What is the Cartesian form of r^2-2r = -theta+tan(theta)- cos^2(theta) r2−2r=−θ+tan(θ)−cos2(θ)?

1 Answer
Sonnhard · Stefan V.
Jul 17, 2018

x^2+y^2-2sqrt(x^2+y^2)=-arctan(y/x)+y/x-x^2/(x^2+y^2)x2+y2−2√x2+y2=−arctan(yx)+yx−x2x2+y2

Explanation:

We are Using

x=r cos(theta)x=rcos(θ)

y=r sin(theta)y=rsin(θ)
theta=arctan(y/x)θ=arctan(yx)
r=sqrt(x^2+y^2)r=√x2+y2

so we get

x^2+y^2-2sqrt(x^2+y^2)=-arctan(y/x)+y/x-x^2/(x^2+y^2)x2+y2−2√x2+y2=−arctan(yx)+yx−x2x2+y2

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