Solução_Calculo_Stewart_6e

F.
428 ¤ CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES TX.10
63. r =3cosθ ⇒ x = r cos θ =3cosθ cos θ, y = r sin θ =3cosθ sin θ ⇒
dy
= dθ −3sin2 θ +3cos 2 θ =3cos2θ =0 ⇒ 2θ = π or 3π ⇔ θ = π or 3π .
2 2 4 4
3
So the tangent is horizontal at √
2
, π and − √ 3 4
2
, 3π 3
same as √
4
2
, − π .
4
dx
= −6sinθ cos θ = −3sin2θ =0 ⇒ 2θ =0or π ⇔ θ =0or π . So the tangent is vertical at (3, 0) and
0, π dθ 2 2 .
65. r =1+cosθ ⇒ x = r cos θ =cosθ (1 + cos θ), y = r sin θ =sinθ (1 + cos θ) ⇒
dy
=(1+cosθ) cosθ − dθ sin2 θ =2cos 2 θ +cosθ − 1=(2cosθ − 1)(cos θ +1)=0 ⇒ cos θ = 1 2
or −1 ⇒
θ = π , π,or 5π ⇒ horizontal tangent at 3
, π
3 3 2 3 , (0,π),and 3 , 5π
2 3 .
dx
= −(1 + cos θ)sinθ − cos θ sin θ = − sin θ (1 + 2 cos θ) =0 ⇒ sin θ =0or cos θ = − 1
dθ 2
⇒
θ =0, π, 2π ,or 4π ⇒ vertical tangent at (2, 0), 1
, 2π
3 3 2 3 ,and 1 , 4π
2 3 .
Note that the tangent is horizontal, not vertical when θ = π,since lim
θ→π
dy/dθ
dx/dθ =0.
67. r =2+sinθ ⇒ x = r cos θ =(2+sinθ) cosθ, y = r sin θ =(2+sinθ) sinθ ⇒
dy
dθ
=(2+sinθ) cosθ +sinθ cos θ =cosθ · 2(1 + sin θ) =0 ⇒ cos θ =0or sin θ = −1 ⇒
θ = π 2 or 3π 2
⇒ horizontal tangent at 3, π 2
dx
and
1,
3π
2
=(2+sinθ)(− sin θ)+cosθ cos θ = −2sinθ − dθ sin2 θ +1− sin 2 θ = −2sin 2 θ − 2sinθ +1
sin θ = 2 ± √ 4+8
= 2 ± 2 √ 3
= 1 − √ √
3 1+ 3
< −1 ⇒
−4 −4 −2 −2
.
⇒
θ 1 =sin −1 − 1 + √ 1
2 2 3 and θ2 = π − θ 1 ⇒ vertical tangent at 3
+ √ 1 3,θ1
2 2 and 3 + √ 1 3,θ2
2 2 .
Note that r(θ 1)=2+sin sin −1 − 1 + √ 1
2 2 3 =2−
1
+ √ 1
2 2 3=
3
+ √ 1
2 2 3.
69. r = a sin θ + b cos θ ⇒ r 2 = ar sin θ + br cos θ ⇒ x 2 + y 2 = ay + bx ⇒
x 2 − bx + 1
2 b2 + y 2 − ay + 1
2 a2 = 1
2 b2 + 1
2 a 2
with center 1
2 b, 1 2 a and radius 1 2
√
a2 + b 2 .
⇒
x − 1 2 b2 + y − 1 2 a2 = 1 4 (a2 + b 2 ), and this is a circle
Note for Exercises 71–76: Maple is able to plot polar curves using the polarplot command, or using the coords=polar option in a regular
plot command. In Mathematica, use PolarPlot. In Derive, change to Polar under Options State. If your graphing device cannot
plot polar equations, you must convert to parametric equations. For example, in Exercise 71, x = r cos θ =[1+2sin(θ/2)] cos θ,
y = r sin θ =[1+2sin(θ/2)] sin θ.
71. r =1+2sin(θ/2). The parameter interval is [0, 4π]. 73. r = e sin θ − 2cos(4θ). The parameter interval is [0, 2π].