Chapter 9.2, Problem 51AYU

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

$r=4\sin \left(5\theta \right)$

To determine

To find: Identify the polar equation $r = 2 \text{sin}\left(3\theta \right)$ and graph it.

Answer to Problem 51AYU

Solution

The graph of $r = 2 \text{cos}\left(3\theta \right)$ is an equation of rose with three petals.

Given:

It is asked to identify and graph the polar equation $r = 2 \text{sin}\left(3\theta \right)$.

Explanation of Solution

Check for symmetry test:

Polar axis:

Replace $\theta$ by $- \theta$. The result is $r = 2 \text{sin}\left(- 3\theta \right) = - 2 \text{sin}\left(3\theta \right)$.

The test fails, so the graph may or may not symmetric with respect to the polar axis.

The line

$\theta = \frac{\pi }{2}$: Replace $\theta$ by $\pi - \theta$. The result is

$\begin{array}{c}r = 2 \text{sin}\left(\pi - 3\theta \right)\\ = 2\left(\text{sin}\pi \text{cos} 3\theta - \text{cos}\pi \text{sin} 3\theta \right)\\ = 2\left(\left(0\right) \text{cos} 3\theta - \left(- 1\right) \text{sin} 3\theta \right)\\ = 2 \text{sin}\left(3\theta \right)\end{array}$

The test satisfied. So the graph is symmetric with respect to the line $\theta = \frac{\pi }{2}$.

The Pole:

Replace $r$ by $- r$. Then the result is $- r = 2 \text{sin}\left(3\theta \right)$, so $r = - 2 \text{sin}\left(3\theta \right)$. This test fails. Replace $\theta$ by $\theta + \pi$. The result is

$r = 2 \text{sin}\left(3\theta + \pi \right) = 2\left(\text{sin}3 \theta \text{cos}\pi + \text{cos} 3\theta \text{sin}\pi \right) = - 2 \text{sin} 3\theta .$

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 line $\theta = \frac{\pi }{2}$, it is enough to find the values from $- \frac{\pi }{2}$ to $\frac{\pi }{2}$.

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

To sketch:

$r = 2 \text{sin}\left(3\theta \right)$.