What the is the polar form of #y = 1/y^3-xy+x^2/y #?

Question

1260 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-18T11:48:23

Polar form is #r^4sin^4theta(1+rcos theta)-r^2cos^2theta*sin^2theta =1#

#y=1/y^3-xy+x^2/y # Multiplying by #y^3# on both sides we get,

#y^4=1-xy^4+x^2*y^2 # or

#y^4+xy^4=1+x^2*y^2 # or

#y^4(1+x)=1+x^2*y^2 # or

#y^4(1+x)-x^2*y^2 =1#

Polar form: #x = r costheta and y = r sin theta and x^2+y^2=r^2#

#r^4sin^4theta(1+rcos theta)-r^4cos^2theta*sin^2theta =1#

Polar form is

#r^4sin^4theta(1+rcos theta)-r^2cos^2theta*sin^2theta =1# [Ans]