Chapter 9.2, Problem 45AYU

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

$r=1+2\sin \theta$

To determine

To find: Identify the polar equation $r = 1 + 2 \text{sin}\theta$ and graph it.

Answer to Problem 45AYU

Solution

The graph of $r = 1 + 2 \text{sin}\theta$ is an equation of Limacon with inner loop.

Given:

It is asked to identify and graph the polar equation $r = 1 + 2 \text{sin}\theta$

Explanation of Solution

Check for symmetry test:

Polar axis:

Replace $\theta$ by $- \theta$. The result is $r = r = r = 1 + 2 \text{sin}\left( - \theta \right) = 1 - 2 \text{sin}\theta$

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

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

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

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

$= 1 + 2 \text{sin}\theta$

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 $- r = 1 + 2 \text{sin}\theta$, so $r = - 1 - 2 \text{sin}\theta$. This test fails. Replace $\theta$ by $\theta + \pi$. The result is

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

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 to $- \frac{\pi }{2} \text{to} \frac{\pi }{2}$.

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

To sketch:

$r = 4 + 2 \text{sin}\theta$