How do you convert $r = 3 θ - csc θ$ $r=3theta - csctheta$ to Cartesian form?

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Answer 1 · 2016-03-14T05:31:38

$\sqrt{x^{2} + y^{2}} = 3 arctan (\frac{y}{x}) - \frac{\sqrt{x^{2} + y^{2}}}{y}$ $sqrt(x^2+y^2)=3arctan(y/x)-sqrt(x^2+y^2)/y$

$(r, θ)$ $(r,theta)$ in polar coordinates is $(r cos θ, r sin θ)$ $(rcostheta,rsintheta)$ in rectangular coordinates and

$(x, y)$ $(x,y)$ in rectangular coordinates is $(\sqrt{x^{2} + y^{2}}, arctan (\frac{y}{x}))$ $(sqrt(x^2+y^2),arctan(y/x))$ in polar coordinates.

Note that $sin θ = \frac{y}{r} = \frac{y}{\sqrt{x^{2} + y^{2}}}$ $sintheta=y/r=y/(sqrt(x^2+y^2)$

Hence $r = 3 θ - csc θ$ $r=3theta-csctheta$ can be written as

$\sqrt{x^{2} + y^{2}} = 3 arctan (\frac{y}{x}) - \frac{\sqrt{x^{2} + y^{2}}}{y}$ $sqrt(x^2+y^2)=3arctan(y/x)-sqrt(x^2+y^2)/y$

How do you convert r=3theta - csctheta r=3θ−cscθ r=3theta - csctheta to Cartesian form?