Chapter 9.2, Problem 53AYU

In Problems 39-62, identify and graph each polar equation.

$r^2=9\cos \left(2\theta \right)$

To determine

To find: Identify the polar equation $r^2 = 9 \text{cos}\left(2\theta \right)$and graph it.

Answer to Problem 53AYU

Solution

The graph of $r^2 = 9 \text{cos}\left(2\theta \right)$ is an equation of Lemniscate.

Given:

It is asked to identify and graph the polar equation $r^2 = 9 \text{cos}\left(2\theta \right)$.

Explanation of Solution

Check for symmetry test:

Polar axis:

Replace $\theta$ by $- \theta$. The result is $r^2 = 9 \text{cos}\left(- 2\theta \right) = 9 \text{cos}\left(2\theta \right)$.

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

$r = 9 \text{cos}\left(\pi - 2\theta \right) = 9\left(\text{cos}\pi \text{cos} 2\theta + \text{sin}\pi \text{sin} 2\theta \right) = 9\left(\left(- 1\right) \text{cos}2 \theta + \left(0\right) \text{sin}2 \theta \right) = - 9 \text{cos}2 \theta$

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 ${\left(- r\right)}^2 = 9 \text{cos}\left(2\theta \right)$, so $r^2 = 9 \text{cos}\left(2\theta \right)$. This test satisfied. Replace $\theta$ by $\theta + \pi$. The result is

$r = 9 \text{cos}\left(2\theta + \pi \right) = 9\left(\text{cos}\pi \text{cos} 2\theta - \text{sin}\pi \text{sin} 2\theta \right) = 9\left(- 1\right) \text{cos}2 \theta = - 9 \text{cos} 2\theta .$

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 $\pi$.

$\theta$	$r^2 = 9 \text{cos}\left(2\theta \right), r = \sqrt{9 \text{cos}\left(2\theta \right)}$
$0$	$\sqrt{9\left(1\right)} = \pm 3$
$\frac{\pi }{6}$	$\sqrt{9 \times \frac{1}{2}} = \pm \frac{3}{\sqrt{2}}$
$\frac{\pi }{4}$	$\sqrt{9\left(0\right)} = 0$
$\frac{\pi }{3}$	$\sqrt{9\left(- \frac{1}{2}\right)} = \pm \frac{3}{\sqrt{2}}$
$\frac{\pi }{2}$	$\sqrt{3\left(- 1\right)} = \pm \sqrt{3}$
$\frac{2\pi }{3}$	$\sqrt{9\left(- \frac{1}{2}\right)} = \pm \frac{3}{\sqrt{2}}$
$\frac{3\pi }{4}$	$\sqrt{9\left(0\right)} = 0$
$\frac{5\pi }{6}$	$\sqrt{9 \times \frac{1}{2}} = \pm \frac{3}{\sqrt{2}}$
$\pi$	$\sqrt{9\left(1\right)} = \pm 3$

To sketch:

$r^2 = 9 \text{cos}\left(2\theta \right)$