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Convert the point $\left(2, \frac{\pi}{3}\right)$ from polar to cartesian coordinates

$\left(2, \frac{\pi}{3}\right) \quad r=2 \quad \theta=\frac{\pi}{3}$

$x=r \cos \theta=2 \cos \left(\frac{\pi}{3}\right)=2 \cdot \frac{1}{2}=1$

$y=r \sin \theta=2 \sin \left(\frac{\pi}{3}\right)=2 \cdot \frac{\sqrt{3}}{2}=\sqrt{3}$

The point $\left(2, \frac{\pi}{3}\right)$ in polar coordinates is the point $(1, \sqrt{3})$ in cartesain  coordinates

Represent the point with cartesian coordinates $(1,-1)$ in terms of polar coordinates.

$({1},-1) \qquad x=1 \quad, \quad y=-1$

$r=\sqrt{x^{2}+y^{2}}=\sqrt{(1)^{2}+(-1)^{2}}=\sqrt{1+1}=\sqrt{2}$

$\tan \theta=\left|\frac{y}{x}\right| \Rightarrow \theta=\tan ^{-1}\left|\frac{y}{x}\right|=\tan ^{-1}\left|\frac{-1}{1}\right|=\tan ^{-1}(1)=\frac{\pi}{4}$

$\theta=\frac{\pi}{4}$

since $x=1\: ,\:y=-1$ is in the 4th quadrant

$\theta=2 \pi-\frac{\pi}{4}=\frac{7 \pi}{4}$

$∴ (r, \theta)=\left(\sqrt{2}, \frac{7 \pi}{q}\right)$

Express the following cartesian equation in terms of polar coordinates $\sqrt{x^{2}+y^{2}}, \sqrt[3]{2 x y}$

we know that $r=\sqrt{x^{2}+y^{2}} \quad, \quad x=r \cos \theta \quad,\quad y=r \sin \theta$

$r=\sqrt[3]{2 x y}$

$r^{3}=(3 \sqrt{2 x y})^{3}=2 x y=2(r \cos \theta)(r \sin \theta)=2 r^{2} \cos \theta \sin a$

$\frac{r^{3}}{r^{2}}=\frac{2 r^{2} \cos \theta \sin \theta}{r^{2}} \rightarrow r=2 \cos \theta \sin \theta$

$r=\sin 2 \theta$

$\sin (2 \theta)=2 \sin \theta \cos \theta$

Change the polar curve $r^{2} \sin (2 \theta)=2$ into rectangular system

$r^{2} \sin (2 \theta)=2$

$\sin (2 \theta)=2 \sin \theta \cos \theta$

$r^{2}(2 \sin \theta \cos \theta)=2$

$\frac{r^{2}(2 \sin \theta \cos \theta)}{2}=\frac{2}{2}$

$r^{2} \sin \theta \cos \theta=1$

$r=\sqrt{x^{2}+y^{2}}$

$r^{2}=x^{2}+y^{2}$

$x=r \cos \theta \rightarrow \cos \theta=\frac{x}{r}$

$y=r \sin \theta \rightarrow \sin \theta=\frac{y}{r}$

$\left(x^{2}+y^{2}\right) \cdot\left(\frac{y}{r}\right) \cdot\left(\frac{x}{r}\right)=1$

$\left(x^{2}+y^{2}\right) \frac{y \cdot x}{r^{2}}=1$

$\left(x^{2}+y^{2}\right) \frac{x \cdot y}{\left(x^{2}+y^{2}\right)}=1$

$x \cdot y=1$