How do you convert (7sqrt3, -7)  into polar coordinates?

Jul 28, 2016

$\left(7 \sqrt{3} , - 7\right) \to \left(14 , - \frac{\pi}{6}\right)$

Explanation:

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

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

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

$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{r = \sqrt{{x}^{2} + {y}^{2}}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

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

here $x = 7 \sqrt{3} \text{ and } y = - 7$

$r = \sqrt{{\left(7 \sqrt{3}\right)}^{2} + {\left(- 7\right)}^{2}} = \sqrt{147 + 49} = \sqrt{196} = 14$

Now $\left(7 \sqrt{3} , - 7\right)$ is in the 4th quadrant and care must be taken to ensure $\theta$ is in the 4th quadrant.

$\theta = {\tan}^{-} 1 \left(- \frac{7}{7 \sqrt{3}}\right) = {\tan}^{-} 1 \left(- \frac{1}{\sqrt{3}}\right) = - \frac{\pi}{6} \text{ 4th quadrant}$

$\Rightarrow \left(7 \sqrt{3} , - 7\right) = \left(14 , - \frac{\pi}{6}\right) = \left(14 , - {30}^{\circ}\right)$