Math 214 — Solutions to Assignment #6
11.3
20. Identify the curve r = tan θ sec θ by finding a Cartesian equation for the curve.
Solution.
x = r cos θ =⇒ cos θ = x/r and tan θ = y/x. So
r = tan θ sec θ =
y r
yr
· = 2 =⇒ y = x2 ,
x x
x
i.e., the curve is a parabola opening upward with vertex (0, 0).
26. Find a polar equation for the curve x 2 − y 2 = 1.
Solution.
Since x = r cos θ and y = r sin θ,
x2 − y 2 = r 2 cos2 θ − r 2 sin2 θ = r 2 (cos2 θ − sin2 θ) = r 2 cos 2θ.
Hence the polar equation is
r 2 cos 2θ = 1
or
r 2 = sec 2θ.
For #34–44, sketch the curve with the given polar equation.
34. r = 1 − 3 cos θ
Solution.
r
4
1
0
π
2
π
3π
2
2π
θ
(4,π )
−2
1
(−2,0)
38. r = 2 cos 3θ
Solution.
Using symmetry about the polar axis,
π
3
2π
3
r
π
6
5π
6
2
0 π
6
π
3
π
2
2π 5 π
3 6
π
θ
−2
4π
3
44. r 2 θ = 1
Solution.
r
r>0
θ
0
r<0
60. Find the slope of the tangent line to the polar curve r = sin 3θ at θ = π/6.
Solution.
The slope of the tangent line is
dy
=
dx
dr
dθ
dr
dθ
sin θ + r cos θ
3 cos 3θ sin θ + sin 3θ cos θ
=
3 cos 3θ cos θ − sin 3θ sin θ
cos θ − r sin θ
3 cos
dy =
dx θ= π
3 cos
6
π
2
π
2
sin π6 + sin π2 cos π6
0+
π
π
π =
cos 6 − sin 2 sin 6
0−
2
√
3
2
1
2
√
= − 3.
11.4
18. Find the area of the region enclosed by one loop of the curve r = 4 sin 3θ.
Solution.
π
3
2π
3
r
π
6
4
0 π
6
π
3
2π 5π
3 6
π
2
π
θ
Using symmetry
A=
Z
π
3
0
1
2
(4 sin 3θ)2 dθ =
2
2
Z
π
6
16 sin2 3θ dθ
0
π6
1
=8
(1 − cos 6θ) dθ = 8 θ − sin 6θ 6
0
0
π 1
4π
− sin π =
.
=8
6
6
3
Z
π
6
24. Find the area of the region that lies inside the first curve and outside the second curve:
r = 1 − sin θ, r = 1.
Solution.
(1,0)
3
Note that the points of intersection are θ = 0, π, 2π (set 1 − sin θ = 1). Using symmetry again,
Z 0
Z
1 0
2
2
[(1 − sin θ) − 1 ] dθ =
A=2·
(−2 sin θ + sin2 θ) dθ
π
2 −π/2
−2
Z 0 1
−2 sin θ + (1 − cos 2θ) dθ
=
π
2
−2
0
1
1 π
1
π
−
= 2−
= 2 cos θ + θ − sin 2θ =2+ .
2
4
2
2
4
π
−
2
30. Find the area of the region that lies inside both curves: r = sin 2θ, r = sin θ.
Solution.
π
3
π
4
To find the points of intersection:
sin 2θ = sin θ =⇒ 2 sin θ cos θ − sin θ = 0 =⇒ sin θ(2 cos θ − 1) = 0
1
π 5π
=⇒ sin θ = 0 or cos θ =
=⇒ θ = 0, π, ,
2
3 3
and the pole. Using symmetry, then,
Z π
Z π
1 3
1 2
2
2
A=2
sin 2θ dθ
sin θ dθ +
2 0
2 π3
Z π
Z π
1 3
1 2
=
(1 − cos 4θ) dθ
(1 − cos 2θ) dθ +
2 0
2 π3
π3 π2 1
1
1
=
θ − sin 2θ + θ − sin 4θ 2
2
4
π
0
3
1 π 1
2π
π 1
π 1
4π
=
− sin
+ − sin 2π − + sin
2 3
2
3
2
4
3
4
3
√ √
√ !
√ !
1 π 3 3
1 π 1
3
1
3
π 3 3
−
=
−
.
=
+
−
= −
2 2
2
2
4
2
2 2
8
4
16
4
32. Same as #30 for the curves r 2 = 2 sin 2θ, r = 1.
Solution.
5 π 12
π4
π 12
The points of intersection are:
2θ =
π 5π 13π 17π
π 5π 13π 17π
,
,
,
=⇒ θ =
,
,
,
.
6 6
6
6
12 12 12 12
Using symmetry,
Z π
Z π
1 12
1 4 2
A=2 2·
2 sin 2θ dθ + 2 ·
1 dθ
π
2 0
2 12
π
12
π4 2
π
π
π
=2
− cos 2θ + θ = 2 − cos + cos 0 + −
2
6
4
12
π
0
12
!
√
√ !
3
2π
6+π−3 3
π √
+1+
=2 −
=2
= 2 + − 3.
2
12
6
3
40. Find all points of intersection of the curves r = cos 3θ and r = sin 3θ.
Solution.
1 3π
2,4
π
3
2π
3
π
6
1 π
,
2 12
1 5π
,
2 12
Since cos 3θ = sin 3θ =⇒ tan 3θ = 1, 3θ = π/4 + nπ and thus θ = π/12 + (nπ)/3. So the
points of intersection are
1 π
√ ,
,
2 12
1 3π
√ ,
2 4
,
5
1 5π
−√ ,
2 12
and the pole.
46. Find the exact length of the polar curve r = e 2θ , 0 ≤ θ ≤ 2π.
Solution.
L=
=
Z
Z
2π
s
r2
0
2π
0
+
√
5e4θ
dr
dθ
2
√ Z
= 5
dθ =
0
2π
e
0
Z
2θ
2π
q
e4θ + (2e2θ )2 dθ
√
√
2π
5 2θ 5 4π
e =
(e − 1).
dθ =
2
2
0
6