Chapter 9.2, Problem 74AYU

To determine

To find: The graph of the polar equation $r=\frac{1}{1-\text{cos}\theta } \left(parabola\right)$.

Answer to Problem 74AYU

Solution:

Graph of the polar equation $r=\frac{1}{1-\text{cos}\theta } \left(parabola\right)$ is

Given:

It is asked to find the graph of the polar equation $r=\frac{1}{1-\text{cos}\theta } \left(parabola\right)$:

Explanation of Solution

Check for symmetry test:

Polar axis:

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

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

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

$=\frac{1}{1+\text{cos}\theta }$

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

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

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 $\pi$.

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

To sketch:

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