Chapter 9.2, Problem 58AYU

To determine

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

Answer to Problem 58AYU

Solution

The graph of $r = 1 - \text{cos}\theta$ is an equation of Limacon without inner loop.

Given:

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

Explanation of Solution

Check for symmetry test:

Polar axis:

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

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

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

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

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

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 $\pi$.

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

To sketch:

$r = 3 + \text{cos}\theta$