How do you convert (r-1)^2= - sin theta costheta +cos^2theta(r−1)2=−sinθcosθ+cos2θ to Cartesian form?

2 Answers
P dilip_k
Dec 1, 2016

We know the relations

x= rcostheta , y =rsintheta and r^2 = x^2+y^2x=rcosθ,y=rsinθandr2=x2+y2

where (x,y) and (r,theta)(x,y)and(r,θ) respectively represent the coordinates of same point in cartesian and in polar form.

Given equation

(r-1)^2=-sinthetacostheta+cos^2theta(r−1)2=−sinθcosθ+cos2θ

=>r^2(r-1)^2=-rsintheta* rcostheta+r^2cos^2theta⇒r2(r−1)2=−rsinθ⋅rcosθ+r2cos2θ

=>(r^2-r)^2=-rsintheta* rcostheta+r^2cos^2theta⇒(r2−r)2=−rsinθ⋅rcosθ+r2cos2θ

=>(x^2+y^2-sqrt(x^2+y^2))^2=-y* x+x^2⇒(x2+y2−√x2+y2)2=−y⋅x+x2

=>(x^2+y^2-sqrt(x^2+y^2))^2=x^2-xy⇒(x2+y2−√x2+y2)2=x2−xy

This equation represents the cartesian form of the given polar equation.

A. S. Adikesavan
Dec 1, 2016

(x^2+y^2)(x^2+y^2-sqrt(x^2+y^2))+y(x+y)=0(x2+y2)(x2+y2−√x2+y2)+y(x+y)=0. Illustrative graph is inserted.

Explanation:

(r-1)^2=r^2-2r+1=-sin theta cos theta + cos^2 theta(r−1)2=r2−2r+1=−sinθcosθ+cos2θ. So,

r^2-2r=-sin theta cos theta -(1-cos^2theta)r2−2r=−sinθcosθ−(1−cos2θ)

=-sin theta cos theta -sin^2theta =-(y(x+y))/r^2=−sinθcosθ−sin2θ=−y(x+y)r2. using

cos theta = x/r and sin theta = y/rcosθ=xrandsinθ=yr, where r = sqrt(x^2+y^2)r=√x2+y2

Cross multiplying and using r=sqrt(x^2+y^2)r=√x2+y2,

(x^2+y^2)(x^2+y^2-2sqrt(x^2+y^2))+y(x+y)=0(x2+y2)(x2+y2−2√x2+y2)+y(x+y)=0

graph{(x^2+y^2)(x^2+y^2-2sqrt(x^2+y^2))+y(x+y)=0 [-5, 5, -2.5, 2.5]}

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