Example 2: Sketching r = sin 2θ
Sketch the curve r = sin 2θ.
Step 1 |
dr/dθ = 2 cos 2θ. |
Step 2 |
r = 0 at θ = 0, , π, . dr/dθ = 0 at θ =  |
Step 3 |
See Figure 7.8.8.
Figure 7.8.8 |
Step 4 |
We take values at intervals of beginning at θ = 0. We can save some time by observing that the values from π to 2π are the same as those from 0 to π.
θ |
r = sin 2θ |
dr/dθ |
tan ψ |
Comments |
0 and π |
0 |
2 |
0 |
crosses origin |
π/8 and 9π/8 |
√2/2 |
√2 |
1/2 |
|r| increasing |
2π/8 and 10π/8 |
1 |
0 |
- |
max |
3π/8 and 11π/8 |
√2/2 |
-√2 |
-1/2 |
|r| decreasing |
4π/8 and 12π/8 |
0 |
-2 |
0 |
crosses origin |
5π/8 and 13π/8 |
-√2/2 |
-√2 |
1/2 |
|r| increasing |
6π/8 and 14π/8 |
-1 |
0 |
- |
min |
7π/8 and 15π/8 |
-√2/2 |
√2 |
-1/2 |
|r| decreasing |
|
Step 5 |
We plot the points and trace out the curve as θ increases from 0 to 2π. Figure 7.8.9 shows the curve at various stages of development. The graph looks like a four-leaf clover.
Figure 7.8.9
|
If r approaches ∞ as θ approaches 0 or π, the curve may have a horizontal asymptote which can be found by computing the limit of y. At θ = π/2 or 3π/2 there may be vertical asymptotes. The method is illustrated in the following example.
|