Chapter 10, Problem 8CR

To determine

The polar equation for the line containing the origin that makes an angle ${30}^{\circ }$ with the positive $x-$ axis.

Answer to Problem 8CR

Solution:

The polar equation of the line containing the origin that makes an angle ${30}^{\circ }$ with the positive $x-$ axis is $\theta =\frac{\pi }{6}$

Explanation of Solution

Given information:

The line passes through the origin and makes an angle ${30}^{\circ }$ with the positive $x-$ axis.

Explanation:

The general equation of the line containing the origin that makes an angle $\theta$ with the positive $x-$ axisis $y=mx$, where $m=\tan \theta$ is the slope

The equation of the line containing the origin that makes an angle ${30}^{\circ }$ with the positive $x-$ axis is $y=mx=\tan \left({30}^{\circ }\right)x=\frac{x}{\sqrt{3}}$.

To convert the rectangular equation into polar equation, substitute polar coordinates $x=r\cos \theta ,y=r\sin \theta$

$\Rightarrow r\sin \theta =\frac{r\cos \theta }{\sqrt{3}}$

$\Rightarrow \sin \theta =\frac{\cos \theta }{\sqrt{3}}$

$\Rightarrow \frac{\sin \theta }{\cos \theta }=\frac{1}{\sqrt{3}}$

$\Rightarrow \tan \theta =\frac{1}{\sqrt{3}}$

Apply ${\tan }^{-1}$ both sides

$\Rightarrow \theta ={\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)$

$\Rightarrow \theta =\frac{\pi }{6}$

Therefore, the polar equation of the line containing the origin that makes an angle ${30}^{\circ }$ with the positive $x-$ axis is $\theta =\frac{\pi }{6}$.