Foci: ( ± 10, 0 )

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 = c 81 ae 3 ⎜⎟ , b −= 45 9

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.

20. This is a vertical ellipse with b = 4 and c = 3.

27. This is a horizontal ellipse with c = 2.

21. This is a horizontal ellipse with a = 5.

25 b

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

= . Since 1 2 a = 14, a 2 = 49 and since vertical, = or b = 2a.

− 49 (4) Suppose the hyperbola is horizontal. 2 = 33 . Thus the x 2 y 2

equation is

2 – 2 = 1 33. 4b This is an hyperbola whose foci are (7, 0) and b

(- 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 − 2 = 1 . Since

Suppose the hyperbola is vertical.

36 and b 2 = c 2 − a 2 = (7 ) 36 13 2 − = –

2 a = 12, a 2 =

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

Thus the equation is

= –3 35. Use implicit differentiation to find the slope:

1 ⎪ a 2 2 2 2 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

0, or y ′ = 2 so the y

2 + 2 = . Since 1 2 a = 26, a 2 = 169 a and since b equation of the tangent line is

+ c 2 , b 2 = 169 (9) − 2 = 88 . Thus the

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

b = 4 and the equation of the arch is + = 1 .

When y = 2,

1, so x =± .

equation of the tangent line is

25 16 2 =− 6 ( The width of the box can at most be y + 6) ( x − 3) or x − 6 y = 9 .

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) 2

The equation of the arch is

0, or

y ′ = 6 so the equation of the

2 2 (2) y 2 4 21

When x = 2,

1, so y =± .

tangent line is ( y − 2) = 6( x − 3) or

6 y = 43 .

The height at a distance of 2 feet to the right of the

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

=− ( 5 y 12) ( x − 5) or

tangent line is

Use implicit differentiation to find the slope: a a

2 x − 2 y y′ = 0 . At the point ( 2, 3) 2 Thus, the length of the latus rectum is b . 6 a

2223 − y ′ =

0, or

so the equation of

3 46. The foci are at (±c, 0)

the tangent line is ( y − 3) =

( x − 2) or

2 b b y 2 = , y =±

41. Use implicit differentiation to find the slope:

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 a 0

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

c = 4357.5 – 4132 = 225.5

x 2 + c 2 225.5 2 y −= 2 0; 2 x + 4 yy ′ = 0

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 .

⎟ x . The equations of the tangent lines are y

Equation of tangent line at x (

When y = , 1, += x =± 3. 2

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 ∫ 1 − 2 dx = 4 ab cos 2 t dt

Equation of tangent line at ( x 0 , y 0 ): =

2 ab ∫ (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

b ⎟ dy 2 a ⎢ y

V =⋅π 2 ∫ 2 0 a ⎜ − ⎜ 1 2 ⎟ =π 2 −

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

2 60. Position the x 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 = 2 + b .

hyperbola. The equation of the tangent line at

xx yy

( x , y ) is 0 – 0

= The equations of the 1.

2 2 asymptotes are y =± x .

=π 2 a b

b ∫ a ⎜ ⎜ 2 –1 ⎟ dx =π b 2 ⎢ − x ⎥

Substitute y = x and y = – x into the

2 ⎡ ( a + b 23/2

equation of the tangent line.

⎡ ( a 2 + b 23/2 ) –3 a 2 a 2 + b 2 + 2 a 3 ⎤ ⎛ bx – ay ⎞

Thus the tangent line intersects the asymptotes at

⎛ ab 2 ab 2 ⎞

V =⋅π 2 ∫ ⎜ 0 b 1– ⎟ 2 dx ⎜

=π 2 b 2 ∫

Observe that bx 2 0 – ay 2 0 = ab 22 .

59. If one corner of the rectangle is at (x, y) the sides

1 ⎛ ab 2 ab 2 ⎞

have length 2x and 2y.

dA b 2 ⎟ dA Thus, the point of contact is midway between the

4 ;0 = when dy y

y 2 − dy

two points of intersection.

61. Add the two equations to get 9 y 2 = 675.

= 0 y =± 53

Substitute y = 53 into either of the two

⎜ ⎜ 1 2 ⎟ ⎟ = 0 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 2 x = a 1 − () 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

uc 22 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

2 2 1 spool at a vertex. Tightly wrap both strings in the a ( a − b )

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.

(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 b

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

where r is the radius of the cylinder. Then

Since c 2 = a 2 + b 2 ,. x = The equation of the

c 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 2 4( 49

4. rotation, translation

This is a circle.

Problem Set 10.3

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 2 –4 2 y –2 x + 2 y += 1 0

This is a circle.

This is a circle.

4 – ⎞ –4 ⎜ ⎛ y ⎟ ⎜ x – ⎞= ⎟ 1

3. 9 2 4 x 2 + y + x 72 – 16 y + 124 = 0 This is a hyperbola.

9( 2 + 8 + + 16) 4( x 2 x y –4 y + 4) = –124 144 16 + +

This is an ellipse.

This is two intersecting lines.

16( x + 6) 2 − 9( y − 5) 2 = 846 12. 4 2 x 2 –4 y + 8 x + y 12 – 6 = 0

This is a hyperbola.

4( x 2 + 2 x +

1) – 4 ⎛ y 2 –3 y + ⎞ =+ 64–9

+ 16) 4( y 2 –4 y + 4) = –160 144 16 + +

4( x + 2 1) – 4 ⎛ y

9( x + 4) + 4( – 2) y = 0 This is a hyperbola.

This is a point.

2 2 13. 4 x – 24 x + 36 = 0 6. 16 x + 9 y + 192 x + 90 y + 1000 = 0 2 4( x –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 +

2 25( 2 2 x + 6 x ++ 9) 9( y − 2 y +=−+ 1) 9 225 9 + 4( x + 4 x + 4) = 16 y − + 32 16 25( x + 3) 2 + 9( y − 1) 2 = 225 4( x + 2) 2 = 16( y − 1)

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 − 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.

Horizontal parabola, p =

Vertex ⎜ − ,1 ⎟ ; Focus is at ⎜ ,1 ⎟ and

⎝ 37. Vertical hyperbola, center (0, 3), 2a = 6, a = 3, 5 ⎠ ⎝ 20 ⎠

29 c = 5, b =

−= 25 9 4

directrix is at x =−

− 38. Vertical ellipse; center (2, 6), a = 8, c = 6,

39. Horizontal parabola, opens to the left

− 10 2

The distance between the vertices is 2a = 6.

Vertex (6, 5), p =

31. 2 16( − 1) 2 x + 25( y + 2) 2 = 400 ( y − 5) =− 4(4)( x − 6) ( x − 1) 2 ( y + 2) 2 ( y − 5) 2 =− 16( x − 6)

25 16

Horizontal ellipse, center (1, –2), a = 5, b = 4,

40. Vertical parabola, opens downward, p = 1

− 25 16 = 3 ( x − 2) 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 = 2

b =− 4 a 2

⎛ An equation for the hyperbola can be written in

Focus is at 3,

and directrix is y = . ( 2) 2 ⎜ 2 ⎟

the form

33. a = 5, b = 4

Substitute (12, 9) into the equation.

( x − 5) 2 ( y − 1) 2 49 144

25 16 a 2 4 − a 2

49(4 − 2 2 a 2 ) 144 − a =

a 2 (4 − a )

34. Horizontal hyperbola, a = 2, c = 3,

4 b 2 = 94 −= 5 a − 197 a + 196 = 0

2 2 2 x 2 − 2) ( y + 1) ( a − 196)( a −= 1) 0

a 2 = 196, 1 2 a =

35. Vertical parabola, opens upward, p = 5 – 3 = 2

Since a < c, a = 1, b = 41 −= 3

x 2 − 2) = 4(2)( y − 3)

y − 2) −

( x − 2) 2 = 8( y − 3) 3

Instructor’s Resource Manual

Section 10.3 609

(2 – )( + 2)–( + 2) 2 u vu v u v = 64

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 θ = cos ( 0.28) − 1 − = 53.13 , we have

11 3 − 4 3 4 4 ( 3 5 u 5 v )( + 96 5 u − 5 v )( 5 u + 5 v ) +

39 ( 5 u + 5 v ) + 240 ( 5 u − 5 v )( + 570 5 u + 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

r 2 = 25 .

x = ay 2 + 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

2 2 centered at (0, 0), 2 p + q = r

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 2 = c 2 − a 2 ⎝

2 − 2 = 1 Kx ⎜

y 2 = b 2 ( 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

original equation is y 2 = Lx , so the conic is

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

(cos 2 u 2 α + sin α ) = d

u=d Thus, the perpendicular distance from the origin is d.

Section 10.3 Instructor’s Resource Manual

+ ⎡ 1 (–)2 2 u v ⎢ ( – )( u vu + ⎤ v ) + ( u + v ) = a

2 + 2( u 2 u – v ) 2 1/ 2 = a 2( 2 u – v ) 2 1/ 2 = a –2 u 2( 2 2 2 u 2 – v ) = a –22 au + 2 u

au

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 ) θ + y (sin ) = ( cos u 2 – cos sin ) ( sin v θ 2 θ θ θ + u θ + v cos sin ) θ θ = u

x (– sin )

2 θ 2 + y (cos ) θ = (– cos sin u θ θ + v sin θ ) ( cos sin + u θ θ + v cos 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 ) 2 cos sin ) + C 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 – cos sin ) θ θ 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 +± (–) 2 AC 2 B ⎜ ⎞ + ⎟ Δ ⎝

second-degree equation can be written in the form

2 + au 2 cv + du + ev + = From Problem 63, f 0.

= then 4ac = 0, so the graph 0,

is a parabola or limiting form.

2 ⎜ + ± a c ⎟ Δ 16 ⎜ + ⎟ −Δ b. If B –4 AC < then 4ac > 0, so the graph 0, ⎝ ⎠ ⎝ a c ⎠

is an ellipse or limiting form.

> 0, then 4ac < 0, so the graph

is a hyperbola or limiting form.

2a ⎜ ⎝ ⎟ ⎠ 2 2 − c a ac c ⎝ ac ⎠

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

66. Ax 2 + Bxy + 2 Cy = can be transformed to 1 68. Δ= 4(25)(1) – 8 2 = 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

2 2 c The area of =

π 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

(–) 2 B u 1 v + ( – )( u vu ++ v ) ( + ) 2 = 1 y = r sin( θφ + ) = r sin cos θ φ + r u cos sin v θ φ

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

> and 0

gt ( () )( dx dt ) =

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. 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 2 = x + 3

1 4–

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.

x y 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

2 = –2sin t

dy 2 (3 t 5 + 7 t 4 –6 t 3 + 10 t 2 –9 t + 3)(1 + t 22 )

dx 2 t 5 4 (1 – ) t 3

dx

26. = 2sin t

27. = 3sec 2 t

5sec tan t t dx

Tangent line: y – 8 = 3(x – 4) or 3x – y – 4 = 0

2 t dx 9

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)

33. = 2sec tan , 2sec t t = 2 t

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 = 65

4 cos t + 4 sin tdt = 2 dt = 2[ ] t ∫ π 0 ∫ 0 0 =π 2

L = ∫ 0 36 t + 9 t dt = 3 ∫ 0 t 4 + t dt

3 ⎡ (4 23/2 + t ) ⎤ = 16 2 8 −

Section 10.4 Instructor’s Resource Manual

L = ∫ ⎜ 1– +

sin 2 + (sec 4 2 −+

tan tdt

= ∫ ⎜ 1 + 2 ⎟ dt = ⎡ – ⎤ t = π /4

t 2 sin 2 ∫ 2 ln 3 ln 3 ∫ t + 2 ∫ t cos tdt

θθ d 0 = ∫ d θ 0

∫ dy

0 1– t 2 ∫ 0 2 b. = 3cos 3 , –3sin 3 θ = θ

1– t

[sin t ] 0 = sin –1

c. The curve in part a goes around the unit circle

dt 1/ 4

once, while the curve in part b goes around

the unit circle three times.

1 2 = 1 2 48. Δ=πΔ S 2 xs ∫

∫ 2 1/ 4 ⎜ t + ⎟ dt ⎝ 2 t 2 ⎠

dt

39 S = ∫ 2 π x ds = ∫ 2 π x

⎝ dt ∫ ⎠

⎟ dt = ⎢ t – ⎥

2 t ⎦ 1/ 4 16 See Section 5.4 of the text

dx

2 44. dy = sech , 2 tanh

2 (1 cos ) sin ∫ 2 π+ t t + cos 2

4 – 4sech 2 + sech t 4 tdt =π 2 (1 cos ) +

t dt =π+ 2[ t 0 sin ] t π 0 =π 4

22 ∫ 3 (2 – sech ) t 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.

( x 2 –4) y dx = [( t + 2 ∫ 3 0 ∫ 1) – 4( t + 4)] dt

π+ 2 (3 sin ) sin 2 cos S 2 ∫

tdt 0

− t ∫ 3 –1 (4 ++− t 2 2 t 15) dt

2 ∫ (3 sin ) + t dt =π 2 [3 – cos ] 2 t π 0 =π 12 t 2 0

t dt ; when y = 1, t = ;

π+ 2 (1 sin ) sin 2 cos t 2 t + tdt ∫ 4

when y = 3, t =.

=π 2 ∫ (1 sin ) + t dt =π 2 [ – cos ] t t 2 0 π =π 4 2 3

1 xy dy = ∫ /4 (sec )(tan ) sec t t t dt

tt 2 + 1 dt 25 ∫ ln 5

3 ln 5 = t ∫

1 y dx = ∫ 0 2 e dt = [2 ] e 0 = 8

S = ∫ 2 –7 π+ ( t 7 )( 1 ++ t 7 ) dt

y = – ⎜ 2 2 ⎟ x 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.

=π 2 ⎡ ⎢ (( t + a ) 2 + 1) 3/2 ⎤

⎥ b. Solve for t when y = 0.

a –2 aa ++ a 1

The time of flight is 0 sin α 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

⎛ 3 = a ⎞ sin ⎛⎞ a t −

t 2 a 2 cos t + a ⎜ ⎟

sin(2 t + t ) 64. x = a 2 cos t − a cos 2 t = a 2 cos −

⎝⎠ 4 = a 2 cos (1 cos ) t − t + a

⎛ 3 a = a ⎞ − ⎛⎞ ⎜ ⎟ sin t ⎜⎟ (sin 2 cos t t + cos 2 sin )

⎜ ⎟ (sin )(1 cos t − t ) + ⎜⎟ sin t = a sin t

t 2 8sin sin 2 t t + 4 sin 2 ) t

dt

63. Consider the following figure similar to the one in

2 the text for Problem 61. 2 ⎛ dy ⎞ = ⎛⎞ a 2 ⎜ 2 dt ⎟ ⎜⎟ (4 cos t − 8 cos cos 2 t t + 4 cos 2 ) t ⎝ ⎠ ⎝⎠ 3

⎛ dx ⎞ 2 ⎛ dy ⎞ 2 ⎛⎞ a 2

⎜ ⎟ + ⎜ ⎟ = ⎜⎟ (8 8sin sin 2 + t t − 8cos cos 2 ) t t ⎝ dt ⎠

= ⎜⎟ (8 16 sin + 2 t cos t − 8 cos 3 t + 8sin 2 t cos ) t

⎝⎠ 3 ⎛⎞ 2 a

(8 24 cos sin + t ⎜⎟ 2 t − 8cos 3 t )

The x - and y-coordinates of the center of the circle

2 dx 2 dy

of radius b are (a + b)cos t and (a + b)sin t

0 ⎜ dt ⎟ + ⎜ ⎟ dt ⎠ 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

b cos ⎛ t ++π=− t ⎞ b cos ⎛

t ⎟ and the

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 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 .

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 ≤≤ 4

b t 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.

75. x = 3 , y =

d. 0 ≤≤ t 8 π t + 1 t 3 + 1

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. 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. r cos

θ ; r sin θ ; 2 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

2 = 36, r=6

y = –1sin π= –

1 tan π = , =

r = ( –2 3 + 2

r = ()() –2 + –2 = r=2 4,

0, – 3 2 2 ⎛ 3 ⎞ ⎛ 1 ⎞ 10 ( 10 ) 10. a. r = ⎜ – ⎟ + ⎜ ⎟ = , r =

b. x = –1cos π= –

y = –1sin π=

Section 10.5 Instructor’s Resource Manual Section 10.5 Instructor’s Resource Manual

d. 2 r 2 = 3 + (–4) 2 = 25, r=5

cos – 3sin θ θ

r =2

3sin – cos θ 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 = 4sin 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

+ ( 1 (1) cos θ −

+ 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

a + b ba and center ⎛ , ⎞ .

2c = 166, c = 83

e == 0.83

39. Recall that the latus rectum is perpendicular to the

axis of the conic through a focus.

π 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 − θ

+ 1 cos ( π 2 0 ) ( 4 − θ 0 )

3 2 cos θ 0 + ( 3 2 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

7. r =

11. r = 1 – 1 sin θ (cardioid)

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. r 2 = –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 = 5cos 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 = 9sin 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)

Since

27. r

θθ ,0 ≥ (spiral of Archimedes)

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.

1 2 cosθ cos θ = 1

− 1 cos θ =+ 1 cos θ

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

12sin 2 θ + 6sin θ −= 6 0 =

1 sin θ

6(2sin θ − 1)(sin θ += 1) 0 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

(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).

cos – ⎜ θ ⎟ = cos θ

1 r = – cos 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 – 3cos θ θ

f. 3 x 2 + 4 y = 2

3 r 2 cos 2 θ + r 4 sin θ = 2 (3cos 2 θ ) r 2 + (4sin ) – 2 θ r = 0

–4sin θ ± 16sin 2 θ +

24 cos 2 θ

6 cos 2 θ –2sin θ ± 4sin 2 θ + 6 cos 2 r θ =

Then tan θ =

r 2 + (2 cos – 4sin ) – 25 θ θ r = 0

a 2 sin cos θ θ − r sin 2 θ = r cos 2 θ

r cos θ + r sin 2 θ = a 2 sin cos θ θ –2 cos θ 4sin θ

(2 cos – 4sin ) θ θ 100

r = a sin 2 θ 2

This is a polar equation for a four-leaved rose. r = – cos θ + 2sin θ ± (cos – 2sin ) θ θ 2 + 25

Section 10.6 Instructor’s Resource Manual

The curve repeats itself after period p if

f ( θ + p ) = f () θ .

50. a. The graph of r =+ 1 sin θ – is the

cos ⎜

⎟ = cos ⎜

rotation of the graph of r = 1 + sin θ by

We need

counter-clockwise about the pole. The graph

1 sin ⎜ θ + ⎟ is the rotation of the

b. I

graph of r = 1 + sin θ by

clockwise about

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

46. r =

1 – 0.5sin 2 θ the graph of r = 1 + cos θ by π 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.

47. r = cos ⎜

51. a. The graph for φ = 0 is the graph for φ ≠ 0

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

graph will pass through the pole and will

48. sin ⎜ ⎟

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 .

2 ∫ (2 cos ) a θ d θ = 2 a cos θθ d 0 ∫ 0

The spiral will wind in the counter-clockwise

= 2 ⎡ + 1 direction. ⎤ a (1 cos 2 ) θθ

2 ∫ (2 cos ) 0

= 1 2 π (4 4 cos + θ + cos 10.7 Concepts Review 2 ∫ θθ ) d

2. ∫ α [ ( )] f θ d θ ⎝

⎢ θ 4sin θ + sin 2 θ ⎥ =π

Problem Set 10.7

1. r = a, a > 0

A = ∫ (5 4 cos ) + θ 2 d θ

(25 40 cos + θ + 16 cos 2 θθ ) d

[25 40 cos + θ + 8(1 cos 2 )] + θθ d

1 2 π = ∫ (33 40 cos + θ + 8cos 2 ) θθ d 2 0

ad θ

a = [33 θ + 40sin θ + 4sin 2 ] θ 0 =π 33

Section 10.7 Instructor’s Resource Manual

5. r = 3 – 3 sin θ

7. r = a(1 + cos θ)

= ∫ (3 – 3sin ) θ d θ 1 2 π

2 0 A = ∫ [ (1 cos )] a + θ 2 d θ 0

(9 – 18sin θ + 9sin 2 θθ ) d a 2 2 π

2 ∫ 0 = ∫ (1 2 cos + θ + cos 2 θθ ) d

∫ ⎢ 9 – 18sin θ + (1 – cos 2 ) θ d θ

2 ∫ 0 ⎢ + 1 2 cos θ + (1 cos 2 ) + θ ⎥ d θ 2

∫ 0 ⎜ – 18sin – cos 2 θ θθ ⎟ d a 2 2 ⎝ 2 2 ⎠ 2 π ⎛ 3 1 ⎞

18cos – sin 2 θ θ

⎢ θ + 2sin θ + sin 2 2 θ ⎥ =

8. r 2 = 6 cos 2 θ

A = ∫ 0 (3 3sin ) + θ 2 d θ

(9 18sin + θ + 9sin 2 θθ ) d A =⋅ 1 2 ∫ = 0 2 ∫ –/4 π 6 cos 2 θθ d 6 ∫ π cos 2 θθ d –/4

= 3[sin 2 ] θ π /4 –/4 π = 6

0 ⎢ + 9 18sin θ + (1 cos 2 ) − θ ⎥ d θ 2 ⎣ 2 ⎦

9. r = 9sin 2 θ

∫ 0 ⎜ + 18sin θ − cos 2 θθ ⎟ d 2 ⎝ 2 2 ⎠

− 9 θ ⎤ 18cos θ sin 2 θ

A =⋅ 2 9sin 2 θθ d = 9 sin 2 θθ d

= 9 [– cos 2 ] θ π /2 0 = 9

Instructor’s Resource Manual

Section 10.7 649

∫ −π cos 2 θθ d /4

A =⋅ 2 ∫ π a cos 2 θθ d = a 2 – 4 cos θ = 0, θ =

–/4 π = a A =⋅ 2 ∫ (2 – 4 cos ) θ d θ

∫ (4 – 16 cos θ + 16 cos θθ )d 0

11. r = 3 – 4 sin θ

[4 – 16 cos θ + 8(1 cos 2 )]d + θθ

(12 16 cos − θ + 8cos 2 )d θθ

12 θ − 16sin θ + 4sin 2 θ ] 0

A =⋅ 2 ∫ 2 –1 3 (3 – 4sin ) θ d θ

2 sin 4 π = /2

(9 – 24sin θ + 16sin ∫ 2

θθ )d

sin –1 3

–1 3 [9 – 24sin θ + 8(1 – cos 2 )]d θθ sin

= ∫ –1 3 (17 – 24sin – 8cos 2 )d θ θθ 2 – 3 cos θ = 0, θ= cos

24 cos – 4sin 2 ] θ θ π /2 1

A =⋅ 2 ∫

θ –1 2 2 (2 – 3cos ) d θ

24 cos – 8sin cos ] θ θ θ π /2

= ∫ –1 2 ⎢ 4 – 12 cos θ + (1 cos 2 ) + θ d θ

= ⎢ θ –12sin θ + sin 2 θ

–12sin θ + sin cos θ θ

Section 10.7 Instructor’s Resource Manual

14. r = 3 cos 2θ

18. r = 3sin , 1 sin θ r =+ θ

A =⋅ 2 (3cos 2 ) θ 2 d θ = 9 cos 2 2 θθ d Solve for the θ -coordinate of the first

2 ∫ 0 ∫ 0 intersection point.

3sin θ =+ 1 sin θ ∫ (1 cos 4 ) + θθ d

π /6 [(3sin ) – (1 sin ) ] θ θ θ

A =⋅ 2 2 d

π /6 (8sin 2 θ – 2sin –1)d θ θ

π /6 (3 – 4 cos 2 – 2sin )d θ θθ = [3 – 2sin 2 θ θ + 2 cos ] θ π /2 π /6 =π

19. r = r 2, 8cos 2 2 = θ

2 π /6 A =⋅ 6 ∫ (4 cos 3 ) θ d θ = 2

0 2 cos 3

= 24 ∫ (1 cos 6 ) + θθ d = θ +

16. r = 2sin 3 θ

Solve for the θ -coordinate of the first intersection point.

A =⋅ 3 ∫ 0 (2sin 3 ) θ 2 d θ = 6 sin 3 2 θθ d θ =

= 3 (1 cos 6 ) d − θθ A =⋅ 4 ∫ (8cos 2 – 4) θ d θ ∫

17. A = ∫ 0 100 d θ – ∫ d 0 =π 49 θ 51

Instructor’s Resource Manual

Section 10.7 651

20. r =− 3 6sin θ 22. r =+ 2 2sin , 2 2 cos θ r =+ θ

Let A 1 be the area inside the large loop and let

[(2 2sin ) + θ A 2 be the area inside the small loop. ∫ −+ (2 2 cos ) ] θ 2 d θ

A =⋅ 1 2 2 1 1 ∫ –/2 π (3 – 6sin ) θ d θ = (8sin θ + 4sin 2 θ – 8cos – 4 cos θ 2 θθ ) d

(9 – 36sin θ + 36sin 2 θθ )d 1 ∫ π –/2 π = ∫ (8sin – 8cos – 4 cos 2 ) θ θ θθ d

–/2 π (27 – 36sin –18cos 2 )d θ θθ = [ − 8cos θ − 8sin θ − 2sin 2 θ ]

36 cos θ −

9sin 2 /6 θ ]

−π /2 =π+ 18 2 23. a.

f () θ = 2 cos , ( ) θ f ′ θ =− 2sin θ

A 2 =⋅ 2 ∫ π (3 – 6sin ) θ 2 /6 d θ (2 cos ) cos θ θ +− ( 2sin ) sin θ m θ =

− (2 cos ) sin θ θ +− ( 2sin ) cos θ θ

[27 2 θ θ θ π 2 + 36 cos – 9sin 2 ]

2 cos θ − 2sin θ

21. r =+ 3 3cos , 3 3sin θ r =+ θ 3 − 3 2 3

b. f ( ) 1 sin , ( ) θ =+ θ f ′ θ = cos θ (1 sin ) cos + θ θ + (cos ) sin θ θ

−+ (1 sin ) sin θ θ + (cos ) cos θ θ cos θ + 2sin cos θ θ

Solve for the θ -coordinate of the intersection

point.

+ 3 3cos θ =+ 3 3sin θ tan θ= 1

c. f () θ = sin 2 , ( ) θ f ′ θ = 2 cos 2 θ

(sin 2 ) cos θ θ + (2 cos 2 ) sin θ θ

− (sin 2 ) sin θ θ + (2 cos 2 ) cos θ θ

A = 1 ∫ [(3 3cos ) – (3 3sin ) ] + 0 θ 2 + θ 2 d θ At π θ = , .

0 (18cos θ + 9 cos 2 θ –18sin – 9sin θ 2 θθ ) d 3 1 3 2 3 () 2 () 2 +− ( 1) () 2 − 4 3

0 (18cos –18sin θ θ + 9 cos 2 ) θθ d − ( )( ) 2 2

= 1 ⎡ ⎢ 18sin θ + 18cos θ + 9 sin 2 ⎤ θ

Section 10.7 Instructor’s Resource Manual Section 10.7 Instructor’s Resource Manual

(4 3cos ) cos = − θ θ + (3sin ) sin θ θ

and 2 cos m 2 θ + cos θ −≠. 10

−− (4 3cos ) sin θ θ + (3sin ) cos θ θ

3 = 3 4 cos 3cos 3sin

4sin θ + 6sin cos θ θ

⎛ a 2 π ⎞⎛ a 4 π ( ⎞

2, a 0,, ,, ) ⎜ ⎟⎜ ⎟

4 cos θ − 3cos 2 θ

4sin θ + 3sin 2 θ

There is no vertical tangent at θ = π since

lim m () θ = 0 (see part (a)).

At θ = , θ → 0

4 1 1 ()() 7 −− 3 25. f ( ) 1 2sin , ( ) θ =− θ f ′ θ =− 2 cos θ

3 3 (1 2sin ) cos − θ θ +− ( 2 cos ) sin θ − θ 4 + 3 − 3 3

()() 2 2 −− (1 2sin ) sin θ θ +− ( 2 cos ) cos θ θ

cos θ − 4sin cos θ θ

24. f () θ = a (1 cos ), ( ) + θ f ′ θ =− a sin θ =

− sin θ + 2sin 2 θ − 2 cos 2 θ

a (1 cos ) cos + θ θ +− ( a sin ) sin θ θ

cos (1 4sin ) θ − θ

− a (1 cos ) sin + θ θ +− ( a sin ) cos θ θ

− sin θ + 2sin 2 θ − 2 cos 2 θ

2 cos 2 θ + cos θ = − = 1

cos θ + cos 2 θ − sin 2 θ

m = 0 when cos θ (1 – 4 sin θ) = 0

− sin θ − 2sin cos θ θ

− sin (1 2 cos ) θ + θ

cos θ =

0, or 1 4sin − θ = 0

θ = ,, θ = θ = sin − 1 ⎛⎞ 1 a. m = 0 when 2 cos θ cos θ 10 ≈ 0.25,

⎜⎟ ≈ 2.89 cos θ = , cos θ =− 1 ⎝⎠ 4

θ =π ,() f θ = 0, so θ =π is the tangent line.

26. Recall from Chapter 5 that L ∫ a ⎜ ⎟ ⎜ ⎟ dt for x and y functions of t and a ≤ t ≤ b. ⎝ dt ⎠

f ′ ( ) cos θ θ − f ( ) sin , ( )sin θ θ = f ′ θ θ + f ( ) cos θ θ

L = ∫ ( ( ) cos f ′ θ θ − f ( ) sin ) θ θ 2 α + ( ( ) sin f ′ θ θ + f ( ) cos ) θ θ 2 d θ

f ( ) (sin θ ] θ + cos 2 θ ) + [ f ′ 2 ( ) (sin θ 2 ] θ + cos 2 θθ ) d =

∫ α [ f () θ ][ + f ′ () θ ] d θ

27. f () θ = a (1 cos ), ( ) + θ f ′ θ =− a sin θ

+ 1 cos θ

L = ∫ [ (1 cos )] a + θ 0 + [– sin ] a θ d θ = a ∫

0 2 2 cos θθ d = 2 a ∫ 0 d θ = 2 2 a ∫ 0 cos d θ 2

2 a ⎢ ∫ 0 cos d θ – ∫ π cos d θ ⎥ = 2 a ⎜ ⎢ 2sin ⎥ − ⎢ 2sin ⎥ ⎟ = 8 a

Instructor’s Resource Manual

Section 10.7 653

28. f = θ /2 ′ θ = () 1 θ e , () f e θ /2

e /2 d θ = ⎡ 5 e /2 2 ⎤ π

ed θ =

0 5( e 1) 4 49.51 ∫ 0 2 ⎣ ⎦ 0

29. If n is even, there are 2n leaves.

1 π /2 n

π /2 n

∫ –/2 π n 2

π /2 n

+ 1 cos 2 n

A = 2 n ∫ ( cos a n θ ) 2 d θ = na 2 2 nd θθ = na 2 θθ d

∫ –/2 π n cos

If n is odd, there are n leaves.

∫ cos nd θθ

30. r = sec θ − 2 cos θ 31. a. Sketch the graph.

Solve for the θ -coordinate when r = 0. sec θ − 2 cos θ = 0 Solve for the θ -coordinate of the intersection.

2a sinθ = 2b cosθ

0 (2 cos ) b 2 d 2 θ θ π 2 ∫ 0

0 ∫ 2 (2 sin ) a θ d θ +

Notice that the loop is produced for −≤≤. θ

2 a ∫ sin 2 θθ d 0 + 2 b 2 cos 2 θθ d

A 0 = (sec θ − θ 2 ∫ 2 cos ) d θ

a 2 0 π (1 – cos 2 ) /2 θθ d + b 2 (1 cos 2 ) + θθ d

2 ∫ (sec θ −+ 4 4 cos θθ ) d –/4 π

= ∫ (sec θ – 2 2 cos 2 ) + θθ d ⎣ 2 ⎦ 0 ⎣ 2 ⎦ θ 0

= tan θ − 2 θ + sin 2 θ

Note that since tan θ = , cos θ =

and sin θ=

Section 10.7 Instructor’s Resource Manual Section 10.7 Instructor’s Resource Manual

32. The area swept from time t 0 to t 1 is

a 2 sin cos θ θ + a 2 cos sin θ θ

θ () t 1 1

∫ θ t rd θ . () 0 2

− a 2 sin sin θ θ + a 2 cos cos θ θ

= By the Fundamental Theorem of Calculus, 2sin cos θ θ

a − b dA So k = where k is the constant angular

At θ = (the pole), 0 m

1 =. 0 2 m momentum.

dt

Let m 2 be the slope of r = b 2 cos θ .

is a constant so A = ( t 1 − t 0 ) .

− b 2 cos sin θ θ − b 2 sin cos θ θ

dt

Equal areas will be swept out in equal time.

cos 2 θ − 2 = θ sin

2sin cos θ θ

At θ = tan − 1 ⎛⎞ b a − ⎜⎟ b , m 2 =− .

⎝⎠ a 2 ab π

At θ = (the pole), m 2 is undefined.

2 Therefore the two circles intersect at right angles.

33. The edge of the pond is described by the equation r = a 2 cos θ . Solve for intersection points of the circles r = ak and r = a 2 cos θ . ak = a 2 cos θ

⎛⎞ k cos θ = , cos θ = − 1 ⎜⎟

2 ⎝⎠ 2 Let A be the grazing area.

∫ cos –1 k ( – 4 cos 2 θθ ) d 2 2 () 2 2 () 2

A =π ( ka ) 2 +⋅ 2 [( ) ka 2 (2 cos ) ] ∫ 2 –1 k − a θ d θ = ka 22 π+ a 2 k 2

cos

π+ a ⎡ k ∫ 2 −

cos − 1 k 2 () 2 2 () 2

⎣ θ − 2 θ − 2sin cos θ θ ⎤ ⎦ cos − 1 2 k () 2

2 ⎢ 2 2 − 1 ⎛⎞ k k 4 − k 2 = ⎤ a ( k −π+− 1) (2 k ) cos

Instructor’s Resource Manual

Section 10.7 655

34. PT = ka − φ a ; φ goes from 0 to k.

38. A =⋅ 4 ∫ 8cos 2 θθ d = 2[4sin 2 ] θ π /4 0 = 8

2 2 ⎣ ⎢ 3 a ⎥ ⎦ 0 f () θ = 8cos 2 , ( ) θ f ′ θ = 8sin 2 θ

A = ∫ 0 ( ka − φ a ) 2 d φ = − ( ka − φ a ) 3 −

8cos 2 θ = 1 ak 23 π /4

8sin 2 6 2 θ

L = 4 ∫ 8cos 2 θ +

The grazing area is

d 0 ≈ θ 14.83

cos 2 θ

35. The untethered goat has a grazing area of π. a 2

From Problem 34, the tethered goat has a grazing

39. r = 4sin ⎜ ⎟ ,0 ≤≤π θ 4

area of a 2 ⎜

π k 2 k 3 π=

2 k 3 +π−π= 3 k 2 6 0 ⎛ 3 θ ⎞

f () θ = 4sin ⎜

2 ⎟ , () f ′ θ = 6 cos ⎝ ⎜ ⎠ 2 ⎟ ⎝ ⎠

Using a numerical method or graphing calculator,

k ≈ 1.26. The length of the rope is approximately

2 1.26a. 2

0 ⎢ 4sin ⎜ 2 ⎟ ⎥ + ⎢ 6 cos ⎜ ⎟ ⎥ d θ ⎣ ⎝ ⎠ ⎦ ⎣ ⎝ 2 ⎠ ⎦

= 2 ∫ + 5 4 cos θθ d ≈ 13.36

+ 16 20 cos 2 ⎜ ⎟ d θ ≈ 63.46

f () θ =+ 2 4 cos , ( ) θ f ′ θ =− 4sin θ

L = 2 ∫ [ + 2 4 cos θ ][ +− 4sin θ ] d θ

4 ∫ + 5 4 cos θθ d ≈ 26.73

10.8 Chapter Review