The Cartesian coordinates of a point are given. a) (2,-2) b) (-1,\sqrt{3}) Find the polar coordinates (r,\theta) of the point, where r is greater than

Jason Farmer 2021-06-05 Answered
The Cartesian coordinates of a point are given.
a) \((2,-2)\)
b) \((-1,\sqrt{3})\)
Find the polar coordinates \((r,\theta)\) 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 \((r,\theta)\) 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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Expert Answer

Mitchel Aguirre
Answered 2021-06-06 Author has 27260 answers
Consider the Cartesian coordinates be (x,y).
The polar coordinates \((r,\theta)\) is defined as,
\(r^2=x^2+y^2\)
\(x=r\cos\theta\)
\(y=r\sin\theta\)
This implies that, \(\tan\theta=\frac{y}{x}\)
a) The Cartesian coordinates be \((2,-2)\).
Calculate r and \(\theta\) as follows:
\(r^2=2^2+(-2)^2\)
\(=8\)
\(r=\pm2\sqrt{2}\)
\(\tan\theta=\frac{y}{x}\)
\(=\frac{-2}{2}\)
\(=-1\)
Thus, the angle \(\theta\) is \(\frac{3\pi}{4},\frac{7\pi}{4}\) as the angle \(0\leq\theta2\pi\)
The polar coordinates of (2,-2) are,
\((2\sqrt{2},\frac{7\pi}{4})\), where r>0
\((-2\sqrt{2},\frac{3\pi}{4})\), where r
b) The Cartesian coordinates be \((-1,\sqrt{3})\).
Calculate r and \(\theta\) as follows:
\(r^2=(-1)^2+(\sqrt{3})^2\)
\(=4\)
\(r=\pm\sqrt{2}\)
\(\tan\theta=\frac{y}{x}\)
\(=\frac{\sqrt{3}}{-1}\)
\(=-\sqrt{3}\)
Thus, the angle \(\theta\) is \(\frac{2\pi}{3},\frac{5\pi}{3}\) as the angle \(0\leq\theta\leq2\pi\)
The polar coordinates of \((-1,\sqrt{3})\) are,
\((2,\frac{2\pi}{3})\), where r>0
\((-2,\frac{5\pi}{3})\), where r<0
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