$\text { Find the polar coordinates, } 0 \leq \theta

Question

$\text { Find the polar coordinates, } 0 \leq \theta < 2 \pi \text { and } r \geq 0 \text {, of the point given in Cartesian coordinates. }$
-$\left( \frac { 1 } { 3 } , \frac { - \sqrt { 3 } } { 3 } \right)$
A) $\left( \frac { 2 } { 3 } , \frac { 4 \pi } { 3 } \right)$
B) $\left( \frac { 2 } { 3 } , \frac { 11 \pi } { 6 } \right)$
C) $\left( \frac { 2 } { 3 } , \frac { 5 \pi } { 3 } \right)$
D) $\left( \frac { 2 } { 3 } , \frac { 7 \pi } { 6 } \right)$

Quizplus · Accepted Answer

To find the polar coordinates, we use the conversion formulas $r=\sqrt{x^2+y^2}$ and $	an 	heta = \frac { y } { x }$. Plugging in the given values, we get $r = \sqrt{\left(\frac{1}{3}ight)^2+\left(- \frac {\sqrt{3}}{3}ight)^2} = \frac{2}{3}$ and $	an 	heta = \frac {- \sqrt{3}/3}{1/3} = -\sqrt{3}$. We know that $	heta$ is in the fourth quadrant ($x>0$ and $y<0$), so we can use the identity $	an 	heta = 	an(	heta + \pi)$ to get $	heta = 	an^{-1}(-\sqrt{3})+\pi = \frac{5\pi}{3}$. Thus, the polar coordinates are $\boxed{	extbf{(C) } \left(\frac{2}{3},\frac{5\pi}{3}ight)}$.