The rectangular coordinates of a point are given. Express the point in polar coordinates with r ≥ 0 and 0° ≤ θ

Question

The rectangular coordinates of a point are given. Express the point in polar coordinates with r ≥ 0 and 0° ≤ θ < 360°.
-$\left( \frac { 1 } { 3 } , \frac { \sqrt { 3 } } { 3 } \right)$
A) $\left( \frac { 2 } { 3 } , 120 ^ { \circ } \right)$
B) $\left( \frac { 1 } { 3 } , 30 ^ { \circ } \right)$
C) $\left( \frac { 2 } { 3 } , 30 ^ { \circ } \right)$
D) $\left( \frac { 2 } { 3 } , 60 ^ { \circ } \right)$

Quizplus · Accepted Answer

To convert rectangular coordinates $\left( \frac { 1 } { 3 } , \frac { \sqrt { 3 } } { 3 } \right)$ to polar coordinates, we use the formulas $r = \sqrt{x^2 + y^2}$ and $\theta = \tan^{-1}\left(\frac{y}{x}\right)$. Here, $r = \sqrt{\left(\frac{1}{3}\right)^2 + \left(\frac{\sqrt{3}}{3}\right)^2} = \sqrt{\frac{1}{9} + \frac{3}{9}} = \sqrt{\frac{4}{9}} = \frac{2}{3}$. For $\theta$, since $x = \frac{1}{3}$ and $y = \frac{\sqrt{3}}{3}$, $\theta = \tan^{-1}\left(\frac{\sqrt{3}/3}{1/3}\right) = \tan^{-1}(\sqrt{3}) = 60^\circ$.

The Rectangular Coordinates of a Point Are Given (13,33)\left( \frac { 1 } { 3 } , \frac { \sqrt { 3 } } { 3 } \right)(31​,33​​)

The Rectangular Coordinates of a Point Are Given $\left( \frac { 1 } { 3 } , \frac { \sqrt { 3 } } { 3 } \right)$