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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

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

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

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

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

\(r^{2} \sin \theta \cos \theta=1\)

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

\(r^{2}=x^{2}+y^{2}\)

\(x=r \cos \theta \rightarrow \cos \theta=\frac{x}{r}\)

\(y=r \sin \theta \rightarrow \sin \theta=\frac{y}{r}\)

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

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

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

\(x \cdot y=1\)

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