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

1 Answer
P dilip_k
Oct 3, 2016

We know the relations

x=rcostheta and y =rsinthetax=rcosθandy=rsinθ

where r and thetaθ are the polar coordinate of a point having rectangular coordinate (x,y)(x,y)
So r^2=x^2+y^2r2=x2+y2

Now the given relation is

(r+1)^2=-sinthetacostheta+costheta(r+1)2=−sinθcosθ+cosθ

r^2(r+1)^2=-(rsintheta)(rcostheta)+r*rcosthetar2(r+1)2=−(rsinθ)(rcosθ)+r⋅rcosθ
=>r^4+2r^3+r^2=-yx+rx⇒r4+2r3+r2=−yx+rx

=>(x^2+y^2)^2+2(x^2+y^2)^(3/2)+(x^2+y^2)=x(x^2+y^2)^(1/2)-xy⇒(x2+y2)2+2(x2+y2)32+(x2+y2)=x(x2+y2)12−xy

This is the rectangular form of the given polar equation.

Related questions
See all questions in Converting Between Systems
Impact of this question
1517 views around the world
You can reuse this answer
Creative Commons License