Jim Lambers
MAT 169
Fall Semester 2009-10
Lecture 38 Examples
These examples correspond to Sections 9.2, 9.3 and 9.4 in the text.
Example (Section 9.3, Exercise 67) Let 𝑃 be any point (except the origin) on the curve 𝑟 = 𝑓 (𝜃).
If 𝜓 is the angle between the tangent line at 𝑃 and the radial line 𝑂𝑃 , show that
tan 𝜓 =
𝑟
𝑑𝑟/𝑑𝜃
[Hint: Observe that 𝜓 = 𝜙 − 𝜃 in the figure on page 504 in the text.]
Solution We have
tan 𝜓 = tan(𝜙 − 𝜃) =
tan 𝜙 − tan 𝜃
.
1 + tan 𝜙 tan 𝜃
Because 𝜙 is the angle that the tangent line makes with the positive 𝑥-axis,
tan 𝜙 =
𝑑𝑦
=
𝑑𝑥
𝑑𝑟
𝑑𝜃
𝑑𝑟
𝑑𝜃
sin 𝜃 + 𝑟 cos 𝜃
cos 𝜃 − 𝑟 sin 𝜃
.
It follows that if we write 𝑟′ = 𝑑𝑟/𝑑𝜃, then
tan 𝜓 =
𝑟′ sin 𝜃+𝑟 cos 𝜃
sin 𝜃
𝑟′ cos 𝜃−𝑟 sin 𝜃 − cos 𝜃
′
𝜃+𝑟 cos 𝜃 sin 𝜃
1 + 𝑟𝑟′ sin
cos 𝜃−𝑟 sin 𝜃 cos 𝜃
,
we can simplify by putting all fractions over a common denominator. Because the common denominators are both equal to cos 𝜃(𝑟′ cos 𝜃 − 𝑟 sin 𝜃), they cancel, and we obtain
tan 𝜓 =
(𝑟′ sin 𝜃 + 𝑟 cos 𝜃) cos 𝜃 − (𝑟′ cos 𝜃 − 𝑟 sin 𝜃) sin 𝜃
.
(𝑟′ cos 𝜃 − 𝑟 sin 𝜃) cos 𝜃 + (𝑟′ sin 𝜃 + 𝑟 cos 𝜃) sin 𝜃
Expanding, and using sin2 𝜃 + cos2 𝜃 = 1, we obtain tan 𝜓 = 𝑟/𝑟′ . □
Example (Section 9.4, Exercise 15) Find the area of the region enclosed by one loop of the curve
𝑟 = sin 2𝜃.
Solution We have 𝑟 = 0 when 𝜃 = 0 and 𝜃 = 𝜋/2, so the interval 0 ≤ 𝜃 ≤ 𝜋/2 corresponds to one
loop of the curve. Therefore, the area of the region enclosed by this loop is
1
2
∫
0
𝜋/2
1
sin 2𝜃 𝑑𝜃 =
2
2
∫
0
𝜋/2
[
]
1
sin 4𝜃 𝜋/2 𝜋
1 − cos 4𝜃
𝑑𝜃 =
𝜃−
= ,
2
4
2
8
0
1
where a double-angle identity was used to rewrite sin2 2𝜃 in such a way that it can be integrated.
□
Example (Section 9.4, Exercise 21) Find the area of the region
that lies inside the curve 𝑟 = 3 cos 𝜃 and outside the curve 𝑟 = 1 + cos 𝜃.
Solution These curves intersect when 3 cos 𝜃 = 1 + cos 𝜃, or cos 𝜃 = 1/2, which is the case when
𝜃 = ±𝜋/3. Therefore, the area 𝐴 of the region between them is given by
∫
1
2
𝐴 =
=
=
=
[3 cos 𝜃]2 − [1 + cos 𝜃]2 𝑑𝜃
−𝜋/3
∫ 𝜋/3
1
2
=
𝜋/3
9 cos2 𝜃 − (1 + 2 cos 𝜃 + cos2 𝜃) 𝑑𝜃
−𝜋/3
𝜋/3
1
2
∫
1
2
∫
1
2
∫
8 cos2 𝜃 − 1 − 2 cos 𝜃 𝑑𝜃
−𝜋/3
𝜋/3
8
−𝜋/3
𝜋/3
1 + cos 2𝜃
− 1 − 2 cos 𝜃 𝑑𝜃
2
3 + 4 cos 2𝜃 − 2 cos 𝜃 𝑑𝜃
−𝜋/3
[
]𝜋/3
1
sin 2𝜃
=
3𝜃 + 4
− 2 sin 𝜃 2
2
−𝜋/3
= 𝜋.
□
Example (Section 9.4, Exercise 23) Find the area of the region that lies inside both of the curves
𝑟 = sin 𝜃 and 𝑟 = cos 𝜃.
Solution These curves intersect when 𝜃 = 𝜋/4. Therefore, if we divide the region that lies inside
both curves with the ray 𝜃 = 𝜋/4, we can obtain its area 𝐴 by computing
𝐴 =
=
=
=
𝜋/4
∫
1
2
[
∫
1 − cos 2𝜃
1 𝜋/2 1 + cos 2𝜃
𝑑𝜃 +
𝑑𝜃
2
2 𝜋/4
2
0
[
]
]
𝜃 sin 2𝜃 𝜋/4
𝜃 sin 2𝜃 𝜋/2
−
+
+
2
4
2
4
0
𝜋/4
sin2 𝜃 𝑑𝜃 +
0
∫
𝜋/4
𝜋 1 𝜋 1
− + −
8 4 8 4
2
1
2
∫
𝜋/2
1
2
𝑐𝑜𝑠2 𝜃 𝑑𝜃
𝜋/4
=
1
(𝜋 − 2).
8
□
3