Jim Lambers
MAT 169
Fall Semester 2009-10
Lecture 37 Examples
These examples correspond to Sections 9.2, 9.3 and 9.4 in the text.
Example (Section 9.2, Exercise 53) A string is wound around a circle and then unwound while
being held taut. The curve traced by the point 𝑃 at the end of the string is called the involute of
the circle. If the circle has radius 𝑟 and center 𝑂 and the initial position of 𝑃 is (𝑟, 0), and if the
parameter 𝜃 is chosen as in the figure (see page 495 of the text), show that parametric equations
of the involute are
𝑥 = 𝑟(cos 𝜃 + 𝜃 sin 𝜃), 𝑦 = 𝑟(sin 𝜃 − 𝜃 cos 𝜃).
Solution For any angle 𝜃, the position vector for the point 𝑃 is the sum of two vectors u and
v, where u is the position vector u for the point 𝑇 on the circle corresponding to 𝜃, and v is
the vector from 𝑇 to 𝑃 . The magnitude of u is 𝑟, the radius of the circle, so its components are
u = ⟨𝑟 cos 𝜃, 𝑟 sin 𝜃⟩.
The magnitude of v is the length of the portion of the string that has been unwound, which is the
length of the arc of the circle that begins at the point (𝑟, 0) and ends after sweeping counterclockwise
(for increasing 𝜃) through 𝜃 radians; this length is 𝑟𝜃.
Because the string is held taut, it is tangent to the circle at 𝑇 , so v and u are perpendicular;
in fact, the angle made by v with the positive 𝑥-axis, when v is translated so that its initial point
is at the origin, is always 90 degrees, or 𝜋/2 radians, less than that of u. That is,
⟨
(
(
𝜋)
𝜋 )⟩
v = 𝑟𝜃 cos 𝜃 −
, 𝑟𝜃 sin 𝜃 −
.
2
2
Using the identities
cos(𝐴 − 𝐵) = cos 𝐴 cos 𝐵 + sin 𝐴 sin 𝐵,
sin(𝐴 − 𝐵) = sin 𝐴 cos 𝐵 − cos 𝐴 sin 𝐵,
with 𝐴 = 𝜃 and 𝐵 = 𝜋/2, we obtain cos(𝜃 − 𝜋/2) = sin 𝜃 and sin(𝜃 − 𝜋/2) = − cos 𝜃. By adding
the components of u and v, we obtain the desired parametric equations for the involute. □
Example (Section 9.3, Exercise 11) Sketch the region in the plane consisting of points whose polar
coordinates satisfy 2 < 𝑟 < 3, 5𝜋/3 ≤ 𝜃 ≤ 7𝜋/3.
Solution A sketch of the region described by the given inequalities is obtained by first sketching
two concentric circles, of radii 2 and 3, and the rays extending from the origin at angles 5𝜋/3 and
7𝜋/3. The region is illustrated in Figure 1. □
Example (Section 9.3, Exercise 13) Identify the curve 𝑟 = 3 sin 𝜃 by finding a Cartesian equation
for the curve.
1
Figure 1: Region satisfying 2 < 𝑟 < 3 and 5𝜋/3 ≤ 𝜃 ≤ 7𝜋/3.
Solution Multiplying both sides by 𝑟 yields 𝑟2 = 3𝑟 sin 𝜃. Using the relations 𝑦 = 𝑟 sin 𝜃 and
𝑥2 + 𝑦 2 = 𝑟2 , we obtain the Cartesian equation
𝑥2 + 𝑦 2 = 3𝑦.
Moving the 3𝑦 to the left side and completing the square yields
𝑥2 + 𝑦 3 − 3𝑦 +
9
9
= ,
4
4
which, upon factoring, becomes
(
)
( )2
3 2
3
𝑥 + 𝑦−
=
.
2
2
2
This is the equation of a circle with center (0, 3/2) and radius 3/2. □
Example (Section 9.3, Exercise 15) Identify the curve 𝑟 = csc 𝜃 by finding a Cartesian equation
for the curve.
2
Solution Applying csc 𝜃 = 1/ sin 𝜃 and multiplying both sides by sin 𝜃 yields 𝑟 sin 𝜃 = 1, which, in
Cartesian coordinates, becomes 𝑦 = 1. This curve is a horizontal line that lies one unit above the
𝑥-axis. □
Example (Section 9.3, Exercise 19) Find a polar equationfor the curve represented by the Cartesian
equation 𝑥2 + 𝑦 2 = 2𝑐𝑥.
Solution Using the relations 𝑥2 + 𝑦 2 = 𝑟2 and 𝑥 = 𝑟 cos 𝜃, we obtain the polar equation 𝑟2 =
2𝑐𝑟 cos 𝜃, which can then be divided through by 𝑟 to obtain a polar equation in standard form,
𝑟 = 2𝑐 cos 𝜃. This curve is a circle with center (𝑐, 0) with radius 𝑐. □
Example (Section 9.4, Exercise 7) Find the area of the region (see page 508 in the text) bounded
by the curve 𝑟 = 4 + 3 sin 𝜃 and the line 𝑥 = 0.
Solution Points on the line 𝑥 = 0 correspond to 𝜃 = −𝜋/2 (for 𝑦 < 0) and 𝜃 = 𝜋/2 (for 𝑦 > 0). It
follows that the area 𝐴 of the region is given by the integral
𝐴 =
=
=
=
=
=
1
2
∫
1
2
∫
𝜋/2
(4 + 3 sin 𝜃)2 𝑑𝜃
−𝜋/2
𝜋/2
16 + 24 sin 𝜃 + 9 sin2 𝜃 𝑑𝜃
−𝜋/2
𝜋/2
1 − cos 2𝜃
𝑑𝜃
2
−𝜋/2
[
(
)]
1
9
sin 2𝜃 𝜋/2
16𝜃 − 24 cos 𝜃 +
𝜃−
2
2
2
−𝜋/2
[
]
9
1
16𝜋 + 𝜋
2
2
41𝜋
.
4
1
2
∫
16 + 24 sin 𝜃 + 9
□
3