Section8.2 Graphs of Polar Equations of Polar Equations... · Section8.2 Graphs of Polar Equations...
17
Section 8.2 Graphs of Polar Equations Graphing Polar Equations The graph of a polar equation r = f (θ), or more generally F (r, θ)=0, consists of all points P that have at least one polar representation (r, θ) whose coordinates satisfy the equation. EXAMPLE: Sketch the polar curve θ =1. Solution: This curve consists of all points (r, θ) such that the polar angle θ is 1 radian. It is the straight line that passes through O and makes an angle of 1 radian with the polar axis. Notice that the points (r, 1) on the line with r> 0 are in the first quadrant, whereas those with r< 0 are in the third quadrant. EXAMPLE: Sketch the following curves: (a) r =2, 0 ≤ θ ≤ 2π. (b) r = θ, 0 ≤ θ ≤ 4π. (c) r = 2 cos θ, 0 ≤ θ ≤ π. 1
Transcript of Section8.2 Graphs of Polar Equations of Polar Equations... · Section8.2 Graphs of Polar Equations...
Section 8.2 Graphs of Polar Equations
Graphing Polar Equations
The graph of a polar equation r = f(θ), or more generally F (r, θ) = 0, consists of all points P thathave at least one polar representation (r, θ) whose coordinates satisfy the equation.
EXAMPLE: Sketch the polar curve θ = 1.
Solution: This curve consists of all points (r, θ) such that the polar angle θ is 1 radian. It is the straightline that passes through O and makes an angle of 1 radian with the polar axis. Notice that the points(r, 1) on the line with r > 0 are in the first quadrant, whereas those with r < 0 are in the third quadrant.
EXAMPLE: Sketch the following curves:
(a) r = 2, 0 ≤ θ ≤ 2π.
(b) r = θ, 0 ≤ θ ≤ 4π.
(c) r = 2 cos θ, 0 ≤ θ ≤ π.
1
EXAMPLE: Sketch the curve r = 2, 0 ≤ θ ≤ 2π.
Solution 1: Since r = 2, it follows that r2 = 4. But r2 = x2 + y2, therefore x2 + y2 = 4 which is a circle ofradius 2 with the center at the origin.
Solution 2: We have
-2 -1 1 2
-2
-1
1
2
r=2, theta=Pi�6
-2 -1 1 2
-2
-1
1
2
r=2, theta=2 Pi�6
-2 -1 1 2
-2
-1
1
2
r=2, theta=3 Pi�6
-2 -1 1 2
-2
-1
1
2
r=2, theta=4 Pi�6
-2 -1 1 2
-2
-1
1
2
r=2, theta=5 Pi�6
-2 -1 1 2
-2
-1
1
2
r=2, theta=6 Pi�6
-2 -1 1 2
-2
-1
1
2
r=2, theta=7 Pi�6
-2 -1 1 2
-2
-1
1
2
r=2, theta=8 Pi�6
-2 -1 1 2
-2
-1
1
2
r=2, theta=9 Pi�6
-2 -1 1 2
-2
-1
1
2
r=2, theta=10 Pi�6
-2 -1 1 2
-2
-1
1
2
r=2, theta=11 Pi�6
-2 -1 1 2
-2
-1
1
2
r=2, theta=12 Pi�6
2
EXAMPLE: Sketch the curve r = θ, 0 ≤ θ ≤ 4π.
Solution: We have
-10 -5 5 10
-10
-5
5
10
r=theta, theta=Pi�3
-10 -5 5 10
-10
-5
5
10
r=theta, theta=2 Pi�3
-10 -5 5 10
-10
-5
5
10
r=theta, theta=3 Pi�3
-10 -5 5 10
-10
-5
5
10
r=theta, theta=4 Pi�3
-10 -5 5 10
-10
-5
5
10
r=theta, theta=5 Pi�3
-10 -5 5 10
-10
-5
5
10
r=theta, theta=6 Pi�3
-10 -5 5 10
-10
-5
5
10
r=theta, theta=7 Pi�3
-10 -5 5 10
-10
-5
5
10
r=theta, theta=8 Pi�3
-10 -5 5 10
-10
-5
5
10
r=theta, theta=9 Pi�3
-10 -5 5 10
-10
-5
5
10
r=theta, theta=10 Pi�3
-10 -5 5 10
-10
-5
5
10
r=theta, theta=11 Pi�3
-10 -5 5 10
-10
-5
5
10
r=theta, theta=12 Pi�3
3
EXAMPLE: Sketch the curve r = 2 cos θ, 0 ≤ θ ≤ π.
Solution 1: Since r = 2 cos θ, it follows that r2 = 2r cos θ. But r2 = x2 + y2 and r cos θ = x, thereforex2 + y2 = 2x. This can be rewritten as (x − 1)2 + y2 = 1 which is a circle of radius 1 with the center at(1, 0).
Solution 2: We have
-0.5 0.5 1.0 1.5 2.0
-1.0
-0.5
0.5
1.0
r=2cosHthetaL, theta=Pi�12
-0.5 0.5 1.0 1.5 2.0
-1.0
-0.5
0.5
1.0
r=2cosHthetaL, theta=2 Pi�12
-0.5 0.5 1.0 1.5 2.0
-1.0
-0.5
0.5
1.0
r=2cosHthetaL, theta=3 Pi�12
-0.5 0.5 1.0 1.5 2.0
-1.0
-0.5
0.5
1.0
r=2cosHthetaL, theta=4 Pi�12
-0.5 0.5 1.0 1.5 2.0
-1.0
-0.5
0.5
1.0
r=2cosHthetaL, theta=5 Pi�12
-0.5 0.5 1.0 1.5 2.0
-1.0
-0.5
0.5
1.0
r=2cosHthetaL, theta=6 Pi�12
-0.5 0.5 1.0 1.5 2.0
-1.0
-0.5
0.5
1.0
r=2cosHthetaL, theta=7 Pi�12
-0.5 0.5 1.0 1.5 2.0
-1.0
-0.5
0.5
1.0
r=2cosHthetaL, theta=8 Pi�12
-0.5 0.5 1.0 1.5 2.0
-1.0
-0.5
0.5
1.0
r=2cosHthetaL, theta=9 Pi�12
-0.5 0.5 1.0 1.5 2.0
-1.0
-0.5
0.5
1.0
r=2cosHthetaL, theta=10 Pi�12
-0.5 0.5 1.0 1.5 2.0
-1.0
-0.5
0.5
1.0
r=2cosHthetaL, theta=11 Pi�12
-0.5 0.5 1.0 1.5 2.0
-1.0
-0.5
0.5
1.0
r=2cosHthetaL, theta=12 Pi�12
4
EXAMPLES:
5
EXAMPLE: Sketch the curve r = 1 + sin θ, 0 ≤ θ ≤ 2π (cardioid).
Solution: We have
-1.5 -1.0 -0.5 0.5 1.0 1.5
-0.5
0.5
1.0
1.5
2.0
r=1+sinHthetaL, theta=Pi�6
-1.5 -1.0 -0.5 0.5 1.0 1.5
-0.5
0.5
1.0
1.5
2.0
r=1+sinHthetaL, theta=2 Pi�6
-1.5 -1.0 -0.5 0.5 1.0 1.5
-0.5
0.5
1.0
1.5
2.0
r=1+sinHthetaL, theta=3 Pi�6
-1.5 -1.0 -0.5 0.5 1.0 1.5
-0.5
0.5
1.0
1.5
2.0
r=1+sinHthetaL, theta=4 Pi�6
-1.5 -1.0 -0.5 0.5 1.0 1.5
-0.5
0.5
1.0
1.5
2.0
r=1+sinHthetaL, theta=5 Pi�6
-1.5 -1.0 -0.5 0.5 1.0 1.5
-0.5
0.5
1.0
1.5
2.0
r=1+sinHthetaL, theta=6 Pi�6
-1.5 -1.0 -0.5 0.5 1.0 1.5
-0.5
0.5
1.0
1.5
2.0
r=1+sinHthetaL, theta=7 Pi�6
-1.5 -1.0 -0.5 0.5 1.0 1.5
-0.5
0.5
1.0
1.5
2.0
r=1+sinHthetaL, theta=8 Pi�6
-1.5 -1.0 -0.5 0.5 1.0 1.5
-0.5
0.5
1.0
1.5
2.0
r=1+sinHthetaL, theta=9 Pi�6
-1.5 -1.0 -0.5 0.5 1.0 1.5
-0.5
0.5
1.0
1.5
2.0
r=1+sinHthetaL, theta=10 Pi�6
-1.5 -1.0 -0.5 0.5 1.0 1.5
-0.5
0.5
1.0
1.5
2.0
r=1+sinHthetaL, theta=11 Pi�6
-1.5 -1.0 -0.5 0.5 1.0 1.5
-0.5
0.5
1.0
1.5
2.0
r=1+sinHthetaL, theta=12 Pi�6
6
EXAMPLE: Sketch the curve r = 1− cos θ, 0 ≤ θ ≤ 2π (cardioid).
Solution: We have
-2.0 -1.5 -1.0 -0.5 0.5
-1.5
-1.0
-0.5
0.5
1.0
1.5r=1-cosHthetaL, theta=Pi�6
-2.0 -1.5 -1.0 -0.5 0.5
-1.5
-1.0
-0.5
0.5
1.0
1.5r=1-cosHthetaL, theta=2 Pi�6
-2.0 -1.5 -1.0 -0.5 0.5
-1.5
-1.0
-0.5
0.5
1.0
1.5r=1-cosHthetaL, theta=3 Pi�6
-2.0 -1.5 -1.0 -0.5 0.5
-1.5
-1.0
-0.5
0.5
1.0
1.5r=1-cosHthetaL, theta=4 Pi�6
-2.0 -1.5 -1.0 -0.5 0.5
-1.5
-1.0
-0.5
0.5
1.0
1.5r=1-cosHthetaL, theta=5 Pi�6
-2.0 -1.5 -1.0 -0.5 0.5
-1.5
-1.0
-0.5
0.5
1.0
1.5r=1-cosHthetaL, theta=6 Pi�6
-2.0 -1.5 -1.0 -0.5 0.5
-1.5
-1.0
-0.5
0.5
1.0
1.5r=1-cosHthetaL, theta=7 Pi�6
-2.0 -1.5 -1.0 -0.5 0.5
-1.5
-1.0
-0.5
0.5
1.0
1.5r=1-cosHthetaL, theta=8 Pi�6
-2.0 -1.5 -1.0 -0.5 0.5
-1.5
-1.0
-0.5
0.5
1.0
1.5r=1-cosHthetaL, theta=9 Pi�6
-2.0 -1.5 -1.0 -0.5 0.5
-1.5
-1.0
-0.5
0.5
1.0
1.5r=1-cosHthetaL, theta=10 Pi�6
-2.0 -1.5 -1.0 -0.5 0.5
-1.5
-1.0
-0.5
0.5
1.0
1.5r=1-cosHthetaL, theta=11 Pi�6
-2.0 -1.5 -1.0 -0.5 0.5
-1.5
-1.0
-0.5
0.5
1.0
1.5r=1-cosHthetaL, theta=12 Pi�6
7
EXAMPLE: Sketch the curve r = 2 + 4 cos θ, 0 ≤ θ ≤ 2π.
Solution: We have
1 2 3 4 5 6
-3
-2
-1
1
2
3
r=2+4cosHthetaL, theta=Pi�6
1 2 3 4 5 6
-3
-2
-1
1
2
3
r=2+4cosHthetaL, theta=2 Pi�6
1 2 3 4 5 6
-3
-2
-1
1
2
3
r=2+4cosHthetaL, theta=3 Pi�6
1 2 3 4 5 6
-3
-2
-1
1
2
3
r=2+4cosHthetaL, theta=4 Pi�6
1 2 3 4 5 6
-3
-2
-1
1
2
3
r=2+4cosHthetaL, theta=5 Pi�6
1 2 3 4 5 6
-3
-2
-1
1
2
3
r=2+4cosHthetaL, theta=6 Pi�6
1 2 3 4 5 6
-3
-2
-1
1
2
3
r=2+4cosHthetaL, theta=7 Pi�6
1 2 3 4 5 6
-3
-2
-1
1
2
3
r=2+4cosHthetaL, theta=8 Pi�6
1 2 3 4 5 6
-3
-2
-1
1
2
3
r=2+4cosHthetaL, theta=9 Pi�6
1 2 3 4 5 6
-3
-2
-1
1
2
3
r=2+4cosHthetaL, theta=10 Pi�6
1 2 3 4 5 6
-3
-2
-1
1
2
3
r=2+4cosHthetaL, theta=11 Pi�6
1 2 3 4 5 6
-3
-2
-1
1
2
3
r=2+4cosHthetaL, theta=12 Pi�6
8
EXAMPLE: Sketch the curve r = cos(2θ), 0 ≤ θ ≤ 2π (four-leaved rose).
Solution: We have
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=cosH2thetaL, theta=Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=cosH2thetaL, theta=2 Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=cosH2thetaL, theta=3 Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=cosH2thetaL, theta=4 Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=cosH2thetaL, theta=5 Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=cosH2thetaL, theta=6 Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=cosH2thetaL, theta=7 Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=cosH2thetaL, theta=8 Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=cosH2thetaL, theta=9 Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=cosH2thetaL, theta=10 Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=cosH2thetaL, theta=11 Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=cosH2thetaL, theta=12 Pi�6
9
EXAMPLE: Sketch the curve r = sin(2θ), 0 ≤ θ ≤ 2π (four-leaved rose).
Solution: We have
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=sinH2thetaL, theta=Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=sinH2thetaL, theta=2 Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=sinH2thetaL, theta=3 Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=sinH2thetaL, theta=4 Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=sinH2thetaL, theta=5 Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=sinH2thetaL, theta=6 Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=sinH2thetaL, theta=7 Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=sinH2thetaL, theta=8 Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=sinH2thetaL, theta=9 Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=sinH2thetaL, theta=10 Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=sinH2thetaL, theta=11 Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=sinH2thetaL, theta=12 Pi�6
10
EXAMPLE: Sketch the curve r = sin(3θ), 0 ≤ θ ≤ π (three-leaved rose).
Solution: We have
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=sinH3thetaL, theta=Pi�12
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=sinH3thetaL, theta=2 Pi�12
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=sinH3thetaL, theta=3 Pi�12
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=sinH3thetaL, theta=4 Pi�12
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=sinH3thetaL, theta=5 Pi�12
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=sinH3thetaL, theta=6 Pi�12
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=sinH3thetaL, theta=7 Pi�12
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=sinH3thetaL, theta=8 Pi�12
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=sinH3thetaL, theta=9 Pi�12
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=sinH3thetaL, theta=10 Pi�12
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=sinH3thetaL, theta=11 Pi�12
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=sinH3thetaL, theta=12 Pi�12
11
EXAMPLE: Sketch the curve r = sin(4θ), 0 ≤ θ ≤ 2π (eight-leaved rose).
Solution: We have
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=sinH4thetaL, theta=Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=sinH4thetaL, theta=2 Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=sinH4thetaL, theta=3 Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=sinH4thetaL, theta=4 Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=sinH4thetaL, theta=5 Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=sinH4thetaL, theta=6 Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=sinH4thetaL, theta=7 Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=sinH4thetaL, theta=8 Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=sinH4thetaL, theta=9 Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=sinH4thetaL, theta=10 Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=sinH4thetaL, theta=11 Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=sinH4thetaL, theta=12 Pi�6
12
EXAMPLE: Sketch the curve r = sin(5θ), 0 ≤ θ ≤ 2π (five-leaved rose).
Solution: We have
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=sinH5thetaL, theta=Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=sinH5thetaL, theta=2 Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=sinH5thetaL, theta=3 Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=sinH5thetaL, theta=4 Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=sinH5thetaL, theta=5 Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=sinH5thetaL, theta=6 Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=sinH5thetaL, theta=7 Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=sinH5thetaL, theta=8 Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=sinH5thetaL, theta=9 Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=sinH5thetaL, theta=10 Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=sinH5thetaL, theta=11 Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=sinH5thetaL, theta=12 Pi�6
13
EXAMPLE: Sketch the curve r = sin(6θ), 0 ≤ θ ≤ 2π (twelve-leaved rose).
Solution: We have
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=sinH6thetaL, theta=Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=sinH6thetaL, theta=2 Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=sinH6thetaL, theta=3 Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=sinH6thetaL, theta=4 Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=sinH6thetaL, theta=5 Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=sinH6thetaL, theta=6 Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=sinH6thetaL, theta=7 Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=sinH6thetaL, theta=8 Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=sinH6thetaL, theta=9 Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=sinH6thetaL, theta=10 Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=sinH6thetaL, theta=11 Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=sinH6thetaL, theta=12 Pi�6
14
EXAMPLE: Sketch the curve r = sin(7θ), 0 ≤ θ ≤ 2π (seven-leaved rose).
Solution: We have
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=sinH7thetaL, theta=Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=sinH7thetaL, theta=2 Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=sinH7thetaL, theta=3 Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=sinH7thetaL, theta=4 Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=sinH7thetaL, theta=5 Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=sinH7thetaL, theta=6 Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=sinH7thetaL, theta=7 Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=sinH7thetaL, theta=8 Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=sinH7thetaL, theta=9 Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=sinH7thetaL, theta=10 Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=sinH7thetaL, theta=11 Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=sinH7thetaL, theta=12 Pi�6
15
EXAMPLE: Sketch the curve r = 1 +1
10sin(10θ), 0 ≤ θ ≤ 2π.
Solution: We have
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=1+sinH10thetaL�10, theta=Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=1+sinH10thetaL�10, theta=2 Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=1+sinH10thetaL�10, theta=3 Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=1+sinH10thetaL�10, theta=4 Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=1+sinH10thetaL�10, theta=5 Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=1+sinH10thetaL�10, theta=6 Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=1+sinH10thetaL�10, theta=7 Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=1+sinH10thetaL�10, theta=8 Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=1+sinH10thetaL�10, theta=9 Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=1+sinH10thetaL�10, theta=10 Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=1+sinH10thetaL�10, theta=11 Pi�6
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
r=1+sinH10thetaL�10, theta=12 Pi�6
16
EXAMPLE: Match the polar equations with the graphs labeled I-VI:
(a) r = sin(θ/2) (b) r = sin(θ/4)
(c) r = sin θ + sin3(5θ/2) (d) r = θ sin θ
(e) r = 1 + 4 cos(5θ) (f) r = 1/√θ
17
