## Calculus (3rd Edition)

(a) $\left( {r,2\pi - \theta } \right)$ (b) $\left( {r,\pi + \theta } \right)$ (c) $\left( {r,\pi - \theta } \right)$ (d) $\left( {r,\frac{\pi }{2} - \theta } \right)$
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) ${r_1} = \sqrt {{x^2} + {{\left( { - y} \right)}^2}} = \sqrt {{x^2} + {y^2}} = r$, ${\theta _1} = {\tan ^{ - 1}}\left( {\frac{{ - y}}{x}} \right)$. Since ${\theta _1}$ is in the fourth quadrant, so ${\theta _1} = 2\pi - {\tan ^{ - 1}}\left( {\frac{y}{x}} \right) = 2\pi - \theta$. The polar coordinates is $\left( {r,2\pi - \theta } \right)$. (b) ${r_2} = \sqrt {{{\left( { - x} \right)}^2} + {{\left( { - y} \right)}^2}} = \sqrt {{x^2} + {y^2}} = r$, ${\theta _2} = {\tan ^{ - 1}}\left( {\frac{{ - y}}{{ - x}}} \right)$. Since ${\theta _2}$ is in the third quadrant, so ${\theta _2} = \pi + {\tan ^{ - 1}}\left( {\frac{y}{x}} \right) = \pi + \theta$. The polar coordinates is $\left( {r,\pi + \theta } \right)$. (c) ${r_3} = \sqrt {{{\left( { - x} \right)}^2} + {y^2}} = \sqrt {{x^2} + {y^2}} = r$, ${\theta _3} = {\tan ^{ - 1}}\left( {\frac{y}{{ - x}}} \right)$. Since ${\theta _3}$ is in the second quadrant, so ${\theta _3} = \pi - {\tan ^{ - 1}}\left( {\frac{y}{x}} \right) = \pi - \theta$. The polar coordinates is $\left( {r,\pi - \theta } \right)$. (d) ${r_4} = \sqrt {{y^2} + {x^2}} = \sqrt {{x^2} + {y^2}} = r$, ${\theta _4} = {\tan ^{ - 1}}\left( {\frac{x}{y}} \right)$, $\tan {\theta _4} = \frac{x}{y} = \frac{1}{{\tan \theta }} = \cot \theta$, Since $\cot \theta = \tan \left( {\frac{\pi }{2} - \theta } \right)$, we have $\tan {\theta _4} = \tan \left( {\frac{\pi }{2} - \theta } \right)$. The polar coordinates is $\left( {r,\frac{\pi }{2} - \theta } \right)$.