calculus one and several variables 10e salas solutions manual ch10

P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU

JWDD027-10 JWDD027-Salas-v1 December 2, 2006 16:41

524 SECTION 10.1

CHAPTER 10

SECTION 10.1

1. y = 12x

2

vertex (0, 0)

focus (0, 12 )

axis x = 0

directrix y = − 12

2. y = − 12x

2

vertex (0, 0)

focus (0,− 12 )

axis x = 0

directrix y = 12

x

y

x

y

3. y = 12 (x− 1)2

vertex (1, 0)

focus (1, 12 )

axis x = 1


4. y = − 12 (x− 1)2

vertex (1, 0)

focus (1,− 12 )

axis x = 1

directrix y = 12

1x

y

1x

y

5. y + 2 = 14 (x− 2)2

vertex (2,−2)

focus (2,−1)

axis x = 2

directrix y = −3

6. y − 2 = 14 (x + 2)2

vertex (−2, 2)

focus (−2, 3)

axis x = −2

directrix y = 1

2x

y

-2x

y



SECTION 10.1 525

7. y = x2 − 4x

vertex (2,−4)

focus (2,− 154 )

axis x = 2


8. y = x2 + x + 1

vertex (− 12 ,

34 )

focus (− 12 , 1)

axis x = − 12

directrix y = 12

2x

y

12

x

y

9.x2

9+

y2

4= 1

center (0, 0)

foci (±√

5, 0)

length of major axis 6

length of minor axis 4

10.x2

4+

y2

9= 1

center (0, 0)

foci (0, ±√

5)



11.x2

4+

y2

6= 1

center (0, 0)

foci (0, ±√

2)

length of major axis 2√

6


12.x2

4+

y2

3= 1

center (0, 0)

foci (±1, 0)


length of minor axis 2√

3



526 SECTION 10.1

13.x2

9+

(y − 1)2

4= 1

center (0, 1)

foci (±√

5, 1)



14. x2 +(y − 3)2

4= 1

center (0, 3)

foci (0, 3 ±√

3)



15.(x− 1)2

16+

y2

64= 1

center (1, 0)

foci (1,±4√

3)



16.(x− 2)2

25+

(y − 3)2

16= 1

center (2, 3)

foci (5, 3) (−1, 3)



17. x2 − y2 = 1

center (0, 0)

transverse axis 2

vertices (±1, 0)

foci (±√

2, 0)

asymptotes y = ±x

18. y2 − x2 = 1

center (0, 0)

transverse axis 2

vertices (0, ±1)

foci (0, ±√

2)

asymptotes y = ±x



SECTION 10.1 527

19.x2

9− y2

16= 1

center (0, 0)

transverse axis 6

vertices (±3, 0)

foci (±5, 0)

asymptotes y = ± 43x

20.x2

16− y2

9= 1

center (0, 0)

transverse axis 8

vertices (±4, 0)

foci (±5, 0)


21.y2

16− x2

9= 1

center (0, 0)

transverse axis 8

vertices (0, ±4)

foci (0, ±5)


22.y2

9− x2

16= 1

center (0, 0)

transverse axis 6

vertices (0, ±3)

foci (0, ±5)


23.(x− 1)2

9− (y − 3)2

16= 1

center (1, 3)

transverse axis 6

vertices (4, 3) and (−2, 3)

foci (6, 3) and (−4, 3)

asymptotes y − 3 = ± 43 (x− 1)

24.(x− 1)2

16− (y − 3)2

9= 1

center (1, 3)

transverse axis 8


foci (6, 3) and (−4, 3)

asymptotes y = ± 34 (x− 1) + 3



528 SECTION 10.1

25.(y − 3)2

4− (x− 1)2

1= 1

center (1, 3)

transverse axis 4

vertices (1, 5) and (1, 1)

foci (1, 3 ±√

5)

asymptotes y − 3 = ±2(x− 1)

26. (x + 1)2 − y2

3= 1

center (−1, 0)

transverse axis 2


foci (1, 0) and (−3, 0)

asymptotes y = ±√

3(x + 1)

27. We can choose the coordinate system so that theparabola has an equation of the form y = αx2, α >0. One of the points of intersection is then the originand the other is of the form (c, αc2). We will assumethat c > 0.

area of R1 =∫ c

0

αx2dx =13αc3 =

13A,

area of R2 = A− 13A = 2

3A.

28. A = A1, B = B1 by the reflection property of parabola;

A = A2, B = B2 by simple geometry. Therefore

A1 = A2, B1 = B2 and C = D = 12π



SECTION 10.1 529

29. The equation of every such parabola takes the form

(x− x0)2 = 4c(y − y0)

This equation can be written

y =(

14c

)x2 −

(x0

2c

)x +

(y0 +

x02

4c

)

vertex(− B

2A,

4AC −B2

4A

), focus

(− B

2A,

4AC −B2 + 14A

), directrix y =

4AC −B2 − 14A

30. Use y =b

a

√a2 − x2.

A =∫ a

0

b

a

√a2 − x2 dx =

b

a

[x

2

√a2 − x2 +

a2

2sin−1 x

a

]a0

=baπ

4.

Also,

xA =∫ a

0

b

ax√a2 − x2 dx = − b

3a

[(a2 − x2)

32

]a0

=ba2

3,

and

yA =∫ a

0

b2

2a2(a2 − x2) dx =

b2

2a2

[a2x− x3

3

]a0

=b2a

3.

Thus, x =4a3π

, and y =4b3π

.

31. By the hint, xy = X2 − Y 2 = 1. In the XY -system a = 1, b = 1, c =√

2. We have center (0, 0),

vertices (±1, 0), foci (±√

2, 0) and asymptotes Y = ±X. Using

x = X + Y and y = X − Y

to convert to the xy-system, we find center (0, 0), vertices (1, 1) and (−1, −1), foci (√

2,√

2) and

(−√

2, −√

2), asymptotes y = 0 and x = 0, transverse axis 2√

2.

32. The points lie on the ellipse: b2x2 + a2y2 = a2b2

b2(a2 cos2 t) + a2(b2 sin2 t) = a2b2(cos2 t + sin2 t) = a2b2

33. 2√π2a4 −A2 /πa

34. Placing the parabola with its vertex at the origin, the equation becomes x2 = 4cy.

When y = 2, x = 2.5, so c =2532

.

Thus the distance from the focus to the center of the mirror is2532

ft.

35. In this case the length of the latus rectum is the width of the parabola at height y = c. With

y = c, 4c2 = x2, and x = ±2c. The length of the latus rectum is thus 4c.



530 SECTION 10.1

36. y =14c

x2 =⇒ dy

dx=

2x4c

=x

2c

At x = ±2c, we thus getdy

dx= ±1

37. A =∫ 2c

−2c

(c− x2

4c

)dx = 2

∫ 2c

0

(c− x2

4c

)dx = 2

[cx− x3

12c

]2c

0

=83c2

x = 0 by symmetry

yA =∫ 2c

−2c

12

(c2 − x4

16c2

)dx =

∫ 2c

0

(c2 − x4

16c2

)dx =

[c2x− x5

80c2

]2c

0

=85c3

y = (85c3)/(

83c2) =

35c

38. V =∫ 2c

0

2πx(c− x2

4c

)dx = 2π

[cx2

2− x4

16c

]2c

0

= 2πc3

yV =∫ 2c

0

πx

(c2 − x4

16c2

)dx = π

[c2x2

2− x6

6 · 16c2

]2c

0

=43πc4

=⇒ y =23c, x = 0 by symmetry

39.kx

p(0)= tan θ =

dy

dx, y =

k

2 p(0)x2 + C

In our figure C = y(0) = 0. Thus the equation of the cable is y = kx2/2p(0), the equation of a

parabola.

40. We’ll work in two dimensions. The directrix of the parabola is perpendicular to the undisturbed light

rays. If the parabola were not there to intercept the rays, the rays would reach the directrix in paths

of the same length (this is an assumption we are making). Also true upon reflection by the parabola

since length of PF=length of PQ.



SECTION 10.1 531

41. Start with any two parabolas γ1, γ2. By moving them we can see to it that they have equations of the

following form:

γ1 : x2 = 4c1y, c1 > 0; γ2 : x2 = 4c2y, c2 > 0.

Now we change the scale for γ2 so that the equation for γ2 will look exactly like the equation for γ1.

Set X = (c1/c2)x, Y = (c1/c2) y. Then

x2 = 4c2y =⇒ (c2/c1)2 X2 = 4c2 (c2/c1)Y =⇒ X2 = 4c1Y.

Now γ2 has exactly the same equation as γ1; only the scale, the units by which we measure distance,

has changed.

42. (a) Take x(t) = a cosh t, y(t) = b sinh t Thenx2

a2− y2

b2= cosh2 t− sinh2 t = 1

range(x) = [a,∞), range(y) = (−∞,∞)

(b) Take x(t) = −a cosh t, y(t) = b sinh t Thenx2

a2− y2

b2= 1,

range(x) = (−∞,−a], range (y) = (−∞,∞)

43. A =2ba

∫ 2a

a

√x2 − a2 dx =

2ba

[x

2

√x2 − a2 − a2

2ln(x +

√x2 − a2

)]2a

a

= [2√

3 − ln (2 +√

3)]ab

44. Let P (x0, y0) be the point of tangency and let l be the tangent line.

slope of l =b2x0

a2y0, slope of F1P =

y0

x0 + c, slope of F2P =

y0

x0 − c

With θ1 as the angle between l and F1P and with θ2 as the angle between F2P and l, we have

tan θ1 =

b2x0

a2y0− y0

x0 + c

1 +b2x0

a2y0

(y0

x0 + c

) and tan θ2 =

y0

x0 − c− b2x0

a2y0

1 +(

y0

x0 − c

)(b2x0

a2y0

)

From the fact that

b2x20 − a2y2

0 = a2b2, (P is on the hyperbola)

it follows readily that tan θ1 = tan θ2.

45. e =√

25 − 16√25

=35

46. e =√

25 − 16√25

=35

47. e =√

25 − 9√25

=45

48. e =√

169 − 144√169

=513

49. E1 is fatter than E2, more like a circle.

50. The ellipse tends to a circle of radius a (Since b approaches a).



532 SECTION 10.1

51. The ellipse tends to a line segment of length 2a.

52. a = 3, c = ea =13· 3 = 1, b =

√a2 − c2 =

√8

Center (0, 0), sox2

9+

y2

8= 1

53. a = 3, c = ea = 23

√2 · 3 = 2

√2, b =

√9 − 8 = 1; x2/9 + y2 = 1

54. The basic equation reads√x2 + y2 = e|x− k|. Square and arrange to

(x +

e2k

1 − e2

)2

+y2

1 − e2=

e2k2

(1 − e2)2

Now set a = ek/(1 − e2) and c = ea. This reduces the equation to

(x + c)2

a2+

y2

a2 − c2= 1

This equation represents an ellipse of eccentricity e = c/a.

55. e = 53 56. e = 5

4

57. e =√

2 58. e = 135

59. The branches of H1 open up less quickly than the branches of H2.

60. The hyperbola tends to the union of two oppositely-directed half lines that begin at the ends of the

transverse axis.

61. The hyperbola tends to a pair of parallel lines separated by the transverse axis.

62. The basic equation reads√x2 + y2 = e|x− k|. Square and rearrange to

(x− e2k

e2 − a

)2

− y2

e2 − 1=

e2k2

(e2 − 1)2

Now set a = ek/(e2 − 1) and c = ea. This reduces the equation to

(x− c)2

a2− y2

c2 − a2= 1

This equation represents a hyperbola of eccentricity e = c/a

63. P (x, y) is on the parabola with directrix l : Ax + By + C = 0 and focus F (a, b) iff d(P, l) = d(P, F )

which happens iff|Ax + By + C|√

A2 + B2=√

(x− a)2 + (y − b)2.

Squaring both sides of this equation and simplifying, we obtain

(Ay −Bx)2 = (2aS + 2AC)x + (2bS + 2BC)y + c2 − (a2 + b2)S

with S = A2 + B2 �= 0.



SECTION 10.2 533

SECTION 10.2

1–8.9. x = 3 cos 1

2π = 0

y = 3 sin 12π = 3

(0, 3)

10. x = 4 cosπ

6= 2

√3

y = 4 sinπ

6= 2(

2√

3, 2)

11. x = − cos (−π) = 1

y = − sin (−π) = 0

(1, 0)

12. x = −1 cosπ

4= −

√2/2

y = −1 sinπ

4= −

√2/2(

−2√

2,−2√

2)

13. x = −3 cos(− 1

3π)

= − 32

y = −3 sin(− 1

3π)

= 32

√3(

− 32 ,

32

√3)

14. x = 2 cos 0 = 2

y = 2 sin 0 = 0

(2, 0)

15. x = 3 cos(− 1

2π)

= 0

y = 3 sin(− 1

2π)

= −3

(0,−3)

16. x = 2 cos 3π = −2

y = 2 sin 3π = 0

(−2, 0)

17. r2 = 02 + 12, r = ±1r = 1: cos θ = 0 and sin θ = 1θ = 1

2π

⎫⎪⎬⎪⎭

[1, 1

2π + 2nπ],

[−1, 3

2π + 2nπ]

18. r2 = 12 + 02, r = ±1r = 1: cos θ = 1 and sin θ = 0θ = 2π

⎫⎪⎬⎪⎭ [1, 2nπ] , [−1, π + 2nπ]

19. r2 = (−3)2 + 02 = 9, r = ±3r = 3: cos θ = −1 and sin θ = 0θ = π

⎫⎪⎬⎪⎭ [3, π + 2nπ], [−3, 2nπ]

20. r2 = 42 + 42, r = ±4√

2r = 4

√2: cos θ = 1√

2and sin θ = 1√

2

θ = 14π

⎫⎪⎬⎪⎭

[4√

2, 14π + 2nπ

],

[−4

√2, 5

4π + 2nπ]



534 SECTION 10.2

21. r2 = 22 + (−2)2 = 8, r = ±2√

2r = 2

√2: cos θ = 1

2

√2, sin θ = − 1

2

√2

θ = 74π

⎫⎪⎬⎪⎭

[2√

2, 74π + 2nπ

],

[−2

√2, 3

4π + 2nπ]

22. r2 = 32 + (3√

3)2, r = ±6r = 6: cos θ = 1

2 and sin θ = −√

32

θ = 53π

⎫⎪⎬⎪⎭

[6, 5

3π + 2nπ],

[−6, 2

3π + 2nπ]

23. r2 =(4√

3)2

+ 42 = 64, r ± 8r = 8: cos θ = 1

2

√3, sin θ = 1

2

r = 16π

⎫⎪⎬⎪⎭

[8, 1

6π + 2nπ],

[−8, 7

6π + 2nπ]

24. r2 = (√

3)2 + 12, r = ±2r = 2: cos θ =

√3

2 and sin θ = − 12

θ = 116 π

⎫⎪⎬⎪⎭

[2, 11

6 π + 2nπ],

[−2, 5

6π + 2nπ]

25. d2 = (x1 − x2)2 + (y1 − y2)2 = (r1 cos θ1 − r2 cos θ2)2 + (r1 sin θ1 − r2 sin θ2)2

= r12 cos2 θ1 − 2r1r2 cos θ1 cos θ2 + r2

2 cos2 θ2

+ r12 sin2 θ1 − 2r1r2 sin θ1 sin θ2 + r2

2 sin2 θ2

= r12 + r2

2 − 2r1r2 (cos θ1 cos θ2 + sin θ1 sin θ2)

= r12 + r2

2 − 2r1r2 cos (θ1 − θ2)

d =√r12 + r22 − 2r1r2 cos (θ1 − θ2)

26. Draw a figure; the result is clear.

27. (a)[12 ,

116 π

](b)

[12 ,

56π]

(c)[12 ,

76π]

28. (a)[3, 5

4π]

(b)[3, 1

4π]

(c)[3, 7

4π]

29. (a)[2, 2

3π]

(b)[2, 5

3π]

(c)[2, 1

3π]

30. (a)[3, 3

4π]

(b)[3, 7

4π]

(c)[3, 1

4π]

31. about the x-axis?: r = 2 + cos (−θ) =⇒ r = 2 + cos θ, yes.

about the y-axis?: r = 2 + cos (π − θ) =⇒ r = 2 − cos θ, no.

about the origin?: r = 2 + cos (π + θ) =⇒ r = 2 − cos θ, no.

32. about the x-axis?: r = cos[2(−θ)] =⇒ r = cos 2θ, yes.

about the y-axis?: r = cos[2(π − θ)] =⇒ r = cos 2θ, yes.

about the origin?: r = cos[2(π + θ)] =⇒ r = cos 2θ, yes.

33. about the x-axis?: r (sin (−θ) + cos (−θ)) = 1 =⇒ r (− sin θ + cos θ) = 1, no.

about the y-axis?: r (sin (π − θ) + cos (π − θ)) = 1 =⇒ r (sin θ − cos θ) = 1, no.

about the origin?: r (sin (π + θ) + cos (π + θ)) = 1 =⇒ r (− sin θ − cos θ) = 1, no.



SECTION 10.2 535

34. about the x-axis?: r sin(−θ) = 1 =⇒ −r sin θ = 1, no.

about the y-axis?: r sin(π − θ) = 1 =⇒ r sin θ = 1 yes.

about the origin?: r sin(π + θ) = 1 =⇒ −r sin θ = 1, no.

35. about the x-axis?: r2 sin (−2θ) = 1 =⇒ −r2 sin 2θ = 1, no.

about the y-axis?: r2 sin (2 (π − θ)) = 1 =⇒ −r2 sin 2θ = 1, no.

about the origin?: r2 sin (2 (π + θ)) = 1 =⇒ r2 sin 2θ = 1, yes.

36. about the x-axis?: r2 cos[2(−θ)] = 1 =⇒ r2 cos 2θ = 1, yes.

about the y-axis?: r2 cos[2(π − θ)] = 1 =⇒ r2 cos 2θ = 1, yes.

about the origin?: r2 cos[2(π + θ)] = 1 =⇒ r2 cos 2θ = 1, yes.

37. x = 2

r cos θ = 2

38. r sin θ = 3

39. 2xy = 1

2 (r cos θ)(r sin θ) = 1

r2 sin 2θ = 1

40. r2 = 9

r = 3

41. x2 + (y − 2)2 = 4

x2 + y2 − 4y = 0

r2 − 4r sin θ = 0

r = 4 sin θ

42. (x− a)2 + y2 = a2

x2 − 2ax + y2 = 0

r2 = 2ar cos θ

r = 2a cos θ

43. y = x

r sin θ = r cos θ

tan θ = 1

θ = π/4

44. x2 − y2 = 4

r2(cos2 θ − sin2 θ) = 4

r2 cos 2θ = 4

[note: division by r okay

since [ 0, 0 ] is on the curve]

45. x2 + y2 + x =√x2 + y2

r2 + r cos θ = r

r = 1 − cos θ

46. y = mx

r sin θ = mr cos θ

θ = a constant

47. (x2 + y2)2 = 2xy

r4 = 2(r cos θ)(r sin θ)

r2 = sin 2θ

48. (x2 + y2)2 = x2 − y2

r4 = r2(cos2 θ − sin2 θ)

r2 = cos 2θ

49. The horizontal line y = 4 50. The vertical line x = 4



536 SECTION 10.2

51. The line y =√

3x 52. θ = ±π3 : the lines y = ±

√3x

53. r = 2 (1 − cos θ)−1

r − r cos θ = 2√x2 + y2 − x = 2

x2 + y2 = (x + 2)2

y2 = 4(x + 1)

a parabola

54. Circle x2 + y2 = 2y

55. r = 6 cos θ

r2 = 6r cos θ

x2 + y2 = 6x

56. The y-axis; x = 0

57. The line y = 2x 58. r = 4 sin(θ + π) = −4 sin θ

r2 = −4r sin θ

x2 + y2 = −4y

x2 + (y + 2)2 = 4

a circle

59. r =4

2 − cos θ2r − r cos θ = 4

2√x2 + y2 − x = 4

4(x2 + y2) = (x + 4)2

3x2 + 4y2 − 8x = 16

an ellipse

60.6

1 + 2 sin θ= r

6 = r + 2r sin θ√x2 + y2 = 6 − 2y

x2 + y2 = 36 − 24y + 4y2

x2 − 3y2 + 24y = 36

a hyperbola

61. r =4

1 − cos θr − r cos θ = 4√

x2 + y2 − x = 4

x2 + y2 = (x + 4)2

y2 = 8x + 16

a parabola

62. r =2

3 + 2 sin θ3r + 2r sin θ = 2

3√x2 + y2 = 2 − 2y

9x2 + 5y2 = 4 − 8y

an ellipse

63.

r = 2a sin θ + 2b cos θ

r2 = 2a r sin θ + 2b r cos θ

x2 + y2 = 2ay + 2bx

(x− b)2 + (y − a)2 = a2 + b2

center: (b, a); radius:√a2 + b2



SECTION 10.3 537

64. r = d + r cos θ 65. 12 (r cos θ + d) = r

r =d

2 − cos θ

66. r = 2(d + r cos θ)

= 2d + 2r cos θ

SECTION 10.3

1. 2. 3.

4. 5. 6.

7. 8. 9.

10. 11. 12.

13. 14. 15.



538 SECTION 10.3

16. 17. 18.

19. 20. 21.

22. 23. 24.

25. 26. 27.

28. 29. 30.



SECTION 10.3 539

31. 32.

33. yes; [1, π] = [−1, 0] and the pair r = −1, θ = 0 satisfies the equation

34. No. 35. yes; the pair r = 12 , θ = 1

2π satisfies the equation

36. yes; [1,− 56π] = [−1, 1

6π], which lies on the curve.

37. [ 2, π] = [−2, 0 ]. The coordinates [−2, 0 ] satisfy the equation r2 = 4 cos θ, and the coordinates

[ 2, π] satisfy the equation r = 3 + cos θ.

38. [2, π/2] = [−2,−π/2] The coordinates [2, π/2] satisfy r2 sin θ = 4, and the coordinates [−2,−π/2]

satisfy r = 2 cos 2θ.

39. (0, 0),(− 1

2 ,12

)40. (0, 1)

41. (1, 0), (−1, 0) 42. (0, 0), (1, 1)



540 SECTION 10.3

43. (0, 0),(

14 ,

14

√3),

(14 ,− 1

4

√3)

44. (0, 0), (0, 1)

45. (0, 0),(±

√3

4 , 34

)46. (0, 0),

(2+

√2

2 , 2+√

22

),

(2−

√2

2 , 2−√

22

)

1

-2 -1.5 -1 -0.5 0.5 1

-1

-0.5

0.5

1

1.5

2

47. (a) The graph of r = 1 + cos(θ − π

3

)is the graph of r = 1 + cos θ rotated counterclockwise π/3

radians; The graph of r = 1 + cos(θ + π

6

)is the graph of r = 1 + cos θ rotated clockwise π/6

radians.

(b) The graph of r = f(θ − α) is the graph of r = f(θ) rotated counterclockwise α radians.

48. (a)

-1.5 1.5

-1.5

1

2 (b) The curves intersect at the pole and at:

[1.172, 0.173], [1.86, 1.036], [0.90, 3.245]

49. (a) (b) The curves intersect at:

the pole and (2, 0), (−1, 0), (−0.25,±0.4330)



SECTION 10.3 541

50. (a)

-2 -1 1 2

-2

-1

1

2 (b) The curves intersect at:

(2, π/6), (2, 5π/6), (2, 3π/2)

51. (a)

-4 -2 2 4

-6

-4

-2

2(b) The curves intersect at the pole and at:

r = 1 − 3 cos θ r = 2 − 5 sin θ

[−2, 0] [2, π]

[3.800, 3.510] [3.800, 3.510]

[2.412, 4.223] [−2.412, 1.081]

[−1.267, 0.713] [−1.267, 0.713]

52. (a) (b) (c) (±0.7484,±0.8651)

53. “Butterfly” curves. The graph for the case k = 2 is:

-2 -1 1 2

-2

-1

1

2

54. An insect? Fly? Mosquito?

-1 1 2 3

-3

-2

-1

1

2

3



542 SECTION 10.3

55. (a) k = 32 (b) k = 5

2

-2 -1 1 2

-1

1

-2 -1 1 2

-1.5

1.5

A petal curve with 2m petals.

56. r = 2 cos(

43 θ

)r = 2 sin

(53 θ

)

-2 -1 1 2

-2

-1

1

2

-2 -1 1 2

-2

-1

1

PROJECT 10.3

1. e =r

d− r cos θ=⇒ r = ed− er cos θ =⇒ r(1 + e cos θ) = ed =⇒ r =

ed

1 + e cos θ

2. (a) Suppose r =ed

1 + e cos θ. Then

r + er cos θ = ed

r = ed− er cos θ

r2 = e2d2 − 2e2dr cos θ + e2r2 cos2 θ

x2 + y2 = e2d2 − 2e2dx + e2x2

(1 − e2)x2 + 2e2dx + y2 = e2d2

(x +

e2d

1 − e2

)2

+y2

1 − e2=

e2d2

(1 − e2)2

(x + c)2 +y2

1 − e2= a2

(a =

ed

1 − e2, c = ea

)

(x + c)2

a2+

y2

(1 − e2)a2= 1

(x + c)2

a2+

y2

a2 − c2= 1



SECTION 10.3 543

(b) From part (a),

x2 + y2 = e2d2 − 2e2dx + e2x2

becomes y2 = d2 − 2dx

so y2 = −4d

2

(x− d

2

)

(c) From part (a), (x +

e2d

1 − e2

)2

+y2

1 − e2=

e2d2

(1 − e2)2

(x− c)2 − y2

e2 − 1= a2

(a =

ed

e2 − 1, c = ea

)

(x− c)2

a2− y2

(e2 − 1)a2= 1

(x− c)2

a2− y2

c2 − a2= 1

3. (a) ellipse: r =8

4 + 3 cos θ=

21 + 3

4 cos θ.

Thus e =34

and34d = 2 =⇒ d =

83.

Rectangular equation:

a =327, c =

247, so

(x + 24

7

)2(327

)2 +y2(247

)2 = 1

(b) hyperbola: r =6

1 + 2 cos θThus e = 2 and 2d = 6 =⇒ d = 3.


a = 2, c = 4, so(x− 4)2

4− y2

12= 1

(c) parabola: r =6

2 + 2 cos θ=

31 + cos θ

Thus e = 1 and d = 3.


y2 = −4(

32

)(x− 3

2

)= −6

(x− 3

2

)

4. r =α

1 − β cos θis the conic section r =

α

1 + β cos θrotated π radians in the

counter clockwise direction:α

1 − β cos θ=

α

1 + β cos (θ − π)



544 SECTION 10.4

r =α

1 − β sin θis the conic section r =

α

1 + β cos θrotated

π

2radians in the clockwise direction:

α

1 − β sin θ=

α

1 + β cos (θ +π

2)

SECTION 10.4

1.

A =∫ π/2

−π/2

12

[a cos θ]2 dθ

= a2

∫ π/2

0

1 + cos 2θ2

dθ

= a2

[θ

2+

sin 2θ4

]π/20

=14πa2

2.A =

∫ π/6

−π/6

12a2 cos2 3θ dθ

=a2

2

[θ

2+

sin 6θ12

]π/6−π/6

=112

πa2

3.A =

∫ π/4

−π/4

12

[a√

cos 2θ]2

dθ

= a2

∫ π/4

0

cos 2θ dθ

= a2

[sin 2θ

2

]π/40

=12a2

4.A =

∫ π/3

−π/3

12a2(1 + 2 cos 3θ + cos2 3θ) dθ

=a2

2

[θ +

23

sin 3θ +θ

2+

sin 6θ12

]π/3−π/3

=12πa2



SECTION 10.4 545

5.A = 2

∫ π

0

12(a2 sin2 θ

)dθ

= a2

∫ π

0

1 − cos 2θ2

dθ

= a2

[θ

2− sin 2θ

4

]π0

=12πa2

6.A = 4

∫ π/2

0

12a2 sin2 2θ dθ

= 2a2

[θ

2− sin 4θ

8

]π/20

=12πa2

7.A =

∫ π/8

0

12

[tan 2θ]2 dθ

=12

∫ π/8

0

(sec2 2θ − 1

)dθ

=12

[12

tan 2θ − θ

]π/80

=14− π

16

8.A =

∫ π/4

0

12(cos2 θ − sin2 θ) dθ

=12

∫ π/4

0

cos 2θ dθ

=[sin 2θ

4

]π/40

=14

9.A =

∫ π/4

0

12

([2 cos θ]2 − [cos θ]2

)dθ

=32

∫ π/4

0

1 + cos 2θ2

dθ

=32

[θ

2+

sin 2θ4

]π/40

=316

π +38



546 SECTION 10.4

10.A =

∫ π/2

0

12(1 + 2 cos θ + cos2 θ − cos2 θ) dθ

=12

∫ π/2

0

(1 + 2 cos θ) dθ

=12

[θ + 2 sin θ]π/20 = 1 +π

4

11.A =

∫ π/4

0

12

[a (4 cos θ − sec θ)]2 dθ

=a2

2

∫ π/4

0

[16 cos2 θ − 8 + sec2 θ

]dθ

=a2

2

∫ π/4

0

[8 (1 + cos 2θ) − 8 + sec2 θ

]dθ

=a2

2[4 sin 2θ + tan θ]π/40 =

52a2

12.A =

∫ π/2

−π/2

12· 14· sec4 θ

2dθ

=18

∫ π/2

−π/2

(1 + tan2 θ

2) sec2 θ

2dθ

=18

[2 tan

θ

2+

23

tan3 θ

2

]π/2−π/2

=23

13.A =

∫ π

0

12

([eθ]2 − [θ]2

)dθ

=12

∫ π

0

(e2θ − θ2

)dθ

= 12

[12e

2θ − 13θ

3]π0

= 112

(3e2π − 3 − 2π3

)

14.A =

∫ π

0

12

[(e2π+θ

)2 − θ2]dθ

=12

[e4π+2θ

2− θ3

3

]π0

=14e6π − 1

4e4π − 1

6π3



SECTION 10.4 547

15.A =

∫ π

0

12

([eθ]2 − [

eθ/2]2)

dθ

=12

∫ π

0

(e2θ − eθ

)dθ

=12

[12e2θ − eθ

]π0

=14(e2π + 1 − 2eπ

)

16.A =

∫ π

0

12

[(e2π+θ

)2 − (eθ)2]

dθ

=12

[e4π+2θ

2− e2θ

2

]π0

=14(e6π − e4π − e2π + 1)

17.A =

∫ 5π/6

π/6

12([ 4 sin θ ]2 − [2]2

)dθ

18.A =

∫ π/2

−π/2

12[(1 + cos θ)2 − (1 − cos θ)2

]dθ

19.A =

∫ π/3

−π/3

12([ 4 ]2 − [ 2 sec θ]2

)dθ



548 SECTION 10.4

20.A = 2

∫ π/2

π/3

12[22 − (4 cos θ)2

]dθ + 2

∫ π

π/2

12· 22 dθ

21.A = 2

{∫ π/3

0

12(2 sec θ)2 dθ +

∫ π/2

π/3

12(4)2 dθ

}

22.A =

∫ 5π/6

π/6

12(1 − 2 sin θ)2 dθ

23.A =

∫ π/3

0

12(2 sin 3θ)2 dθ

24.A =

∫ 5π/3

π/3

12[(2 − cos θ)2 − (1 + cos θ)2

]dθ



SECTION 10.4 549

25.A = 2

{∫ π/6

0

12

(sin θ)2 dθ +∫ π/2

π/6

12

(1 − sin θ)2 dθ

}

26.A = 2

∫ π/12

0

12(5 cos 6θ)2 dθ

27.A = π − 8

∫ π/4

0

12

(cos 2θ)2 dθ

28.A =

∫ π/4

0

12(2a sin θ)2 dθ +

∫ π/2

π/4

12(2a cos θ)2 dθ

29.

-1 1 2

-1

1

2A = 2

∫ π/6

0

12 [2 sin θ]2 dθ + 1

12π

= 4∫ π/6

0

sin2 θ dθ + 112π

= 2[θ − 1

2 sin 2θ]π/60

+ 112π = 5

12π − 12

√3



550 SECTION 10.4

30. The curves intersect at the point [a, π/6] in the first quadrant. By symmetry:

A = 4∫ π/6

0

12

[2a2 cos 2θ − a2

]dθ = 2

[a2 sin 2θ − a2θ

]π/60

= a2√

3 − 13 πa

2

31. The area of one petal of the curve r = a cos 2nθ is given by:

2∫ π/4n

0

12 (a cos 2nθ)2 dθ = a2

∫ π/4n

0

cos2 2nθ dθ

= a2

∫ π/4n

0

(12

+cos 4nθ

2

)dθ

= a2

[12θ +

sin 4nθ8n

]π/4n0

=π a2

8n

The total area enclosed by r = a cos 2nθ isπ a2

2.

The area of one petal of the curve r = a sin 2nθ is given by:

A = 2∫ π/4n

0

12 (a sin 2nθ)2 dθ = a2

∫ π/4n

0

(12

+cos 4nθ

2

)dθ =

π a2

8n

and the total area enclosed by the curve isπ a2

2.

32. Since there are 2n + 1 petals, total area = (2n + 1)· (area of one petal). So

A = (2n + 1) · 2∫ π/[2(2n+1)]

0

a2

2cos2[(2n + 1)θ] dθ = (2n + 1)a2

[θ

2+

sin[2(2n + 1)θ]4(2n + 1)

]π/[2(2n+1)]

0

= (2n + 1) · πa2

4(2n + 1)=

π

4a2, independent of n.

33. Let P = {α = θ0, θ1, θ2, . . . , θn = β} be a partition of the interval [α, β]. Let θ∗i be the midpoint of

[θi−1, θi] and let r∗i = f(θ∗i ). The area of the i th “triangular” region is 12 (r∗i )Δθi, where Δθi = θi − θi−1,

and the rectangular coordinates of its centroid are(approximately)(

23 r

∗i cos θ∗i ,

23 r

∗i sin θ∗i

).

The centroid (xp, yp) of the union of the triangular regions satisfies the following equations

xp Ap =13

(r∗1)3 cos θ1Δθ1 +

13

(r∗2)3 cos θ2Δθ2 + · · · + 1

3(r∗n)3 cos θnΔθn

yp Ap =13

(r∗1)3 sin θ1Δθ1 +

13

(r∗2)3 sin θ2Δθ2 + · · · + 1

3(r∗n)3 sin θnΔθn

As ‖P‖ → 0, the union of the triangular regions tends to the region Ω and the equations above tend

to

xA =∫ β

α

13r3 cos θ dθ

y A =∫ β

α

13r3 sin θ dθ

The result follows from the fact that A =∫ β

α

12r2 cos θ dθ.



SECTION 10.4 551

34. A = 14πr

2; ρ(θ) = r; by symmetry, x = y.

xA =13

∫ π/2

0

r3 cos θ dθ =r3

3

[sin θ

]π/20

=r3

3; x = y =

4r3π

35. Since the region enclosed by the cardioid r = 1 + cos θ is symmetric with respect to the x-axis, y = 0.

To find x :

A =∫ 2π

0

r2 dθ =∫ 2π

0

(1 + cos θ)2 dθ

=∫ 2π

0

(1 + 2 cos θ + cos2 θ) dθ

=∫ 2π

0

(32

+ 2 cos θ12

cos 2θ)

dθ

=[32

+ 2 sin θ +14

sin 2θ]2π

0

= 3π

and23

∫ 2π

0

r3 cos θ dθ =23

∫ 2π

0

(1 + cos θ)3 cos θ dθ

=23

∫ 2π

0

(cos θ + 3 cos2 θ + 3 cos3 θ + cos4 θ

)dθ

=23

∫ 2π

0

(158

+ 4 cos θ + 2 cos 2θ +18

cos 4θ − 3 sin2 θ cos θ)

dθ

=23

[158

θ + 4 sin θ + sin 2θ +132

sin 4θ − sin3 θ

]2π

0

=52π

Thus x =5π/23π

=56.

36.∫ 2π

0

r2 dθ =∫ 2π

0

(4 + 4 sin θ + sin2 θ) dθ =[4θ − 4 cos θ +

θ

2− sin 2θ

4

]2π

0

= 9π

x =19π

· 23

∫ 2π

0

r3 cos θ dθ =2

27π

∫ 2π

0

(2 + sin θ)3 cos θ dθ = 0

y =2

27π

∫ 2π

0

(2 + sin θ)3 sin θ dθ =2

27π· 51

8· 2π =

1718

37. A =∫ 2π

0

12 [2 + cos θ]2 dθ = 1

2

∫ 2π

0

(4 + 4 cos θ + cos2 θ

)dθ = 1

2

[4θ + 4 sin θ + 1

2θ + 14 sin 2θ

]2π0

= 92π

38. r = 2 cos 3θ is a petal curve with 3 petals. The area of one petal is given by:

A = 2∫ π/6

0

12 (2 cos 3θ)2 dθ = 4

∫ π/6

0 cos2 3θ dθ = 2∫ π/6

0 (1 + cos 6θ) dθ = 2[θ + 1

6 sin 6θ]π/60

= 13 π

The total area enclosed by the is π.



552 SECTION 10.4

39.

-1 1 2

-1

1

2

A = 6∫ π/9

0

12

[(4 cos 3θ)2 − 4

]dθ

= 3∫ π/9

0

[16 cos2 3θ − 4

]dθ

= 3∫ π/9

0

(4 + 8 cos 6θ) dθ

= 3[4θ + 4

3 sin 6θ]π/90

= 43π + 2

√3

40. The curves intersect at [0.6667, 1.2310] in the first quadrant

By symmetry, the area is given by:

A = 2∫ 1.2310

0

12

[4 cos2 θ − (1 − cos θ)2

]dθ

=∫ 1.2310

0

[3 cos2 θ + 2 cos θ − 1

]dθ

∼= 2.9725

-2 -1 1 2

-1

-0.5

0.5

1

41. (a) y2 = x2

(a− x

a + x

)(b) Let a = 2

r2 sin2 θ = r2 cos2 θ(a− r cos θa + r cos θ

)

sin2 θ(a + r cos θ) = cos2 θ(a− r cos θ)

r cosθ = a cos 2θ

r = a cos 2θ sec θ

(c) A =∫ 5π/4

3π/4

12a2 cos2 2θ sec2 θ dθ

= 2∫ 5π/4

3π/4

cos2 2θ sec2 θ dθ (a = 2)

= 2∫ 5π/4

3π/4

(2 cos2 θ − 1

)2cos2 θ

dθ

= 2∫ 5π/4

3π/4

(4 cos2 θ − 4 + sec2 θ

)dθ

= 2∫ 5π/4

3π/4

(−2 + 2 cos 2θ + sec2 θ

)dθ

= 2 [−2θ + sin 2θ + tan θ]5π/43π/4 = 8 − 2π



SECTION 10.5 553

42. (a) (x2 + y2)2 = ax2y =⇒ r4 = ar2 cos2 θr sin θ =⇒ r = a sin θ cos2 θ

(b) same for all values of a,

with different scale

(c) A =∫ π/2

0

12(2 sin θ)2 cos4 θ dθ

= 2∫ π/2

0

(cos4 θ − cos6 θ) dθ

= 2 · π

32=

π

16

SECTION 10.5

1. 4x = (y − 1)2 2. 2x + 3y = 13

3. y = 4x2 + 1, x ≥ 0 4. y = (x + 1)3 − 5

5. 9x2 + 4y2 = 36 6. x = (y − 2)2 + 1

7. 1 + x2 = y2 8. (x− 2)2 + y2 = 1

9. y = 2 − x2, −1 ≤ x ≤ 1 10. y = 4 − x2, x > 0

11. 2y − 6 = x, −4 ≤ x ≤ 4 12. 1 + y2 = x2

13. y = x− 1 14. x = 6 − 3y

15. xy = 1 16. y = x2



554 SECTION 10.5

17. 2x + y = 11 18. 1 + y2 = x2

19. x = sin 12πy 20. x2 + 4y2 = 4

21. 1 + x2 = y2

22. x(t) = f(t) cos t, y(t) = f(t) sin t, t ∈ [α, β]

23. (a) x (t) = − sin 2πt, y (t) = cos 2πt (b) x (t) = sin 4πt, y (t) = cos 4πt

(c) x (t) = cos 12πt, y (t) = sin 1

2πt (d) x (t) = cos 32πt, y (t) = − sin 3

2πt

24. (a) x(t) = 3 cos 2πt, y(t) = −4 sin 2πt (b) x(t) = 3 sin 2πt, y(t) = 4 cos 2πt

(c) x(t) = −3 cos 4πt, y(t) = −4 sinπt (d) x(t) = 3 cos 12πt, y(t) = 4 sin 1

2πt

25. x (t) = tan 12πt, y (t) = 2

26. Any continuous function unbounded on (0, 1) will do. For example, f(t) =1t

27. x (t) = 3 + 5t, y (t) = 7 − 2t 28. x(t) = 2 + 4t, y(t) = 6 − 3t



SECTION 10.5 555

29. x (t) = sin2 πt, y (t) = − cosπt 30. x(t) = 4(1 − t)2, y(t) = 2(1 − t)

31. x (t) = (2 − t)2, y (t) = (2 − t)3 32. x(t) = (t + 1)3, y(t) = (t + 1)2

33.∫ d

c

y(t)x′(t) dt =∫ d

c

f(x(t))x′(t) dt =∫ b

a

f(x) dx = area below C

34.∫ d

c

x(t)y(t)x′(t) dt =∫ d

c

x(t)f(x(t))x′(t) dt =∫ b

a

xf(x) dx = xA

∫ d

c

12

[y(t)]2 x′(t) dt =∫ d

c

12

[f(x(t))]2 x′(t) dt =∫ d

c

12

[f(x)]2 dx = yA

35.∫ d

c

π [y(t)]2 x′(t) dt =∫ d

c

π [f(x(t))]2 x′(t) dt =∫ b

a

π [f(x)]2 dx = Vx

∫ d

c

2πx(t)y(t)x′(t) dt =∫ d

c

2πx(t)f(x(t))x′(t) dt =∫ b

a

2πxf(x) dx = Vy

36.∫ d

c

πx(t) [y(t)]2 x′(t) dt =∫ d

c

πx(t) [f(x(t))]2 x′(t) dt =∫ b

a

πx [f(x)]2 dx = xVx = y Vy

37. A =∫ 2π

0

x(t) y′(t) dt

= r2

∫ 2π

0

(1 − cos t) dt

= r2 [t− sin t]2π0 = 2πr2

38. xA =∫ 2π

0

r t r(1 − cos t) r dt = r3

∫ 2π

0

(t− t cos t) dt = r3

[t2

2− t sin t− cos t

]2π

0

= 2r3π2

=⇒ x =2r3π2

2πr2= πr

yA =∫ 2π

0

12[r2(1 − cos t)2

]r dt =

r3

2

∫ 2π

0

(1 − 2 cos t + cos2 t

)dt =

r3

2

[3t2

− 2 sin t +sin 2t

2

]2π

0

=3πr3

2=⇒ y =

3πr3/22πr2

=34r

39. (a) Vx = 2πyA = 2π(

34r) (

2πr2)

= 3π2r3

(b) Vy = 2πxA = 2π (πr) 2πr2 = 4π3r3

40. (a) xVx =∫ 2π

0

πr t r2(1 − cos t)2r dt = πr4

∫ 2π

0

(t− 2t cos t + t cos2 t) dt = 3π3r4

=⇒ x = πr, y = 0

(b) yVy =∫ 2π

0

πr t r2(1 − cos t)2r dt = 3π3r4 =⇒ y =34r, x = 0



556 SECTION 10.5

41. x (t) = −a cos t, y(t) = b sin t t ∈ [ 0, π]

42. (a) A = 2∫ π

0

y(t)x′(t) dt = 2∫ π

0

(b sin t)(a sin t) dt = 2ab[t

2− sin 2t

4

]π0

= πab

(b) xA =∫ π

0

(− cos t)(b sin t)(a sin t) dt = −a2b

∫ π

0

sin2 t cos t dt = 0 =⇒ x = 0

yA =∫ π

0

12b2 sin2 t(a sin t) dt =

ab2

2

∫ π

0

(1 − cos2 t) sin t dt =23ab2

=⇒ y =23

ab2

πab/2=

4b3π

43. (a) Equation for the ray: y + 2x = 17, x ≥ 6.

Equation for the circle: (x− 3)2 + (y − 1)2 = 25.

Simultaneous solution of these equations gives the points of intersection: (6, 5) and (8, 1).

(b) The particle on the ray is at (6, 5) when t = 0. However, when t = 0 the particle on the circle is

at the point (−2, 1). Thus, the intersection point (6, 5) is not a collision point.

The particle on the ray is at (8, 1) when t = 1. Since the particle on the circle is also at (8, 1) when

t = 1, the intersection point (8, 1) is a collision point.

44. (a) Equation of ellipse:(x− 2)2

9+

(y − 3)2

49= 1

Equation of parabola: y = − 715

(x− 1)2 +15715

Solving simultaneously gives the points of intersection (2, 10) and (5, 3)

(b) Particle 2 is at (2, 10) at t = 0, but particle 1 is at (−1, 3) at t = 0, so no collision

at (2, 10).

Both particles are at (5, 3) when t = 1, so (5, 3) is a collision point.

45. If x(r) = x(s) and r �= s, then

r2 − 2r = s2 − 2s

r2 − s2 = 2r − 2s

(1) r + s = 2.

If y(r) = y(s) and r �= s, then

r3 − 3r2 + 2r = s3 − 3s2 + 2s(r3 − s3

)− 3

(r2 − s2

)+ 2 (r − s) = 0

(2)(r2 + rs + s2

)− 3 (r + s) + 2 = 0.

Simultaneous solution of (1) and (2) gives r = 0 and r = 2. Since (x(0), y(0)) = (0, 0) = (x(2), y(2)),

the curve intersects itself at the origin.



SECTION 10.5 557

46. x(r) = x(s) and r �= s =⇒ cos r(1 − 2 sin r) = cos s(1 − 2 sin s)

y(r) = y(s) and r �= s =⇒ sin r(1 − 2 sin r) = sin s(1 − 2 sin s)

Both equations can be satisfied simultaneously with r �= s in [0, π] only for r =π

6, s =

5π6

, or

r =5π6, r =

π

6So the curve intersects itself at (0, 0).

47. Suppose that r, s ∈ [ 0, 4 ] and r �= s.

x(r) = x(s) =⇒ sin 2πr = sin 2πs.

y(r) = y(s) =⇒ 2r − r2 = 2s− s2 =⇒ 2(r − s) = r2 − s2 =⇒ 2 = r + s.

Now we solve the equations simultaneously:

sin 2πr = sin [2π (2 − r)] = − sin 2πr

2 sin 2πr = 0

sin 2πr = 0.

Since r ∈ [ 0, 4 ], r = 0, 12 , 1, 3

2 , 2, 52 , 3, 7

2 , 4.

Since s ∈ [ 0, 4 ] and r �= s and r + s = 2, we are left with r = 0, 12 ,

32 , 2. Note that

(x(0), y(0)) = (0, 0) = (x(2), y(2)) and(x(

12

), y

(12

))=(0, 3

4

)=(x(

32

), y

(32

)).

The curve intersects itself at (0, 0) and(0, 3

4

).

48.

x(r) = x(s) and r �= s =⇒

⎧⎪⎪⎨⎪⎪⎩

r3 − 4r = s3 − 4s

r3 − s3 = 4(r − s)

r2 + rs + s2 = 4

y(r) = y(s) and r �= s =⇒

⎧⎪⎪⎨⎪⎪⎩

r3 − 3r2 + 2r = s3 − 3s2 + 2s

r3 − s3 − 3(r2 − s2) + 2(r − s) = 0

(r2 + rs + s2) − 3(r + s) + 2 = 0

Solving simultaneously gives r = 0, s = 2, so curve intersect itself at (0, 0)

49. x = 2t, y = 4t− t2; 0 ≤ t ≤ 6: From the first equation, t = 12x. Substituting this into the second

equation, we get: y = 2x− 14x

2, a parabola. The limits on t imply that 0 ≤ x ≤ 12. The particle

moves along the parabola y = 2x− 14x

2 from the point (0, 0) to the point (12,−12).

50. The particle starts at the upper right and ends at the lower right.

-8 -6 -4 -2 2 x

-2

-1

1

2

y



558 SECTION 10.5

51. x = cos(t2 + t), y = sin(t2 + t) =⇒ x2 = cos2(t2 + t), y2 = sin2(t2 + t).

Since cos2(t2 + t) + sin2(t2 + t) = 1, we have x2 + y2 = 1 the unit circle.

The particle starts at the point (1, 0) and moves around the unit circle in the counterclockwise

direction.

52. x = cos(ln t), y = sin(ln t) =⇒ x2 + y2 = 1 the unit circle.

The particle starts at the point (1, 0) and moves around the unit circle in the counterclockwise direction.

53. x(θ) = cos θ(a− b sin θ), y(θ) = sin θ(a− b sin θ)

(a) a = 1, b = 2 (b) a = 2, b = 2 (c) a = 2, b = 1

(d) The curves are limacons; the curve has an inner loop if a < b and no loop if a > b.

PROJECT 10.5

1. x”(t) = 0 =⇒ x′(t) = C; x′(0) = v0 cos θ =⇒ x′(t) = v0 cos θ.

Integrating again, x(t) = (v0 cos θ)t + x0 (since x(0) = x0)

Similarly, since y”(t) = −g =⇒ y′(t) = −gt + C; y′(0) = v0 sin θ =⇒ y′(t) = −gt + v0 sin θ.

Integrating again, y(t) = − 12gt

2 + (v0 sin θ)t + y0 (since y(0) = y0)

2. From the first equation, we get t =1

v0 cos θ[x(t) − x0].

Substituting this into the second equation gives the desired result.

3. (a) Set x0 = 0, y0 = 0, g = 32.

parametric equations: x = (v0 cos θ)t, y = −16t2 + (v0 sin θ)t

rectangular equation: y = − 16v0

2(sec2 θ)x2 + (tan θ)x.

(b) To find the range, set y = 0 and solve for x, x �= 0:

−16v20

sec2 θ x2 + tan θ x = 0 =⇒ x =v20

16sin θ cos θ

(c) y(t) = 0 (and t �= 0) when t =v0

16sin θ

(d) The range 116v0

2 sin θ cos θ = 132v0

2 sin 2θ is maximal when θ = 14π for then sin 2θ = 1.

(e) Set x =132

v02 sin 2θ = b, θ =

12

arcsin(

32bv0

2

).

4. (a) maximum height — approximately 9,000 ft. range — approximately 61,000 ft.



SECTION 10.6 559

SECTION 10.6

1. x′(1) = 1, y′(1) = 3, slope 3, point (1, 0); tangent: y = 3(x− 1)

2. x′(2) = 4, y′(2) = 1, slope = 14 , point (4, 7); tangent: y − 7 = 1

4 (x− 4)

3. x′(0) = 2, y′(0) = 0, slope 0, point (0, 1); tangent: y = 1

4. x′(1) = 2, y′(1) = 4, slope = 2, point (1, 1); tangent: y − 1 = 2(x− 1)

5. x′(1/2) = 1, y′(1/2) = −3, slope − 3, point(

14 ,

94

); tangent: y − 9

4 = −3(x− 1

4

)6. x′(1) = −1, y′(1) = 2, slope = −2, point (1, 2); tangent: y − 2 = −2(x− 1)

7. x′(π

4

)= −3

4

√2, y′

(π4

)=

34

√2, slope − 1, point

(14

√2 ,

14

√2)

;

tangent: y − 14

√2 = −

(x− 1

4

√2)

8. x′(0) = 1, y′(0) = −3, slope = −3, point (1, 3); tangent: y − 3 = −3(x− 1)

9. x(θ) = cos θ (4 − 2 sin θ), y(θ) = sin θ (4 − 2 sin θ), point (4, 0)

x′(θ) = −4 sin θ − 2(cos2 θ − sin2 θ

), y′(θ) = 4 cos θ − 4 sin θ cos θ

x′(0) = −2, y′(0) = 4, slope − 2, tangent: y = −2 (x− 4)

10. x(θ) = 4 cos 2θ cos θ, y(θ) = 4 cos 2θ sin θ, point (0,−4)

x′(θ) = −8 sin 2θ cos θ − 4 cos 2θ sin θ, y′(θ) = −8 sin 2θ sin θ + 4 cos 2θ cos θ

x′(π2 ) = 4, y′(π2 ) = 0, m = 0; tangent: y = −4.

11. x(θ) =4 cos θ

5 − cos θ, y(θ) =

4 sin θ

5 − cos θ, point

(0,

45

)

x′(θ) =−20 sin θ

(5 − cos θ)2, y′(θ) =

4 (5 cos θ − 1)(5 − cos θ)2

x′(π

2

)= −4

5, y′

(π2

)= − 4

25, slope

15, tangent: y − 4

5=

15x

12. x(θ) =5 cos θ

4 − cos θ, y(θ) =

5 sin θ

4 − cos θ, point

(5√

38 −

√3,

58 −

√3

)

x′(θ) =−20 sin θ

(4 − cos θ)2, y′(θ) =

5(4 cos θ − 1)(4 − cos θ)2

, slope1 − 2

√3

2

tangent: y − 58 −

√3

=

(1 − 2

√3

2

)(x− 5

√3

8 −√

3

)



560 SECTION 10.6

13. x(θ) =cos θ (sin θ − cos θ)

sin θ + cos θ, y(θ) =

sin θ (sin θ − cos θ)sin θ + cos θ

, point (−1, 0)

x′(θ) =sin θ cos 2θ + 2 cos θ

(sin θ + cos θ)2, y′(θ) =

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

x′(0) = 2, y′(0) = −1, slope − 12 , tangent y = − 1

2 (x + 1)

14. x(θ) =sin θ + cos θsin θ − cos θ

cos θ, y(θ) =sin θ + cos θsin θ − cos θ

sin θ, point (0, 1)

x′(θ) =−2 cos θ

(sin θ − cos θ)2− sin θ + cos θ

sin θ − cos θsin θ, y′(θ) =

−2 sin θ

(sin θ − cos θ)2+

sin θ + cos θsin θ − cos θ

cos θ

x′(π/2) = −1, y′(π/2) = −2, slope 2; tangent: y − 1 = 2x

15. x(t) = t, y(t) = t3 16. x(t) = t3, y(t) = t

x′(0) = 1, y′(0) = 0, slope 0 x′(0) = 0, y′(0) = 1

tangent y = 0 tangent x = 0

17. x(t) = t5/3, y(t) = t 18. x(t) = t, y(t) = t5/3

x′(0) = 0, y′(0) = 1, slope undefined x′(0) = 1, y′(0) = 0

tangent x = 0 tangent y = 0

19. x′(t) = 3 − 3t2, y′(t) = 1

x′(t) = 0 =⇒ t = ±1; y′(t) �= 0

(a) none

(b) at (2, 2) and (−2, 0)



SECTION 10.6 561

20. x′(t) = 2t− 2, y′(t) = 3t2 + 12

x′(t) = 0 =⇒ t = 1

y′(t) �= 0 for all t

(a) no horizontal tangents

(b) at (−1, 13) 2 4x

10

20

y

21. curve traced once completely with t ∈ [ 0, 2π)

x′(t) = −4 cos t, y′(t) = −3 sin t

x′(t) = 0 =⇒ t =π

2,

3π2

;

y′(t) = 0 =⇒ t = 0, π

(a) at (3, 7) and (3, 1)

(b) at (−1, 4) and (7, 4)

22. x′(t) = 2 cos 2t, y′(t) = cos t

x′(t) = 0 =⇒ t =π

4,

3π4

5π4

7π4

y′(t) = 0 =⇒ t =π

2,

3π2

(a) at (0, ±1)

(b) at (±1, ±√

22 )

23. x′(t) = 2t− 2, y′(t) = 3t2 − 6t + 2

x′(t) = 0 =⇒ t = 1

y′(t) = 0 =⇒ t = 1 ± 13

√3

(a) at(− 2

3 , ± 29

√3)

(b) at (−1, 0)

24. x′(t) = 5 sin t, y′(t) = cos t

x′(t) = 0 =⇒ t = 0, π

y′(t) = 0 =⇒ t =π

2,

3π2

(a) at (2, 2) and (2, 4)

(b) at (−3, 3) and (7, 3)

25. curve traced completely with t ∈ [ 0, 2π)

x′(t) = − sin t, y′(t) = 2 cos 2t

x′(t) = 0 =⇒ t = 0, π

y′(t) = 0 =⇒ t =π

4,

3π4,

5π4,

7π4

(a) at(± 1

2

√2, ±1

)(b) at (±1, 0)



562 SECTION 10.6

26. x′(t) = 2 cos t, y′(t) = 5 cos t

x′(t) and y′(t) are near zero separately.

(a) none, (b) none.

27. First, we find the values of t when the curve passes through (2, 0).

y(t) = 0 =⇒ t4 − 4t2 = 0 =⇒ t = 0, ±2.

x(−2) = 2, x(0) = 2, x(2) = −2.

The curve passes through (2, 0) at t = −2 and t = 0.

x′(t) = −1 − π

2sin

πt

4, y′(t) = 4t3 − 8t.

At t = −2, x′(−2) =π

2− 1, y′(t) = −16, tangent: y =

322 − π

(x− 2).

At t = 0, x′(0) = −1, y′(0) = 0, tangent: y = 0.

28. Passes through (0, 1) at t = 1 and t = −1

x′(t) = 3t2 − 1, y′(t) = sinπ

2t +

π

2t cos

π

2t;

at t = 1 : x′(1) = 2, y′(1) = 1, tangent: y − 1 =x

2at t = −1 : x′(−1) = 2, y′(−1) = −1, tangent: y − 1 = −x

2

29. The slope of OP is tan θ1. The curve r = f(θ) can be parameterized by setting

x(θ) = f(θ) cos θ, y(θ) = f(θ) sin θ.

Differentiation gives

x′(θ) = −f(θ) sin θ + f ′(θ) cos θ, y′(θ) = f(θ) cos θ + f ′(θ) sin θ.

If f ′(θ1) = 0, then

x′(θ1) = −f(θ1) sin θ1, y′(θ1) = f(θ1) cos θ1.

Since f(θ1) �= 0, we have

m =y′(θ1)x′(θ1)

= −cot θ1 = − 1slope of OP

.

30. x(θ) = (a− cos θ) cos θ, y(θ) = (a− cos θ) sin θ goes through (0, 0) when θ = ± cos−1 a

x′(θ) = −a sin θ + 2 cos θ sin θ, y′(θ) = a cos θ + sin2 θ − cos2 θ

At θ = cos−1 a, sin θ =√

1 − a2 =⇒ x′(cos−1 a) = a√

1 − a2, y′(cos−1 a) = 1 − a2

=⇒ m1 =√

1 − a2

a

At θ = − cos−1 a, sin θ = −√

1 − a2 =⇒ x′(− cos−1 a) = −a√

1 − a2, y′(− cos−1 a) = 1 − a2

=⇒ m2 = −√

1 − a2

a

We want m1 = − 1m2

:√

1 − a2

a=

a√1 − a2

=⇒ a =√

22



SECTION 10.6 563

31. x′(t) = 3t2, y′(t) = 2t 32. x′(t) = 3t2, y′(t) = 5t4

33. x′(t) = 5t4, y′(t) = 3t2 34. x′(t) = 3t2, y′(t) = 6t2

35. x′(t) = 2t, y′(t) = 2t

ray: y = x + 1, x ≥ 0

36.d2y

dx2=

d

dt

(dy

dx

)· dtdx

=d

dt

[y′(t)x′(t)

]· 1x′(t)

=x′(t)y′′(t) − y′(t)x′′(t)

[x′(t)]3

37. By (9.7.5),d2y

dx2=

(− sin t)(− sin t) − (cos t)(− cos t)(− sin t)3

=−1

sin3 t. At t =

π

6,

d2y

dx2= −8.

38.d2y

dx2=

3t2 · 0 − 1 · 6t(3t2)3

= − 29t5

. At t = 1,d2y

dx2= −2

9

39. By (9.7.5),d2y

dx2=

(et) (e−t) − (−e−t) (et)(et)3

= 2e−3t. At t = 0,d2y

dx2= 2.



564 SECTION 10.7

40.d2y

dx2=

− sin 2t cos t + 2 sin t cos 2tsin3 2t

At t =π

4,

d2y

dx2= −

√2

2

41.d2y

dx2= cot3 t 42. the “curve” is the straight line x + y = 1

43. tangent line: y − 2 = − 163

(x− 1

8

)44. tangent line: y − 4

√3

4 +√

3=

−11 +

√3

(x− 4

4 +√

3

)

SECTION 10.7

1. L =∫ 1

0

√1 + 22 dx =

√5 2. L =

∫ 1

0

√1 + 32 dx =

√10

3. L =∫ 4

1

√√√√1 +

[32

(x− 4

9

)1/2]2

dx =∫ 4

1

32√x dx =

[x3/2

]41

= 7

4. L =∫ 44

0

√1 +

(32x1/2

)2

dx =∫ 44

0

√1 +

94x dx =

827

[(1 +

94x)3/2

]44

0

= 296

5. L =∫ 3

0

√1 +

(12√x− 1

2√x

)2

dx =∫ 3

0

(12√x +

12√x

)dx =

[13x3/2 + x1/2

]3

0

= 2√

3

6. L =∫ 2

1

√1 + (x− 1) dx =

23

[x3/2

]21

=23(2√

2 − 1) ∼= 1.22

7. L =∫ 1

0

√1 +

[x (x2 + 2)1/2

]2dx =

∫ 1

0

(x2 + 1

)dx =

[13x3 + x

]1

0

=43

8. L =∫ 4

2

√1 + x2(x2 − 2) dx =

∫ 4

2

(x2 − 1) dx =[x3

3− x

]4

2

=503

9. L =∫ 5

1

√1 +

[12

(x− 1

x

)]2

dx =∫ 5

1

12

(x +

1x

)dx =

[12

(12x2 + lnx

)]5

1

= 6 +12

ln 5 ∼= 6.80

10. L =∫ 4

1

√1 +

(x

4− 1

x

)2

dx =∫ 4

1

(x

4+

1x

)dx =

[x2

8+ lnx

]4

1

=158

+ ln 4 ∼= 3.26



SECTION 10.7 565

11. L =∫ 8

1

√1 +

[12(x1/3 − x−1/3

)]2

dx =∫ 8

1

12

(x1/3 + x−1/3

)dx =

12

[34x4/3 +

32x2/3

]8

1

=638

12. L =∫ 2

1

√1 +

(x4

2− 1

2x−4

)2

dx =∫ 2

1

(x4

2+

x−4

2

)dx =

12

[x5

5− 1

3x3

]2

1

=779240

13. L =∫ π/4

0

√1 + tan2 x dx =

∫ π/4

0

secx dx = [ln |secx + tanx|]π/40 = ln(1 +

√2)∼= 0.88

14. L =∫ 1

0

√1 + x2 dx =

∫ π/4

0

sec3 u du =12

[secu tanu + ln | secu + tanu|]π/40

=12

[√2 + ln(1 +

√2)]∼= 1.15

15. L =∫ 2

1

√1 +

(√x2 − 1

)2

dx =∫ 2

1

x dx =[12x2

]2

1

=32

16. L =∫ ln 2

0

√1 + sinh2 x dx =

∫ ln 2

0

coshx dx = [sinhx]ln 20 =

34

17. L =∫ 1

0

√1 +

[√3 − x2

]2dx =

∫ 1

0

√4 − x2 dx =

∫ π/6

0

4 cos2 u du

(x = 2 sinu)∧

= 2∫ π/6

0

(1 + cos 2u) du = 2[u + 1

2 sin 2u]π/60

= 13π + 1

2

√3

18. L =∫ π/2

π/6

√1 + cot2 x dx =

∫ π/2

π/6

cscx dx = [ln | cscx− cotx|]π/2π/6 = − ln(2 −√

3) ∼= 1.32

19. v(t) =√

(2t)2 + 22 = 2√t2 + 1

initial speed = v (0) = 2, terminal speed = v(√

3)

= 4

s =∫ √

3

0

2√t2 + 1 dt = 2

∫ π/3

0

sec3 u du = 2[12

secu tanu +12

ln |secu + tanu|]π/30

(t = tanu) (by parts)

= 2√

3 + ln(2 +

√3) ∼= 4.78

20. v(t) =√

1 + t2 initial speed = v(0) = 1, terminal speed = v(1) =√

2

s =∫ 1

0

√1 + t2 dt =

12

[√2 + ln(1 +

√2)]∼= 1.15



566 SECTION 10.7

21. v(t) =√

(2t)2 + (3t2)2 dt = t(4 + 9t2

)1/2initial speed = v(0) = 0, terminal speed = v(1) =

√13

s =∫ 1

0

t(4 + 9t2

)1/2dt =

[127

(4 + 9t2

)3/2]1

0

=127

(13√

13 − 8)

22. v(t) =√

9a2 cos4 t sin2 t + 9a2 sin4 t cos2 t = 3a cos t sin t

initial speed = v(0) = 0, terminal speed = v(π/2) = 0

s =∫ π/2

0

3a cos t sin t dt =[3a sin2 t

2

]π/20

=32a

23. v(t) =√

[et cos t + et sin t]2 + [et cos t− et sin t]2 =√

2 et

initial speed = v(0) =√

2, terminal speed =√

2 eπ

s =∫ π

0

√2 et dt =

[√2 et

]π0

=√

2 (eπ − 1)

24. v(t) =√

(t cos t)2 + (t sin t)2 = t initial speed = v(0) = 0, terminal speed = v(π) = π

s =∫ π

0

t dt =[t2

2

]π0

=π2

2

25. L =∫ 2π

0

√[x′(θ)]2 + [y′(θ)]2 dθ =

∫ 2π

0

√a2(1 − cos θ)2 + a2 sin2 θ dθ

= a

∫ 2π

0

√2(1 − cos θ) dθ = 2a

∫ 2π

0

sinθ

2dθ = −4a

[cos

θ

2

]2π

0

= 8a

26.√

[x′(θ)]2 + [y′(θ)]2 =√

(−2a sin θ + 2a sin 2θ)2 + (2a cos θ − 2a cos 2θ)2 = 2a√

2 − 2 cos θ

L =∫ 2π

0

2a√

2 − 2 cos θ dθ = 2a∫ 2π

0

2 sinθ

2dθ = 4a

[−2 cos

θ

2

]2π

0

= 16a

27.(a) L =∫ 2π

0

√(−3a sin θ − 3a sin 3θ)2 + (3a cos θ − 3a cos 3θ)2 dθ

= 3a∫ 2π

0

√sin2 θ + 2 sin θ sin 3θ + sin2 3θ + cos2 θ − 2 cos θ cos 3θ + cos2 3θ dθ

= 3a∫ 2π

0

√2(1 − cos 4θ) dθ = 6a

∫ 2π

0

|sin 2θ| dθ

= 24a∫ π/2

0

sin 2θ dθ = −12a [cos 2θ]π/20 = 24a

(b) The result follows from the identities: cos 3θ = 4 cos3 θ − 3 cos θ; sin 3θ = 3 sin θ − 4 sin3 θ



SECTION 10.7 567

28.√

[x′(θ)]2 + [y′(θ)]2 =√

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

1 + θ2

L =∫ 2π

0

√1 + θ2 dθ =

[12θ√

1 + θ2 +12

ln |θ +√

1 + θ2|]2π

0

= π√

1 + 4π2 +12

ln(2π +√

1 + 4π2)

29. L = circumference of circle of radius 1 = 2π

30. L = half circumference of circle of radius 3 = 3π

31. L =∫ 4π

0

√[eθ]2 + [eθ]2 dθ =

∫ 4π

0

√2 eθdθ =

[√2 eθ

]4π0

=√

2(e4π − 1

)

32. L =∫ 2π

−2π

√(aeθ)2 + (aeθ)2 dθ =

∫ 2π

−2π

√2aeθ dθ = a

√2(e2π − e−2π) (a > 0)

33. L =∫ 2π

0

√[e2θ]2 + [2e2θ]2 dθ =

∫ 2π

0

√5 e2θdθ =

[12

√5 e2θ

]2π

0

=12

√5(e4π − 1

)

34. L = 2∫ π

0

√(1 + cos θ)2 + (− sin θ)2 dθ = 2

∫ π

0

√2 + 2 cos θ dθ = 4

∫ π

0

cosθ

2dθ = 8

35. L =∫ π/2

0

√(1 − cos θ)2 + sin2 θ dθ =

∫ π/2

0

√2 − 2 cos θ dθ

=∫ π/2

0

(2 sin

12θ

)dθ =

[−4 cos

12θ

]π/20

= 4 − 2√

2

36. L =∫ π/4

0

√(2a sec θ)2 + (2a sec θ tan θ)2 dθ =

∫ π/4

0

2a sec2 θ dθ =[2a tan θ

]π/40

= 2a

37. s =∫ 1

0

√[1

1 + t2

]2

+[ −t

1 + t2

]2

dt =∫ 1

0

dt√1 + t2

=∫ π/4

0

secu du = [ln |sec u + tanu|]π/40 = ln(1 +

√2)

(t = tanu)

initial speed = v (0) = 1, terminal speed = v (1) = 12

√2

38. s =∫ 2π

0

√(− sin t)2 + (1 − cos t)2 dt =

∫ 2π

0

√2 − 2 cos t dθ =

[−4 cos

t

2

]2π

0

= 8

initial speed = v (0) = 0, terminal speed = v (2π) = 0

39. c = 1; the curve y = ex is the curve y = lnx reflected in the line y = x



568 SECTION 10.7

40. By the hint:L =

∫ 4

1

√1 +

(32x1/2

)2

dx =∫ 4

1

√1 +

94x dx =

827

[(1 +

94x

)3/2]4

1

=127

[80√

10 − 13√

13]∼= 7.63

41. coordinates of the midpoint:(

12 , − 7

2

)42. coordinates of the midpoint (approx):

(3.2911, 5.9705)

43. L =∫ b

a

√1 + sinh2 x dx =

∫ b

a

√cosh2 x dx =

∫ b

a

coshx dx = A

44. L =∫ e

1

√1 +

4x2

dx ∼= 2.6625

45.f(x) = sinx− x cosx

f ′(x) = x sinx

L =∫ π

0

√1 + x2 sin2 x dx ∼= 4.6984

46. L =∫ π/3

0

√(2e2t cos 2t− 2e2t sin 2t)2 + (2e2t sin 2t + 2e2t cos 2t)2 dt =

√8∫ π/3

0

e2t dt ∼= 10.0699

47. x = t2, y = t3 − t

x′ = 2t y′ = 3t2 − 1

L =∫ 1

−1

√(2t)2 + (3t2 − 1)2 dt ∼= 2.7156

48. (a) (b) L = 2∫ 1

0

√(x′(t))2 + (y′(t))2 dt

= 2∫ 1

0

√(3 − 6t3

(t3 + 1)2

)2

+(

6t− 3t4

(t3 + 1)2

)2

dt

= 3.5485



SECTION 10.7 569

49. r = 1 − cos θ, r′ = sin θ

L =∫ π

0

√(1 − cos θ)2 + sin2 θ dθ = 4

50. r = sin 5θ is a petal curve with 5 petals. The length of one petal is given by:

L =∫ π/5

0

√sin2 5θ + 25 cos2 5θ dθ ∼= 2.101

The length of the entire curve is: 10.505 (approx.). The curve is traversed twice for 0 ≤ θ ≤ 2π.

Thus the length of that curve is 2(10.505) ∼= 21.01.

51. (a) L =∫ 2π

0

√a2 sin2 t + b2 cos2 t dt = 4

∫ π/2

0

√a2(1 − cos2 t) + b2 cos2 t dt

= 4a∫ π/2

0

√1 − e2 cos2 t dt, where e =

√a2 − b2

a

(b) L = 4∫ π/2

0

√25 − 9 cos2 t dt ∼= 28.3617

52. The line segment that joins Pi−1 to Pi has length√(xi − xi−1)2 + [f(xi) − f(xi−1)]2

This can be written√1 +

[f(xi) − f(xi−1)

xi − xi−1

]2

(xi − xi−1) =

√1 +

[f(xi) − f(xi−1)

xi − xi−1

]2

Δxi

The mean value theorem gives the existence of x∗i in [xi−1, xi] such that

f(xi) − f(xi−1)xi − xi−1

= f ′(x∗i ).

The length of Pi−1Pi can then be written√1 + [f ′(x∗

i )]2 Δxi

The length of the polygonal approximation is the sum of such expressions from i = 1 to i = n.

53.√

1 + [f(x)]2 =√

1 + tan2 [α(x)] = | sec [α(x)]|

54. Balancing the vertical forces we have

k

∫ x

0

√1 + [f ′(t)]2 dt = p(x) sin θ. (weight=vertical pull at x)

Balancing the horizontal forces we have

p(0) = p(x) cos θ. (pull at 0 = horizontal pull at x)



570 SECTION 10.8

Together these equations give

k

∫ x

0

√1 + [f ′(x)]2 dt = p(0) tan θ = p(0)f ′(θ).

Differentiating with respect to x, we have

k√

1 + [f ′(x)]2 = p(0)f ′′(x)

f ′′(x)√1 + [f ′(x)]2

= b. (set k/p(0) = b)

Integration gives

ln(f ′(x) +

√1 + [f ′(x)]2

)= bx + C.

In the coordinate system of the figure, we have f ′(0) = 0. Therefore C = 0 and

ln(f ′(x) +

√1 + [f ′(x)]2

)= bx

Exponentiating both sides we have

f ′(x) +√

1 + [f ′(x)]2 = ebx√1 + [f ′(x)]2 = ebx − f ′(x)

1 + [f ′(x)]2 = ebx − 2ebxf ′(x) + [f ′(x)]2

2ebxf ′(x) = e2bx − 1

f ′(x) =12(ebx − e−bx) = sinh bx

A final integration gives

f(x) =1b

cosh bx + K.

Adjusting the coordinate system vertically so that f(0) = 1/b we have

f(x) =1b

cosh bx

Now set a = 1/b

SECTION 10.8

1. L = length of the line segment = 1

(x, y) =(

12 , 4

)(the midpoint of the line segment)

Ax = lateral surface area of cylinder of radius 4 and side 1 = 8π.

2. L =∫ 1

0

√1 + 22 dx =

√5

xL =∫ 1

0

x√

1 + 22 dx =√

52

, x =12

yL =∫ 1

0

2x√

1 + 22 dx =√

5, y = 1

Ax = 2πyL = 2π√

5



SECTION 10.8 571

3.L =

∫ 3

0

√1 +

(43

)2

dx =(

53

)3

= 5

xL =∫ 3

0

x

√1 +

(43

)2

dx =53

[12x2

]3

0

=152, x =

32

yL =∫ 3

0

43x

√1 +

(43

)2

dx =(

43

)(152

)= 10, y = 2

Ax = 2πyL = 2π(2)(5) = 20π

4. L =∫ 5

0

√1 +

(−12

5

)2

dx = 5

√1 +

14425

= 13

xL =∫ 5

0

x

√1 +

14425

dx =252

√1 +

14425

, x =52

yL =∫ 5

0

(12 − 12

5x

)√1 +

14425

dx =125

· 252

√1 +

14425

, y = 6

Ax = 2πyL = 156π

5. L =∫ 2

0

√(3)2 + (4)2 dt = (2)(5) = 10

xL =∫ 2

0

3t√

(3)2 + (4)2 dt = 15[12t2]2

0

= 30, x = 3

yL =∫ 2

0

4t√

(3)2 + (4)2 dt = 20[12t2]2

0

= 40, y = 4

Ax = 2πyL = 2π(4)(10) = 80π

6. L =18

circumference of circle of radius 5 =54π

xL =∫ π/4

0

x(θ)√x′(θ)2 + y′(θ)2 dθ =

∫ π/4

0

5 cos θ · 5 dθ =25√

22

, x =10√

2π

yL =∫ π/4

0

y(θ)√x′(θ)2 + y′(θ)2 dθ =

∫ π/4

0

5 sin θ · 5 dθ = 25

(1 −

√2

2

), y =

10π

(2 −√

2)

Ax = 2πyL = 25π(2 −√

2)

7. L =∫ π/6

0

√4 sin2 t + 4 cos2 t dt = 2

(π6

)=

13π

xL =∫ π/6

0

2 cos t√

4 sin2 t + 4 cos2 t dt = 4 [sin t ]π/60 = 2, x =6π

yL =∫ π/6

0

2 sin t√

4 sin2 t + 4 cos2 t dt = 4 [− cos t]π/60 = 4 − 2√

3, y = 6(2 −

√3)/π

Ax = 2πyL = 2π(6(2 −√

3 )/π ) 13π = 4π(2 −

√3 )



572 SECTION 10.8

8.L =

∫ π/2

0

√(−3 cos2 t sin t)2 + (3 sin2 t cos t)2 dt =

∫ π/2

0

3 cos t sin t dt =[3 sin2 t

2

]π/20

=32

xL =∫ π/2

0

cos3 t · 3 cos t sin t dt =[−3

5cos5 t

]π/20

=35, x =

25

yL =∫ π/2

0

sin3 t · 3 cos t sin t dt =[35

sin5 t

]π/20

=35, y =

25

Ax = 2πyL =65π

9. x(t) = a cos t, y = a sin t; t ∈ [ 13π,23π]

L =∫ 2π/3

π/3

√a2 sin2 t + a2 cos2 t dt =

13πa

by symmetry x = 0

yL =∫ 2π/3

π/3

a sin t√a2 sin2 t + a2 cos2 t dt = a2

∫ 2π/3

π/3

sin t dt

= a2 [− cos t]2π/3π/3 = a2, y = 3a/π

Ax = 2πyL = 2πa2

10. L =∫ π

0

√(1 + cos θ)2 + sin2 θ dθ =

∫ π

0

√2 + 2 cos θ dθ

=∫ π

0

2

√1 + cos θ

2dθ =

∫ π

0

2 cos12θ dθ = 4

xL =∫ π

0

cos θ(1 + cos θ)(

2 cos12θ

)dθ =

∫ π

0

(cos2

12θ − sin2 1

2θ

)(2 cos2

12θ

)(2 cos

12θ

)dθ

= 4∫ π

0

(1 − 2 sin2 1

2θ

)(1 − sin2 1

2θ

)(cos

12θ

)dθ

= 4∫ π

0

(cos

12θ − 3 sin2 1

2θ cos

12θ + 2 sin4 1

2θ cos

12θ

)dθ =

165

; x =45

yL =∫ π

0

sin θ(1 + cos θ)(

2 cos12θ

)dθ

=∫ π

0

(2 sin

12θ cos

12θ

)(2 cos2

12θ

)(2 cos

12θ

)dθ

= 8∫ π

0

(cos4

12θ sin

12θ

)dθ =

165

; y =45

Ax = 2πyL =325π

11. Ax =∫ 2

0

23πx3

√1 + x4 dx =

19π[(1 + x4)3/2

]20

=19π(17

√17 − 1) ∼= 24.1179

12. Ax =∫ 2

1

2π√x

√1 +

14x

dx =∫ 2

1

π√

4x + 1 dx = π

[16(4x + 1)3/2

]2

1

=16π(27 − 5

√5)



SECTION 10.8 573

13. Ax =∫ 1

0

12πx3

√1 +

916

x4 dx =427

π

[(1 +

916

x4

)3/2]1

0

=61432

π

14. Ax =∫ 4

0

2π · 3√x

√1 +

94x

dx =∫ 4

0

3π√

4x + 9 dx =π

2

[(4x + 9)3/2

]40

= 49π

15. Ax =∫ π/2

0

2π cosx√

1 + sin2 x dx =∫ 1

0

2π√

1 + u2 du

∧u = sinx

= 2π[12u

√1 + u2 + 1

2 ln(u +

√1 + u2

)]10

= π[√

2 + ln(1 +

√2)]

∧(8.5.1)

16. Ax =∫ 0

−1

4π√

1 − x

√1 +

11 − x

dx = 4π∫ 0

−1

√2 − x dx = −8

3

[(2 − x)3/2

]0−1

=83π(3

√3 − 2

√2)

17. Ax =∫ π/2

0

2π(eθ sin θ)√

[eθ cos θ − eθ sin θ]2 + [eθ sin θ + eθ cos θ]2 dθ

= 2π√

2∫ π/2

0

e2θ sin θ dθ

= 2π√

2[15

(2e2θ sin θ − e2θ cos θ

)]π/20

= 25

√2π (2eπ + 1)

(by parts twice)

18. Ax =∫ ln 2

0

2π coshx√

1 + sinh2 x dx

= 2π∫ ln 2

0

cosh2 x dx

=12π

∫ ln 2

0

(e2x + 2 + e−2x) dx =12π

[12e2x + 2x− 1

2e−2x

]ln 2

0

=116

π(15 + 16 ln 2)

19. (a) A =∫ 2π

0

y(θ)x′(θ) dθ [see (9.6.4)]

=∫ 2π

0

a2(1 − cos θ)2 dθ

= a2

∫ 2π

0

(1 − 2 cos θ + cos2 θ) dθ

= a2

∫ 2π

0

(32− 2 cos θ +

12

cos 2θ)

dθ

= a2

[32θ − 2 sin θ +

14

sin 2θ]2π

0

= 3π a2



574 SECTION 10.8

(b) A =∫ 2π

0

2π y(θ)√

[x′(θ)]2 + [y′(θ)]2 dθ (9.9.2)

=∫ 2π

0

2π a(1 − cos θ)√a2(1 − cos θ)2 + a2 sin2 θ dθ

= 2π a2

∫ 2π

0

(1 − cos θ)√

2 − 2 cosθ dθ

= 4π a2

∫ 2π

0

(1 − cos θ) sinθ

2dθ

= 4π a2

∫ 2π

0

(2 sin

θ

2− 2 cos2

θ

2sin

θ

2

)dθ

= 4π a2

[−4 cos

θ

2

]2π

0

+16π a2

3[cos3(θ/2)

]2π0

=64π a2

3

20. (a) The graph with a = 1;

−2−4 2 4 x

−2

−4

2

4y

(b) A = 2∫ π

0

y(θ)x′(θ) dθ

= 2∫ π

0

(3 sin θ − sin 3θ)(−3 sin θ − 3 sin 3θ) dθ

= −6∫ π

0

(3 sin2 θ + 2 sin θ sin 3θ − sin2 3θ) dθ

= 6π

(b) A =∫ π

0

2πy(θ)√x′(θ)2 + y′(θ)2 dθ = 2π

∫ π

0

(3a sin θ − a sin 3θ)3a√

2 − 2 cos 4θ dθ

21. A = 12θs2

2 − 12θs1

2

= 12 (θs2 + θs1)(s2 − s1)

= 12 (2πR + 2πr)s = π(R + r)s

22. (a) x =1

π4 r

2 − π4 a

2·(− 4a

3π· π

4a2 +

4r3π

· π4r2

)=

43π

(r3 − a3

r2 − a2

)=

43π

(r2 + ar + a2

r + a

)

y = x

(b) lima→r

43π

(r2 + ar + a2

r + a

)=

2rπ

; = x = y =2rπ.

23. (a) The centroids of the 3, 4, 5 sides are the midpoints(

32 , 0

), (3, 2),

(32 , 2

).

(b) x(3 + 4 + 5) = 32 (3) + 3(4) + 3

2 (5), 12x = 24, x = 2

y(3 + 4 + 5) = 0(3) + 2(4) + 2(5), 12y = 18, y =32



SECTION 10.8 575

(c) A = 13 (3)(4) = 6

xA =∫ 3

0

x

(43x

)dx =

∫ 3

0

43x2dx =

49[x3]30

= 12, x = 2

yA =∫ 3

0

12

(43x

)2

dx =∫ 3

0

89x2dx =

827

[x3]30

= 8, y =43

(d) x (4 + 5) = 3(4) +32(5), 9x =

392, x =

136

y (4 + 5) = 2(4) + 2(5), 9y = 18, y = 2

(e) Ax = 2π(2)(5) = 20π

24. Set y(t) = t, x(t) =12

[t√t2 − 1 + ln |t−

√t2 − 1|

], t ∈ [2, 5]

x′(t) =√t2 − 1 y′(t) = 1

A =∫ 5

2

2πt√

(√t2 − 1)2 + 12 dt =

∫ 5

2

2πt2 dt =[2π

t3

3

]5

2

= 78π

25. Set y(t) = t, x(t) =1

6a2

(t3 +

3a4

t

)t ∈ [a, 3a]

x′(t) =1

6a2

(3t2 − 3a4

t2

), y′(t) = 1

A =∫ 3a

a

2πt

√1

36a4

(3t2 − 3a4

t2

)2

+ 1 dt = 2π∫ 3a

a

t

6a2

(3t2 +

3a4

t2

)dt

=π

a2

∫ 3a

a

(t3 +

a4

t

)dt =

π

a2

[t4

4+ a4 ln 4

]3a

a

= πa2(20 + ln 3)

26. Ax = 2πyL = 2π(b)(2πa) = 4π2ab

27. (a) No: f ′(x) = −x/√r2 − x2 is not defined at x = ±r. We can however begin with a smaller

interval [−r + ε, r − ε], integrate, and take the limit as ε → 0.

(b) A =∫ 2π

0

2πr√

(sin t)2 dt = 2πr∫ 2π

0

| sin t| dt = 8πr

C is not simple. The curve [in this case the line segment that joins (1, r) to (−1, r)] is traced

out twice.

27. The band can be obtained by revolving about the x-axis the graph of the function

f(x) =√r2 − x2, x ∈ [ a, b ].

A straightforward calculation shows that the surface area of the band is 2πr(b− a).



576 SECTION 10.8

28. C revolved about the y-axis generates a surface of area

Ay = 2πxL = 2πxr(θ2 − θ1).

C revolved about the x-axis generates a surface of area

Ax = 2πxyL = 2πyr(θ2 − θ1).

By our solution to Exercise 27.

Ay = 2πr2(sin θ2 − sin θ1)

and

Ax = 2πr2(cos θ1 − cos θ2).

Therefore

x = r

(sin θ2 − sin θ1

θ2 − θ1

), y = r

(cos θ1 − cos θ2

θ2 − θ1

).

29. (a) Parameterize the upper half of the ellipse by

x(t) = a cos t, y(t) = b sin t; t ∈ [ 0, π ].

Here √[x′(t)]2 + [y′(t)]2 =

√a2 sin2 t + b2 cos2 t =

√a2 − (a2 − b2) cos2 t,

which, with c =√a2 − b2, can be written

√a2 − c2 cos2 t. Therefore,

A =∫ π

0

2πb sin t√a2 − c2 cos2 t dt = 4πb

∫ π/2

0

sin t√a2 − c2 cos2 t dt.

Setting u = c cos t, we have du = −c sin t and

A = −4πbc

∫ 0

c

√a2 − u2 du =

4πbc

[u

2

√a2 − u2 +

a2

2sin−1

(ua

)]c0

= 2πb2 +2πa2b

csin−1

( ca

)= 2πb2 +

2πabe

sin−1 e

where e is the eccentricity of ellipse: e = c/a.

(b) Parameterize the right half of the ellipse by

x(t) = a cos t, y(t) = b sin t; t ∈[−1

2π,

12π

].

Again√

[x′(t)]2 + [y′(t)]2 =√a2 − c2 cos2 t where c =

√a2 − b2.

Therefore

A =∫ π/2

−π/2

2πa cos t√a2 − c2 cos2 t dt.

Set u = c sin t. Then du = c cos t dt and

A =2πac

∫ c

−c

√b2 + u2 du =

2πac

[u

2

√b2 + u2 +

b2

2ln∣∣∣u +

√b2 + u2

∣∣∣]c−c



SECTION 10.8 577

Routine calculation gives

A = 2πa2 +πb2

eln∣∣∣∣1 + e

1 − e

∣∣∣∣.30. Set s′(t) =

√[x′(t)]2 + [y′(t)]2. Let P = {t0, t1, · · · , tn} be a partition of [c,d]. The partition

breaks up the surface into n surfaces of revolution with areas

Ai =∫ ti

ti−1

2πy(t)s′(t) dt

and centroids

xi = x(t∗i ) with t∗i ∈ [ti−1, ti].

xA = x1A1 + · · · + xnAn

= x(t∗1)∫ t1

t0

2πy(t)s′(t) dt + · · · + x(t∗n)∫ tn

tn−1

2πy(t)s′(t) dt

∼= x(t∗1)[2πy(t∗1)s

′(t∗1)Δt1] + · · · + x(t∗n)[2πy(t∗n)s′(t∗n)Δtn]

= 2πx(t∗1)y(t∗1)s

′(t∗1)Δt1 + · · · + 2πx(t∗n)y(t∗n)s′(t∗n)Δtn

As ‖P‖ → 0, the expression on the right tends to∫ d

c

2πx(t)y(t)s′(t) dt =∫ d

c

2πx(t)y(t)√

[x′(t)]2 + [y′(t)]2 dt.

31. Such a hemisphere can be obtained by revolving about the x-axis the curve

x(t) = r cos t, y(t) = r sin t; t ∈ [ 0, 12π ].

Therefore, xA =∫ π/2

0

2π(r cos t)(r sin t)√r2 sin2 t + r2 cos2 t dt

=∫ π/2

0

2πr3 sin t cos t dt = πr3[sin2 t

]π/20

= πr3.

A = 2πr2; x = xA/A = 12r.

The centroid lies on the midpoint of the axis of the hemisphere.

32. The cone can be generated by revolving about the x-axis the graph of the function

f(x) = (r/h)x, x ∈ [0, h].

Formula 9.9.8 gives

xA =∫ h

0

2πf(x)√

1 + [f ′(x)]2 dx =2πrh2

√h2 + r2

∫ h

0

x2 dx =23πrh

√h2 + r2.

A = πr√h2 + r2; x = xA/A = 2

3h

The centroid of the surface lies on the axis of the cone at a distance of 23h from the vertex of the

cone.



578 SECTION 10.8

33. Such a surface can be obtained by revolving about the x-axis the graph of the function

f(x) =(R− r

h

)x + r, x ∈ [0, h].

Formula (9.9.8) gives

xA =∫ h

0

2πxf(x)√

1 + [f ′(x)]2 dx

=2πh

√h2 + (R− h)2

∫ h

0

[(R− r

h

)x2 + rx

]dx

=π

3

√h2 + (R− r)2 (2R + r)h

A = π(R + r)s = π(R + r)√h2 + (R− r)2 and x =

xA

A=(

2R + r

R + r

)h

3.

The centroid of the surface lies on the axis of the cone(

2R + r

R + r

)h

3units from the base of radius r.

PROJECT 10.8

1. Referring to the figure we have

x(θ) = OB −AB = Rθ −R sin θ = R (θ − sin θ)

y(θ) = BQ−QC = R−R cos θ = R (1 − cos θ).

2. (a) x′(θ) = R(1 − cos θ), y′(θ) = R sin θ.

The arches end at θ = 2nπ, and x′(2nπ) = y′(2nπ) = 0

(b) A =∫ 2π

0

y(θ)x′(θ) dθ = R2

∫ 2π

0

(1 − cos θ)2 dθ = 3πR2

(c)

L =∫ 2π

0

√[R (1 − cos θ)]2 + [R sin θ]2 dθ

= R

∫ 2π

0

√2 − 2 cos θ dθ

= R

∫ 2π

0

√4 sin2

(θ

2

)dθ = 2R

∫ 2π

0

sinθ

2dθ = 4R

[− cos

θ

2

]2π

0

= 8R



SECTION 10.8 579

3. (a) x = πR by symmetry

yA =∫ 2π

0

12[y(θ)]2x′(θ) dθ

=∫ 2π

0

12R2(1 − cos θ)2[R(1 − cos θ)] dθ

=12R3

∫ 2π

0

(1 − 3 cos θ + 3 cos2 θ − cos3 θ

)dθ =

52πR3

A = 3πR2 (by Exercise 2b) y =(

52πR

3)/(3πR2

)= 5

6R

(b) Vx = 2πyA = 2π(

56R

)(3πR2

)= 5π2R3

(c) Vy = 2πxA = 2π(πR)(3πR2) = 6π3R3

4.

5. (a)dy

dx=

y′(φ)x′(φ)

=R sinφ

R (1 + cosφ)=

2 sin 12φ cos 1

2φ

2 cos2 12φ

= tan12φ; α =

12φ

(b)s =

∫ φ

0

√[x′(t)]2 + [y′(t)]2 dt =

∫ φ

0

√[R(1 + cos t)]2 + [R sin t]2 dt

= R

∫ φ

0

√2 + 2 cos t dt = R

∫ φ

0

2 cos12t dt = 4R

[sin

12t

]φ0

= 4R sin12φ = 4R sinα

Now note that the tangent at (x(φ), y(φ)) has slope

m =y′(φ)x′(φ)

=sinφ

1 + cosφ=

sinφ(1 − cosφ)1 − cos2 φ

=1 − cosφ

sinφ=

2 sin2 φ2

2 sin φ2 cos φ

2

= tanφ

2

So the inclination of the tangent isφ

2= α

6. (a) Already shown more generally in Example 6 of Section 9.8.

(b) Combining d2s/dt2 = −g sinα with s = 4R sinα, we have

d2s

dt2= − g

4Rs.

This is simple harmonic motion with angular frequency ω = 12

√g/R (see Section 18.6)

and period T = 2π/ω = 4π√R/g.



580 REVIEW EXERCISES

7.d2s

dt2= −g sinα = −g · 2R√

π2R2 + 4R2= − 2g√

π2 + 4

Sinceds

dt= 0 and s = R

√π2 + 4 when t = 0, integrating twice gives

s = − g√π2 + 4

t2 + R√π2 + 4

s = 0 at t2 =1gR(π2 + 4), so t =

√R(π2 + 4)/g

REVIEW EXERCISES

1. The equation can be written x2 = 4(y + 1), a parabola.

vertex (0,−1)

focus (0, 0)

axis y-axis

directrix y = −2 − 2 2x

− 1

1

2

3

y

2. The equation can be written y2

4 − (x−5)2

16 = 1, a hyperbola.

center (5, 0)

foci (5,±2√

5)

vertices (5,±2)

asymptotes x = 5 ± 2y

length of the transverse axis: 4.

− 5 5 10x

− 4

4

y

3. The equation can be written (x+3)2

1 + y2

14

= 1, an ellipse.

center (−3, 0)

foci (−3 ± 3√

32 , 0)

length of the major axis: 2

length of the minor axis: 1− 3

x

y

4. The equation can be written (y−1)2

9 + (x−1)2

4 = 1, an ellipse.

center (1, 1)

foci (1, 1 ±√

5)


length of the minor axis: 4

2x

1

y



REVIEW EXERCISES 581

5. The equation can be written (y+1)2

4 − (x−1)2

9 = 1, a hyperbola.

center (1,−1)

foci (1,−1 ±√

13)

vertices (1, 1), (1,−3)

asymptotes: y = −1 ± 23 (x− 1)

length of the transverse axis: 4

1x

1

y

6. The equation can be written (x− 5)2 = 8(y − 2), a parabola.

vertex (5, 2)

focus (5, 4)

directrix: y = 0

axis: x = 5 5x

2

y

7. The equation can be written (x−1)2

25 + (y+2)2

9 = 1, an ellipse.

center (1,−2)

foci (5,−2), (−3,−2)


length of the minor axis: 6

1x

− 2

y

8. The equation can be written (x+√

3)2

3 − (y+√

3)2

2 = 1, a hyperbola.

center (−√

3,−√

3)

foci (±√

5 −√

3,−√

3)

vertices (−2√

3,−√

3) and (0,−√

3)

asymptotes: y = −√

3 ±√

2√3(x +

√3)

length of the transverse axis: 2√

3

−3x

−3

y

9. x = −4 cos π3 = −2, y = −4 sin π

3 = −2√

3; (−2,−2√

3)

10. x =√

2 cos (− 5π4 ) = −1, y =

√2 sin (− 5π

4 ) = 1; (−1, 1).

11. r = 4, θ = 3π/2; [4, 32π + 2nπ], n = 0,±1,±2, · · ·; [−4, π

2 + 2nπ], n = 0,±1,±2, · · ·

12. r =√

(1/4) + (1/4) =√

22 , θ = 5π

4 ;[√

22 , 5

4π + 2nπ], n = 0,±1,±2, · · ·;[

−√

22 , 1

4π + 2nπ], n = 0,±1,±2, · · ·

13. r =√

16 = 4, θ = −π6 ; [4,−π

6 + 2nπ], n = 0,±1,±2, · · ·; [−4, 56π + 2nπ], n = 0,±1,±2, · · ·

14. r = 2, θ = 23π;

[2, 2

3π + 2nπ], n = 0,±1,±2, · · ·;

[−2, 5

3π + 2nπ], n = 0,±1,±2, · · ·

15. Set y = r sin θ and x = r cos θ : r sin θ = r2 cos2 θ =⇒ r = sec θ tan θ.




16. Set y = r sin θ and x = r cos θ : r2 cos2 θ + r2 sin2 θ − 2r cos θ = 0 =⇒ r = 2 cos θ

17. Set y = r sin θ and x = r cos θ:

r2 cos2 θ + r2 sin2 θ − 4r cos θ + 2r sin θ =⇒ r = 4 cos θ − 2 sin θ.

18. Set y = r sin θ and x = r cos θ :

r2 sin2 θ(1 − r2 cos2 θ) = r4 cos4 θ

sin2 θ − r2 cos2 θ sin2 θ = r2 cos4 θ

sin2 θ = r2 cos2 θ

r = ± tan θ

19. x = 5

20. r + 4 cos θ = 0, r2 + 4r cos θ = 0 =⇒ x2 + y2 + 4x = 0

21. Multiplication by r gives r2 = 3r cos θ + 4r sin θ =⇒ x2 + y2 − 3x− 4y = 0.

22. The equation can be written r2 cos 2θ = 1 Since cos 2θ = cos2 θ − sin2 θ, we have x2 − y2 = 1.

23.

-2 2

-2

-4

24.

-3 3

-1

1

25.

-1

26.

-1 1

-2

-3




27.

-1 1

2 28.

2

intersections: [√

22 , 1

6π], [√

22 , 5

6π], the pole intersections: [√

3, π6 ], the pole

29.

A =∫ 2π

0

12(2(1 − cos θ))2 dθ = 2

∫ 2π

0

(1 − 2 cos θ + cos2 θ) dθ

= 2∫ 2π

0

(1 − 2 cos θ +

1 + cos 2θ2

)θ

= 2[32θ − 2 sin θ +

14

sin 2θ]2π

0

= 6π-4 -2

-2

2

30. A = 2∫ π

2

0

12(4 sin 2θ) dθ

= 4[− 1

2cos 2θ

]π/20

= 4

-1 1

-1

1

31. One petal is formed when θ ranges from −π/8 to π/8. By symmetry, the area of one petal is

A = 2∫ π/8

0

12(4 cos2 4θ) dθ = 4

∫ π/8

0

1 + cos 8θ2

dθ == 4[12θ +

116

sin 8θ]π/80

=π

4

32. A =∫ π/2

0

12[sin2 θ − (1 − cos θ)2] dθ =

12

∫ π/2

0

(2 cos θ − 1 − cos 2θ) dθ

= 12

[2 sin θ − θ − 1

2 sin 2θ]π/20

= 1 − π4




33.∫ π

4

0

12(2 sin θ)2dθ +

∫ 34π

π4

12(sin θ + cos θ)2dθ =

∫ π4

0

(1 − cos 2θ)dθ +∫ 3

4π

π4

12(1 + sin 2θ)dθ

=π

4− 1

2+

π

4=

π

2− 1

2

34. Since x2 =1t2

, we have

y = t2 + 1 =1x2

+ 1 =x2 + 1x2

.

-2 -1 1 2x

1

5

10

y

35. y = cos 2t = 1 − 2 sin2 θ

y = 1 − 2x2, 0 ≤ x ≤ 1.

1x

-1

1

y

36. cosh2 t− sinh2 = 1 =⇒ x2 − y2 = 1, graph: the right branch of the hyperbola since x > 0.

37. y = x2 − 2x + 1 = (x− 1)2, x ≥ 1, graph: the right half of the parabola.

38. Since y = t2 + 4t + 8 = (t + 2)2 + 4,

y = x2 + 4, x ≤ 2.

2x

4

y

39. x(t) = 1 + (5 − 1)t = 1 + 4t, y(t) = 4 + (6 − 4)t = 4 + 2t, t ∈ [0, 1].

40. x(t) = 2 + (−2 − 2)t = 2 − 4t, y(t) = −1 + (3 − [−1])t = −1 + 4t, t ∈ [0, 1].

41. x(t) = 3 cos(−t + π2 ) = 3 sin t, y(t) = 2 sin(−t + π

2 ) = 2 cos t.

42. x(t) = 3t− 1, x′(t) = 3; y(t) = 9t2 − 3t, y′(t) = 18t− 3.

At t = 1 : x(1) = 2, y(1) = 6, x′(1) = 3, y′(1) = 15; tangent line: y − 6 = 5(x− 2) or y = 5x− 4.

43. x(t) = 3et, x′(t) = 3et; y(t) = 5e−t, y′(t) = −5e−t.

At t = 0 : x(0) = 3, y(0) = 5, x′(0) = 3, y′(0) = −5; tangent line: y − 5 = − 53 (x− 3) or

5x + 3y = 30.




44. x(θ) = r cos θ = 2 sin 2θ cos θ, y(θ) = r sin θ = 2 sin 2θ sin θ;

x′(θ) = 4 cos 2θ cos θ − 2 sin 2θ sin θ, y′(θ) = 4 cos 2θ sin θ + 2 sin 2θ cos θ;

at θ = π/4 : x(π/4) =√

2, x′(π/4) = −√

2; y(π/4) =√

2, y′(π/4) =√

2

tangent line: y −√

2 =√

2−√

2(x−

√2) = −(x−

√2) or x + y = 2

√2.

45. (a) For a horizontal tangent, set y′(t) = 0 : 3t2 + 3t− 6 = 0 =⇒ t = −2, 1.

The curve has a horizontal tangent at: (x(−2), y(−2)) = (0, 10) and (x(1), y(1)) = (3,−7/2).

(b) For a vertical tangent, set x′(t) = 0 (provided y′(t) �= 0 simultaneously):

x′(t) = 2t + 2 =⇒ t = −1.

Since y′(−1) �= 0, the curve has a vertical tangent at (x(−1), y(−1)) = (−1, 13/2).

46. (a) For a horizontal tangent, set y′(t) = 0 : 2 cos t− 2 cos 2t = (2 cos t + 1)(cos t− 1) = 0.

The curve has a horizontal tangent at the points where t = 2π3 + 2nπ, −2π

3 + 2nπ, 2nπ,

n = 0,±1,±2, · · ·.(b) for a vertical tangent, set x′(t) = 0 : −2 sin t + 2 sin 2t = −2 sin t(1 − 2 cos t) = 0.

the curve has a vertical tangent at the points where t = ±π3 + 2nπ, nπ, n = 0,±1,±2, · · ·.

47. x = 3t2, x′ = 6t, x′′ = 6; y = 4t3, y′ = 12t2, y′′ = 24t

dy

dx=

12t2

6t= 2t;

d2y

dx2=

6t(24t) − 12t2(6)(6t)3

=13t

48. Setdx

dt= a, and

d2y

dt2= b. Then x(t) = at + C1 and y(t) = 1

2bt2 + C2t + C3 where C1, C2, C3

are arbitrary constants. In rectangular coordinates, the equation will have the form

y = Ax2 + Bx + C, a parabola

49. L =∫ 5/9

0

√1 + (y′(x))2dx =

∫ 5/9

0

√1 +

94xdx =

1927

.

50. L =∫ 1/2

0

√1 + [y′(x)]2 dx =

∫ 1/2

0

√1 +

4x2

(1 − x2)2dx =

∫ 1/2

0

1 + x2

1 − x2dx

=∫ 1/2

0

[2

1 − x2− 1

]dx =

[ln

(1 + x)(1 − x)

− x]1/20

= ln 3 − 12

51. L =∫ π

2

0

√sin2 t + 4 sin2 t cos2 t dt =

∫ π2

0

2 sin t

√14

+ cos2 t dt

= 2∫ 1

0

√14

+ u2 du (substitution u = cos t)

= 2

[u

2

√14

+ u2 +18

ln

(u +

√14

+ u2

)]1

0

=√

52

+14

ln(√

5 + 2)




52. L =∫ 4

0

√t2 + 6t + 9 dt =

∫ 4

0

(t + 3) dt =[12t2 + 3t

]4

0

= 20.

53. By symmetry L =∫ 2π

0

√(1 − sin θ)2 + cos2 θ dθ = 2

∫ π

0

√(1 − sin θ)2 + cos2 θ dθ.

L =∫ 2π

0

√(1 − sin θ)2 + cos2 θdθ = 2

∫ π

0

√2(1 − sin θ) dθ

= 2∫ π

0

√4 cos2(θ/2) dθ = 4

∫ π

0

cos(θ/2) dθ = 8[sin (θ/2)

]π0

= 8

54. L =∫ √

5

0

√θ4 + 4θ2 dθ =

∫ √5

0

θ√θ2 + 4 dθ =

13

[(θ2 + 4)3/2

]√5

0=

193

55.∫ 24

0

2π(2√x)

√1 +

1xdx = 4π

∫ 24

0

√1 + x dx =

8π3

[(1 + x)3/2

]240

=992π

3

56. Let u =√t + 1. Then

∫ 8

3

2πt√t + 1 dt =

∫ 3

2

2π(u2 − 1)2u2 du = 4π∫ 3

2

(u4 − u2)du =215215

π.

57. y =x3

6+

12x∫ 3

1

2π(x3

6+

12x

)√1 +

(x2

2− 1

2x2

)2

dx = 2π∫ 3

1

(x3

6+

12x

)(x2

2+

12x2

)dx

= 2π∫ 3

1

(x5

12+

x

4+

x

12+

14x3

)dx =

2497π108

58. 3x2 + 4y2 = 3a2 =⇒ y =√

32

√a2 − x2

A = 2∫ a

0

2π ·√

32

√a2 − x2

√1 +

34

x2

a2 − x2dx =

√3π

∫ a

0

√4a2 − x2 dx

=√

3π[12x√

4a2 − x2 + 2a2 arctanx√

4a2 − x2

]a0

=√

33

π2a2 +32πa2

59. Assume that f ′(x) > 0. Then (f−1)′(y) =1

f ′(x), where y = f(x), dy = f ′(x) dx.

The length of the graph of f−1 is given by:

∫ f(b)

f(a)

√1 + [(f−1)′(y)]2 dy =

∫ f(b)

f(a)

√1 + 1/[f ′(x)]2 dy

=∫ b

a

1f ′(x)

√(f ′(x))2 + 1 f ′(x) dx

=∫ b

a

√1 + (f ′(x))2 dx




60. Solve the differential equations

x′ = x y′ = 2y

with initial conditions x(0) = 4, y(0) = 2. We get x(t) = 4et, y(t) = 2e2t.

Initial speed: ν0 = 16√

2; terminal speed: ν1 =√

16e2 + 16e4 = 4e√

1 + e2.

s =∫ 1

0

√16e2t + 16e4t dt = 4

∫ 1

0

et√

1 + e2t dt

= 2[e√

1 + e2 + ln(e +

√1 + e2

)− 1 − ln

(1 +

√2)]

61. Ifx

a= cos3 θ and

y

a= sin3 θ, then

(xa

)2/3

+(ya

)2/3

= cos2 θ + sin2 θ = 1 and x2/3 + y2/3 = a2/3.

62. Assume a ≥ 0. By symmetry,

L = 4∫ π/2

0

√(3a cos2 θ sin θ)2 + (3a sin2 θ cos θ)2 dθ

= 4∫ π/2

0

3a cos θ sin θ dθ = 12a[12

sin2 θ

]π/20

= 6a

63. By symmetry, x = y.

xL

4=∫ π/2

0

a cos3 θ(3a cos θ sin θ) dθ = −3a2

5

[cos5 θ

]π2

0=

3a2

5

Therefore, (x, y) =(

2a5,2a5

)

64. (a) The speed μ satisfies

|μ|2 = (x′(t))2 + (y′(t))2

= (4 − π cosπt)2 + (4 − π sinπt)2 = 32 + π − 8π(cosπt + sinπt)

= 32 + π − 8π√

2 cos(πt− π

4)

The particle has minimum speed√

32 + π − 8√

2π at t = 1/4;

the particle has maximum speed√

32 + 9π at t = 1.

(b)y′(1/4)x′(1/4)

= 1.

calculus one and several variables 10e salas solutions manual ch10

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