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 𝑟 tan 𝜓 = 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, 𝑑𝑦 tan 𝜙 = 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
′ 𝑟
′ sin 𝜃 + 𝑟 cos 𝜃 cos 𝜃 − 𝑟 sin 𝜃
1 + 𝑟
′ 𝑟
′ sin cos 𝜃 𝜃
+
− 𝑟 𝑟
− cos sin 𝜃 𝜃 sin 𝜃 cos 𝜃 sin cos 𝜃 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
′
( 𝑟 ′ sin 𝜃 + 𝑟 cos 𝜃 ) cos 𝜃 − ( 𝑟
′ cos 𝜃 − 𝑟 sin 𝜃 ) cos 𝜃 + ( 𝑟 ′ cos 𝜃 − 𝑟 sin 𝜃 ) sin 𝜃 sin 𝜃 + 𝑟 cos 𝜃 ) sin 𝜃
.
Expanding, and using sin
2 𝜃 + cos 2 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 2 𝜃 .
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
∫ 𝜋/ 2 sin
2
2 𝜃 𝑑𝜃 =
0
1 ∫ 𝜋/ 2
2
0
1 − cos 4 𝜃 𝑑𝜃 =
2
1 [ 𝜃 −
4 sin 4 𝜃 ] 𝜋/ 2
2
0
= 𝜋
8
,
1

where a double-angle identity was used to rewrite sin
2
□
2 𝜃 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
𝐴 =
=
=
1
2
∫ 𝜋/ 3
− 𝜋/ 3
[3 cos 𝜃 ]
2
− [1 + cos 𝜃 ]
2 𝑑𝜃
1
2
∫ 𝜋/ 3
− 𝜋/ 3
9 cos
2 𝜃 − (1 + 2 cos 𝜃 + cos
2 𝜃 ) 𝑑𝜃
1
2
∫ 𝜋/ 3
− 𝜋/ 3
8 cos
2 𝜃 − 1 − 2 cos 𝜃 𝑑𝜃
=
=
1
2
∫ 𝜋/ 3
− 𝜋/ 3
8
1 + cos 2 𝜃
2
− 1 − 2 cos 𝜃 𝑑𝜃
1 ∫ 𝜋/ 3
2
− 𝜋/ 3
3 + 4 cos 2 𝜃 − 2 cos 𝜃 𝑑𝜃
=
1
2
= 𝜋.
[
3 𝜃 + 4 sin 2 𝜃
2
− 2 sin 𝜃
] 𝜋/ 3
− 𝜋/ 3
□
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 =
=
=
=
1
2
∫ 𝜋/ 4 sin
2
0 𝜃 𝑑𝜃 +
1 ∫ 𝜋/ 2 𝑐𝑜𝑠
2 𝜃 𝑑𝜃
2 𝜋/ 4
1 ∫ 𝜋/ 4
2
0
1 − cos 2 𝜃 𝑑𝜃 +
2
1
2
∫ 𝜋/ 2 𝜋/ 4
1 + cos 2 𝜃 𝑑𝜃
2
[ 𝜃 𝜋
8
2
−
−
1
4 sin 2 𝜃 ] 𝜋/ 4
+
4
0
+ 𝜋
8
−
1
4
[ 𝜃
2
+ sin 2 𝜃 ] 𝜋/ 2
4 𝜋/ 4
2

□
=
1
8
( 𝜋 − 2) .
3