , 0). Thus, AB = 2 p + x 0 – x 0 = 2. p points is a parabola by definition.
39. Let P 1 and P 2 denote (, x 1 y 1 ) and ( x 2 , y 2 ), dy δ x
respectively. PP 12 = PF 1 + PF 2 since the focal
dx H
chord passes through the focus. By definition of a 2 δ x
parabola, PF 1 =+ p x 1 and PF 2 =+ p x 2 . 2 H
Thus, the length of the chord is
y (0) = 0 implies that C = 0. y =
PP 12 = x 1 + x 2 + 2. p 2 H
L =++ p p 2 p = 4 p This is an equation for a parabola.
42. a. AT () 1 is the area of the trapezoid formed by
2 (a, 0), P, Q, (b, 0) minus the area the two trapezoids formed by (a, 0), P, 2 (, ) cc , (c, 0) and by (c, 0), (, ) cc ,
a + bbc − c − a b − a
Q , (b, 0). Observe that since c =
()() 2 2 (–) b a AT ()
c. Using reasoning similar to part b, AT ( n ) =
d. Area =
[ a + b ]–() AS =
Section 10.1 Instructor’s Resource Manual
Using a CAS, we find that the largest vertical distance between the catenary and the parabola is roughly 67 feet.
44. a. Since the vertex is on the positive y-axis and
10.2 Concepts Review
the parabola crosses the x-axis, its equation is
of the form: x 2 = 4( py − k ) , where k is the
y -coordinate of the vertex; that is
x 2 = 4( py − 630) . Since the point (315, 0) x 2 y is on the parabola, we have 2
(315) = p 4 (0 630) or 4 − p =
Thus the parabola has the equation
3. foci
x 2 =− 157.5( y − 630) 4. to the other focus; directly away from the other
focus
b. Solving for y , we get
1 Problem Set 10.2
1. Horizontal ellipse The catenary for the Gateway Arch is
y = 758 128 cosh −
. 2. Horizontal hyperbola
y = 758 128cosh − x
3. Vertical hyperbola
4. Horizontal hyperbola
5. Vertical parabola
6. Vertical parabola
7. Vertical ellipse
8. Horizontal hyperbola
c. Because of symmetry, we can focus on the x 2 largest vertical distance between the graphs 2 y
= 1; horizontal ellipse
for positive x.
Let fx () = y Arch − y parabola . That is,
a = 4, b = 2, c = 23
fx () = ⎜ 758 128 cosh −
Vertices: (±4, 0)
Foci: ( ± 2 3, 0 )
Instructor’s Resource Manual
Section 10.2 597
10. – = 1; horizontal hyperbola
= vertical ellipse 1;
Foci: ( ± 2 5, 0 )
Foci: ( ± 0, 6 )
Asymptotes: y =± x
= horizontal ellipse 1;
11. – x= 1; vertical hyperbola 4 9
Foci: ( ± 21, 0 )
Vertices: (0, ±2)
Foci: ( ± 0, 13 )
Asymptotes: y =± x
= horizontal hyperbola 1;
b 10, 2, 14 = c =
12. + 1; horizontal ellipse ) =
Vertices: ( ± 10, 0
Foci: ( ± 14, 0
a b 7, 2, 3 c = )
Vertices: ( ± 7, 0 ) Asymptotes: y =± x
Foci: ( ± 3, 0 )
Section 10.2 Instructor’s Resource Manual
22. This is a horizontal hyperbola with a = 4 and
= horizontal hyperbola 1;
Foci: ( ± 10, 0 )
23. This is a vertical hyperbola with a = 4 and c = 5.
Asymptotes: y =± x b =
24. This is a vertical hyperbola with a = 3. ⎛⎞ 3 9 81 45
17. This is a horizontal ellipse with a = 6 and c = 3.
25. This is a horizontal hyperbola with a = 8.
−= 36 9 27 1 The asymptotes are 1 y =± x , so b= or b = 4.
18. This is a horizontal ellipse with c = 6.
19. This is a vertical ellipse with c = 5.
27. This is a horizontal ellipse with c = 2.
20. This is a vertical ellipse with b = 4 and c = 3.
21. This is a horizontal ellipse with a = 5.
25 b 2
28. This is a horizontal hyperbola with c = 4.
Instructor’s Resource Manual
Section 10.2 599
29. The asymptotes are y =± x . If the hyperbola is
1 32. This is an ellipse whose foci are (4, 0) and (- 4, 0)
2 and whose major diameter has length 2 a = 14.
b 1 Since the foci are on the x-axis, it is the major axis horizontal = or a = 2b. If the hyperbola is
of the ellipse so the equation has the form
a 2 14, a = 49 and since vertical, = or b = 2a.
2 + 2 = . Since 1 2 =
− 49 (4) Suppose the hyperbola is horizontal. 2 a b c b = 33 . Thus the x 2 y 2
equation is
33. This is an hyperbola whose foci are (7, 0) and
(- 7, 0) and whose axis is the x-axis. So the
b 2 = –5 x 2 y 2
This is not possible.
equation has the form 2 −
= 1 . Since
Suppose the hyperbola is vertical.
a 2 12, a = 36 and b = c − a = 2 (7 ) 36 − = 13
Thus the equation is
34. This is an hyperbola whose foci are (0, 6) and
(0, - 6) and whose axis is the y-axis. So the
– x= 1
equation has the form 2 − 2 = 1 . Since
Thus the equation is
35. Use implicit differentiation to find the slope:
b ⎪⎭
x + y y′ = 0 . At the point
equation of the tangent line is
36. Use implicit differentiation to find the slope:
31. This is an ellipse whose foci are (0, 9) and (0, - 9)
and whose major diameter has length 2 a = 26. 1 x y + y′ = . At the point 0
Since the foci are on the y-axis, it is the major axis
of the ellipse so the equation has the form
− y ′ = 0 , or y ′ = 2 so the y
2 + 2 = . Since 1 2 a = 26, a = 169 and since
equation of the tangent line is
a 2 = b 2 + c 2 , b 2 = 169 (9) − 2 ( y + 2) = 2( x − 3 2) or 2 x −= y 8 = . 88 . Thus the 2 y 2 x
equation is
Section 10.2 Instructor’s Resource Manual
37. Use implicit differentiation to find the slope:
43. Let the y-axis run through the center of the arch
x + y y′ = 0 . At the point
and the x-axis lie on the floor. Thus a = 5 and
x 27 2 9 y 2
b = 4 and the equation of the arch is + = 1 .
When y = 2,
1, so x =± .
equation of the tangent line is
6 The width of the box can at most be
6 53 ≈ 8.66 ft.
38. Use implicit differentiation to find the slope:
44. Let the y-axis run through the center of the arch
and the x-axis lie on the floor.
x − y y′ = 0 . At the point ( 3, 2) x 2 y 2 2
The equation of the arch is
y ′ = 0 , or
y ′ = 6 so the equation of the
(2) 2 y 2 4 21
When x = 2,
1, so y =± .
tangent line is ( y − 2) = 6( x − 3) or
25 16 5 The height at a distance of 2 feet to the right of the
6 x − 6 y = 43 .
center is
≈ 3.67 ft.
39. Use implicit differentiation to find the slope:
2 x + 2 y y′ = 0 . At the point (5,12)
45. The foci are at (±c, 0).
10 24 y ′ = 0 , or y ′ =−
so the equation of the
c = a 12 2 – b 2
tangent line is ( y − 12) =− ( x − 5) or
40. Use implicit differentiation to find the slope:
2 2 x − 2 y y′ = 0 . At the point ( 2, 3) 2 Thus, the length of the latus rectum is b . a
22 − 23 y ′ = 0 , or
so the equation of
3 46. The foci are at (±c, 0)
the tangent line is ( y − 3) =
( x − 2) or
41. Use implicit differentiation to find the slope:
2 , y =±
y y′ = 0 . At the point
2 b 2 Thus, the length of the latus rectum is
y ′ = 0 , or y ′ = . The tangent line is 0 a
13 horizontal and thus has equation
y = 13 .
47. a = 18.09, b = 4.56,
c = (18.09) 2 − (4.56) 2 ≈ 17.51
42. Use implicit differentiation to find the slope: The comet’s minimum distance from the sun is
x + y y′ = . At the point 0 (7, 0) 18.09 – 17.51 ≈
0.58 AU.
48. a −= c c 0.13, , = ae a (1 −= e ) 0.13,
+ 0 y ′ = 0 , or is undefined. y ′ . The tangent line
is vertical and thus has equation x = 7 .
a += c a (1 +≈ e ) 1733(1 0.999925) + ≈ 3466 AU
Instructor’s Resource Manual
Section 10.2 601
49. a – c = 4132; a + c = 4583
2a = 8715; a = 4357.5
54. The slope of the line is
50. (See Example 5) Since a += c 49.31 and
2 y into the equation of the 2 a =
a −= c 29.65 , we conclude that
Substitute x
78.96, 2 c = 19.66 and so
ellipse.
a = 39.48, c = 9.83 . Thus
2 y 2 + 2 y 2 −= 2 0; y =±
b = a − c = 1462.0415 ≈ 38.24 . So the
major diameter = 2 a = 78.96 and the minor
The tangent lines are tangent at ⎜ − 1,
⎟ and
diameter = 2 b = 76.48 .
⎜ 1, x − y ⎟ . The equations of the tangent lines are =
Equation of tangent line at ( x 0 , y 0 ): 2 2 2 2
When y = , 1, += x =± 3.
dx
The points of tangency are ⎜ 3, ⎟ and
Let x = a sin t then dx = a cos t d t. Then the limits ⎛
π ⎜ − 3, ⎟ are 0 and .
a x 2 π /2
A 4 b ∫ 2 − 0 1 2 dx = 4 ab ∫ 0 cos t dt
Equation of tangent line at ( x 0 , y 0 ): 2 ab ∫ 0 (1 cos 2 ) t dt = 2 ⎢ ab t + ⎣
When y = –6,
= x 1, 2 2. =± b 2
The points of tangency are
2 2, – 6 and
V =⋅π 2 ∫ 0 a ⎜ − ⎜ 1 2 ⎟ ⎟ dy =π 2 a ⎢ y − ⎥
Substitute x =−
into the equation of the
hyperbola. 98
y 2 − 7 y 2 − 35 = 0, y =± 3
9 The coordinates of the points of tangency are
( − 7, 3 ) and ( −. 7, 3 )
Section 10.2 Instructor’s Resource Manual
60. Position the x -axis on the axis of the hyperbola
57. y =± b − 1
such that the equation 2 – 2 = describes the 1
The vertical line at one focus is x = a 2 + b 2 .
hyperbola. The equation of the tangent line at
2 xx
( x , y ) is 0 – yy 0 = The equations of the 1.
V =π
∫ 0 b –1 ⎟ 0
a 2 dx a 2 b ⎜ 2 a ⎟
2 asymptotes are y =± x .
2 ⎡ x 3 =π ⎤
b ∫ a ⎜ ⎜ 2 –1 dx =π ⎟ ⎟ b ⎢ 2 − x ⎥
Substitute y = x and y = – x into the
⎡ ( a 2 + b 2 23/2 )
equation of the tangent line.
Thus the tangent line intersects the asymptotes at
=π 2 b ∫ ⎜ 1– 2 ⎟ dx =π 2 b 2 x −
⎥ =π ab ⎜ bx 0 + ay 0 bx 0 + ay 0 ⎟
Observe that bx 0 – ay 0 = 22 ab .
59. If one corner of the rectangle is at (x, y) the sides
1 ⎛ ab 2 ab 2 ⎞
have length 2x and 2y.
b 2 ⎝ ⎟ ⎠ dA Thus, the point of contact is midway between the
4 ;0 = when dy y dy
two points of intersection.
61. Add the two equations to get 2 9 y = 675.
2 y 3 y − 2 = 0 y =± 53
Substitute y = 53 into either of the two
equations and solve for x
⇒ x = ±6
The point in the first quadrant is
y = 0 or y =±
The Second Derivative Test shows that y =
is
a maximum.
b x = a 1 − ()
b 2 Therefore, the rectangle is a 2 by b 2 .
Instructor’s Resource Manual
Section 10.2 603
62. Substitute x = 6 – 2y into x 2 + 4 y 2 = 20. AP
y 2 –3 y += 2 0 Thus the curve is the right branch of the horizontal (y – 1)(y – 2) = 0
y = 1 or y = 2 hyperbola with a = , so b =⎜ 1– ⎟ c .
uc
v 2 x ⎟ = 4 or x = 2 ⎝ ⎠ The points of intersection are (4, 1) and (2, 2).
The equation of the curve is
() v 2
2 = 1. ⎜ x ≥ ⎟ v 2 1– u c 2 ⎝
69. Let P(x, y) be the location of the explosion.
3 AP = 3 BP + 12 AP – BP = 4
Thus, P lies on the right branch of the horizontal hyperbola with a = 2 and c = 8, so b = 2 15.
64. If the original path is not along the major axis, the Since BP = CP , the y-coordinate of P is 5. ultimate path will approach the major axis.
65. Written response. Possible answer: the ball will
= lim ⎝
follow a path that does not go between the foci.
x →∞ ⎢
66. Consider the following figure.
Observe that 2( α + β) = 180°, so α + β = 90°. The
72. x = a cos , sin ty = a t − b sin t = ( ab − ) sin t
ellipse and hyperbola meet at right angles.
cos t = , sin t =
a ab −
67. Possible answer: Attach one end of a string to F x 2 y and attach one end of another string to F ′ . Place a 2
spool at a vertex. Tightly wrap both strings in the
same direction around the spool. Insert a pencil
Thus the coordinates of R at time t lie on an
through the spool. Then trace out a branch of the
ellipse.
hyperbola by unspooling the strings while keeping both strings taut.
Section 10.2 Instructor’s Resource Manual
73. Let (x, y) be the coordinates of P as the ladder
slides. Using a property of similar triangles,
75. The equations of the hyperbolas are – = 1
76. Position the x-axis on the plane so that it makes the angle φ with the axis of the cylinder and the
Square both sides to get
y -axis is perpendicular to the axis of the cylinder.
2 or bx + 22 ay = 22 ab or
(See the figure.)
74. Place the x-axis on the axis of the hyperbola such x 2 y 2
that the equation is 2 – 2 = 1. One focus is at a b
(c , 0) and the asymptotes are y =± x . The
equations of the lines through the focus,
If P(x, y) is a point on C, ( sin ) x φ 2 + y 2 = r 2
perpendicular to the asymptotes, are
y =± ( – ). x c Then solve for x in x 2 y b 2
where r is the radius of the cylinder. Then
Since c = a + b ,. x = The equation of the
2 When a < 0, the conic is an ellipse. When a > 0,
directrix nearest the focus is x =
, so the line
the conic is a hyperbola. When a = 0, the graph is
c two parallel lines. through a focus and perpendicular to an asymptote intersects that asymptote on the directrix nearest
the focus.
Instructor’s Resource Manual
Section 10.2 605
7. y 2 –5–4–6 x y = 0
10.3 Concepts Review
This is a parabola.
2 2 2. 14; ellipse 8. 4 x + 4 y + x 8 – 28 – 11 y = 0
4. rotation, translation
This is a circle.
Problem Set 10.3
9. 3 x 2 + 3 y 2 –6 x + 12 y + 60 = 0 3( x 2 2 –2 x ++ 1) 3( y 2 2 + 4 y + 4) = –60 3 12 ++
1. x + y –2 x + 2 y += 1 0 3( – 1) x 2 + 3( y + 2) 2 2 = –45
( x –2 x ++ 1) ( y 2 + 2 y += 1) –1 1 1 ++ This is the empty set. ( – 1) x 2 + ( y + 1) 2 = 1
10. 4 x –4 y –2 x + 2 y += 1 0
This is a circle.
This is a circle.
⎜ x – 4 ⎟ ⎞= 1
3. 9 2 x 2 + 4 y + x 72 – 16 y + 124 = 0 This is a hyperbola. 9( 2 x + 8 x + 16) + 4( 2 y –4 y + 4) = –124 144 16 + +
11. 4 x 2 –4 y 2 + 8 x + y 12 – 5 = 0 9( x + 4) 2 + 4( – 2) 2 y = 36
This is an ellipse.
4( x 2 + 2 x +
1) – 4 ⎜ y 2 –3 y + ⎟ =+ 54–9
This is two intersecting lines.
16( x + 6) 2 − 9( y − 5) 2 = 846 12. 4 x 2 –4 2 y + 8 x + y 12 – 6 = 0
This is a hyperbola.
4( x 2 + 2 x +
1) – 4 ⎛ ⎜ y 2 –3 y + ⎞ ⎟ =+ 64–9
5. 9 x 2 + 4 y 2 + x 72 – 16 y + 160 = 0 ⎝
9( x 2 + 8 x + 16) + 4( y 2 –4 y + 4) = –160 144 16 + +
2 x 3 + 1) – 4 ⎛
9( x + 4) + 4( – 2) y = 0 This is a hyperbola.
This is a point.
13. 4 2 2 2 x – 24 x + 36 = 0
6. 16 x + 9 y + 192 x + 90 y + 1000 = 0 4( x 2 –6 x + 9) = –36 36 +
This is a line.
16( x + 6) 2 + 9( y + 5) 2 = –199
This is the empty set.
Section 10.3 Instructor’s Resource Manual
14. 4 x 2 – 24 x + 35 = 0 19. ( x + 2) 2 = 8( y − 1) 4( x 2 –6 x + 9) = –35 36 +
4( – 3) x 2 = 1 This is two parallel lines.
Instructor’s Resource Manual
Section 10.3 607
23. x 2 + 4 y 2 − 2 x + 16 y += 1 0 26. x 2 − 4 y 2 − 14 x − 32 y − 11 = 0
( x 2 − 2 x ++ 1) 4( y 2 + 4 y + 4) =−++ 1 1 16 ( x 2 − 14 x +
x 2 + 6 x ++ 9) 9( 2 y − 2 y +=−+ 1) 9 225 9 + 4( x + 4 x + 4) = 16 y − + 32 16
Section 10.3 Instructor’s Resource Manual
29. 2 y 2 − 4 y − 10 x = 0 36. An equation for the ellipse can be written in the
( x − 2) 2 ( y − 3) 2
2( y − 2 y += 1) 10 x + 2 form
=. 1
( y − 1) 2
= 5 ⎜ x + ⎟ Substitute the points into the equation.
Therefore, a = 4 and b = 2.
( y − Horizontal parabola, 2 = 3)
Vertex ⎜ − ,1 ⎟ ; Focus is at
⎜ ,1
20 ⎟ ⎝ and ⎠
37. Vertical hyperbola, center (0, 3), 2a = 6, a = 3,
c = 5, b =
−= 25 9 4
directrix is at x =− 29 .
2 − 38. Vertical ellipse; center (2, 6), a = 8, c = 6,
= 39. Horizontal parabola, opens to the left
a 9, 3 =
− 10 2
The distance between the vertices is 2a = 6.
Vertex (6, 5), p =
31. 2 + 16( 2 x − 1) 25( y + 2) = 400 ( y − 5) =− 4(4)( x − 6)
( x − 1) 2 ( y + 2) 2 ( y − 5) 2 =− 16( x − 6)
25 16
40. Vertical parabola, opens downward, p = 1
Horizontal ellipse, center (1, –2), a = 5, b = 4,
− 25 16 = 3 ( x − 2) =− 4( y − 6)
Foci are at (–2, –2) and (4, –2).
41. Horizontal ellipse, center (0, 2), c = 2
Since it passes through the origin and center is at
Vertical parabola, opens downward, vertex
42. Vertical hyperbola, center (0, 2), c = 2,
3, , 1 p =
b =− 4 a
⎛ An equation for the hyperbola can be written in
Focus is at ⎜ 3, ⎟ and directrix is y = .
( y − 2) 2 x 2
2 the form
33. a = 5, b = 4
Substitute (12, 9) into the equation.
( x − 5) 2 ( y − 1) 2 49 144
2 = 1 25 16 a 4 − a
49(4 − a ) 144 − a = a (4 − a )
34. Horizontal hyperbola, a = 2, c = 3,
4 b 2 = 94 −= 5 a − 197 a + 196 = 0 ( x − 2) 2 ( y + 1) 2 2 ( 2 a − 196)( a −= 1) 0
196, 1 2 a =
35. Vertical parabola, opens upward, p = 5 – 3 = 2
Since a < c, a = 1, b = 41 −= 3
( 2 x − 2) = 4(2)( y − 3)
( y − 2) −
( 2 x − 2) = 8( y − 3) 3
Instructor’s Resource Manual
Section 10.3 609
Section 10.3 Instructor’s Resource Manual
– x 2 + 7 xy – y 2 –62–62 x y = 0
(–) u v 2 1 + 3 ( – )( u vu ++ v ) ( u + v ) 2 + (–)( u v ++ u v ) = 13
− 3 3 Since 0 ≤ 2 θπ ≤ , sin 2 θ is positive, so cos 2θ is negative; using a 3-4-5
4 v right triangle, we conclude cos 2 θ = − . Thus
. Rotating through the angle
cos − 1 ( 0.8) − = 71.6 , we have
= This is a hyperbola in standard position 1.
4 36 in the uv-system; its axis is the v-axis, and a = 2, b =. 6
Instructor’s Resource Manual
Section 10.3 611
50. A = 11, B = 96, C = 39, D = 240, E = 570, F = 875 y
Since 0 ≤ 2 θπ ≤ , cos 2 θ is negative; using a 7-24-25 right triangle, we
conclude cos 2 θ=− .
Thus sin θ =
Rotating through the angle = cos 1 θ − ( 0.28) − = 53.13 , we have
11 3 4 3 4 ( 4 5 u − 3 5 v )( + 96 5 u − 5 v )( 5 u + 5 v ) +
39 4 3 ( 2 5 u + 5 v ) + 240 3 ( 5 u − 4 v )( + 570 5 4 5 u + 3 5 v ) =− 875
or
3 u 2 − v 2 + 24 u + 6 v =− 35
3( u 2 +8u+16) ( − v 2 -6 v +
This is a hyperbola in standard position in the uv-system; its axis is the u-
axis, its center is (,) uv =− ( 4, 3) and a =
51. 34 x 2 + 24 xy + 41 y 2 + 250 y = –325
(3 – 4 ) u v 2 + (3 – 4 )(4 u v u + 3) v + (4 u + 3) v 2 + 50(4 u + 3) v = –325
50 u 2 + 25 v 2 + 200 u + 150 v = –325
This is an ellipse in standard position in the uv-system, with major axis
parallel to the v-axis. Its center is ()( uv , = − − and 2, 3 ) a =, 2 b = 2 .
Section 10.3 Instructor’s Resource Manual
The graph consists of the two parallel lines u = − and 2 u =. 3
53. a. If C is a vertical parabola, the equation for C
c. If C is a circle, an equation for C can be can be written in the form y = ax 2 + bx +. c written in the form ( x − h ) 2 + ( y − k ) 2 = r 2 .
Substitute the three points into the equation.
Substitute the three points into the equation.
2=a–b+c
Solve the system to get a = 1, b = –1, c = 0.
(3 − h ) 2 +− (6 k ) 2 = r 2
5 5 Solve the system to get h = , k = , and
b. If C is a horizontal parabola, an equation for
C can be written in the form
x = ay + by + . Substitute the three points c 2 2
into the equation.
54. Let (p, q) be the coordinates of P. By
1 properties of similar triangles and since
Solve the system to get a = , b = –1, c = 0.
4 KP
and α = . Solve
. Since P(p, q) is
a point on a circle of radius r centered at (0, 0), p 2 + q 2 = r 2
equation for C.
Instructor’s Resource Manual
Section 10.3 613
2 55. 2 y = Lx + Kx 56. Parabola: horizontal parabola, opens to the right, ⎛
2 p = c – a, y 2 = 4( c − ax )( − a )
Kx ⎜ + x +
Hyperbola: horizontal hyperbola, b = c − a
2 − 2 = 1 Kx ⎜
y 2 = b ( x 2 − a 2 ) Now show that y 2
If K < –1, the conic is a vertical ellipse. If
K = –1, the conic is a circle. If –1 < K < 0,
(hyperbola) is greater than y 2 (parabola).
the conic is a horizontal ellipse. If K = 0, the
2 b 2 c 2 original equation is 2 y = Lx , so the conic is 2 − a
a horizontal parabola. If K > 0, the conic is a
horizontal hyperbola.
( c + ac )( − a )
( x + ax )( − a )
If 1 −< K < (a horizontal ellipse) the length 0 ( c + ax )( + a )
2 ( c − ax )( − a ) Section 10.2)
(2 )(2 ) a a
of the latus rectum is (see problem 45,
From general considerations, the result for a
c + a > 2a and x + a > 2a since c > a and x > a vertical ellipse is the same as the one just
except at the vertex.
obtained.
For K =− 1 (a circle) we have
If K = 0 (a horizontal parabola) we have
Lx y ; = 4 xp ; = , and the latus
rectum is
2 Lp = 2 L = L .
4 If K > 0 (a horizontal hyperbola) we can use
the result of Problem 46, Section 10.2. The
2b 2
length of the latus rectum is
, which is
equal to L .
57. x = u cos α – v sin α y = u sin α + v cos α (u cos α – v sin α) cos α + (u sin α + v cos α) sin α = d
u (cos 2 α + sin 2 α ) = d
u=d Thus, the perpendicular distance from the origin is d.
Section 10.3 Instructor’s Resource Manual
2 The corresponding curve is a parabola with x > 0 and y > 0.
59. x = u cos θ – v sin θ; y = u sin θ + v cos θ x (cos ) + y (sin ) = ( cos u 2 – cos sin ) ( sin v + u θ 2 θ θ θ θ θ + v cos sin ) θ θ = u
x (– sin ) θ + y (cos ) = (– cos sin u + v sin 2 ) ( cos sin + u + v cos θ 2 θ θ θ θ θ θ ) = v Thus, u = x cos θ + y sin θ and v = –x sin θ + y cos θ.
60. u = 5 cos 60 °− 3sin 60 °=−
v = –5sin 60 °− 3cos 60 °= –
61. Rotate to eliminate the xy-term. x 2 + 14 xy + 49 y 2 = 100
(–7) u v ; y =
(7 u + v )
(–7) u v 2 14 + 49 ( – 7 )(7 u v u ++ v ) (7 u + v ) 2 = 100
50 u 2 = 100 u 2 = 2 u =± 2 Thus the points closest to the origin in uv-coordinates are ( 2, 0 ) and ( – 2, 0 . )
() 2 = or x = () –2 = – 52 5 52 5
() 72 = or y = –7 2 52 = 5 ( ) – 52 5
The points closest to the origin in xy-coordinates are ⎜ ,
and –,– .
Instructor’s Resource Manual
Section 10.3 615
62. x = u cos θ – v sin θ y = u sin θ + v cos θ
Ax 2 = Au ( cos – sin ) θ v θ 2 = Au ( 2 cos 2 θ –2 uv cos sin θ θ + v 2 sin 2 θ ) Bxy = Bu ( cos – sin )( sin θ v θ u θ + v cos ) θ = Bu ( 2 cos sin θ θ + uv (cos 2 θ – sin 2 θ )– v 2 cos sin ) θ θ Cy 2 = Cu ( sin θ + v cos ) θ 2 = Cu ( 2 sin 2 θ + 2 uv cos sin θ θ + v 2 cos 2 θ )
Ax 2 + Bxy + Cy 2 = ( cos A 2 θ + B cos sin θ θ + C sin 2 θ ) u 2 + (–2 cos sin A θ θ + B (cos 2 θ – sin 2 θ ) + C 2 cos sin ) θ θ uv
+ ( sin A 2 θ – B cos sin θ θ + C cos 2 θ ) v 2 Thus, a = A cos 2 θ + B cos sin θ θ + C sin 2 θ and c = A sin 2 θ – B cos sin θ θ + C cos 2 θ .
a += c A (cos 2 θ + sin 2 θ ) + B (cos sin θ θ – cos sin ) θ θ + C (sin 2 θ + cos 2 θ ) =+ AC
63. From Problem 62, a = A cos 2 θ + B cos sin θ θ + C sin 2 θ ,
b = –2 cos sin A θ θ + B (cos 2 θ – sin 2 θ ) 2 cos sin , + C θ θ and
b 2 –4 ac = ( B 2 –4 AC ) cos 4 θ + 2( B 2 –4 AC ) cos 2 θ sin 2 θ + ( B 2 –4 AC ) sin 4 θ = ( B 2 –4 AC )(cos 2 θ )(cos 2 θ + sin 2 θ )( + B 2 –4 AC )(sin 2 θ )(cos 2 θ + sin 2 θ ) = ( B 2 –4 AC )(cos 2 θ + sin 2 θ ) = B 2 –4 AC
64. By choosing an appropriate angle of rotation, the
c. ⎛ ⎜ AC +± (– AC ) + B ⎞ ⎟
second-degree equation can be written in the form
2 + au 2 cv + du + ev + = From Problem 63, f 0.
2 = then 4ac = 0, so the graph
is a parabola or limiting form.
b. If B –4 AC < then 4ac > 0, so the graph 0,
⎠ Δ 16 ⎝ a c ⎠
is an ellipse or limiting form.
c. If B –4 AC > 0, then 4ac < 0, so the graph
is a hyperbola or limiting form.
65. a. From Problem 63, –4 ac = B 2 –4 AC = Δ or –
b. From Problem 62, a + c = A + C.
The two values are
and .
Section 10.4 Instructor’s Resource Manual
2 + 2 = can be transformed to
66. 2 Ax + Bxy Cy 1 68. Δ= 4(25)(1) – 8 = 36
2 au 2 + cv = Since 4 1. ac = Δ > the graph is an 0,
Since c < a, = ⎛ ++
2 ⎜ 2 AC (– AC ) + B ⎞ ⎟
ellipse or a limiting form. += ( AC + ) > 0,
so a > 0 and c > 0. Thus, the graph is an ellipse (or
The area of au + 2 cv = is
Thus, the distance between the foci is 26 – 8 10
67. cot 2 θ = 0, θ =
4 and the area is .
(–) u v
69. From Figure 6 it is clear that
v = r sin and φ u = r cos φ .
Also noting that y = r sin( θφ + ) leads us to
1 2 B + 1 ++ + 2 =
y = r sin( θφ + ) = r sin cos + r cos sin
(–) u v
( – )( u vu v )
2 2 2 = ( r cos φ )( sin θ )( + r sin φ )( cos θ )
2 + B 2 2– B
u 2 v = 1 = u sin θ + v cos θ
a. The graph is an ellipse if
> and 0
2 10.4 Concepts Review
0, B > so –2 < B <2.
1. simple; closed; simple
2 + B 2– B
b. The graph is a circle if
, so
2. parametric; parameter
B = 0.
3. cycloid
c. The graph is a hyperbola if
2 4. dy dt dx dt = gt ( () )( ) ft ′ ()
Problem Set 10.4
2 + B 1. a.
d. The graph is two parallel lines if
b. Simple; not closed
Instructor’s Resource Manual
Section 10.4 617 Section 10.4 617
b. Simple; not closed
b. Simple; not closed
b. Simple; not closed
b. Simple; not closed
c. t = x + 3 y = 2 x + 6
Section 10.4 Instructor’s Resource Manual Section 10.4 Instructor’s Resource Manual
b. Not simple; not closed
b. Simple; not closed
b. Simple; not closed
Instructor’s Resource Manual
Section 10.4 619 Section 10.4 619
c. t = x 2 + 2
t = 4– y
1 2 b. Simple; closed
b. Simple; not closed
c. t = x 2 + 3
t = 4– y 2
4 b. Simple; closed
Section 10.4 Instructor’s Resource Manual
b. Not simple; closed
c. sin 2 r x = b. Not simple; closed
x +y=9
b. Not simple; closed
b. Not simple; closed
Instructor’s Resource Manual
Section 10.4 621
9 9 x +y=9
19. a.
b. Not simple; not closed
y = 2(cos 2 2 θ 2 – sin θ )
b. Not simple; not closed
y = –8sin 2 θ cos 2 θ dy ′
Section 10.4 Instructor’s Resource Manual
= – = 2sin t
= –2sin t
dy 2 (3 t 5 + 7 t 4 –6 t 3 + 10 t 2 –9 t + 3)(1 + t 22 )
26. = 2sin t
27. = 3sec t dx 2 8 t 5
dt
dy =
5sec tan t t dx
Tangent line: y – 8 = 3(x – 4) or 3x – y – 4 = 0
dx 2
Instructor’s Resource Manual
Section 10.4 623
Tangent line: y += 1 2 ⎜ x + ⎟ or 2x – y + 2 = 0
Tangent line: y −=− ( x − or 2)
x + 6y – 4 = 0
dx
33. = 2sec tan , 2sec 2 t t = t dt
L = ∫ 0 49 + dt = 13 ∫ 0 dt = 13[ ] t 3 0 = 3 13
Tangent line: y +
L = ∫ –3 14 + dt = 5 ∫ –3 dt = 5[ ] t 3 –3 = 65
4 cos ∫ 2 0 t + 4 sin 2 tdt = 2 ∫ π 0 dt = 2[ ] t 0 =π 2
L = ∫ 0 36 t + 9 t dt = 3 ∫ t 4 t dt 0 2 +
3 ⎡ (4 + t 23/2 ) ⎤ = 16 2 − 8
Section 10.4 Instructor’s Resource Manual
− cos t = sec t − cos t ∫
L = ∫ sin t + 4 2 (sec t −+ 2 2 1 ⎛ 1 ⎞ 0 2 cos t dt )
1 1 + t 2 + dt =
⎜ 1 + ⎟ dt
tan tdt
= ∫ ⎜ 1 + 2 = ⎟ ⎡ dt ⎢ t – ⎤ ⎥ = π /4
∫ dy
0 1– t 2 ∫ 0
dt dx
1– t 2 b. = 3cos 3 , –3sin 3 θ = θ d θ d θ
c. The curve in part a goes around the unit circle
+ ⎜ 4 t – + 4 ⎟ dt
once, while the curve in part b goes around
the unit circle three times.
4 t ++
dt
48. Δ=πΔ S 2 xs
39 S = ∫ 2 π x ds = ∫ 2 π a x a ⎜ dt ⎟ + ⎜ ⎟ dt = ∫ ⎜ 2 t + 2 ⎟ dt = ⎢ t – ⎥ = ⎝ ⎠ ⎝ dt ⎠
2 t ⎦ 1/ 4 16 See Section 5.4 of the text
sech , 2 tanh t = t 49. = – sin , cos t = t
sech t + 4 tanh tdt
S ∫ π+ 2 (1 cos ) sin t t + 0 cos tdt
4 – 4sech 2 t + sech 4 tdt =π 2 (1 cos ) +
t dt =π+ 2[ t 0 sin ] t π 0 =π 4
3 = 3 ∫ (2 – sech ) 2 t 2 dt = ∫ (2 – sech ) 2 t dt
= [2 – tanh ] t t 3 –3 =
12 – 2 tanh 3
Instructor’s Resource Manual
Section 10.4 625 Section 10.4 625
dy
55. dx = dt; when x = 0, t = –1; when x = 1, t = 0.
∫ 0 ∫ –1 [( t + 1) – 4( t + 4)] dt
S = π+
∫ 2 2 (3 sin ) sin t t + 2 cos tdt
–1 (4 − t 3 ++− t 2 2 t 15) dt 2 π
=π 2 2 ∫ 2 (3 sin ) +
56. dy = sec 2 t dt ; when y = 1, t = ;
∫ 4 π+ 2 (1 sin ) sin t + cos tdt
2 when y = 3, t =π =. 2
∫ (1 sin ) + t dt =π 2 [ – cos ] t t 0 0 =π 4 3
1 xy dy = ∫ /4 (sec )(tan ) sec t t 2 t dt
S = ∫ 2 π ⎜ t 3/2 ⎞ ⎟ t + dt
23 57. dx = 2 2 e dt t 4 π
= ∫ tt 2 + 1 dt 25 ln 5
y dx =
2 t e dt = [2 ] e t 3 ln 5 ∫ 1 ∫ 0 0 = 8
S = ∫ 2 π+ ( t 7 )( 1 ++ t 7 ) –7 dt
y = – ⎜ 2 ⎟ x 2 2 + (tan ) α x
⎝ v 0 cos α ⎠
S = ∫ 2( π+ t a )( t + a ) 2 + 1 dt
This is an equation for a parabola.
1 a =π 2 ⎡ (( t + a ⎢ 2 ) + 1) 3/2 ⎤
b. Solve for t when y = 0.
– ( a –2 aa ++ a 1 ) ⎥
α The time of flight is 0 sin seconds.
Section 10.4 Instructor’s Resource Manual Section 10.4 Instructor’s Resource Manual
be the range as a function of α.
Let the wheel roll along the x-axis with P initially
π The range is the largest possible when α =.
61. The x- and y-coordinates of the center of the circle of radius b are (a – b)cos t and (a – b)sin t, respectively. The angle measure (in a clockwise
direction) of arc BP is t . The horizontal change
from the center of the circle of radius b to P is
b cos – ⎜ ⎜ t – t ⎟ ⎟ = b cos ⎛
⎞ t and the vertical
⎛ ab change is sin − − − =− ⎞
b ⎟ ⎟ b sin ⎜
ab −
Therefore, x = ( ab − ) cos t +
b cos
t and ⎝ b
Let the wheel roll along the x-axis with P initially
x = OM = ON − MN = at − b sin t y = MP = RN = NC + CR =− a b cos t 62. From Problem 61,
a Substitute b =.
⎛⎞ a x = ⎜ 4 ⎟ cos t + ⎜⎟ cos(3 ) t ⎝ ⎠ ⎝⎠ 4
= ⎜ ⎟ cos t + ⎜⎟ (cos 2 cos t t − sin 2 sin ) t t
⎛⎞ a
cos t + (cos 3 t – sin 2 t cos – 2 sin t ⎜ 2 ⎟ ⎜⎟ t cos ) t
2 a 3 3 ⎜ 4 ⎟ (cos )(1 sin t − t ) + ⎜⎟ cos t = a cos t
sin(3 ) t
Instructor’s Resource Manual
Section 10.4 627 Section 10.4 627
⎜ ⎟ sin t − ⎜⎟ (sin 2 cos t t +
⎜ ⎟ (sin )(1 cos t − t ) + ⎜⎟ sin t = a sin t
x 2/3 + y 2/3 = a 2/3 cos 2 t + a 2/3 sin 2 t = a 2/3 2 ⎛ 2 dx ⎞ = ⎛⎞ a 2 ⎜ 2 dt ⎟ ⎜⎟ (4 sin t + 8sin sin 2 t t + 4 sin 2 ) t
63. Consider the following figure similar to the one in
2 the text for Problem 61. 2 ⎛ dy ⎞ ⎛⎞ a
+ dt ⎟ ⎜ ⎟ = ⎜⎟ (8 8sin sin 2 + t t − 8 cos cos 2 ) t ⎝ t ⎠ ⎝ dt ⎠ ⎝⎠ 3
= ⎛⎞ a (8 16 sin + ⎜⎟ 2 t cos t − 8 cos 3 t + 8sin 2 t cos ) t
⎝⎠ 3 ⎛⎞ 2 a
⎜⎟ 3 8cos t ) ⎝⎠ 3
The x - and y-coordinates of the center of the circle
2 dx 2 dy
∫ 0 ⎜ dt ⎟ ⎜ ⎟ dt
of radius b are (a + b)cos t and (a + b)sin t
⎝ dt ⎠
respectively. The angle measure (in a counter-
a 2/3 π
= a ∫ 0 + 8 24 cos t − 32 cos t dt
clockwise direction) of arc PB is t . The
horizontal change from the center of the circle of
Using a CAS to evaluate the length, L =
16 a
. radius b to P is
⎛ a a + b b cos
⎜ t ++π=− t ⎞ ⎟ b cos ⎛
t ⎞ and the
b ⎜ b ⎝ ⎟ ⎠ ⎝ ⎠ vertical change is
b sin ⎜ t ++π=− t ⎟ b sin ⎜
t ⎟ . Therefore,
Section 10.4 Instructor’s Resource Manual
The curve touches a horizontal border six
= times and touches a vertical border twice.
a − c cos t = a 2 ⎜ 1 −
cos 2 t ⎟
⎝ Note that the curve is traced out five times. ⎠
P = 4 ∫ 0 ⎜ ⎟ + ⎜ ⎟ dt
0 1 − e cos t dt The curve touches a horizontal border 18 times and touches a vertical border four times.
b. P = 4 ∫ 0 1 −
cos 2 t
dt 68. This is a closed curve even thought the graph does
16 not look closed because the graph retraces itself.
0 16 – cos 2 t dt ≈ 6.1838 1
(The answer is near 2 π because it is slightly smaller than a circle of radius 1 whose perimeter is 2 π ).
0 16 – cos tdt ≈ 6.1838
The curve touches a horizontal border twice and touches a vertical border twice.
The curve touches a horizontal border five times and touches a vertical border three times.
Instructor’s Resource Manual
Section 10.4 629 Section 10.4 629
sin bt =± 1 or bt = π ( odd) k ; that is,
π , where k = 1, 3, … ,4 b − 1 .
2 b Hence, H = 2b.
b. The graph touches a vertical side if
cos at =± 1 or at = n π ( an integer) n ; that is,
t n = π , where n = 0,1, … ,2 a − 1 .
d. x = cos 2 t ; y = sin 3 t
Hence, V = 2a.
1 c. If t
0 yields a corner, then (see a. and b.) then
at 0 = n π , bt 0 = π so that
2 == n
. Thus corners can only
occur if u is even and v is odd. Assume that
is the case, write u = 2 r , and assume we have
e. x = cos 6 t ; y = sin 9 t a corner at t 0 ; then t 0 = π and
at 0 =
ak
π = n π , ( an integer). n Thus
ak
2 rwk = rk =
is an integer; hence v is a
2 b 2 vw
factor of rk. But v and r have no factors in common, so v must be a factor of k. Conclusion: k is an odd multiple of v. Thus
corners occur at
π where
f. x = cos12 t ; y = sin18 t
m = vvv ,3,5, … , (4 w − 1) v ; therefore C = 2w.
Now if we count corner contacts as half horizontal and half vertical, the ratio of vertical contacts to horizontal contacts is given by
70. Consider the curve defined parametrically by x = cos at , y = sin bt t , ∈ [0, 2 ) π ; we assume a
and b are integers. This graph will be contained in
the box with sides x =± 1, y =± 1 . Let H be the
number of times the graph touches a horizontal side, V the number of times it touches a vertical side, and C the number of times it touches a corner (right now, C is included in H and V). Finally, let w = the greatest common divisor of a and b;
write a =⋅ uw , b = ⋅ . Note: vw
in lowest
u terms is . v
Section 10.4 Instructor’s Resource Manual
Given a parameterization of the form x = cos
f −1 (t) and y = sin f(t), the point moves around the curve (which is a circle of radius 1) at a
speed of ft ′ (). The point travels clockwise
b.
1 around the circle when f(t) is decreasing and counterclockwise when f(t) is increasing.
Note that in part d, only part of the circle will
0.5 be traced out since the range of f(t) = sin t is [–1, 1].
The curve traced out is the graph of y = x 2
0.5 1 The curve traced out is the graph of y = x 2
The curve traced out is the graph of y =− x 2 for 16 −≤≤. x 0
Instructor’s Resource Manual
Section 10.4 631 Section 10.4 631
a = 3, b = 1
The curve traced out is the graph of y = x 2
for 0 ≤≤ x 32 a = 5, b = 2
All of the curves lie on the graph of y =± x 2 ,
but trace out different parts because of the parameterization
is the reduced fraction of .
c. 0 ≤≤ t 4 π
The length of the t-interval is 2q π . The number of times the graph would touch the circle of radius
a during the t-interval is p. If
is irrational, the
b curve is not periodic.
d. 0 ≤≤ t 8 π t + 1 t + 1
75. x = 3 , y = 3
Let = where
is the reduced fraction
a of . The length of the t-interval is 2q π.
When x > 0, t > 0 or t < –1.
When x < 0, –1 < t < 0.
The number of times the graph would touch
When y > 0, t > –1. When y < 0, t < –1.
the circle of radius a during the t-interval is p. Therefore the graph is in quadrant I for t > 0,
quadrant II for –1 < t < 0,
If is irrational, the curve is not periodic.
b quadrant III for no t, and quadrant IV for t < –1.
Section 10.4 Instructor’s Resource Manual
10.5 Concepts Review
1. infinitely many
2 2. r cos θ ; r sin θ ; r
3. circle; line
Problem Set 10.5
Instructor’s Resource Manual
Section 10.5 633
7. a. x = 1cos π= 0 d. x = –2 2 cos π= 0
y = 1sin π= 1 y = –2 2 sin π= –2 2
b. x = –1cos π= –
4 2 9. a. r = () 33 + 3 = 36, r=6
y = –1sin π= –
4 2 tan θ =
b. r = ( –2 3 ) + 2 = 16, r=4
c. r = ()() –2 + –2 = r=2 4,
0, – 3 2 ) ⎛ 3 ⎞ ⎛ 1 ⎞ 10 10. a. 10 r 2 = ⎜ – ⎟ + ⎜ ⎟ = , r =
b. x = –1cos π= –
y = –1sin π=
Section 10.5 Instructor’s Resource Manual
2 2 2 c. r = 0 + (–2) = 4, r=2
2 d. r 3 (–4) = 25, r=5
cos – 3sin θ θ
r =2
3sin θ – cos θ
Instructor’s Resource Manual
Section 10.5 635
19. r cos θ+3=0
cos ( θ −π )
21. r sin θ–1=0
27. r = 4 sin θ
y –1=0
r = 2(2) cos ⎜ θ −
y =1
, circle
22. r 2 – 6 cos – 4 sin r θ r θ += 9 0 x 2 + y 2 –6–4 x y += 9 0
( x 2 –6 x ++ 9) ( y 2 –4 y + 4) = –9 9 4 ++ ( – 3) x 2 + ( – 2) y 2 = 4
23. r = 6, circle
28. r =− 4 cos θ r = 2(2) cos ( θ −π ) , circle
cos ( θ − π 2 )
, line
Section 10.5 Instructor’s Resource Manual
2 6 r , ellipse = + 1 (1) cos θ − π , parabola
1 + () 2 cos ( θ − 2 )
+ 1 2 cos( θ −π () )
() hyperbola
36. 3cos − π ( θ
, line
Instructor’s Resource Manual
Section 10.5 637
37. By the Law of Cosines,
b. The length of the major diameter is
a 2 = r 2 + c 2 − 2 rc cos( θα − ) (see figure below).
4 The length of the minor diameter is
1– + e
4 41. a + c = 183, a – c = 17
This is an equation of a circle with radius
2a = 200, a = 100
2c = 166, c = 83
⎛ and center ba ⎞
e == 0.83
39. Recall that the latus rectum is perpendicular to the
axis of the conic through a focus.
ed c = ea = (0.0167)92.9 = 1.55143
1 + e cos 2
Perihelion = a – c ≈
91.3 million miles
Thus the length of the latus rectum is 2ed.
43. Let sun lie at the pole and the axis of the parabola
40. a. The point closest to the pole is at θ 0 .
lie on the pole so that the parabola opens to the
ed ed left. Then the path is described by the equation
1 + e cos(0) 1 + e r d =
The point furthest from the pole is at θ 0 +π . θ equation and solve for d.
. Substitute (100, 120 ) + into the
d = 50 The closest distance occurs when θ =° 0 .
= 25 million miles
+ 1 cos 0 °
Section 10.5 Instructor’s Resource Manual
44. a. 4 =
+ 1 cos π ( 2 − θ
+ 1 cos ( 4 − θ 0 )
3 2 cos θ 0 + ( 32 − ) 8 sin θ 0 −= 2 0
4.24 cos θ 0 −
3.76 sin θ 0 −= 2 0
b. A graph shows that a root lies near 0.5. Using Newton’s Method, θ 0 ≈ 0.485 .
d =+ 4 4 sin θ 0 ≈ 5.86
c. The closest the comet gets to the sun is r =
Instructor’s Resource Manual
Section 10.5 639
3. r sin θ+4=0
10.6 Concepts Review
sin θ Since sin( −=− θ ) sin θ , test 2 is passed. The
1. limaçon
2. cardioid other two tests fail so the graph has only y-axis symmetry.
3. rose; odd; even
4. spiral
Problem Set 10.6
Changing θ →− θ or r → − yields an r
equivalent set of equations. Therefore all 3 tests
cos θ
are passed. Since cos( −= θ ) cos θ , the graph is symmetric about the x-axis. The other symmetry tests fail.
2. ( – 3) r ⎜ θ – ⎞= 0
5. r = 2 cos θ
Since cos( −= θ ) cos θ , the graph is symmetric r = 3 or θ =
about the x-axis. The other symmetry tests fail. θθ = 0 defines a line through the pole. Since a line forms an angle of π radians, changing
θ →+ πθ results in an equivalent set of equations, thus passing test 3. The other two tests fail so the graph has only origin symmetry.
6. r = 4 sin θ Since sin( −=− θ ) sin θ , the graph is symmetric
about the y-axis. The other symmetry tests fail.
Section 10.6 Instructor’s Resource Manual
11. r = 1 – 1 sin θ (cardioid)
7. r =
1 – cos θ
Since sin( πθ − ) = sin θ , the graph is symmetric
Since cos( −= θ ) cos θ , the graph is symmetric
about the y-axis. The other symmetry tests fail.
about the x-axis. The other symmetry tests fail.
12. r =
2 – 2 sin θ (cardioid)
8. r =
1 sin θ
Since sin( πθ − ) = sin θ , the graph is symmetric
Since sin( πθ − ) = sin θ , the graph is symmetric
about the y-axis. The other symmetry tests fail.
about the y-axis. The other symmetry tests fail.
9. r = 3 – 3 cos θ (cardioid)
13. r = 1 – 2 sin θ (limaçon)
Since cos( −= θ ) cos θ , the graph is symmetric
Since sin( πθ − ) = sin θ , the graph is symmetric
about the x-axis. The other symmetry tests fail.
about the y-axis. The other symmetry tests fail.
10. r = 5 – 5 sin θ (cardioid) Since sin( πθ − ) = sin θ , the graph is symmetric
14. r = 4 – 3 cos θ (limaçon)
about the y-axis. The other symmetry tests fail.
Since cos( −= θ ) cos θ , the graph is symmetric about the x-axis. The other symmetry tests fail.
Instructor’s Resource Manual
Section 10.6 641
15. r = 2 – 3 sin θ (limaçon)
19. 2 r = –9 cos 2 θ (lemniscate)
Since sin( πθ − ) = sin θ , the graph is symmetric
r =± 3 – cos 2 θ
about the y-axis. The other symmetry tests fail.
Since cos( 2 ) − θ = cos 2 θ and
cos(2( πθ − )) = cos(2 π − 2) θ = cos( 2 ) − θ = cos 2 θ the graph is symmetric about both axes and the
origin.
16. r = 5 – 3 cos θ (limaçon) Since cos( −= θ ) cos θ , the graph is symmetric
about the x-axis. The other symmetry tests fail.
20. r 2 =− 16 cos 2 θ (lemniscate)
r =±− 4 cos 2 θ Since cos( 2 ) − θ = cos 2 θ and
cos(2( πθ − )) = cos(2 π − 2) θ = cos( 2 ) − θ = cos 2 θ the graph is symmetric about both axes and the
origin.
17. r 2 = 4 cos 2 θ (lemniscate) r =± 2 cos 2 θ
Since cos( 2 ) − θ = cos 2 θ and cos(2( πθ − )) = cos(2 π − 2) θ = cos( 2 ) − θ = cos 2 θ the graph is symmetric about both axes and the
origin.
21. r = 5 cos 3 θ (three-leaved rose) Since cos( 3 ) − θ = cos(3 ) θ , the graph is symmetric about the x-axis. The other symmetry tests fail.
18. r 2 = 9 sin 2 θ (lemniscate)
r =± 3 sin 2 () θ
Since sin(2( πθ + )) = sin(2 π + 2) θ = sin 2 θ , the
22. r = 3sin 3 θ (three-leaved rose)
graph is symmetric about the origin. The other
Since sin( 3 ) − θ =− sin(3 ) θ , the graph is
symmetry tests fail. symmetric about the y-axis. The other symmetry tests fail.
Section 10.6 Instructor’s Resource Manual
23. r = 6sin 2 θ (four-leaved rose)
27. r = θθ ,0 ≥ (spiral of Archimedes)
Since
sin(2( πθ − )) = sin(2 π − 2) θ
No symmetry. All three tests fail.
= sin( 2 ) − θ =− sin(2 ) θ and sin( 2 ) − θ =− sin(2 ) θ , the graph is symmetric about both axes and the origin.
28. r = θθ 2, 0 ≥ (spiral of Archimedes) No symmetry. All three tests fail.
24. r = 4 cos 2 θ (four-leaved rose) Since cos( 2 ) − θ = cos 2 θ and
cos(2( πθ − )) = cos(2 π − 2) θ = cos( 2 ) − θ = cos 2 θ the graph is symmetric about both axes and the
origin.
29. r = e θ ,0 θ ≥ (logarithmic spiral) No symmetry. All three tests fail.
25. r = 7 cos 5 θ (five-leaved rose) Since cos( 5 ) − θ = cos 5 θ , the graph is symmetric
about the x-axis. The other symmetry tests fail.
30. r = e θ /2 ,0 θ ≥ (logarithmic spiral) No symmetry. All three tests fail.
26. r = 3sin 5 θ (five-leaved rose) Since sin( 5 ) − θ =− sin 5 θ , the graph is
symmetric about the y-axis. The other symmetry
31. r = ,0 θ > (reciprocal spiral)
θ No symmetry. All three tests fail.
tests fail.
Instructor’s Resource Manual
Section 10.6 643
32. r =− ,0 θ > (reciprocal spiral)
35. r = 3 3 cos , 3sin θ r = θ
θ No symmetry. All three tests fail.
3 3 cos θ = 3sin θ
(0, 0) is also a solution since both graphs include the pole.
Note that r = –5 is equivalent to r = 5. −= 5
cos θ = 0 37. 6 r = 6sin , θ r =
(0, 0) is also a solution since both graphs include the pole.
Section 10.6 Instructor’s Resource Manual
12 sin 2 θ + 6 sin θ −= 6 0 = 1 sin θ 6(2 sin θ − 1)(sin θ += 1) 0 2
sin θ
, sin θ =− 1 r sin θ 2
θ = ,, θ = θ = 40. Consider the following figure.
4 cos 2 θ = ( 2 2 sin θ ) xr = ar − br cos θ
2 2 ( x − ar ) =− bx − 4 8sin θ = 8sin θ 22 22
(0, 0) is also a solution since both graphs includes the pole.
39. Consider r = cos θ .
if b < a
The graph is clearly symmetric with respect to the y-axis.
Substitute (r, θ) by (–r, –θ).
2 Substitute (r, θ) by (r, π – θ)
Instructor’s Resource Manual
Section 10.6 645 Section 10.6 645
This is the equation of a lemniscate.
r (sin θ – 3 cos θ) = 2
42. Consider the following figure.
sin θ – 3cos θ
f. 3 x 2 + 4 y = 2
3 r 2 cos 2 θ + r 4 sin θ = 2 (3cos 2 ) r θ 2 + (4 sin ) – 2 θ r = 0
Then tan θ =
This is a polar equation for a four-leaved rose. r = – cos θ + 2 sin θ ± (cos – 2 sin ) 2 θ θ + 25
Section 10.6 Instructor’s Resource Manual
49. r =+ 1 3cos ⎜⎟
The curve repeats itself after period p if
f ( θ + p ) = f () θ .
50. a. The graph of r =+ 1 sin ⎜ θ –
rotation of the graph of r = 1 + sin θ by
We need
counter-clockwise about the pole. The graph
of r =+ 1 sin ⎛
45. a. VII
⎜ θ + ⎟ is the rotation of the
b. I
graph of r = 1 + sin θ by
clockwise about 3
c. VIII the pole.
d. III
b. r = 1 – sin θ = 1 + sin(θ – π )
e. V
The graph of r = 1 + sin θ is the rotation of the graph of r = 1 – sin θ by π about the
2 ⎠ The graph of r = 1 + sin θ is the rotation of
h. IV
46. r =
1 – 0.5sin 2 θ the graph of r = 1 + cos π θ by counter- 2
clockwise about the pole.
d. The graph of r = f( θ) is the rotation of the graph of r = f( θ – α) by a clockwise about the pole.
51. a. The graph for φ = 0 is the graph for φ ≠ 0
47. r = cos ⎜
rotated by φ counterclockwise about the pole.
b. As n increases, the number of “leaves”
increases.
c. If a > b , the graph will not pass through the pole and will not “loop.” If b < a , the
48. r = sin ⎛ 5 ⎞
graph will pass through the pole and will
have 2n “loops” (n small “loops” and n large “loops”). If a = b , the graph passes
through the pole and will have n “loops.” If
ab ≠ 0, n > 1, and φ = 0 , the graph will be
symmetric about θ = k , where k = 0, n − 1.
Instructor’s Resource Manual
Section 10.7 647
52. The number of loops is 2n.
2. r = 2a cos θ, a > 0
53. The spiral will unwind clockwise for c < 0. The spiral will unwind counter-clockwise for c > 0.
54. This is for c = 4 π.
0 (2 cos ) a θ d θ = 2 a cos d
The spiral will wind in the counter-clockwise
a ∫ 0 (1 cos 2 ) θθ d a ⎢ θ + sin 2 θ =π a
(2 cos ) θ ∫ 2 + 0 d θ
0 (4 4 cos + θ + cos 2 θθ ) 2 d
10.7 Concepts Review
2. ∫ α [ ( )] f θ d θ ⎝
3. ∫ 0 (2 + 2 cos ) θ d θ
4. r = 5 + 4 cos θ
4. f( θ) = 0
Problem Set 10.7
1. r = a, a > 0
∫ 2 (5 4 cos ) + θ d θ
(25 40 cos + +
16 cos ∫ 2 0 θ θθ ) d
2 ∫ 0 [25 40 cos + θ + 8(1 cos 2 )] + θθ d
= ∫ 0 (33 40 cos + θ + 8 cos 2 ) θθ d 2 1
ad θ =π 0 2 a = [33 θ + 40 sin θ + 4 sin 2 ] θ 0 =π 33
Section 10.7 Instructor’s Resource Manual
5. r = 3 – 3 sin θ
7. r = a(1 + cos θ)
A = ∫ 0 (3 – 3sin ) θ d θ 1 2 π
A = ∫ 0 [ (1 cos )] a + θ 2 d θ
0 (9 – 18sin θ + 9 sin θθ ) d a 2 2 2 π =
0 (1 2 cos + θ + cos ∫ 2 θθ ) d
0 ⎢ 9 – 18sin θ + (1 – cos 2 ) 2 θ 2 ⎥ d θ a 2 ⎣ 2 ⎦ π ⎡
∫ 0 ⎢ + 1 2 cos θ + (1 cos 2 ) + θ ⎥ d θ
– 18sin θ – cos 2 θθ d
2 ∫ 0 ⎜ + 2 cos θ + cos 2 θθ ⎟ d ⎡ 1 27 9 ⎤ 27 ⎝ 2 2 ⎠
(3 3sin ) 2 + d
(9 18sin θ
9 sin θθ ) d A =⋅ 2 ∫ –/4
∫ 0 ⎢ + 9 18sin θ + (1 cos 2 ) − θ ⎥ d θ 2 ⎣ 2 ⎦
A =⋅ 2 ∫ 0 9 sin 2 θθ d = 9
2 ∫ 0 sin 2 θθ d
[– cos 2 ] θ π /2 0 = 9
Instructor’s Resource Manual
Section 10.7 649