How do you change the rectangular coordinate (-6, 6sqrt3) into polar coordinates?

Aug 25, 2016

$\left(- 6 , 6 \sqrt{3}\right) \to \left(12 , \frac{2 \pi}{3}\right)$

Explanation:

To convert from $\textcolor{b l u e}{\text{rectangular to polar coordinates}}$

That is $\left(x , y\right) \to \left(r , \theta\right)$

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder}}$

$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{r = \sqrt{{x}^{2} + {y}^{2}}} \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ and } \textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\theta = {\tan}^{-} 1 \left(\frac{y}{x}\right)} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

here x = - 6 and y $= 6 \sqrt{3}$

$\Rightarrow r = \sqrt{{6}^{2} + {\left(6 \sqrt{3}\right)}^{2}} = \sqrt{36 + 108} = \sqrt{144} = 12$

Now $\left(- 6 , 6 \sqrt{3}\right)$ is in the 2nd quadrant so we must ensure that $\theta$ is in the 2nd quadrant.

$\theta = {\tan}^{-} 1 \left(\frac{6 \sqrt{3}}{-} 6\right) = {\tan}^{-} 1 \left(- \sqrt{3}\right)$

$= - \frac{\pi}{3} \text{ in 4th quadrant}$

$\Rightarrow \theta = \left(\pi - \frac{\pi}{3}\right) = \frac{2 \pi}{3} \text{ in 2nd quadrant}$

$\Rightarrow \left(- 6 , 6 \sqrt{3}\right) \to \left(12 , \frac{2 \pi}{3}\right)$