James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

Section 10.3 Polar Coordinates 661
¨=1
1
O
(_1, 1)
(_2, 1)
FIGURE 7
(1, 1)
(2, 1)
(3, 1)
x
Example 5 Sketch the polar curve − 1.
SOLUtion This curve consists of all points sr, d such that the polar angle is
1 radian. It is the straight line that passes through O and makes an angle of 1 radian
with the polar axis (see Figure 7). Notice that the points sr, 1d on the line with r . 0
are in the first quadrant, whereas those with r , 0 are in the third quadrant.
Example 6
(a) Sketch the curve with polar equation r − 2 cos .
(b) Find a Cartesian equation for this curve.
SOLUTION
(a) In Figure 8 we find the values of r for some convenient values of and plot the corresponding
points sr, d. Then we join these points to sketch the curve, which appears
to be a circle. We have used only values of between 0 and , since if we let increase
beyond , we obtain the same points again.
n
FIGURE 8
Table of values and
graph of r − 2 cos
r − 2 cos
0 2
y6 s3
y4 s2
y3 1
y2 0
2y3 21
3y4 2s2
5y6 2s3
22
π ”0, ’
2
π
”1, ’ 3
2π
”_1, ’ 3
π
”œ„, 2 ’ 4
π
”œ„, 3 ’ 6
(2, 0)
5π
”_ œ„, 3 ’
3π
6
”_ œ„, 2 ’ 4
(b) To convert the given equation to a Cartesian equation we use Equations 1 and 2.
From x − r cos we have cos − xyr, so the equation r − 2 cos becomes r − 2xyr,
which gives
Completing the square, we obtain
2x − r 2 − x 2 1 y 2 or x 2 1 y 2 2 2x − 0
sx 2 1d 2 1 y 2 − 1
which is an equation of a circle with center s1, 0d and radius 1.
n
y
P
Figure 9 shows a geometrical illustration
that the circle in Example 6 has the
equation r − 2 cos . The angle OPQ is
a right angle (Why?) and so ry2 − cos .
O
¨
r
2
Q
x
FIGURE 9
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