Example 5 from Algebra and Trigonometry, 10.4 Polar Coordinates: Graphs

Sketching the Graph of a One-Loop Limaçon

Graph the equation $r = 4 - 3 \sin θ .$

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Solution
First, testing the equation for symmetry, we find that it fails all three symmetry tests, meaning that the graph may or may not exhibit symmetry, so we cannot use the symmetry to help us graph it. However, this equation has a graph that clearly displays symmetry with respect to the line $θ = \frac{π}{2},$ yet it fails all the three symmetry tests. A graphing calculator will immediately illustrate the graph’s reflective quality.
Next, we find the zeros and maximum, and plot the reflecting points to verify any symmetry. Setting $r = 0$ results in $θ$ being undefined. What does this mean? How could $θ$ be undefined? The angle $θ$ is undefined for any value of $\sin θ > 1.$ Therefore, $θ$ is undefined because there is no value of $θ$ for which $\sin θ > 1.$ Consequently, the graph does not pass through the pole. Perhaps the graph does cross the polar axis, but not at the pole. We can investigate other intercepts by calculating $r$ when $θ = 0.$
$\begin{array}{l} r (0) = 4 - 3 \sin (0) \\ r = 4 - 3 \cdot 0 = 4 \end{array}$
So, there is at least one polar axis intercept at $(4, 0) .$

Next, as the maximum value of the sine function is 1 when $θ = \frac{π}{2},$ we will substitute $θ = \frac{π}{2}$ into the equation and solve for $r .$ Thus, $r = 1.$
Make a table of the coordinates similar to Table 5.

$θ$ $0$ $\frac{π}{6}$ $\frac{π}{3}$ $\frac{π}{2}$ $\frac{2 π}{3}$ $\frac{5 π}{6}$ $π$ $\frac{7 π}{6}$ $\frac{4 π}{3}$ $\frac{3 π}{2}$ $\frac{5 π}{3}$ $\frac{11 π}{6}$ $2 π$

$r$ 4 2.5 1.4 1 1.4 2.5 4 5.5 6.6 7 6.6 5.5 4

Table 5

The graph is shown in Figure 10.

Figure 10 One-loop limaçon

Your turn

The book's Try It right after this example: same idea, new numbers. Only the answer is given.

Sketch the graph of $r = 3 - 2 \cos θ .$

How did it go?

$θ$	$0$	$\frac{π}{6}$	$\frac{π}{3}$	$\frac{π}{2}$	$\frac{2 π}{3}$	$\frac{5 π}{6}$	$π$	$\frac{7 π}{6}$	$\frac{4 π}{3}$	$\frac{3 π}{2}$	$\frac{5 π}{3}$	$\frac{11 π}{6}$	$2 π$
$r$	4	2.5	1.4	1	1.4	2.5	4	5.5	6.6	7	6.6	5.5	4

Solution

Your turn

Solution

Your turn