What is the Cartesian form of r = -sintheta+4csc^2theta r=−sinθ+4csc2θ?

1 Answer
Salvatore I.
Nov 4, 2016

(x^2+y^2)(y^2-4*root2(x^2+y^2))+y^3=0(x2+y2)(y2−4⋅2√x2+y2)+y3=0

Explanation:

Let's rewrite the given function in polar coordinates exclusively in term of sin(theta)sin(θ) so that it turns into R=-sin(theta)+4/sin^2(theta)R=−sin(θ)+4sin2(θ)
By recalling that x=R*cos(theta)x=R⋅cos(θ) and y=R*sin(theta)y=R⋅sin(θ), it follows that sin(theta)=y/Rsin(θ)=yR and R=root2(x^2+y^2)R=2√x2+y2.

By replacing both of them it into the given function we get
root2(x^2+y^2)=-y/root2(x^2+y^2)+4(x^2+y^2)/y^22√x2+y2=−y2√x2+y2+4x2+y2y2

After some simple algebraic manipulations we get
the implicit form of the curve on the x-y plane

(x^2+y^2)y^2+y^3-4(x^2+y^2)root2(x^2+y^2)=0(x2+y2)y2+y3−4(x2+y2)2√x2+y2=0

in other terms

(x^2+y^2)(y^2-4*root2(x^2+y^2))+y^3=0(x2+y2)(y2−4⋅2√x2+y2)+y3=0

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