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

Question

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

Quizplus · Accepted Answer

The polar coordinates $r$ and $\theta$ can be found from the Cartesian coordinates $(x, y)$ using the formulas $r = \sqrt{x^2 + y^2}$ and $\theta = \arctan\left(\frac{y}{x}\right)$. For the point $\left(\frac{1}{6}, \frac{\sqrt{3}}{6}\right)$, $r = \sqrt{\left(\frac{1}{6}\right)^2 + \left(\frac{\sqrt{3}}{6}\right)^2} = \sqrt{\frac{1}{36} + \frac{3}{36}} = \sqrt{\frac{4}{36}} = \frac{1}{3}$. The angle $\theta$ is found by recognizing that the point lies on the line $y = \sqrt{3}x$, which corresponds to a $60^\circ$ or $\frac{\pi}{3}$ angle in standard position. Therefore, the correct polar coordinates are $\left(\frac{1}{3}, \frac{\pi}{3}\right)$.