Sketching Polar Curves Examples
So we have looked at various families of polar curves, however, there are tons of families of curves and it is not reasonable to memorize them all and their properties, so let's attempt to graph some polar curves.
Example 1
Graph the polar curve defined by the equation $r = \theta$ on the interval [0, 2π]
Let's first graph some values to see what is happening with the polar coordinates:
| θ | r | (r, θ) |
|---|---|---|
| 0 | 0 | (0, 0) |
| π/2 | π/2 | (π/2, π/2) |
| π | π | (π, π) |
| 3π/2 | 3π/2 | (3π/2, 3π/2) |
| 2π | 2π | (2π, 2π) |
So as θ increases, the the distance away from the origin also increases on the entire interval [0, 2π]. Hence, we obtain somewhat of a spiral that originates at the origin or point (0, 0). The graph we obtain is as follows:
Example 2
Graph the polar curve defined by the equation $r = \cos \theta + 1$ on the interval [0, 2π]
Once again let's find some values to see what is happening:
| θ | r | (r, θ) |
|---|---|---|
| 0 | 2 | (2, 0) |
| π/4 | √2/2 + 1 ≈ 1.707 | (1.707, π/4) |
| π/2 | 1 | (1, π/2) |
| 3π/4 | -√2/2 + 1 ≈ 0.292 | (0.292, 3π/4) |
| π | 0 | (0, π) |
| 5π/4 | -√2/2 + 1 ≈ 0.292 | (0.292, 5π/4) |
| 3π/2 | 1 | (1, 3π/2) |
| 7π/4 | √2/2 + 1 ≈ 1.707 | (1.707, 3π/4) |
| 2π | 2 | (2, 2π) |
From this table, we should see that as θ increases, the distance away from the origin for a point P tracing the curve decreases on the interval (0, π) and increases again on the interval (π, 2π). Also notice that the curve starts at coordinates (2, 0), intersects the vertical polar axis at point (1, π/2), intersects the horizontal polar axis at (0, π) and intersects the vertical polar axis once again at (1, 3π/2). Hence we obtain the following graph from this information:
Example 3
Graph the polar curve defined by the equation $r = 2\sin \theta - 1$ on the interval [0, 2π].
Once again we will plot some points on the curve:
| θ | r | (r, θ) |
|---|---|---|
| 0 | -1 | (-1, 0) |
| π/6 | 0 | (0, π/6) |
| π/4 | √2 - 1 ≈ 0.41 | (0.41, π/4) |
| π/2 | 1 | (1, π/2) |
| 3π/4 | √2 - 1 ≈ 0.41 | (0.41, 3π/4) |
| 5π/6 | 0 | (0, 5π/6) |
| π | -1 | (-1, π) |
| 5π/4 | -√2 - 1 ≈ -2.41 | (-2.41, 5π/4) |
| 3π/2 | -3 | (-3, 3π/2) |
| 7π/4 | -√2 - 1 ≈ -2.41 | (-2.41, 7π/4) |
| 2π | -1 | (-1, 2π) |
Starting at θ = 0, the curve begins to be traced on the negative portion of the polar axis. The distance from the origin for the point P tracing the curve out decreases on the interval (0, π/6) where the sign of r shades. Also notice that the position of the point tracing the curve is going to be in quadrant 3 from (0, π/6) as opposed to quadrant 2:
Then from the interval (π/6, π/2), r becomes positive and the curve continues in quadrant 1 with the distance away from the origin increasing:
Now from the interval (π/2, 5π/6), r begins to decrease:
And on the interval (5π/6, π), r becomes negative once again with the distance from the origin increasing.
Now on the interval (π, 3π/2) r remains negative but starts in quadrant 1 instead of quadrant 3 since r is negative. The distance away from the origin continues to increase:
Then from the interval (3π/2, 2π), r is still negative but the distance away from the origin decreases and we have sketched the entire curve:
