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

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Answer 1 · 2016-12-01T02:50:02

We know the relations

#x= rcostheta , y =rsintheta and r^2 = x^2+y^2#

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

Given equation

#(r-1)^2=-sinthetacostheta+cos^2theta#

#=>r^2(r-1)^2=-rsintheta* rcostheta+r^2cos^2theta#

#=>(r^2-r)^2=-rsintheta* rcostheta+r^2cos^2theta#

#=>(x^2+y^2-sqrt(x^2+y^2))^2=-y* x+x^2#

#=>(x^2+y^2-sqrt(x^2+y^2))^2=x^2-xy#

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

Answer 2 · 2016-12-01T04:35:34

#(x^2+y^2)(x^2+y^2-sqrt(x^2+y^2))+y(x+y)=0#. Illustrative graph is inserted.

#(r-1)^2=r^2-2r+1=-sin theta cos theta + cos^2 theta#. So,

#r^2-2r=-sin theta cos theta -(1-cos^2theta)#

#=-sin theta cos theta -sin^2theta =-(y(x+y))/r^2#. using

#cos theta = x/r and sin theta = y/r#, where #r = sqrt(x^2+y^2)#

Cross multiplying and using #r=sqrt(x^2+y^2)#,

#(x^2+y^2)(x^2+y^2-2sqrt(x^2+y^2))+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]}