Chapter 9.2, Problem 59AYU

To determine

To find: Identify the polar equation $r = 1 - 3 \text{cos} \theta$ and graph it.

Answer to Problem 59AYU

Solution

The graph of $r = 1 - 3 \text{cos} \theta$ is an equation of Limacon with inner loop.

Given:

It is asked to identify and graph the polar equation $r = 1 - 3 \text{cos} \theta$.

Explanation of Solution

Check for symmetry test:

Polar axis:

Replace $\theta$ by $- \theta$. The result is $r = 1 - 3 \text{cos}\left(- \theta \right) = 1 - 3 \text{cos} \theta$

The test satisfied. So the graph is symmetric with respect to the polar axis.

The line $\theta =\frac{\pi }{2}$:

Replace $\theta$ by $\pi -\theta$. The result is

$r = 1 - 3 \text{cos} \left(\pi - \theta \right) = 1 - 3 \left(\text{cos} \pi \text{cos} \theta + \text{sin} \pi \text{sin} \theta \right) = 1 - 3\left(\left(- 1\right) \text{cos} \theta + \left(0\right) \text{sin} \theta \right) = 1 + 3 \text{cos} \theta$

The test fails, so the graph may or may not symmetric with respect to the line $\theta =\frac{\pi }{2}$.

The Pole:

Replace $r$ by $- r$. Then the result is $- r = 1 - 3 \text{cos} \theta$, so $r = - 1 + 3 \text{cos} \theta$. This test fails. Replace $\theta$ by $\theta + \pi$. The result is

$r = 1 - 3 \text{cos} \left(\theta + \pi \right) = 1 - 3 \left(\text{cos} \pi \text{cos} \theta - \text{sin} \pi \text{sin} \theta \right) = 1 - 3\left(- 1\right) \text{cos} \theta = 1 + 3 \text{cos} \theta .$

This test also fails, so the graph may or may not be symmetric with respect to the pole.

Let’s identify points on the graph by assigning values to the angle $\theta$ and calculating the corresponding values of $r$.

As the polar equation is symmetric about the polar axis, it is enough to find the values from 0 to $\pi$.

$\theta$	$r = 1 - 3 \text{cos} \theta$
$0$	$1 - 3\left(1\right) = - 2$
$\frac{\pi }{6}$	$1 - 3 \frac{\sqrt{3}}{2}$
$\frac{\pi }{4}$	$1 - 3 \frac{1}{\sqrt{2}}$
$\frac{\pi }{3}$	$1 - 3 \times \frac{1}{2} = - \frac{1}{2}$
$\frac{\pi }{2}$	$1 - 3\left(0\right) = 1$
$\frac{2\pi }{3}$	$1 - 3 \times \left(- \frac{1}{2}\right) = \frac{5}{2}$
$+\frac{3\pi }{4}$	$1-3\times \left(-\frac{1}{\sqrt{2}}\right)=\frac{\sqrt{2}+3}{\sqrt{2}}$
$\frac{5\pi }{6}$	$1-3\times \left(-\frac{\sqrt{3}}{2}\right)=\frac{2+3\sqrt{3}}{2}$
$\pi$	$1 - 3\left(- 1\right) = 4$

To sketch:

$r = 1 - 3 \text{cos} \theta$