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

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Answer 1 · 2016-10-03T16:19:42

We know the relations

#x=rcostheta and y =rsintheta#

where r and #theta# are the polar coordinate of a point having rectangular coordinate #(x,y)#
So #r^2=x^2+y^2#

Now the given relation is

#(r+1)^2=-sinthetacostheta+costheta#

#r^2(r+1)^2=-(rsintheta)(rcostheta)+r*rcostheta#
#=>r^4+2r^3+r^2=-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#

This is the rectangular form of the given polar equation.