421

(1)

70. Continued Multiply by x.

n n x x

x n n

( )

+ =

−

= ∞

∑

1 2

1 3

1

Replace x by 1

x. n n

x

x x

n n

( )

( ) ,

+ = − ⎛ ⎝⎜

⎞ ⎠⎟

=

− >

= ∞

∑

1

2

1 1 2

1 1

1 3

2

3

(b) Solvex x to get for

x x x

=

− ≈ >

2

1 2 769 1

2

3

( ) . .

71. (a) Computing the coefficients,

f

f x x f

f x

( )

( ) ( ) , ( )

( ) (

1 1

2

1 1 1

4 2

2

=

′ = − + ′ = −

′′ =

− _so

xx f

f x x

+ ′′ =

′′′ = − +

− −

1 1

2 1 8

6 1

3

4

) , ( )

!

( ) ( ) ,

so

so ′′′′ = − = −₊

f

f n

n n

n

( ) ! ,

! ( )

.

( )

1 3

1 16 1

2 1 In general

Soof x x x n x

n n

( )= − − +( − ) + + −( ) ( − ) +

+

1 2

1 4

1

8 1

1 2

2

1

(b)Ratio test for absolute convergence: lim

n

n n

n n x

x

→∞ + +

+

−

− =

− 1

2

2 1

1 2

1

2

1

i

x

−

< ⇒ − < < 1

2 1 1 3.

The series converges absolutely onn (−1 3, ). At the series is

which diverges by th

x

n

= −

= ∞

∑

1 1

2

0

, ,

ee th-term testn . At the series is

which diverges

x n

n

= −

= ∞

∑

3 1 1

2

0

, ( ) ,

b

by the th-term testn . The interval of convergence is (−1, 3).

(c) P x₃ x x

2

1 2

1 4

1 8 ( )= − − +( − )

P₃

2

0 5 1

2 0 5 1

4

0 5 1

8 0 65265

( . )= − . − +( . − ) = . 72. (a) Ratio test for absolute convergence:

lim( ) lim

n

n n

n

n _n

n x

n x

n n

x x

x

→∞

+

+ _→∞

+

= + =

1 2

2 1

2 2

2

1

1 i

<< ⇒ − < <1 2 x 2

The series converges absolutely on (−2, 2).

The series diverges at both endpoints by the nth-term

test, since lim lim ( ) .

n n

n

n n

→∞ ≠0and →∞−1 ≠0

The interval of convergence is (−2, 2).

(b)The series converges at −1 and forms an alternating

series:− + − +1 + − +

2 2 4

3 8

4

16 ( 1) 2 .

n n n

The nth-term of this series decreases in absolute value to 0, so the truncation error after 9 terms is less than the absolute value of the10thterm. Thus error < 10 <

210 0 01. . 73. (a) P x₁( )= − +1 2x

(b) P x₂ 1 2x 3x2

2 ( )= − + −

(c) P x₃ 1 2x 3x2 x3

2 2 3 ( )= − + − +

(d) P₃ 0 7 1 2 0 7 3 2 3

2 0 7 2 3 0 7 ( . )= − + ( . )− ( . ) + ( . )

Chapter 10 Parametric, Vector, and Polar

Functions

Section 10.1

Parametric Functions

(pp. 531–537)

Exploration 1 Investigating Cycloids

1.

[0, 20] by [–1, 8]

2. x=2naπfor any integer n.

3. a>0and1−cost≥0soy≥0.

4. An arch is produced by one complete turn of the wheel. Thus, they are congruent.

5. The maximum value of yis 2a and occurs when

x=(2n+1)aπ for any integer n.

6. The function represented by the cycloid is periodic with period 2aπ, and each arch represents one period of the graph. In each arch, the graph is concave down, has an absolute maximum of 2a at the midpoint, and an absolute minimum of 0 at the two endpoints.

Quick Review 10.1 1. t x

y t x x

= −

= + = − + = +

1

2 3 2( 1) 3 2 1

2. t x

y t x x

=

= − = ⎛

⎝⎜ ⎞

⎠⎟ − = − 3

54 3 54

3 3 2 3

3

(2)

3. x t

y t

t t

x y

2 2

1 1 = =

+ =

+ = sin cos

cos sin

4. y t t t

y x

= =

=

sin2 2sin cos 2

5. x y

x

y x

2 2

2

1 1 = =

= + = +

tan sec

sec tan

θ θ θ

6. x y

x y

2 2 2

2 2

1 1

= = +

csc cot

cot

θ θ

θ

7. x y

y x

2 2

2

2 2 1

2 1

=

= = −

= −

cos

cos cos

θ

θ θ

8. x y

y x

2 2

2 1 2

1 2 =

= = −

= − sin

cos sin

θ

θ θ

9. x y

y x

2 2

2 2 2

2

1

1 0

=

= = −

= − ≤ ≤

cos

sin cos

, ( )

θ

θ θ

θ π

10. x y

y x

2 2

2 2 2

2

1

1 2

=

= = −

= − − ≤ ≤

cos

sin cos

, ( )

θ

θ θ

π θ π

Section 10.1 Exercises 1. Yes, y is a function of x.

y=2x−9

(3, –3)

(9, 9)

1 10

10

1

–3

0 x

y

2. Yes, y is a function of x. y= x +

2 ₇

4

(1, 2) (3, 4)

1 5

1

0 x

y

4

3. Yes, y is a function of x.

y= x2−1

(1, 0) (2, 1.732)

3 3

1

0 x

y

4. No, y is not a function of x. x2 y

2

4 1

+ =

(0, 2)

(0, –2)

3 1

–3 –1 x

y

5. Yes, y is a function of x. y= −1 2x2

(0, 1)

(1, -1) 2 2

–2 0

–2 x

(3)

6. No, y is not a function of x. x=sin (3y)

(0, 0) (0, 3.14159)

2 4

–2 x

y

7. (a) dy

dx

t

t t

= −2 = − 4

1 2 sin

cos tan

8. (a) dy

dx

t t

= −

− =

3

3 sin sin

(b) d y

dx d dt

t 2

2

3 0 =

( )

−sin = 9. (a) dy

dx t t

t

t t

=

− + = − + = − +

− −

3 3 1

9 9

3 3

3

1 2

( )

/

(b) d y

dx d

dt t

t t

2

2 1 2 3 2

3 3 1

3 =

− + ⎛ ⎝

⎜ ⎞_⎠⎟

− +₍ ₎−/ = − /

10. (a) dy

dx t

t t

=

− = − 1 1 2

/ /

(b)d y

dx d dt t

t t

2

2 2

2

1

= −

− =

( ) / 11. (a) dy

dx t t

= − 3

2 3

2

(b) d y

dx d dt

t t t

t t

t 2

2

3

2 3

6 18

2 3

= −

⎛ ⎝⎜

⎞ ⎠⎟

− =

− −

( )

12. (a) dy

dx t t

= − +

2 1

(b) d y

dx d dt

t t

2

2 3

2 1

2 3

4

2 1

= − + ⎛ ⎝⎜

⎞ ⎠⎟

− =( + )

13. (a)dy

dx

t t

=sec tan = sec2 sin

(b) d y

dx d d t t

t t

2

2 2

3

= sin =

sec cos

14. (a) dy

dx

t

t t

= −

− =

2 2

sin ( )

sin cos

(b) d y

dx d

d t t

t 2

2

2 1

=

− =

( cos ) sin

15. (a) dy

dx t

t

= ⎛ ⎝⎜ ⎞⎠⎟ ₌ 4 1

1/ 4

(b) d y

dx d dt

t 2

2

4

1 0

= ( )= / 16. (a) dy

dx t t

t t

=5 =

1 5

5 5

e e /

(b) d y

dx d dt t

t t t

t

t t

2

5

2 5 5

5

1 25 5

= ( )= +

/ e

e e

17. (a)

x y

1

(b) (0.5, –0.25)

(c) We seek to minimize y as a function of t, so we compute

dy/dt= 2t+ 1, which is negative for – 2≤ t < –0.5 and positive for –0.5 < t≤ 2. There is a relative minimum at t= –0.5, where (x, y) = (0.5, –0.25).

18. (a)

x y

1

(b) (–1, 6)

(c) We seek to minimize x as a function of t, so we compute

dx/dt=2t+ 2, which is negative for –2 ≤ t < –1 and positive for –1 < t ≤ 3. There is a relative minimum at

t= –1, where (x, y)= (–1, 6). (b) d y

dx

dy dt dx dt

d

dt t

t 2

2

1 2 4

1

= ′ =

− ⎛

⎝⎜ ⎞⎠⎟ ₌− /

/

tan cos

2 2 4

1 8

2

3

sec

cos sec

t

(4)

19. (a)

x y

1

(b) (2, 0)

(c) We seek to maximize x as a function of t, so we compute

dx/dt=2 cos t, which is positive for 0≤ <t π/ and2 negative for π/2< ≤t π.There is a relative maximum at

t=π/ ,2 where (x, y) = (2, 0). 20. (a)

x y

1 (b) (0, 2)

(c) We seek to minimize yas a function of t, so we compute

dy dt/ =2sec tant t,which is negative for − ≤ <1 t 0 and positive for 0< ≤t 1. There is a relative minimum at t=0, where (x, y) = (0, 2).

21. (a) y

x

(b) (0, 1)

(c) We seek to maximize y as a function of t, so we compute

dy dt/ = −2sin( ),2t which is positive for 1 5. ≤ <t π and negative for π < ≤t 4 5. . There is a relative maximum at t=π, where (x, y) = (0,.1).

22. (a)

x y

(b) ( (In50),In(400))≈(3.912, 5.991)

(c) We seek to maximize x as a function of t, so we compute

dx dt/ =1/ ,t which is positive for all t > 0. There is an endpoint maximum at t= 10, where ( , )x y =( (In50), In(400)).

23. (a)dy d

dt t

dy t

t t x

= − + = = = ±

= + ±

( sin ) cos

cos cos( )

1

0 1

2 1

xx y y

=

= − + ± = −

− 2

1 1

2 0

2 0 2 2

sin( ) ,

( , )and( , ) (b) dx d

dx t

t t x

= +

= − = − = ±

= + ±

( cos ) sin sin

cos( 2

0 1

2 1))

,

sin( )

( , ) ( , )

x y y

=

= − + ± = −

− −

1 3

1 1

1

1 1 and 3 1

24. (a)dy d

dt t

dy t

t

= = ≠

(tan ) sec sec

2 2

(5)

24. Continued (b) dx d

dt t

= (sec )

dx t t t

t t

t t x

= =

= = ±

= ±

sec tan sin cos sin

cos sec(

2

0 1

1 1 1 1

1 0

1 0 1 0

) , tan( ) ( , ) ( , )

x y y

= −

= ±

= − and 25. (a)dy d

dt t t

= (3−4)

dy t

t t

x x

y

= −

= ±

= ± =

= ±

3 4

0 3 4

2 3

2 2

3 0 845 3 155

2

2 2

. , .

3

3 4

2 3 3 079 3 079 0 845 3

3

⎛ ⎝⎜

⎞ ⎠⎟ −

± ⎛ ⎝⎜

⎞ ⎠⎟ = −

−

y . , .

( . , .. )

( . , . )

079 3 155 3 079

and

(b) dx d

dt t

= (2− )

dx= −

≠ − 1

0 1

Nowhere 26. (a)dy d

dt t

= (1 3+ sin )

dy t

t t x x y

= = = ±

= − + ±

= − = +

3

0 3

1

2 3 1

2 1 3

cos cos

cos( ) sinn( ) ,

( , ) ( , )

± = −

− − −

1 2 4

2 2 2 4

y

and (b) dx d

dt t

= (− +2 3cos )

dx t

t t

x x y

= − = − = = − + = − = +

3

0 3

0

2 3 0

5 1 1 3

sin sin

cos( ) ,

ssin( )

( , ) ,

0 1

1 1 5 1

y=

−

and ( )

27. S=

∫

( sin )− t2+(cos )t2dt 0

2π

S dt

S t

S

=

= = −

=

∫

1

2 0

2

0 2

π

π _π

π

28. S=

∫

( cos )3 t2+ −( 3sin )t2dt 0

π

S dt

S t

S

=

= = −

=

∫

9

3 3 0

3

0

π π

29.S t t t t

t

= − + +

+

∫

(( sin sin cos )

( cos

/

8 8 8

8

2 0

2

π

−−8cost+8tsin ) )t2 1 2/ dt

S t t t t dt

S t dt

= +

=

∫

/ (( cos ) ( sin ) )/

/

8 8

8

2 2 1 2

0 2

π π

∫∫

= = −

=

S t

S

42 0

0

2 ₂

2

π

π π

/

30. S= ((−18sin3tcos23t)2+(18sin23tcos ) )3t2 1 2/ 0

2ππ

∫

dt

S=12

31. S=

∫

(( 2t+3)2+ +(1 t) )2 / dt 0

3 1 2

S t t dt

S

= + +

=

∫

(2 )1 2/

0 3

4 4

21 2

32. S=

∫

(( 8t+8)2+(2t+1) )2 /dt 0

2 1 2

S t t dt

S

= + +

=

∫

(4 12 9)/ 10

2 1 2

0 2

33. S=

∫

(( )t2 2+( ) )t 2 1 2/dt 0

1

S t t dt

S

= +

= −

∫

( )

/

4 2

0

1 1 2

2 2 1 3

34. S

t t t t

=

+ −

⎛

⎝⎜ ⎞⎠⎟ + −

⎛ ⎝

⎜⎜ 1 ⎞_⎠⎟⎟

2 2

0 sec tan cos ( sin )

π//3

∫

dt

S t t t

t

= + +

+ ⎛

⎝

⎜sin (cos_(sin (sin₎ ) )⎞_⎠⎟

/ 2 2 2

2 0

3 1

1

π

∫∫

=

dt

S ln 2

35. (a) x′= −2sin2t y, ′=2cos2t, so Length=

∫

/ (−2sin2)2+( cos2 2)2

0 2

t t

π

=

∫

2 =

0 2

dt π

π

(6)

35. Continued

(b) x′ =πcosπt y, ′ = −πsinπt, so

Length= + −

−

∫

( cosπ πt)2 ( πsinπt dt)2

1 2 1 2

= =

−

∫

π dt π.

1 2 1 2

36. x′= −3sin ,t y′=4cos , sot

Length=

∫

(−3sin )2+(4 cos )2

0 2

t t dt

π

which using NINT evaluates to ≈ 22.103. 37. In the first integral, replace t with x. Then dx

dt becomes dx

dx=1.

38. Parameterize the curve as x=g y y( ), =y c, ≤ ≤y d. The parameter is y itself, so replace t with y in the general formula. Then dy

dt becomes dy dy=1.

39. dx

dt =a(1−cos )t

(Note: integrate with respect to x from 0 to 2aπ; integrate with respect to t from 0 to 2π.)

Area=

∫

a y dx 0

2 π

= − −

= − +

∫

a t a t dt

a t t

( cos ) ( cos ) cos cos

1 1

1 2

0 2

2 2

π

( ))

sin sin

dt

a t t t t a

0 2

2

0 2

2 2

1

4 2 3

π

∫

= ⎡ − + +

⎣⎢

⎤ ⎦⎥ =

2 2

40. dx

dt =a(1−cos ),t so

Volume=

∫

ππ⎡⎣a(1−cos )t ⎤⎦2a(1−cos )t dt 0

2

= − + −

= −

∫

π

a t t t dt

a t

3 2 3

0 2

3

1 3 3

3

( cos cos cos )

siin sin

sin sin

t t t

t t

+ + ⎡

⎣⎢

−⎛_⎝⎜ − ⎞_⎠⎟⎤ ⎦ ⎥ 3 2

3

4 2

1 3

3

0 2 2

2 3

5

π

= a

41. x t b t

y a b t a b

= −

= − < <

a sin and

cos (0 )

42. x t b t

y a b t a b a

= −

= − < <

a sin and

cos ( 2 )

43. S=

∫

((3 2− cos )t 2+( sin ) )2 t 2 1 2dt 0

2π

S t t t dt

S

= − + +

= −

∫

( cos cos sin )

( c

9 12 4 4

13 12

2 2 1 2

0 2π

o os ) .

t dt S

1 2 0

2

21 010

π

∫

=

44. S=

∫

((2 3− cos )t2+( sin ))3 t 1 2dt 0

2π

S t t t dt

S

= − + +

= −

∫

( cos cos sin )

(

4 12 9 9

13 12

2 2 1 2

0 2π

ccos ) .

t dt S

1 2 0

2

21 010

π

∫

=

45. False. Indeed, y may not even be a function of x. (See Example 1.)

46. True. The ordered pairs (x, f(x)) and (t, f(t)) are exactly the same.

47. B. 48. C. dy

dx

t t

= − = −

− < csc cot

cos sin

sin ,

1 1

4 0

2

2π

decreassing

d y dx

d

dt t

t

t t

t 2

2

2 2

1 =

− ⎛

⎝⎜ sin ⎞⎠⎟₌ cos

(csc cot )

cos =

1

4

sin t

1 4

0

4

sin ,

π > concave up

49. C. x=ln ( )t =ln ( )y y=ex

50. D.

51. (a) QP has length t, so P can be obtained by starting at Q

and moving t sin t units right and t cos t units downward.

(If either quantity is negative, the corresponding direction is reversed.) Since Q=(cos , sin ), thet t

coordinates of P are

x=cost+tsintandy=sint−tcos .t y

Q _t_{sin t}

t t

t tcos t

P(x, y)

(1,0) x

(b) x′ =tcos ,t y′ =tsin , sot

Length=

_∫

( cos )t t2+( sin )t t2dt=

_∫

t dt

0 2 0

2π π

=2π2

52. All distances are a times as big as before. (a) x=a(cott+tsin ),t y=a(sint−tcos )t

(7)

52. Continued

For exercises 35 38− ,x′ =v₀cosθ and y′ =v₀sinθ−32t, and y=0 t=0 t=v

16

0

for or sinθ. The maximum height is attained in mid-flight at t=v0

32 sin

. θ

To find the path length, evaluate vsin / (v₀cos )2 (v₀sin t dt)2

0 16

32

0

θ θ

θ

+ −

∫

using NINT. To find the maximum height, calculate

y_max=(v sin )⎛v sin v sin ⎝⎜

⎞ ⎠⎟−

⎛ ⎝⎜

⎞ ⎠

0 θ 0₃₂ 16 0₃₂

θ θ

⎟⎟

2

. 53. (a) The projectile hits the ground when y= 0.

y t t

t t

= ° − =

= = ° ≈

( sin )

sin .

150 20 16 0

0 75

8 20 3

or 2206

150 20 150 20 32

′ = ° ′ = ° −

x cos ,y sin t

Length=

° + ° −

( cos ) ( sin )

(

150 20 2 150 20 32 2

0 7

t dt 5

5sin20°)/8

∫

which, using NINT, evaluates to ≈ 461.749 ft (b) The maximum height of the projectile occurs when

′ =

y 0,

sot= ° y⎛ ° ft

⎝⎜ ⎞⎠⎟≈ 75

16 20

75

16 20 41 125

sin , sin .

54. (a) ≈ 641.236 ft (b) 5625

64 ≈87 891. ft 55. (a) ≈ 840.421 ft

(b) 16 875

64 263 672 ,

.

≈ ft

56. (a) It is not necessary to use NINT.

Length=

∫

(150 32− ) =⎡_⎣150 −162⎤_⎦

0 75 8 0

75 8

t dt t t

=5625=

8 703 125. ft (b) 5625

16 =351 5625.

57. S=

∫

2 y − t2+ t2 1 2dt

0 2

π

(( sin ) cos )/

S y dt

S yt

=

= = =

∫

2 1

2 2 2 2 8

0 2

0

2 2

π

π π π π

π π

( )

( )( ) 58. S=

∫

2 y t−1 2 2+ t1 2 2 1 2dt

0 2

π (( / ) ( / ) )/

S y

t t dt

S

= ⎛_⎝⎜ + ⎞_⎠⎟

= ≈

∫

2 1

2 2 1 25 14 214

0

2 1 2

π π

/

( )( . ) .

59. S=

∫

2 y 2t2+ 12 1 2dt 0

3

π (( ) ( ) )/

S y t dt

S

= +

=

∫

=

2 4 1

2 4 7 105 178 561

2 1 2

0 3

π π

( )

( )( . ) .

/

60. S y

t t t t

=

+ −

⎛

⎝⎜ ⎞⎠⎟ + −

⎛ ⎝

⎜⎜ ⎞_⎠⎟

2 1

2

π

sec tan cos ( sin ) ⎟⎟

∫

1 2

0 3

/ /

dt

π

S y t t

t t dt

=

+ +

⎛ ⎝

⎜ ⎞_⎠⎟

2

1

2 2

1 2

0 π

sin cos

(sin ) sin

/

ππ

π π

3

2 5 1

∫

= =

S (. )( )

Section 10.2

Vectors in the Plane

(pp. 538–547)

Quick Review 10.2 1. (5 1− )2+ −(3 2)2= 17 2. 3 2

5 1 1 4 − − =

3. Solve 3 5 3

1 4

5 2 −

− = =

b b

; .

4. Slope of PQ RQ

= − 1 , so 3 2 5 1

5 3

3 11

−

− = − −−b andb= .

5. Slope of AB=Slope of CD, so 3 0 1 0

3 0 5 −

− = −−a and a=4.

6. Slope of AB₌Slope of CD, so 5 1 3 1

2 6 8 −

− = −− =

b

and b=6.

7. v t d dt t t

( )= ( sin )

v t t t t

a t d

dt t t t

a t

( ) sin cos ( ) (sin cos ) ( ) co

= +

=2 sst−tsint

8. x t( )=

∫

v t dt( ) =

∫

(3t2−12t dt)

x t t t C

x C

C X

( )

( ) ( )

( )

= − +

= − + =

= = −

3 2

3 0

3

6

0 0 6 0 40

40

4 4 66 4( )2+40=8 9. x t( ) =

∫

v t dt( ) =

∫

(3t2−12t dt)

0 4 0

4

x t( )=⎡_⎣t3− t2⎤_{⎦ =} 0 4

6 32

10. ( cos2 2)2 ( 3sin )3 2

0 2

t + − t dt

∫

π

= +

=

∫

4 2 9 3

15 289

2 2

0 2

cos sin

.

t t dt

(8)

Section 10.2 Exercises 1. (2, 3) (0, 0)− = 2, 3 2. (0, 0) (2, 3)− = − −2, 3 3. (2, 1) (1, 3)− − = 1,−4

4. P= − +4 2 − = − 2

3 1

2 1 1

, ( , )

(−1 −, ) (0, 0)1 = −1,1 5. v= 22+22= 8

tanθ= 2,θ= °

2 45

6. v= − 22+ 22=2 tanθ= − 2, θ= °

2 135

7. v= 32+ =12 2 tanθ= 1 ,θ= °

3 30

8. v= − + −22 ( 2 3)2=4 tanθ= −2 3,θ= °

2 240

9. v= − +52 02=5 tanθ= , θ

− = °

0

5 180

10. v= 02+42=4 cosθ=0,θ= °

4 90

11. x=4cos(180)= −4

y=4sin(180)=0 −4 0,

12. x=6cos(270)=0

y=6sin(270)= −6 0,−6

13. x=5cos(100)= −0 868.

y=5sin(100)=4 924. −0 868 4 924. , .

14. x=13cos(200)= −12 216.

y=13sin(200)= −4 446. −12 216. ,−4 446.

15. x= ⎛

⎝⎜ ⎞⎠⎟=

3 2 180

4 3

cos π

π

y= ⎛

⎝⎜ ⎞⎠⎟=

3 2 180

4 3

sin π

π

3 3,

16. x= ⎛

⎝⎜ ⎞ ⎠⎟=

2 3 180

6 3

cos π

π

y= ⎛

⎝⎜ ⎞⎠⎟=

2 3 180

6 3

sin π

π

3, 3

17. (a) 3 3 3 2( ), (− ) = 9, −6 (b) 92+ −( 6)2= 117=3 13 18. (a) − − −2( 2), 2 5( ) = 4,−10

(b) 42+ −( 10)2= 116=2 29 19. (a) 3+ − − + =( 2), 2 5 1 3,

(b) 12+32= 10

20. (a) 3− − − − =( 2), 2 5 5,−7 (b) 52+ −( 7)2= 74 21. (a) 2u= 2 3 2

( )

,

( )

−2 = 6,−4

3 3 v

u v

= − = −

− = − − − − = −

3 2 3 5 6 15

2 6 6 4 15 12 1

( ), ( ) ,

( ), , 99

(b) 122+ −( 19)2= 505 22. (a) −2u= −2 3( ),− −2( 2) = −6 4,

5 5 2 5 5 10 25

2 5 6 10 4 25 1

v

u v

= − = −

− + = − + − + = −

( ), ( ) ,

( ), 66 29,

(b) (−16)2+292= 1097

23. (a) 3 5

3 5 3

3

5 2

9 5

6 5 u= ( ), (− ) = ,− 4

5 4

5 2

4 5 5

8 5 4 3

5 4 5

9 5

8 5 v

u v

= − = −

+ = + −⎛ ⎝⎜ ⎞⎠⎟

( ), ( ) ,

,−− +6 =

5 4

1 5

14 5 ,

(b) 1 5

14 5

197 5

2 2

⎛ ⎝⎜

⎞ ⎠⎟ + ⎛⎝⎜

(9)

24. (a) − 5 = − − − = − 13

5 13 3

5 13 2

15 13

10 13

u ( ), ( ) ,

12 13

12 13 2

12 13 5

24 13

60 13 5

13 12 13 v

u

= − = −

− +

( ), ( ) ,

v

v= − + −⎛

⎝⎜ ⎞⎠⎟ + = − 15

13 24 13

10 13

60

13 3

70 13

, ,

(b) (− ) + ⎛ ⎝⎜

⎞ ⎠⎟ =

3 70

13

6421 13

2 2

25.Initial velocity is 70° north of east:

325 cos70°, sin70° ≈ 111 157 305 400. , . . Wind velocity is 130° north of east:

40 cos130°, sin130° ≈ −25 712 30 642. , . . Add the two vectors to get ≈ 85 445 336 042. , . . The speed is the magnitude, ≈346 735. mph. The direction is tan .

. .

− ⎛

⎝⎜

⎞

⎠⎟≈ °

1 336 042

85 445 75 734 north of east, or ≈14 266. ° east of north.

26. x=4cos135= −2 2

y=4sin135=2 2

The true velocity is 2−2 2 2 2, , so the true angle is

θ = ⎛ −

⎝

⎜ ⎞_⎠⎟ = °

−

cos1 2 2 2 .

4 106 3 and the true speed is (2 2 2− )2+(2 2)2=2 95. mph

27. v dr t dt

d

dt t t t t

= ( )= 32, 23 = 6,62

a dv t dt

d

dt t t t

= ( )= 6 6, 2 = 6 12,

28. v dr t dt

d

dt t t t t

= ( )= sin2 2, cos = 2cos2,−2sin

a dv t dt

d

dt t t t t

= ( )= 2cos2,−2sin = −4sin2,−2cos

29. v dr t dt

d

dt t t

t t t t t

= ( )= _e−_,_e− = _e− − _e−_,−_e−

a dv t dt

d

dt t t

t t t t t t

= ( )= _e− − _e−_,−_e− = −₂_e + _e−_, _e−

30. v dr t dt

d

dt t t t t

= ( )= 2cos , sin3 2 4 = −6sin , cos3 8 4

a dv t dt

d

dt t t t

= ( )= −6sin , cos3 8 4 = −18cos ,3 −32sin44t

31. v dr t dt

d

dt t t t t

t t t

= = + −

= +

( )

sin , cos

cos ,

2 ₂ 2 ₂

2 2 2 2 ++2sin2t a dv t

dt d

dt t t t t

t

= = + +

= − ( )

cos , sin

sin ,

2 2 2 2 2 2

2 4 2 2++4cos2t

32. v dr t dt

d

dt t t t t t t t t t

= ( )= sin , cos = sin + cos , cos − siint

a dv t dt

d

dt t t t t t t

t t

= = + −

= −

( )

sin cos , cos sin

cos s

2 iin ,t −2sint−tcost

33. v dr t dt

d

dt t t t t

= ( )= cos , sin3 2 = −3sin , cos3 2 2

a dv t dt

d

dt t t t t

= ( )= −3sin ,3 2cos2 = −9cos ,3 −4sin2

[–1.6, 1.6] by [–1.1, 1.1] 0≤t≤ 2p

34. v dr t dt

d

dt t t t t

= ( )= sin4, cos3 = 4cos4,−3sin3

a dv t dt

d

dt t t t

= ( )= 4cos4,−3sin3 = −16sin4,−9cos3tt

[–1.6, 1.6] by [–1.1, 1.1] 0≤t≤ 2p

35. (a) v t d

dt t t t

( )= sin4 cos , sin2

= −

=

4 4 4 2 2

2 2 0

4

cos cos sin sin , cos ,

:

t t t t t

t π

Speed 22 2 (b)

[–4, 4] by [–1.2, 1.2] 0≤t≤ 2p

(c) To the right 36. (a) v t d

dt

t t t t

(10)

36. Continued (b) lim /

/ lim lim

t t

t t _t

dy dt dx dt

→∞ →∞

− − _→∞

= +

− =

e e

e22

2

1 1 1

t t

+ − = e (c) For any t,

x y t t t t

t t t

2 2 2 2

2 ₂ 2 2 ₂

+ = + − −

= + + − −

− −

−

( ) ( )

(

e e e e

e e e ++

=

−

e 2 4

t₎

(d) The velocity at t=0 is 0 2, .

[–9, 9] by [–6, 6] 0≤t≤ 3

37. (a) 32 2 1

0 3

t − t dt + t dt

∫

,

∫

cosπ

= − +

= + + =

t3 t2 t t

0 3

1

2 18 6 3 20 9

, sin

, ,

π π

(b) (32 2)2 (1 cos )2

0 3

t − t + + t dt

∫

π

≈19 343.

38. (a) 2 4 4 2

0 3

πcos πtdt, πsin πtdt

∫

= −

= + + =

1

2 4 2 2

7 0 2 0 7 2

0 3

sin , cos

, ,

πt πt

(b) (2 cos4 )2 (4 sin2 )2

0 3

π πt + π πt dt

∫

≈28 523.

39. (a)

∫

(t+1)−1dt,

∫

(t+2)− dt 0

3 ₂

0 3

= + − +

= + −

−

ln ( ) , ( )

ln , .

t 1 t 2

3 4 1 7

0 3

1 0 3

(b)

∫

((t+1) )−1 2+ +((t 2) )−2 2dt 0

3

≈1 419.

40. (a) e e

0 3

0 t₋_{t dt} t₊_{t dt}

∫

,

∫

3

= − +

= + + =

et t et t

2

0 3

2

0 3

2 2

1 14 586 1 23 586 15 586 ,

. , . . ,,24 586.

(b)

∫

(et−t)2+(et+t dt)2

0 3

≈27 791.

41. The parametric equations are x= − +t3 t2 2 and

y= +t sin(π πt) +6.

[0, 23] by [5, 10] 0≤t≤ 3

42. The parametric equations are x=1 t +

2sin(4π ) 7 and

y= −2cos(2πt)+4.

(Note: The particle traverses the Figure-8 three times, finishing where it started.)

[6, 8] by [1.5, 6.5] 0≤t≤ 3

43. (a) v t d

dt t t

( )= 5cos , sin

6 3 6

π π

v t t t

v t t

( ) sin , cos

( ) sin

= −

= ⎛− ⎝⎜

5

6 6

1

2 6

5

6 6

π π π π

π π ⎞⎞ ⎠⎟ + ⎛⎝⎜

⎞ ⎠⎟

= ≈

2 2

1

2 6

7 12 2 399

π π

π

cos

( ) .

t

v t

(b) a t d

dt t t

( )= −5 sin , cos

6 6

1

2 6

π π π π

a t t t

a t

t

( ) cos , sin

( )

= − −

= −

=

5

72 6 12 6

5 72

2 2

π π π π

π _,,_{− π}2 ₃

24

(c) x2 t

2

5 6 = ⎛_⎝⎜ cosπ ⎞_⎠⎟

y t

x y

2

2 2

3 6

25 9 1

= ⎛ ⎝⎜ ⎞⎠⎟

+ =

sinπ

44. (a) v t d

dt t t

( )= 〈secπ , tanπ 〉

v t t t t

v t

t

( ) sec tan , sec

( ) ,

/

= =

=

π π π π π

π π

2 1 4

2 2

(11)

44. Continued (b) x2=(secπt)2

y t

x y

2 2

2 2 ₁

= − =

(tanπ )

(c) The upper part of the right branch:

[–1, 3.7] by [–1, 3.1] 0≤t< 1/2

45. (a) v t d dt

t t

( )= − ,

+ +

1 1

2 1

2

2 2

v t t

t

t t

( )

( ) , ( )

= − +

− + 4 1

2 2 1

2 2

2

2 2

(b) No. The x-component of velocity is zero only if t= 0, while the y-component of velocity is zero only if t= 1. At no time will the velocity be 0 0, .

(c) lim , ,

t t t

t t

→∞

−

+ + = −

1 1

2

1 1 0

2

2 2

46. (a) v t d

dt t t

( )= sin , cos2

v t( )= cos ,t −2sin2t

(b) v t( )= 0 0, t= t=

2

3 2 and

where π π

(c) cos2t= −1 2sin2t

y= −1 2x2

(d)

[–2, 2] by [–1.5, 1.5] 0≤t≤ 2π

47. (a) d

dt t t

t t

e sin ,e cos

= 〈 + − 〉

= −

e e e e

e e

t t t t

t t

t t t t

m t

sin cos , cos sin

cos siin

sin cos

t

t t

t

e +e _=π = −

2

1

(b) 〈e1sin ( )1+e1cos ( ),1 e1cos( )1 −e1sin ( )1〉

= 〈 − 〉

( )

= + − =

3 756 0 817

3 7562 0 8172 3 8

, , .

( , ) ( . ) .

v t 444

(c) ( sin cos ) ( cos sin )

.

et t+et t + et t−et t dt

≈

∫

2 2

0 1

2 4300

48. (a) d

dt〈 −t t 〉

2 3

3 2 3 ,

= 〈 〉

= +

= ≈

2 2

2 4 2 4

1088 32 985

2

2 2 2

t t

v v

,

( ( )) ( ( ) ) .

(b) (( ) ( ) ) .

/

2 2

46 062

2 2 2 1 2

0 4

t + t dt

≈

∫

(c) t x

y x

dy

dx x

= +

= + 3 2 3 3

3

3 2

1 2

( )

/

49. (a) .

3 2

3 2 3 942

2 2

4

2 2 4

+ +

= + − ≈

∫

( sin( ))

cos( ) .

t dt

t t

(b) y− =y₁ m x( −x₁)

y− = − x

+ −

5 6

2 sin4( 3)

(c)Speed = (2+sin )42+ −( 6)2≈6 127.

(d) _〈 ₊ ₊ _〉

≈ 〈− −

8 16 2 2 16 7 8 16

7 661 50

cos , ( sin ) ( ) cos . , .2205〉

50. (a) y− =y₁ m x( −x₁)

y− =5 3 4 x−

8 4

cos

sin ( )

(b) Speed = ( cos )3 42+(sin )82≈2 196.

(c) ( cos( )) (sin( )) .

3 2 741

2 2 3 2

0 1

t + t dt

≈

∫

(d) 4 3 5 3 2

2 3 2

3

+ +

⎛

⎝⎜

∫

sin( )t dt,

∫

cos( )t dt⎞⎠⎟

=⎛_⎝⎜ + − + ⎞_⎠⎟

≈

4 3 5 3

4 004

2

3 ₂

2 3

( cos( )) , cos( ) ( . ,

t

5 5 724. )

51. False. For example, uand –1(u) have opposite directions. 52. False. For example, 〈 3 0, 〉 + 〈0 1, 〉 = 〈 3 1, 〉, which has a

direction angle of 30°. 53. E. d

dt〈 +t t+ 〉 = t t+

2 ₁ ₂ ₃ ₂ 1

2 3

, ln( ) ,

d

dt 2t t t

1

2 3 2

4 2 32

, ,

( )

+ = − +

54. D. 〈 +4

∫

2 7+

∫

3 〉

0 2 0

2

cos( )t dt, sin( )t dt

〈 + + 〉

= 〈 〉

4 0 461 7 0 452 4 461 7 452

. , .

(12)

55. B. x y x 1 1 2

7 40 5 36

7 40 4 50

4 140 3 0

= =

= = −

cos .

sin .

cos . 66

4 140 2 57

5 36 3 06 4 50 2 57

2

2 2

y = =

− + +

=

sin .

( . . ) ( . . )

7 7 435. 56. B. dx

dt =2cos2t

dy

dt = −5sin5t

Speed = ( cos )2 42+ −( 5sin10)2=3 018.

57. The velocity vector is −x, 1−x2 , which has slope

− 1−x2

x . The acceleration vector is d

dt x d

dt x

dx dt

x x

dx dt

(− ),

( )

− = − , −

−

1 2

1

2

= −

x x

x

, ,

2

1

which has slope x

x

1− 2

. Since the slopes are negative reciprocals of each other, the vectors are orthogonal.

58. The position vector is 〈cos , sin ,t t〉 which has slope tan t. The velocity vector is 〈−sin , cos ,t t〉 which has slope

− 1

tant. The acceleration vector is〈−cos ,t −sint〉,which

has slope tan t.

The velocity slope is the negative reciprocal of the position and acceleration slopes, so velocity is orthogonal to position and to acceleration.

59. (a) t t

t t

t

− = −

− = − =

3 3

2 4

3

2 1

2

Since t= 2 also solves (t−3) =3t− ,

2 2

2 _{the particles}

collide when t= 2.

(b)First particle: v₁( )2 =〈1,−2〉, so the direction unit vector is 1

5 2

5

,− .

Second particle: v₂ 3 2

3 2

( )t = , , so the direction unit

vector is 1 2

1 2

, .

60. (a)Referring to the figure, look at the circular arc from the point where t= 0 to the point “m”. On the one hand, this arc has length given by r₀θ, but it also has length given by vt. Setting these two quantities equal gives the result.

(b) v( )t vsinvt, cos

r v

vt r

= −

0 0

and

a( )t v cos , sin cos

r vt r

v r

vt r

v r

= − 2 − = −

0 0

2

0 0

2

0 2

0

,, sinvt

r₀

(c)From part (b), a( )t v r( ).

r t

= −⎛ ⎝⎜

⎞ ⎠⎟

0 2

So, by Newton’s

second law,F= − ⎛ r ⎝⎜

⎞ ⎠⎟

m v

r₀ 2

. Substituting for F in the law of gravitation gives the result.

(d)Set vt

r₀=2π and solve for vT.

(e)Subsitute 2πr0

T for v in v r 2

0

=GM and solve for T2.

2

4 1

4

0 2

2 0

2 2

0 3

π

r T

GM r r T

GM r

T GM

r

T

⎛ ⎝⎜

⎞ ⎠⎟ =

=

2

2 2

0 3

4 = π

GMr

61. Let u = a b, be one of the vectors. It has slope b

a, so the

perpendicular vector v must have slope −a

b. Thus

v= kb,−ka for some nonzero scalar k, and the dot product is uiv= a b, , kb,−ka =kab+ −

(

kab

)

=0. 62. (a)The diagram shows, by vector addition, that v + w = u,

so w = u– v.

(b) This is just the Law of Cosines applied to the triangle, the sides of which are the magnitudes of the vectors. (c)By the HMT Rule, w=〈u₁−v₁,u₂−v₂〉.So

u2 v2 w2 ₁2 ₂2 ₁2 ₂2

1 1

2

+ − = + + +

− − +

( ) ( )

[( ) (

u u v v

u v u₂₂ ₂ 2

1 2

2 2

1 2

2 2

1 2

1 1 1

2 2

2

−

= + + +

− − + + −

v

u u v v

u u v v u

) ]

2 2

2

2 2 2

2

1 1 2 2

u v v

u v u v

+ ⎡

⎣ ⎤⎦

= +

= ( + )

(d) From part (b), w2=u2+v2−2u vcos ,θ so u2+v2−w2=2u vcosθ.

From part (c), u2+v2−w2=2

(

u v_{1 1}+u v_{2 2}

)

. Substituting, we get 2

(

u v_{1 1}+u v_{2 2}

)

=2u vcos ,θ

(13)

Section 10.3

Polar Functions (pp. 548–559)

Exploration 1 Graphing Polar Curves

Parametrically

See text page 551. Quick Review 10.3 1. x=4cos30=2 3

y=4 sin30=2 〈2 3 2, 〉

2. A=πr2 30 =π 2 = π

360 6

30

360 3

( )

3. A= r ⎛

⎝⎜ ⎞⎠⎟= =

π π

π π π

2 2

8 1 2

1

16 ( )8 4 4. x2+y2=25

5. Graph y=⎛ −x

⎝ ⎜4 ₃ ⎞_⎠⎟

2 1 2/

and y= −⎛ −x

⎝ ⎜4 ₃ ⎞_⎠⎟

2 1 2/

6. dy

dx dy dt dx dt

d dy t

d dx t

= / = =

/

/ ( sin ) / ( cos )

cos 5

3

5 tt

t t

−3 = − 5 3

sin cot

7. −5 = 3cot( )2 0 763. 8. x=3cost=0

t

y t

y y

= =

= ⎛

⎝⎜ ⎞⎠⎟= ±

( )

= ±

− π

π 2 5 5

2 5 1

5

0 5

sin sin

( , ) annd ( , )0 5 9. y t

t

x t

x x

= =

= = ±

= ± −

5 0

0 3

3 0 3 1

3 3 0

sin cos

cos( ) ( ) ( , ))and 3 0( , )

10. ( cos )3 2 ( sin )5 2 12 763.

0 t + t dt≈

∫

π

Section 10.3 Exercises 1.

[–3, 3] by [–2, 2]

(a) x= ⎛

⎝⎜ ⎞⎠⎟= 2 cos

4 1

π

y= ⎛

⎝⎜ ⎞ ⎠⎟=

( )

2 sin

4 1

1, 1 π

(b) x=1 cos (0)=1

y=1 sin (0)=0 (1, 0)

(c) x= ⎛

⎝⎜ ⎞⎠⎟= 0 cos

2 0w

π

y= ⎛

⎝⎜ ⎞⎠⎟=

( )

0 sin

2 0

0, 0 π

(d) x= − ⎛

⎝⎜ ⎞⎠⎟= − 2 cos

4 1

π

y= − ⎛

⎝⎜ ⎞⎠⎟= − − −

(

)

2 sin

4 1

1, 1 π

2.

[–9, 9] by [–6, 6]

(a) x= − ⎛

⎝⎜ ⎞ ⎠⎟= 3 cos 5

6 3 3

2 π

y= − ⎛

⎝⎜ ⎞ ⎠⎟= − −

⎛ ⎝

⎜ ⎞_⎠⎟ 3 sin 5

6 3 2 3 3

, 3 π

2 2

(b) x= ⎛

⎝⎜ ⎞ ⎠⎟ ⎛

⎝⎜

⎞ ⎠⎟=

−

5 cos tan 4

3 3

1

y= ⎛

⎝⎜ ⎞⎠⎟ ⎛

⎝⎜

⎞ ⎠⎟=

( )

−

5 sin tan 4

3 4

3, 4

1

(c) x y

= − =

( )

1 cos 7 1 1, 0

π π 1sin7 0

(d) x= ⎛

⎝⎜ ⎞ ⎠⎟= − 2 3 cos 2

3 3

π

y= ⎛

⎝⎜ ⎞⎠⎟= −

(

)

2 3 sin 2

3 3

(14)

3.

[–6, 6] by [–4, 4]

(a) r= − + =12 12 2

θ π π

π =

− ⎛ ⎝⎜

⎞ ⎠⎟= − −

⎛

⎝⎜ ⎞⎠⎟

−

tan 1

1 and

1 5

4 3

4

2 5

4 2

3 ,

, , ππ

4 ⎛ ⎝⎜ ⎞⎠⎟

(b) r= 1+ −

( )

3 2= ±2

θ π π

π π

= ⎛−

⎝

⎜ ⎞_⎠⎟ = −

− ⎛ ⎝⎜

⎞

⎠⎟ −

−

tan ,

, ,

1 3

2 3 2 2

3 and 2 33

⎛ ⎝⎜

⎞ ⎠⎟

(c) r= 02+32= ±3

θ π π

π π

= ⎛

⎝⎜ ⎞⎠⎟ ⎛

⎝⎜ ⎞ ⎠⎟

⎛ ⎝⎜

−

tan 3

0 =2, 5

2 3,

2 and 3, 5

2

1

⎞⎞ ⎠⎟

(d) r= −

( )

12+02= ±1

θ π

π

= ⎛

⎝⎜ ⎞⎠⎟= −

(

)

( )

−

tan ,

,

1 0

1 0 and 1, 4.

[–9, 9] by [–6, 6]

(a) r= −

( )

32+ −( 1)2= ±2

θ π π

π π

= −

− ⎛ ⎝⎜

⎞ ⎠⎟= −

⎛ ⎝⎜

⎞ ⎠⎟

−

tan ,

,

1 1

3 6

7 6 2

6 and 2, 7

6 ⎛⎛ ⎝⎜

⎞ ⎠⎟

(b) r= 32+42= ±5 θ

π

= ⎛

⎝⎜ ⎞⎠⎟

− + ⎛

⎝⎜ ⎞ ⎠⎟ ⎛

⎝⎜

⎞ ⎠⎟

−

tan 4

3

5, tan 4

3 and 5

1

1 _{,, tan} 4

3

1

− ⎛

⎝⎜ ⎞ ⎠⎟ ⎛

⎝⎜

⎞ ⎠⎟

(c) r= 0 +2

( )

−22= ±2

θ= ⎛− π π π π

⎝⎜ ⎞⎠⎟= − ⎛⎝⎜ − ⎞⎠⎟

−

tan 2

0 and 2,

3

1

2 3

2 2 2

, , ,

2 2 ⎛ ⎝⎜ ⎞⎠⎟

(d) r= 22+02= ±2

θ π

π

= ⎛

⎝⎜ ⎞⎠⎟

( ) (

)

−

tan 0

2 = 0, 2 2, 0 and 2, 2

1

5.

[–6, 6] by [–4, 4] 0≤u≤ 2p 6.

[–6, 6] by [–4, 4] 0≤u≤ 2p 7.

[–6, 6] by [–4, 4] 0≤u≤ 2p 8.

[–6, 6] by [–4, 4] 9.

[–6, 6] by [–4, 4] 10.

(15)

11. Cardioid

[–3, 3] by [–2, 2] 0≤u≤ 2p 12. Cardioid

[–6, 6] by [–4, 4] 0≤u≤ 2p 13. Rose

[–3, 3] by [–2, 2] 0≤u≤p 14. Rose

[–4.5, 4.5] by [–3, 3] 0≤u≤ 2p 15. Limaçon

[–4.5, 4.5] by [–3, 3] 0≤u≤ 2p 16. Limaçon

[–3, 3] by [–2, 2] 0≤u≤ 2p

17. Lemniscate

[–3, 3] by [–2, 2] –p / 4 ≤u≤p / 4 18. Lemniscate

[–1.5, 1.5] by [–1, 1] 0≤u≤p / 2 19. Circle

[–6, 6] by [–4, 4] 0≤u≤p 20. Circle

[–4.5, 4.5] by [–3, 3] 0≤u≤p 21. r=4 cscθ

1 4

4 =

=r

y

sin

,a horizontal lineθ 22. r= −3secθ

1 3

3 = − = −r

x

cos

,a verticle lineθ 23. x+ =y 1, a line

(slope = –1, y-intercept = 1) 24. r2=x2+y2=1,

(16)

25. r=

− 5

2 sinθ cosθ

1 5

2

2 5 2

= −

−r = r =

y x y

sinθ cosθ

, a line (slope , -inntercept=5) 26. r

r xy xy

2

2 2

1 sin

( sin cos ) ,

θ

θ θ

= = =

= a hyperbolaa 27. π2_cos2θ₌_r2_sinθ

x2=y2, the union of two lines:y= ±x

28. r r

x y x

x x y

x y

2

2 2

4 4

4 4 4 0

2 = − + = −

+ + − + =

+ + =

cos

( )

θ

4

4,a circle center( = −( 2 0, ),radius=2) 29. r r

x y r

x y

2

2 2

8

8 16 16 0

4 16

=

+ − + − =

+ − =

sin sin

( ) ,

θ θ

a ccircle center( = −( 2 0, ),radius=2) 30. r

r r r

x y x y

x

= +

+ = +

−

2 2

2

2 2

cos sin

(

θ θ

1

1 1 2

2 2

) ( ) ,

( ( , ),

+ − =

= =

y

a circle center radius )) 31. It is a parabola.

[–6, 6] by [–4, 4] 0≤u≤ 2p 32. It is a parabola.

[–6, 6] by [–4, 4] 0≤u≤ 2p 33. It is a parabola.

[–6, 6] by [–4, 4] 0≤u≤ 2p

34. It is a parabola.

[–6, 6] by [–4, 4] 0≤u≤ 2p 35. It is a hyperbola.

[–6, 6] by [–4, 4] 0≤u≤ 2p 36. It is an ellipse.

[–6, 6] by [–4, 4] 0≤u≤ 2p

37. It is an ellipse.

[–6, 6] by [–4, 4] 0≤u≤ 2p 38. It is a hyperbola.

[–6, 6] by [–4, 4] 0≤u≤ 2p 39. r= − +1 sinθ

dy dx

d d

d d dy dx

= − +

− +

=

θ θ θ

( sin ) sin ( sin ) cos (

1 1 2

2 1

2 2

sin ) cos

cos sin sin

θ θ

θ θ θ

θ θ π

−

− −

= − =

At 0: 1

(17)

40. r dy dx d d d d dy d = = cos

(cos sin ) (cos cos ) 2

2 2 θ

θ θ θ

xx=

−

− −

cos cos sin sin sin cos cos sin

θ θ θ θ

θ θ θ

2 2 2

2 2 2θθ

θ θ π θ π θ π At = = − = = 0 2 0 2 0 : : : : undefined At At At undeefined 41. r= −2 3sinθ

dy dx d d d d dy dx = − − =

θ θ θ

( sin ) sin ( sin ) cos

( 2 3 2 3

2 2 6

3 2 3

2 0 2

2

−

− − −

sin ) cos sin ( sin ) cos ( , ) :

θ θ

θ θ θ

At 3 3 1 2 0 2 2 3 5 3 2 0 At At At − ⎛ ⎝⎜ ⎞ ⎠⎟

( )

⎛ ⎝⎜ ⎞ ⎠⎟ , : , : , : π π π

42. r=3 1( −cos )θ

dy dx d d d d dy =

(

−

)

−

(

)

θ θ θ

3 1 3 1

( cos ) sin ( cos ) cos

d dx= −

+ +

− ⎛

⎝

6 3 3

3 2 1

1 5 3

2

cos cos

sin ( cos ) . ,

θ θ

π

At_⎜⎜ ⎞

⎠⎟ ⎛ ⎝⎜ ⎞ ⎠⎟ : . , : ( , ) : undefined At At und

4 5 2

3 0

6 π

π eefined

At 3 3

2 1

, π : ⎛ ⎝⎜

⎞ ⎠⎟

43. 1 2 4 2

2 0

2

+

(

)

∫

π cosθ dθ

=

(

+ +

)

= + +

∫

12 16 4 4

2 9

2 0

2

cos cos

sin cos sin

θ θ θ

π

d

0 0

18 0 18

0 2 ⎡⎣ ⎤⎦ = − = π π π

44. 1 2 2 2

2 0

2

+

(

)

∫

π sinθ dθ

=

(

+ +

)

= −

(

+

)

−

∫

12 4 4 4

4 3 2 2 0 2 sin sin sin cos

θ θ θ

π

d

((

)

⎡⎣ ⎤⎦

= − = 0

2

6 0 6

π

π π

45. 1 4 1 2 2 2 0 2

cos θ θ

π

(

)

∫

d =⎡ + ⎣⎢ ⎤ ⎦⎥ = − =

sin2 cos2 2 32

8 0 8

0 2

θ θ θ

π π

π

46. 1

2 2 4

2 0

2

( sin θ) θ

π d

∫

=⎡ − ⎣⎢ ⎤ ⎦⎥ = − = 1

4 4 4

2 0 2

0 2

(sin θ θ θ)

π π

π

cos 4

47. 1

2 4 2

0

4 _{( cos} θ θ₎

π d

∫

= ⎡ ⎣⎢ ⎤ ⎦⎥ = 1

2( sin4 2θ) ₀ 2

π/4

48. 1

2 2 3

0 ( sin θ θ)

π d

∫

=⎡ − ⎣⎢ ⎤ ⎦⎥ = 1

2( 2cos3θ) ₀ 4

π

49. 1 2 3 2

2 0

2

( − cos )

∫

π θ dθ

= − +

= −

∫

12 9 4 4

1

2 2 4

2 0

2

( cos cos )

( sin cos sin

θ θ θ

θ θ π d θθ θ π π π + ⎡ ⎣⎢ ⎤ ⎦⎥ = − = 11

11 0 11

0 2

)

50. 1

2 2 1

2 0

2 3

( sin )

/ θ θ π − ⎛ ⎝⎜ ⎞⎠⎟

∫

d = + + = − − −

∫

12 4 4 1

2 2

2 0

2

3 _{( sin} _sin ₎

( (sin ) cos

θ θ θ

θ θ π d 3 3 2 0 544 0 2 3

θ) π

. ⎡ ⎣⎢ ⎤ ⎦⎥ ≈

51. 1

2 2 2

2 2

(( sin ) ( cos ) )

/ /

θ θ θ

(18)

52. 1 2 2

1

2 2 1

2 2

0

0 ( sin )θ θ (( sin )θ ) θ

π π

d −

∫

− d

∫

= −_⎡⎣

(

−

)

_{⎤⎦ +}⎡ + ⎣⎢

⎤ ⎦⎥ =

sin cosθ θ θ sin cosθ θ θ π

π π

0

2 2

3 3

3 2 −

53. 1

2 2 1 4

1 2 4 2 1

2 2

(( ( −cos )) − ) − ( −( ( −cos ) ))

−

θ dθ _π θ dθ

//2 /2

π π

∫

0

2

=1_⎡⎣ − + _{⎤⎦ − −} −

2 4 3

1

2 2

0 2

sin cosθ θ sinθ θ π ⎡⎣ (sin cosθ θ θ))⎤⎦

= −

−π π

π

/2 /2

5 8

54. 1

2 2 1 2 1

2 2

(( ( cos )) ( ( cos ) ))

/ /

+ − +

−

∫

π_π θ θ dθ

= ⎡⎣8 ⎤⎦₋ =6 −16

2 2

sin

/ /

θ π_π π

55. 1

2 4 2 1

2 2

2

( ( ( sin )) )

/ /

− −

−

∫

π_π θ dθ

=_⎡⎣(sin − ) cos − _⎤⎦₋ = −

/ /

θ 4 θ θ π_π 8 π

2 2

56. 1

2 4 2 4

2 0 (( cos θ) ) θ

π

−

∫

/6 d

=_⎡⎣ + _⎤⎦

= +

2 2 2

4 3 8

3

0

(sin θcos θ θ) π

π/6

57.

[–3, 3] by [–2, 2]

0≤u≤ 2pfor the cardioid 0≤u≤pfor the circle

1

2 3 1

1 2 4

2 2

(( cos ) ( cos ) ) ( sin co

θ θ θ

θ

π π

− +

=

−

∫

d

/3 /3

ssθ sinθ θ

π π

π

− +

⎡ ⎣⎢

⎤ ⎦⎥

= −

2 3

/3 /3

58.

[–6, 6] by [–4, 4] 0≤u≤ 2p 1

2 2 2 1

4

2 2

0 (( ) ( ( sin )) )

(sin ) cos

− −

=_⎡⎣ − −

∫

θ θ

θ θ θ

π

d

⎤⎤⎦

=4 858 0

π

.

59.

[–3, 3] by [–2, 2] 0≤u≤p A

Slop

=

(

)

= −⎡ ⎣⎢

⎤ ⎦⎥ =

∫

12 2 3

2 3

3

2 0

0

sin cos ( )

θ θ

θ _π

π

d

ee=

=

d d

θ θ θ

θ θ θ θ π

( sin sin ) ( sin cos ) sin

2 3

6

/4

θθ θ θ θ

θ θ θ θ

cos cos sin

cos cos sin sin

3 2 3

6 3 2 3

+ − ⎡

⎣

⎢ ⎤

⎦⎦ ⎥

=

= θ π/4

1 2

60. (a) 1 5

3

2 0

3 4

+ − ⎛

⎝⎜ ⎤_⎦⎥

∫

/ y y dy

= + +

⎛

⎝⎜ ⎞⎠⎟ ₊ + ₋

⎡

⎣ ⎢ ⎢ ⎢ ⎢

⎤

⎦ ⎥ ⎥ ⎥ ⎥ ln

/

y y

y y y

2

2 2

0 3

1 2

5 6

4 4

0 347 = . (b) x=rcosθ

y r

x y

r

= = + − =

−

(

)

=

= sin

cos sin

c θ

θ θ

2 2

2 2 2

2

1 1

o

os2θ−sin2θ (c) Letα =tan−1

( )

3 5/ .

Then the area is 1

2

1

2 2

0 _cos _θ _sin _θ θ.

α

− ⎛

⎝⎜ ⎞⎠⎟

∫

d

61. True. Polar coordinates determine a unique point. 62. False. Integrating from 0 to 2πtraverses the curve twice,

giving twice the area. The correct upper limit of integration is π.