What the is the polar form of y = x^2y-x/y^2 +xy^2 y=x2y−xy2+xy2?

1 Answer
James
May 20, 2018

color(blue)[r^2=(-costheta)/[sin^3theta-r^2*sin^3theta*cos^2theta-r^2costheta*sin^4theta]]r2=−cosθsin3θ−r2⋅sin3θ⋅cos2θ−r2cosθ⋅sin4θ

Explanation:

Note that

color(red)[y=r*sintheta]y=r⋅sinθ

color(red)[x=r*costheta]x=r⋅cosθ

y = x^2y-x/y^2 +xy^2y=x2y−xy2+xy2

(r*sintheta)=(r*costheta)^2*(r*sintheta)-(r*costheta)/(r*sintheta)^2+(r*costheta)(r*sintheta)^2(r⋅sinθ)=(r⋅cosθ)2⋅(r⋅sinθ)−r⋅cosθ(r⋅sinθ)2+(r⋅cosθ)(r⋅sinθ)2

rsintheta=r^3sintheta*cos^2theta-(costheta)/(rsin^2theta)+r^3costheta*sin^2thetarsinθ=r3sinθ⋅cos2θ−cosθrsin2θ+r3cosθ⋅sin2θ

[rsintheta-r^3sintheta*cos^2theta-r^3costheta*sin^2theta]/1=-(costheta)/(rsin^2theta)rsinθ−r3sinθ⋅cos2θ−r3cosθ⋅sin2θ1=−cosθrsin2θ

r^2sin^3theta-r^4*sin^3theta*cos^2theta-r^4costheta*sin^4theta=-costhetar2sin3θ−r4⋅sin3θ⋅cos2θ−r4cosθ⋅sin4θ=−cosθ

color(blue)[r^2=(-costheta)/[sin^3theta-r^2*sin^3theta*cos^2theta-r^2costheta*sin^4theta]]r2=−cosθsin3θ−r2⋅sin3θ⋅cos2θ−r2cosθ⋅sin4θ

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