 # The Cartesian coordinates of a point are given. a) (2,-2) mattgondek4 2021-09-21 Answered
The Cartesian coordinates of a point are given.
a) $\left(2,-2\right)$
b) $\left(-1,\sqrt{3}\right)$
Find the polar coordinates $\left(r,\theta \right)$ of the point, where r is greater than 0 and 0 is less than or equal to $\theta$, which is less than $2\pi$
Find the polar coordinates $\left(r,\theta \right)$ of the point, where r is less than 0 and 0 is less than or equal to $\theta$, which is less than $2\pi$
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Consider the Cartesian coordinates be (x,y).
The polar coordinates $\left(r,\theta \right)$ is defined as,
${r}^{2}={x}^{2}+{y}^{2}$
$x=r\mathrm{cos}\theta$
$y=r\mathrm{sin}\theta$
This implies that, $\mathrm{tan}\theta =\frac{y}{x}$
a) The Cartesian coordinates be $\left(2,-2\right)$.
Calculate r and $\theta$ as follows:
${r}^{2}={2}^{2}+{\left(-2\right)}^{2}$
$=8$
$r=±2\sqrt{2}$
$\mathrm{tan}\theta =\frac{y}{x}$
$=\frac{-2}{2}$
$=-1$
Thus, the angle $\theta$ is $\frac{3\pi }{4},\frac{7\pi }{4}$ as the angle $0\le \theta 2\pi$
The polar coordinates of (2,-2) are,
$\left(2\sqrt{2},\frac{7\pi }{4}\right)$, where r>0
$\left(-2\sqrt{2},\frac{3\pi }{4}\right)$, where r<0
b) The Cartesian coordinates be $\left(-1,\sqrt{3}\right)$.
Calculate r and $\theta$ as follows:
${r}^{2}={\left(-1\right)}^{2}+{\left(\sqrt{3}\right)}^{2}$
$=4$
$r=±\sqrt{2}$
$\mathrm{tan}\theta =\frac{y}{x}$
$=\frac{\sqrt{3}}{-1}$
$=-\sqrt{3}$
Thus, the angle $\theta$ is $\frac{2\pi }{3},\frac{5\pi }{3}$ as the angle $0\le \theta \le 2\pi$
The polar coordinates of $\left(-1,\sqrt{3}\right)$ are,
$\left(2,\frac{2\pi }{3}\right)$, where r>0
$\left(-2,\frac{5\pi }{3}\right)$, where r<0