Answer
(a) $\left( {r,\theta } \right) = \left( {1,0} \right)$
(b) $\left( {r,\theta } \right) = \left( {2\sqrt 3 ,\frac{\pi }{6}} \right)$
(c) $\left( {r,\theta } \right) = \left( {2\sqrt 2 ,\frac{{3\pi }}{4}} \right)$
(d) $\left( {r,\theta } \right) = \left( {2,\frac{{2\pi }}{3}} \right)$
Work Step by Step
Use the conversion formula from rectangular coordinates to polar coordinates given by
$r = \sqrt {{x^2} + {y^2}} $, ${\ \ \ }$ $\theta = {\tan ^{ - 1}}\left( {\frac{y}{x}} \right)$.
(a) Since $\left( {x,y} \right) = \left( {1,0} \right)$, we have
$r = \sqrt {1 + 0} $, ${\ \ \ }$ $\theta = {\tan ^{ - 1}}\left( {\frac{0}{1}} \right)$.
So, $\left( {r,\theta } \right) = \left( {1,0} \right)$.
(b) Since $\left( {x,y} \right) = \left( {3,\sqrt 3 } \right)$, we have
$r = \sqrt {9 + 3} $, ${\ \ \ }$ $\theta = {\tan ^{ - 1}}\left( {\frac{{\sqrt 3 }}{3}} \right)$.
So, $\left( {r,\theta } \right) = \left( {2\sqrt 3 ,\frac{\pi }{6}} \right)$.
(c) Since $\left( {x,y} \right) = \left( { - 2,2} \right)$, we have
$r = \sqrt {4 + 4} $, ${\ \ \ }$ $\theta = {\tan ^{ - 1}}\left( {\frac{2}{{ - 2}}} \right)$.
So, $\left( {r,\theta } \right) = \left( {2\sqrt 2 ,\frac{{3\pi }}{4}} \right)$.
(d) Since $\left( {x,y} \right) = \left( { - 1,\sqrt 3 } \right)$, we have
$r = \sqrt {1 + 3} $, ${\ \ \ }$ $\theta = {\tan ^{ - 1}}\left( {\frac{{\sqrt 3 }}{{ - 1}}} \right)$.
So, $\left( {r,\theta } \right) = \left( {2,\frac{{2\pi }}{3}} \right)$.
