Section 8.2 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 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
EXAMPLE: Sketch the curve r = 2, 0 ≤ θ ≤ 2π. Solution 1: Since r = 2, it follows that r2 = 4. But r2 = x2 + y 2 , therefore x2 + y 2 = 4 which is a circle of radius 2 with the center at the origin. Solution 2: We have r=2, theta=Pi6
-2
r=2, theta=2 Pi6 2
2
1
1
1
1
-1
2
-2
-2
1
-1
-1
-2
-2
-2
r=2, theta=5 Pi6 2
2
1
1
1
1
-1
2
-2
1
-1
2
-2
1
-1
-1
-1
-1
-2
-2
-2
r=2, theta=8 Pi6 2
2
1
1
1
1
2
-2
1
-1
2
-2
1
-1
-1
-1
-1
-2
-2
-2
r=2, theta=11 Pi6 2
2
1
1
1
1
2
-2
1
-1
2
-2
1
-1
-1
-1
-1
-2
-2
-2
2
2
r=2, theta=12 Pi6
2
-1
2
r=2, theta=9 Pi6
2
-1
2
r=2, theta=6 Pi6
2
r=2, theta=10 Pi6
-2
2
-1
r=2, theta=7 Pi6
-2
1
-1
-1
r=2, theta=4 Pi6
-2
r=2, theta=3 Pi6
2
2
EXAMPLE: Sketch the curve r = θ, 0 ≤ θ ≤ 4π. Solution: We have r=theta, theta=Pi3
-10
r=theta, theta=2 Pi3
10
10
10
5
5
5
5
-5
10
-10
-10
5
-5 -5
-10
-10
-10
r=theta, theta=5 Pi3 10
10
5
5
5
5
-5
10
-10
5
-5
10
-10
5
-5
-5
-5
-5
-10
-10
-10
r=theta, theta=8 Pi3 10
10
5
5
5
5
10
-10
5
-5
10
-10
5
-5
-5
-5
-5
-10
-10
-10
r=theta, theta=11 Pi3 10
10
5
5
5
5
10
-10
5
-5
10
-10
5
-5
-5
-5
-5
-10
-10
-10
3
10
r=theta, theta=12 Pi3
10
-5
10
r=theta, theta=9 Pi3
10
-5
10
r=theta, theta=6 Pi3
10
r=theta, theta=10 Pi3
-10
10
-5
r=theta, theta=7 Pi3
-10
5
-5
-5
r=theta, theta=4 Pi3
-10
r=theta, theta=3 Pi3
10
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 + y 2 and r cos θ = x, therefore x2 + y 2 = 2x. This can be rewritten as (x − 1)2 + y 2 = 1 which is a circle of radius 1 with the center at (1, 0). Solution 2: We have r=2cosHthetaL, theta=Pi12
r=2cosHthetaL, theta=2 Pi12
r=2cosHthetaL, theta=3 Pi12
1.0
1.0
1.0
0.5
0.5
0.5
0.5
-0.5
1.0
1.5
2.0
0.5
-0.5
1.0
1.5
2.0
0.5
-0.5
-0.5
-0.5
-0.5
-1.0
-1.0
-1.0
r=2cosHthetaL, theta=4 Pi12
r=2cosHthetaL, theta=5 Pi12 1.0
1.0
0.5
0.5
0.5
0.5
1.0
1.5
2.0
0.5
-0.5
1.0
1.5
2.0
0.5
-0.5
-0.5
-0.5
-0.5
-1.0
-1.0
-1.0
r=2cosHthetaL, theta=7 Pi12
r=2cosHthetaL, theta=8 Pi12 1.0
1.0
0.5
0.5
0.5
0.5
1.0
1.5
2.0
0.5
-0.5
1.0
1.5
2.0
0.5
-0.5
-0.5
-0.5
-0.5
-1.0
-1.0
-1.0
1.0
1.5
1.0
1.5
r=2cosHthetaL, theta=10 Pi12
r=2cosHthetaL, theta=11 Pi12
r=2cosHthetaL, theta=12 Pi12
1.0
1.0
1.0
0.5
0.5
0.5
0.5
-0.5
1.0
1.5
2.0
0.5
-0.5
1.0
1.5
2.0
0.5
-0.5
-0.5
-0.5
-0.5
-1.0
-1.0
-1.0
4
2.0
2.0
r=2cosHthetaL, theta=9 Pi12
1.0
-0.5
1.5
r=2cosHthetaL, theta=6 Pi12
1.0
-0.5
1.0
1.0
1.5
2.0
2.0
EXAMPLES:
5
EXAMPLE: Sketch the curve r = 1 + sin θ, 0 ≤ θ ≤ 2π (cardioid). Solution: We have r=1+sinHthetaL, theta=Pi6
-1.5
-1.0
r=1+sinHthetaL, theta=2 Pi6
2.0
2.0
2.0
1.5
1.5
1.5
1.0
1.0
1.0
0.5
0.5
0.5
0.5
-0.5
1.0
1.5
-1.5
-1.0
0.5
-0.5
1.5
1.5
1.0
1.0
1.0
0.5
0.5
0.5
0.5
1.0
1.5
-1.5
-1.0
0.5
-0.5
1.0
1.5
-1.5
-1.0
0.5
-0.5
-0.5
r=1+sinHthetaL, theta=8 Pi6
2.0
1.5
1.5
1.5
1.0
1.0
1.0
0.5
0.5
0.5
1.0
1.5
-1.5
-1.0
0.5
-0.5
1.0
1.5
-1.5
-1.0
0.5
-0.5
-0.5
r=1+sinHthetaL, theta=11 Pi6
2.0
1.5
1.5
1.5
1.0
1.0
1.0
0.5
0.5
0.5
-0.5
1.0
1.5
-1.5
-1.0
0.5
-0.5 -0.5
6
1.5
r=1+sinHthetaL, theta=12 Pi6
2.0
0.5
1.0
-0.5
2.0
-0.5
1.5
r=1+sinHthetaL, theta=9 Pi6
2.0
0.5
1.0
-0.5
2.0
-0.5
1.5
r=1+sinHthetaL, theta=6 Pi6
1.5
-0.5
1.0
-0.5
2.0
r=1+sinHthetaL, theta=10 Pi6
-1.0
-1.0
2.0
-0.5
-1.5
-1.5
2.0
r=1+sinHthetaL, theta=7 Pi6
-1.0
1.5
r=1+sinHthetaL, theta=5 Pi6
-0.5
-1.5
1.0
-0.5
r=1+sinHthetaL, theta=4 Pi6
-1.0
0.5
-0.5
-0.5
-1.5
r=1+sinHthetaL, theta=3 Pi6
1.0
1.5
-1.5
-1.0
0.5
-0.5 -0.5
1.0
1.5
EXAMPLE: Sketch the curve r = 1 − cos θ, 0 ≤ θ ≤ 2π (cardioid). Solution: We have r=1-cosHthetaL, theta=Pi6
-2.0
-1.5
-1.0
r=1-cosHthetaL, theta=2 Pi6 1.5
1.5
1.0
1.0
1.0
0.5
0.5
0.5
0.5
-0.5
-1.5
-1.0
-1.5
-1.0
-1.5
-1.0
-1.0
0.5
-0.5
-2.0
-1.5
-1.0
0.5
-0.5 -0.5
-1.0
-1.0
-1.0
-1.5
-1.5
-1.5
r=1-cosHthetaL, theta=5 Pi6
r=1-cosHthetaL, theta=6 Pi6
1.5
1.5
1.5
1.0
1.0
1.0
0.5
0.5
0.5
0.5
-0.5
-2.0
-1.5
-1.0
0.5
-0.5
-2.0
-1.5
-1.0
0.5
-0.5
-0.5
-0.5
-0.5
-1.0
-1.0
-1.0
-1.5
-1.5
-1.5
r=1-cosHthetaL, theta=8 Pi6
r=1-cosHthetaL, theta=9 Pi6
1.5
1.5
1.5
1.0
1.0
1.0
0.5
0.5
0.5
0.5
-0.5
-2.0
-1.5
-1.0
0.5
-0.5
-2.0
-1.5
-1.0
0.5
-0.5
-0.5
-0.5
-0.5
-1.0
-1.0
-1.0
-1.5
-1.5
-1.5
r=1-cosHthetaL, theta=10 Pi6
-2.0
-1.5
-0.5
r=1-cosHthetaL, theta=7 Pi6
-2.0
-2.0
-0.5
r=1-cosHthetaL, theta=4 Pi6
-2.0
r=1-cosHthetaL, theta=3 Pi6
1.5
r=1-cosHthetaL, theta=11 Pi6
r=1-cosHthetaL, theta=12 Pi6
1.5
1.5
1.5
1.0
1.0
1.0
0.5
0.5
0.5
0.5
-0.5
-2.0
-1.5
-1.0
0.5
-0.5
-2.0
-1.5
-1.0
0.5
-0.5
-0.5
-0.5
-0.5
-1.0
-1.0
-1.0
-1.5
-1.5
-1.5
7
EXAMPLE: Sketch the curve r = 2 + 4 cos θ, 0 ≤ θ ≤ 2π. Solution: We have r=2+4cosHthetaL, theta=Pi6
r=2+4cosHthetaL, theta=2 Pi6
r=2+4cosHthetaL, theta=3 Pi6
3
3
3
2
2
2
1
1
1
1
2
3
4
5
6
1
2
3
4
5
6
1
-1
-1
-1
-2
-2
-2
-3
-3
-3
r=2+4cosHthetaL, theta=4 Pi6
r=2+4cosHthetaL, theta=5 Pi6 3
3
2
2
2
1
1
1
2
3
4
5
6
1
2
3
4
5
6
1
-1
-1
-1
-2
-2
-2
-3
-3
-3
r=2+4cosHthetaL, theta=7 Pi6
r=2+4cosHthetaL, theta=8 Pi6 3
3
2
2
2
1
1
1
2
3
4
5
6
1
2
3
4
5
6
1
-1
-1
-1
-2
-2
-2
-3
-3
-3
r=2+4cosHthetaL, theta=10 Pi6
r=2+4cosHthetaL, theta=11 Pi6 3
3
2
2
2
1
1
1
2
3
4
5
6
1
2
3
4
5
6
1
-1
-1
-1
-2
-2
-2
-3
-3
-3
8
5
6
2
3
4
5
6
2
3
4
5
6
r=2+4cosHthetaL, theta=12 Pi6
3
1
4
r=2+4cosHthetaL, theta=9 Pi6
3
1
3
r=2+4cosHthetaL, theta=6 Pi6
3
1
2
2
3
4
5
6
EXAMPLE: Sketch the curve r = cos(2θ), 0 ≤ θ ≤ 2π (four-leaved rose). Solution: We have r=cosH2thetaL, theta=Pi6
-1.0
r=cosH2thetaL, theta=2 Pi6
1.0
1.0
1.0
0.5
0.5
0.5
0.5
-0.5
1.0
-1.0
-1.0
0.5
-0.5 -0.5
-1.0
-1.0
-1.0
r=cosH2thetaL, theta=5 Pi6 1.0
1.0
0.5
0.5
0.5
0.5
-0.5
1.0
-1.0
0.5
-0.5
1.0
-1.0
0.5
-0.5
-0.5
-0.5
-0.5
-1.0
-1.0
-1.0
r=cosH2thetaL, theta=8 Pi6 1.0
1.0
0.5
0.5
0.5
0.5
1.0
-1.0
0.5
-0.5
1.0
-1.0
0.5
-0.5
-0.5
-0.5
-0.5
-1.0
-1.0
-1.0
r=cosH2thetaL, theta=11 Pi6 1.0
1.0
0.5
0.5
0.5
0.5
1.0
-1.0
0.5
-0.5
1.0
-1.0
0.5
-0.5
-0.5
-0.5
-0.5
-1.0
-1.0
-1.0
9
1.0
r=cosH2thetaL, theta=12 Pi6
1.0
-0.5
1.0
r=cosH2thetaL, theta=9 Pi6
1.0
-0.5
1.0
r=cosH2thetaL, theta=6 Pi6
1.0
r=cosH2thetaL, theta=10 Pi6
-1.0
1.0
-0.5
r=cosH2thetaL, theta=7 Pi6
-1.0
0.5
-0.5
-0.5
r=cosH2thetaL, theta=4 Pi6
-1.0
r=cosH2thetaL, theta=3 Pi6
1.0
EXAMPLE: Sketch the curve r = sin(2θ), 0 ≤ θ ≤ 2π (four-leaved rose). Solution: We have r=sinH2thetaL, theta=Pi6
-1.0
r=sinH2thetaL, theta=2 Pi6
1.0
1.0
1.0
0.5
0.5
0.5
0.5
-0.5
1.0
-1.0
-1.0
0.5
-0.5 -0.5
-1.0
-1.0
-1.0
r=sinH2thetaL, theta=5 Pi6 1.0
1.0
0.5
0.5
0.5
0.5
-0.5
1.0
-1.0
0.5
-0.5
1.0
-1.0
0.5
-0.5
-0.5
-0.5
-0.5
-1.0
-1.0
-1.0
r=sinH2thetaL, theta=8 Pi6 1.0
1.0
0.5
0.5
0.5
0.5
1.0
-1.0
0.5
-0.5
1.0
-1.0
0.5
-0.5
-0.5
-0.5
-0.5
-1.0
-1.0
-1.0
r=sinH2thetaL, theta=11 Pi6 1.0
1.0
0.5
0.5
0.5
0.5
1.0
-1.0
0.5
-0.5
1.0
-1.0
0.5
-0.5
-0.5
-0.5
-0.5
-1.0
-1.0
-1.0
10
1.0
r=sinH2thetaL, theta=12 Pi6
1.0
-0.5
1.0
r=sinH2thetaL, theta=9 Pi6
1.0
-0.5
1.0
r=sinH2thetaL, theta=6 Pi6
1.0
r=sinH2thetaL, theta=10 Pi6
-1.0
1.0
-0.5
r=sinH2thetaL, theta=7 Pi6
-1.0
0.5
-0.5
-0.5
r=sinH2thetaL, theta=4 Pi6
-1.0
r=sinH2thetaL, theta=3 Pi6
1.0
EXAMPLE: Sketch the curve r = sin(3θ), 0 ≤ θ ≤ π (three-leaved rose). Solution: We have r=sinH3thetaL, theta=Pi12
-1.0
r=sinH3thetaL, theta=2 Pi12
1.0
1.0
1.0
0.5
0.5
0.5
0.5
-0.5
1.0
1.0
-1.0
0.5
-0.5 -0.5
-1.0
-1.0
-1.0
r=sinH3thetaL, theta=5 Pi12 1.0
1.0
0.5
0.5
0.5
0.5
-0.5
1.0
-1.0
0.5
-0.5
1.0
-1.0
0.5
-0.5
-0.5
-0.5
-0.5
-1.0
-1.0
-1.0
r=sinH3thetaL, theta=8 Pi12 1.0
1.0
0.5
0.5
0.5
0.5
1.0
-1.0
0.5
-0.5
1.0
-1.0
0.5
-0.5
-0.5
-0.5
-0.5
-1.0
-1.0
-1.0
r=sinH3thetaL, theta=11 Pi12 1.0
1.0
0.5
0.5
0.5
0.5
1.0
-1.0
0.5
-0.5
1.0
-1.0
0.5
-0.5
-0.5
-0.5
-0.5
-1.0
-1.0
-1.0
11
1.0
r=sinH3thetaL, theta=12 Pi12
1.0
-0.5
1.0
r=sinH3thetaL, theta=9 Pi12
1.0
-0.5
1.0
r=sinH3thetaL, theta=6 Pi12
1.0
r=sinH3thetaL, theta=10 Pi12
-1.0
0.5
-0.5 -0.5
r=sinH3thetaL, theta=7 Pi12
-1.0
-1.0
-0.5
r=sinH3thetaL, theta=4 Pi12
-1.0
r=sinH3thetaL, theta=3 Pi12
1.0
EXAMPLE: Sketch the curve r = sin(4θ), 0 ≤ θ ≤ 2π (eight-leaved rose). Solution: We have r=sinH4thetaL, theta=Pi6
-1.0
r=sinH4thetaL, theta=2 Pi6
1.0
1.0
1.0
0.5
0.5
0.5
0.5
-0.5
1.0
-1.0
-1.0
0.5
-0.5 -0.5
-1.0
-1.0
-1.0
r=sinH4thetaL, theta=5 Pi6 1.0
1.0
0.5
0.5
0.5
0.5
-0.5
1.0
-1.0
0.5
-0.5
1.0
-1.0
0.5
-0.5
-0.5
-0.5
-0.5
-1.0
-1.0
-1.0
r=sinH4thetaL, theta=8 Pi6 1.0
1.0
0.5
0.5
0.5
0.5
1.0
-1.0
0.5
-0.5
1.0
-1.0
0.5
-0.5
-0.5
-0.5
-0.5
-1.0
-1.0
-1.0
r=sinH4thetaL, theta=11 Pi6 1.0
1.0
0.5
0.5
0.5
0.5
1.0
-1.0
0.5
-0.5
1.0
-1.0
0.5
-0.5
-0.5
-0.5
-0.5
-1.0
-1.0
-1.0
12
1.0
r=sinH4thetaL, theta=12 Pi6
1.0
-0.5
1.0
r=sinH4thetaL, theta=9 Pi6
1.0
-0.5
1.0
r=sinH4thetaL, theta=6 Pi6
1.0
r=sinH4thetaL, theta=10 Pi6
-1.0
1.0
-0.5
r=sinH4thetaL, theta=7 Pi6
-1.0
0.5
-0.5
-0.5
r=sinH4thetaL, theta=4 Pi6
-1.0
r=sinH4thetaL, theta=3 Pi6
1.0
EXAMPLE: Sketch the curve r = sin(5θ), 0 ≤ θ ≤ 2π (five-leaved rose). Solution: We have r=sinH5thetaL, theta=Pi6
-1.0
r=sinH5thetaL, theta=2 Pi6
1.0
1.0
1.0
0.5
0.5
0.5
0.5
-0.5
1.0
-1.0
-1.0
0.5
-0.5 -0.5
-1.0
-1.0
-1.0
r=sinH5thetaL, theta=5 Pi6 1.0
1.0
0.5
0.5
0.5
0.5
-0.5
1.0
-1.0
0.5
-0.5
1.0
-1.0
0.5
-0.5
-0.5
-0.5
-0.5
-1.0
-1.0
-1.0
r=sinH5thetaL, theta=8 Pi6 1.0
1.0
0.5
0.5
0.5
0.5
1.0
-1.0
0.5
-0.5
1.0
-1.0
0.5
-0.5
-0.5
-0.5
-0.5
-1.0
-1.0
-1.0
r=sinH5thetaL, theta=11 Pi6 1.0
1.0
0.5
0.5
0.5
0.5
1.0
-1.0
0.5
-0.5
1.0
-1.0
0.5
-0.5
-0.5
-0.5
-0.5
-1.0
-1.0
-1.0
13
1.0
r=sinH5thetaL, theta=12 Pi6
1.0
-0.5
1.0
r=sinH5thetaL, theta=9 Pi6
1.0
-0.5
1.0
r=sinH5thetaL, theta=6 Pi6
1.0
r=sinH5thetaL, theta=10 Pi6
-1.0
1.0
-0.5
r=sinH5thetaL, theta=7 Pi6
-1.0
0.5
-0.5
-0.5
r=sinH5thetaL, theta=4 Pi6
-1.0
r=sinH5thetaL, theta=3 Pi6
1.0
EXAMPLE: Sketch the curve r = sin(6θ), 0 ≤ θ ≤ 2π (twelve-leaved rose). Solution: We have r=sinH6thetaL, theta=Pi6
-1.0
r=sinH6thetaL, theta=2 Pi6
1.0
1.0
1.0
0.5
0.5
0.5
0.5
-0.5
1.0
-1.0
-1.0
0.5
-0.5 -0.5
-1.0
-1.0
-1.0
r=sinH6thetaL, theta=5 Pi6 1.0
1.0
0.5
0.5
0.5
0.5
-0.5
1.0
-1.0
0.5
-0.5
1.0
-1.0
0.5
-0.5
-0.5
-0.5
-0.5
-1.0
-1.0
-1.0
r=sinH6thetaL, theta=8 Pi6 1.0
1.0
0.5
0.5
0.5
0.5
1.0
-1.0
0.5
-0.5
1.0
-1.0
0.5
-0.5
-0.5
-0.5
-0.5
-1.0
-1.0
-1.0
r=sinH6thetaL, theta=11 Pi6 1.0
1.0
0.5
0.5
0.5
0.5
1.0
-1.0
0.5
-0.5
1.0
-1.0
0.5
-0.5
-0.5
-0.5
-0.5
-1.0
-1.0
-1.0
14
1.0
r=sinH6thetaL, theta=12 Pi6
1.0
-0.5
1.0
r=sinH6thetaL, theta=9 Pi6
1.0
-0.5
1.0
r=sinH6thetaL, theta=6 Pi6
1.0
r=sinH6thetaL, theta=10 Pi6
-1.0
1.0
-0.5
r=sinH6thetaL, theta=7 Pi6
-1.0
0.5
-0.5
-0.5
r=sinH6thetaL, theta=4 Pi6
-1.0
r=sinH6thetaL, theta=3 Pi6
1.0
EXAMPLE: Sketch the curve r = sin(7θ), 0 ≤ θ ≤ 2π (seven-leaved rose). Solution: We have r=sinH7thetaL, theta=Pi6
-1.0
r=sinH7thetaL, theta=2 Pi6
1.0
1.0
1.0
0.5
0.5
0.5
0.5
-0.5
1.0
-1.0
-1.0
0.5
-0.5 -0.5
-1.0
-1.0
-1.0
r=sinH7thetaL, theta=5 Pi6 1.0
1.0
0.5
0.5
0.5
0.5
-0.5
1.0
-1.0
0.5
-0.5
1.0
-1.0
0.5
-0.5
-0.5
-0.5
-0.5
-1.0
-1.0
-1.0
r=sinH7thetaL, theta=8 Pi6 1.0
1.0
0.5
0.5
0.5
0.5
1.0
-1.0
0.5
-0.5
1.0
-1.0
0.5
-0.5
-0.5
-0.5
-0.5
-1.0
-1.0
-1.0
r=sinH7thetaL, theta=11 Pi6 1.0
1.0
0.5
0.5
0.5
0.5
1.0
-1.0
0.5
-0.5
1.0
-1.0
0.5
-0.5
-0.5
-0.5
-0.5
-1.0
-1.0
-1.0
15
1.0
r=sinH7thetaL, theta=12 Pi6
1.0
-0.5
1.0
r=sinH7thetaL, theta=9 Pi6
1.0
-0.5
1.0
r=sinH7thetaL, theta=6 Pi6
1.0
r=sinH7thetaL, theta=10 Pi6
-1.0
1.0
-0.5
r=sinH7thetaL, theta=7 Pi6
-1.0
0.5
-0.5
-0.5
r=sinH7thetaL, theta=4 Pi6
-1.0
r=sinH7thetaL, theta=3 Pi6
1.0
EXAMPLE: Sketch the curve r = 1 +
1 sin(10θ), 0 ≤ θ ≤ 2π. 10
Solution: We have r=1+sinH10thetaL10, theta=Pi6
-1.0
1.0
1.0
0.5
0.5
0.5
0.5
-0.5
1.0
0.5
-0.5
1.0
-1.0
0.5
-0.5
-0.5
-0.5
-1.0
-1.0
-1.0
r=1+sinH10thetaL10, theta=5 Pi6 1.0
1.0
0.5
0.5
0.5
0.5
-0.5
1.0
-1.0
0.5
-0.5
1.0
-1.0
0.5
-0.5
-0.5
-0.5
-0.5
-1.0
-1.0
-1.0
r=1+sinH10thetaL10, theta=8 Pi6 1.0
1.0
0.5
0.5
0.5
0.5
1.0
-1.0
0.5
-0.5
1.0
-1.0
0.5
-0.5
-0.5
-0.5
-0.5
-1.0
-1.0
-1.0
r=1+sinH10thetaL10, theta=11 Pi6 1.0
1.0
0.5
0.5
0.5
0.5
1.0
-1.0
0.5
-0.5
1.0
-1.0
0.5
-0.5
-0.5
-0.5
-0.5
-1.0
-1.0
-1.0
16
1.0
r=1+sinH10thetaL10, theta=12 Pi6
1.0
-0.5
1.0
r=1+sinH10thetaL10, theta=9 Pi6
1.0
-0.5
1.0
r=1+sinH10thetaL10, theta=6 Pi6
1.0
r=1+sinH10thetaL10, theta=10 Pi6
-1.0
-1.0
-0.5
r=1+sinH10thetaL10, theta=7 Pi6
-1.0
r=1+sinH10thetaL10, theta=3 Pi6
1.0
r=1+sinH10thetaL10, theta=4 Pi6
-1.0
r=1+sinH10thetaL10, theta=2 Pi6
1.0
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)
17
√ r = 1/ θ
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
EXAMPLE: Sketch the curve r = 2, 0 ≤ θ ≤ 2π. Solution 1: Since r = 2, it follows that r2 = 4. But r2 = x2 + y 2 , therefore x2 + y 2 = 4 which is a circle of radius 2 with the center at the origin. Solution 2: We have r=2, theta=Pi6
-2
r=2, theta=2 Pi6 2
2
1
1
1
1
-1
2
-2
-2
1
-1
-1
-2
-2
-2
r=2, theta=5 Pi6 2
2
1
1
1
1
-1
2
-2
1
-1
2
-2
1
-1
-1
-1
-1
-2
-2
-2
r=2, theta=8 Pi6 2
2
1
1
1
1
2
-2
1
-1
2
-2
1
-1
-1
-1
-1
-2
-2
-2
r=2, theta=11 Pi6 2
2
1
1
1
1
2
-2
1
-1
2
-2
1
-1
-1
-1
-1
-2
-2
-2
2
2
r=2, theta=12 Pi6
2
-1
2
r=2, theta=9 Pi6
2
-1
2
r=2, theta=6 Pi6
2
r=2, theta=10 Pi6
-2
2
-1
r=2, theta=7 Pi6
-2
1
-1
-1
r=2, theta=4 Pi6
-2
r=2, theta=3 Pi6
2
2
EXAMPLE: Sketch the curve r = θ, 0 ≤ θ ≤ 4π. Solution: We have r=theta, theta=Pi3
-10
r=theta, theta=2 Pi3
10
10
10
5
5
5
5
-5
10
-10
-10
5
-5 -5
-10
-10
-10
r=theta, theta=5 Pi3 10
10
5
5
5
5
-5
10
-10
5
-5
10
-10
5
-5
-5
-5
-5
-10
-10
-10
r=theta, theta=8 Pi3 10
10
5
5
5
5
10
-10
5
-5
10
-10
5
-5
-5
-5
-5
-10
-10
-10
r=theta, theta=11 Pi3 10
10
5
5
5
5
10
-10
5
-5
10
-10
5
-5
-5
-5
-5
-10
-10
-10
3
10
r=theta, theta=12 Pi3
10
-5
10
r=theta, theta=9 Pi3
10
-5
10
r=theta, theta=6 Pi3
10
r=theta, theta=10 Pi3
-10
10
-5
r=theta, theta=7 Pi3
-10
5
-5
-5
r=theta, theta=4 Pi3
-10
r=theta, theta=3 Pi3
10
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 + y 2 and r cos θ = x, therefore x2 + y 2 = 2x. This can be rewritten as (x − 1)2 + y 2 = 1 which is a circle of radius 1 with the center at (1, 0). Solution 2: We have r=2cosHthetaL, theta=Pi12
r=2cosHthetaL, theta=2 Pi12
r=2cosHthetaL, theta=3 Pi12
1.0
1.0
1.0
0.5
0.5
0.5
0.5
-0.5
1.0
1.5
2.0
0.5
-0.5
1.0
1.5
2.0
0.5
-0.5
-0.5
-0.5
-0.5
-1.0
-1.0
-1.0
r=2cosHthetaL, theta=4 Pi12
r=2cosHthetaL, theta=5 Pi12 1.0
1.0
0.5
0.5
0.5
0.5
1.0
1.5
2.0
0.5
-0.5
1.0
1.5
2.0
0.5
-0.5
-0.5
-0.5
-0.5
-1.0
-1.0
-1.0
r=2cosHthetaL, theta=7 Pi12
r=2cosHthetaL, theta=8 Pi12 1.0
1.0
0.5
0.5
0.5
0.5
1.0
1.5
2.0
0.5
-0.5
1.0
1.5
2.0
0.5
-0.5
-0.5
-0.5
-0.5
-1.0
-1.0
-1.0
1.0
1.5
1.0
1.5
r=2cosHthetaL, theta=10 Pi12
r=2cosHthetaL, theta=11 Pi12
r=2cosHthetaL, theta=12 Pi12
1.0
1.0
1.0
0.5
0.5
0.5
0.5
-0.5
1.0
1.5
2.0
0.5
-0.5
1.0
1.5
2.0
0.5
-0.5
-0.5
-0.5
-0.5
-1.0
-1.0
-1.0
4
2.0
2.0
r=2cosHthetaL, theta=9 Pi12
1.0
-0.5
1.5
r=2cosHthetaL, theta=6 Pi12
1.0
-0.5
1.0
1.0
1.5
2.0
2.0
EXAMPLES:
5
EXAMPLE: Sketch the curve r = 1 + sin θ, 0 ≤ θ ≤ 2π (cardioid). Solution: We have r=1+sinHthetaL, theta=Pi6
-1.5
-1.0
r=1+sinHthetaL, theta=2 Pi6
2.0
2.0
2.0
1.5
1.5
1.5
1.0
1.0
1.0
0.5
0.5
0.5
0.5
-0.5
1.0
1.5
-1.5
-1.0
0.5
-0.5
1.5
1.5
1.0
1.0
1.0
0.5
0.5
0.5
0.5
1.0
1.5
-1.5
-1.0
0.5
-0.5
1.0
1.5
-1.5
-1.0
0.5
-0.5
-0.5
r=1+sinHthetaL, theta=8 Pi6
2.0
1.5
1.5
1.5
1.0
1.0
1.0
0.5
0.5
0.5
1.0
1.5
-1.5
-1.0
0.5
-0.5
1.0
1.5
-1.5
-1.0
0.5
-0.5
-0.5
r=1+sinHthetaL, theta=11 Pi6
2.0
1.5
1.5
1.5
1.0
1.0
1.0
0.5
0.5
0.5
-0.5
1.0
1.5
-1.5
-1.0
0.5
-0.5 -0.5
6
1.5
r=1+sinHthetaL, theta=12 Pi6
2.0
0.5
1.0
-0.5
2.0
-0.5
1.5
r=1+sinHthetaL, theta=9 Pi6
2.0
0.5
1.0
-0.5
2.0
-0.5
1.5
r=1+sinHthetaL, theta=6 Pi6
1.5
-0.5
1.0
-0.5
2.0
r=1+sinHthetaL, theta=10 Pi6
-1.0
-1.0
2.0
-0.5
-1.5
-1.5
2.0
r=1+sinHthetaL, theta=7 Pi6
-1.0
1.5
r=1+sinHthetaL, theta=5 Pi6
-0.5
-1.5
1.0
-0.5
r=1+sinHthetaL, theta=4 Pi6
-1.0
0.5
-0.5
-0.5
-1.5
r=1+sinHthetaL, theta=3 Pi6
1.0
1.5
-1.5
-1.0
0.5
-0.5 -0.5
1.0
1.5
EXAMPLE: Sketch the curve r = 1 − cos θ, 0 ≤ θ ≤ 2π (cardioid). Solution: We have r=1-cosHthetaL, theta=Pi6
-2.0
-1.5
-1.0
r=1-cosHthetaL, theta=2 Pi6 1.5
1.5
1.0
1.0
1.0
0.5
0.5
0.5
0.5
-0.5
-1.5
-1.0
-1.5
-1.0
-1.5
-1.0
-1.0
0.5
-0.5
-2.0
-1.5
-1.0
0.5
-0.5 -0.5
-1.0
-1.0
-1.0
-1.5
-1.5
-1.5
r=1-cosHthetaL, theta=5 Pi6
r=1-cosHthetaL, theta=6 Pi6
1.5
1.5
1.5
1.0
1.0
1.0
0.5
0.5
0.5
0.5
-0.5
-2.0
-1.5
-1.0
0.5
-0.5
-2.0
-1.5
-1.0
0.5
-0.5
-0.5
-0.5
-0.5
-1.0
-1.0
-1.0
-1.5
-1.5
-1.5
r=1-cosHthetaL, theta=8 Pi6
r=1-cosHthetaL, theta=9 Pi6
1.5
1.5
1.5
1.0
1.0
1.0
0.5
0.5
0.5
0.5
-0.5
-2.0
-1.5
-1.0
0.5
-0.5
-2.0
-1.5
-1.0
0.5
-0.5
-0.5
-0.5
-0.5
-1.0
-1.0
-1.0
-1.5
-1.5
-1.5
r=1-cosHthetaL, theta=10 Pi6
-2.0
-1.5
-0.5
r=1-cosHthetaL, theta=7 Pi6
-2.0
-2.0
-0.5
r=1-cosHthetaL, theta=4 Pi6
-2.0
r=1-cosHthetaL, theta=3 Pi6
1.5
r=1-cosHthetaL, theta=11 Pi6
r=1-cosHthetaL, theta=12 Pi6
1.5
1.5
1.5
1.0
1.0
1.0
0.5
0.5
0.5
0.5
-0.5
-2.0
-1.5
-1.0
0.5
-0.5
-2.0
-1.5
-1.0
0.5
-0.5
-0.5
-0.5
-0.5
-1.0
-1.0
-1.0
-1.5
-1.5
-1.5
7
EXAMPLE: Sketch the curve r = 2 + 4 cos θ, 0 ≤ θ ≤ 2π. Solution: We have r=2+4cosHthetaL, theta=Pi6
r=2+4cosHthetaL, theta=2 Pi6
r=2+4cosHthetaL, theta=3 Pi6
3
3
3
2
2
2
1
1
1
1
2
3
4
5
6
1
2
3
4
5
6
1
-1
-1
-1
-2
-2
-2
-3
-3
-3
r=2+4cosHthetaL, theta=4 Pi6
r=2+4cosHthetaL, theta=5 Pi6 3
3
2
2
2
1
1
1
2
3
4
5
6
1
2
3
4
5
6
1
-1
-1
-1
-2
-2
-2
-3
-3
-3
r=2+4cosHthetaL, theta=7 Pi6
r=2+4cosHthetaL, theta=8 Pi6 3
3
2
2
2
1
1
1
2
3
4
5
6
1
2
3
4
5
6
1
-1
-1
-1
-2
-2
-2
-3
-3
-3
r=2+4cosHthetaL, theta=10 Pi6
r=2+4cosHthetaL, theta=11 Pi6 3
3
2
2
2
1
1
1
2
3
4
5
6
1
2
3
4
5
6
1
-1
-1
-1
-2
-2
-2
-3
-3
-3
8
5
6
2
3
4
5
6
2
3
4
5
6
r=2+4cosHthetaL, theta=12 Pi6
3
1
4
r=2+4cosHthetaL, theta=9 Pi6
3
1
3
r=2+4cosHthetaL, theta=6 Pi6
3
1
2
2
3
4
5
6
EXAMPLE: Sketch the curve r = cos(2θ), 0 ≤ θ ≤ 2π (four-leaved rose). Solution: We have r=cosH2thetaL, theta=Pi6
-1.0
r=cosH2thetaL, theta=2 Pi6
1.0
1.0
1.0
0.5
0.5
0.5
0.5
-0.5
1.0
-1.0
-1.0
0.5
-0.5 -0.5
-1.0
-1.0
-1.0
r=cosH2thetaL, theta=5 Pi6 1.0
1.0
0.5
0.5
0.5
0.5
-0.5
1.0
-1.0
0.5
-0.5
1.0
-1.0
0.5
-0.5
-0.5
-0.5
-0.5
-1.0
-1.0
-1.0
r=cosH2thetaL, theta=8 Pi6 1.0
1.0
0.5
0.5
0.5
0.5
1.0
-1.0
0.5
-0.5
1.0
-1.0
0.5
-0.5
-0.5
-0.5
-0.5
-1.0
-1.0
-1.0
r=cosH2thetaL, theta=11 Pi6 1.0
1.0
0.5
0.5
0.5
0.5
1.0
-1.0
0.5
-0.5
1.0
-1.0
0.5
-0.5
-0.5
-0.5
-0.5
-1.0
-1.0
-1.0
9
1.0
r=cosH2thetaL, theta=12 Pi6
1.0
-0.5
1.0
r=cosH2thetaL, theta=9 Pi6
1.0
-0.5
1.0
r=cosH2thetaL, theta=6 Pi6
1.0
r=cosH2thetaL, theta=10 Pi6
-1.0
1.0
-0.5
r=cosH2thetaL, theta=7 Pi6
-1.0
0.5
-0.5
-0.5
r=cosH2thetaL, theta=4 Pi6
-1.0
r=cosH2thetaL, theta=3 Pi6
1.0
EXAMPLE: Sketch the curve r = sin(2θ), 0 ≤ θ ≤ 2π (four-leaved rose). Solution: We have r=sinH2thetaL, theta=Pi6
-1.0
r=sinH2thetaL, theta=2 Pi6
1.0
1.0
1.0
0.5
0.5
0.5
0.5
-0.5
1.0
-1.0
-1.0
0.5
-0.5 -0.5
-1.0
-1.0
-1.0
r=sinH2thetaL, theta=5 Pi6 1.0
1.0
0.5
0.5
0.5
0.5
-0.5
1.0
-1.0
0.5
-0.5
1.0
-1.0
0.5
-0.5
-0.5
-0.5
-0.5
-1.0
-1.0
-1.0
r=sinH2thetaL, theta=8 Pi6 1.0
1.0
0.5
0.5
0.5
0.5
1.0
-1.0
0.5
-0.5
1.0
-1.0
0.5
-0.5
-0.5
-0.5
-0.5
-1.0
-1.0
-1.0
r=sinH2thetaL, theta=11 Pi6 1.0
1.0
0.5
0.5
0.5
0.5
1.0
-1.0
0.5
-0.5
1.0
-1.0
0.5
-0.5
-0.5
-0.5
-0.5
-1.0
-1.0
-1.0
10
1.0
r=sinH2thetaL, theta=12 Pi6
1.0
-0.5
1.0
r=sinH2thetaL, theta=9 Pi6
1.0
-0.5
1.0
r=sinH2thetaL, theta=6 Pi6
1.0
r=sinH2thetaL, theta=10 Pi6
-1.0
1.0
-0.5
r=sinH2thetaL, theta=7 Pi6
-1.0
0.5
-0.5
-0.5
r=sinH2thetaL, theta=4 Pi6
-1.0
r=sinH2thetaL, theta=3 Pi6
1.0
EXAMPLE: Sketch the curve r = sin(3θ), 0 ≤ θ ≤ π (three-leaved rose). Solution: We have r=sinH3thetaL, theta=Pi12
-1.0
r=sinH3thetaL, theta=2 Pi12
1.0
1.0
1.0
0.5
0.5
0.5
0.5
-0.5
1.0
1.0
-1.0
0.5
-0.5 -0.5
-1.0
-1.0
-1.0
r=sinH3thetaL, theta=5 Pi12 1.0
1.0
0.5
0.5
0.5
0.5
-0.5
1.0
-1.0
0.5
-0.5
1.0
-1.0
0.5
-0.5
-0.5
-0.5
-0.5
-1.0
-1.0
-1.0
r=sinH3thetaL, theta=8 Pi12 1.0
1.0
0.5
0.5
0.5
0.5
1.0
-1.0
0.5
-0.5
1.0
-1.0
0.5
-0.5
-0.5
-0.5
-0.5
-1.0
-1.0
-1.0
r=sinH3thetaL, theta=11 Pi12 1.0
1.0
0.5
0.5
0.5
0.5
1.0
-1.0
0.5
-0.5
1.0
-1.0
0.5
-0.5
-0.5
-0.5
-0.5
-1.0
-1.0
-1.0
11
1.0
r=sinH3thetaL, theta=12 Pi12
1.0
-0.5
1.0
r=sinH3thetaL, theta=9 Pi12
1.0
-0.5
1.0
r=sinH3thetaL, theta=6 Pi12
1.0
r=sinH3thetaL, theta=10 Pi12
-1.0
0.5
-0.5 -0.5
r=sinH3thetaL, theta=7 Pi12
-1.0
-1.0
-0.5
r=sinH3thetaL, theta=4 Pi12
-1.0
r=sinH3thetaL, theta=3 Pi12
1.0
EXAMPLE: Sketch the curve r = sin(4θ), 0 ≤ θ ≤ 2π (eight-leaved rose). Solution: We have r=sinH4thetaL, theta=Pi6
-1.0
r=sinH4thetaL, theta=2 Pi6
1.0
1.0
1.0
0.5
0.5
0.5
0.5
-0.5
1.0
-1.0
-1.0
0.5
-0.5 -0.5
-1.0
-1.0
-1.0
r=sinH4thetaL, theta=5 Pi6 1.0
1.0
0.5
0.5
0.5
0.5
-0.5
1.0
-1.0
0.5
-0.5
1.0
-1.0
0.5
-0.5
-0.5
-0.5
-0.5
-1.0
-1.0
-1.0
r=sinH4thetaL, theta=8 Pi6 1.0
1.0
0.5
0.5
0.5
0.5
1.0
-1.0
0.5
-0.5
1.0
-1.0
0.5
-0.5
-0.5
-0.5
-0.5
-1.0
-1.0
-1.0
r=sinH4thetaL, theta=11 Pi6 1.0
1.0
0.5
0.5
0.5
0.5
1.0
-1.0
0.5
-0.5
1.0
-1.0
0.5
-0.5
-0.5
-0.5
-0.5
-1.0
-1.0
-1.0
12
1.0
r=sinH4thetaL, theta=12 Pi6
1.0
-0.5
1.0
r=sinH4thetaL, theta=9 Pi6
1.0
-0.5
1.0
r=sinH4thetaL, theta=6 Pi6
1.0
r=sinH4thetaL, theta=10 Pi6
-1.0
1.0
-0.5
r=sinH4thetaL, theta=7 Pi6
-1.0
0.5
-0.5
-0.5
r=sinH4thetaL, theta=4 Pi6
-1.0
r=sinH4thetaL, theta=3 Pi6
1.0
EXAMPLE: Sketch the curve r = sin(5θ), 0 ≤ θ ≤ 2π (five-leaved rose). Solution: We have r=sinH5thetaL, theta=Pi6
-1.0
r=sinH5thetaL, theta=2 Pi6
1.0
1.0
1.0
0.5
0.5
0.5
0.5
-0.5
1.0
-1.0
-1.0
0.5
-0.5 -0.5
-1.0
-1.0
-1.0
r=sinH5thetaL, theta=5 Pi6 1.0
1.0
0.5
0.5
0.5
0.5
-0.5
1.0
-1.0
0.5
-0.5
1.0
-1.0
0.5
-0.5
-0.5
-0.5
-0.5
-1.0
-1.0
-1.0
r=sinH5thetaL, theta=8 Pi6 1.0
1.0
0.5
0.5
0.5
0.5
1.0
-1.0
0.5
-0.5
1.0
-1.0
0.5
-0.5
-0.5
-0.5
-0.5
-1.0
-1.0
-1.0
r=sinH5thetaL, theta=11 Pi6 1.0
1.0
0.5
0.5
0.5
0.5
1.0
-1.0
0.5
-0.5
1.0
-1.0
0.5
-0.5
-0.5
-0.5
-0.5
-1.0
-1.0
-1.0
13
1.0
r=sinH5thetaL, theta=12 Pi6
1.0
-0.5
1.0
r=sinH5thetaL, theta=9 Pi6
1.0
-0.5
1.0
r=sinH5thetaL, theta=6 Pi6
1.0
r=sinH5thetaL, theta=10 Pi6
-1.0
1.0
-0.5
r=sinH5thetaL, theta=7 Pi6
-1.0
0.5
-0.5
-0.5
r=sinH5thetaL, theta=4 Pi6
-1.0
r=sinH5thetaL, theta=3 Pi6
1.0
EXAMPLE: Sketch the curve r = sin(6θ), 0 ≤ θ ≤ 2π (twelve-leaved rose). Solution: We have r=sinH6thetaL, theta=Pi6
-1.0
r=sinH6thetaL, theta=2 Pi6
1.0
1.0
1.0
0.5
0.5
0.5
0.5
-0.5
1.0
-1.0
-1.0
0.5
-0.5 -0.5
-1.0
-1.0
-1.0
r=sinH6thetaL, theta=5 Pi6 1.0
1.0
0.5
0.5
0.5
0.5
-0.5
1.0
-1.0
0.5
-0.5
1.0
-1.0
0.5
-0.5
-0.5
-0.5
-0.5
-1.0
-1.0
-1.0
r=sinH6thetaL, theta=8 Pi6 1.0
1.0
0.5
0.5
0.5
0.5
1.0
-1.0
0.5
-0.5
1.0
-1.0
0.5
-0.5
-0.5
-0.5
-0.5
-1.0
-1.0
-1.0
r=sinH6thetaL, theta=11 Pi6 1.0
1.0
0.5
0.5
0.5
0.5
1.0
-1.0
0.5
-0.5
1.0
-1.0
0.5
-0.5
-0.5
-0.5
-0.5
-1.0
-1.0
-1.0
14
1.0
r=sinH6thetaL, theta=12 Pi6
1.0
-0.5
1.0
r=sinH6thetaL, theta=9 Pi6
1.0
-0.5
1.0
r=sinH6thetaL, theta=6 Pi6
1.0
r=sinH6thetaL, theta=10 Pi6
-1.0
1.0
-0.5
r=sinH6thetaL, theta=7 Pi6
-1.0
0.5
-0.5
-0.5
r=sinH6thetaL, theta=4 Pi6
-1.0
r=sinH6thetaL, theta=3 Pi6
1.0
EXAMPLE: Sketch the curve r = sin(7θ), 0 ≤ θ ≤ 2π (seven-leaved rose). Solution: We have r=sinH7thetaL, theta=Pi6
-1.0
r=sinH7thetaL, theta=2 Pi6
1.0
1.0
1.0
0.5
0.5
0.5
0.5
-0.5
1.0
-1.0
-1.0
0.5
-0.5 -0.5
-1.0
-1.0
-1.0
r=sinH7thetaL, theta=5 Pi6 1.0
1.0
0.5
0.5
0.5
0.5
-0.5
1.0
-1.0
0.5
-0.5
1.0
-1.0
0.5
-0.5
-0.5
-0.5
-0.5
-1.0
-1.0
-1.0
r=sinH7thetaL, theta=8 Pi6 1.0
1.0
0.5
0.5
0.5
0.5
1.0
-1.0
0.5
-0.5
1.0
-1.0
0.5
-0.5
-0.5
-0.5
-0.5
-1.0
-1.0
-1.0
r=sinH7thetaL, theta=11 Pi6 1.0
1.0
0.5
0.5
0.5
0.5
1.0
-1.0
0.5
-0.5
1.0
-1.0
0.5
-0.5
-0.5
-0.5
-0.5
-1.0
-1.0
-1.0
15
1.0
r=sinH7thetaL, theta=12 Pi6
1.0
-0.5
1.0
r=sinH7thetaL, theta=9 Pi6
1.0
-0.5
1.0
r=sinH7thetaL, theta=6 Pi6
1.0
r=sinH7thetaL, theta=10 Pi6
-1.0
1.0
-0.5
r=sinH7thetaL, theta=7 Pi6
-1.0
0.5
-0.5
-0.5
r=sinH7thetaL, theta=4 Pi6
-1.0
r=sinH7thetaL, theta=3 Pi6
1.0
EXAMPLE: Sketch the curve r = 1 +
1 sin(10θ), 0 ≤ θ ≤ 2π. 10
Solution: We have r=1+sinH10thetaL10, theta=Pi6
-1.0
1.0
1.0
0.5
0.5
0.5
0.5
-0.5
1.0
0.5
-0.5
1.0
-1.0
0.5
-0.5
-0.5
-0.5
-1.0
-1.0
-1.0
r=1+sinH10thetaL10, theta=5 Pi6 1.0
1.0
0.5
0.5
0.5
0.5
-0.5
1.0
-1.0
0.5
-0.5
1.0
-1.0
0.5
-0.5
-0.5
-0.5
-0.5
-1.0
-1.0
-1.0
r=1+sinH10thetaL10, theta=8 Pi6 1.0
1.0
0.5
0.5
0.5
0.5
1.0
-1.0
0.5
-0.5
1.0
-1.0
0.5
-0.5
-0.5
-0.5
-0.5
-1.0
-1.0
-1.0
r=1+sinH10thetaL10, theta=11 Pi6 1.0
1.0
0.5
0.5
0.5
0.5
1.0
-1.0
0.5
-0.5
1.0
-1.0
0.5
-0.5
-0.5
-0.5
-0.5
-1.0
-1.0
-1.0
16
1.0
r=1+sinH10thetaL10, theta=12 Pi6
1.0
-0.5
1.0
r=1+sinH10thetaL10, theta=9 Pi6
1.0
-0.5
1.0
r=1+sinH10thetaL10, theta=6 Pi6
1.0
r=1+sinH10thetaL10, theta=10 Pi6
-1.0
-1.0
-0.5
r=1+sinH10thetaL10, theta=7 Pi6
-1.0
r=1+sinH10thetaL10, theta=3 Pi6
1.0
r=1+sinH10thetaL10, theta=4 Pi6
-1.0
r=1+sinH10thetaL10, theta=2 Pi6
1.0
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)
17
√ r = 1/ θ
