How do you convert # r=-2 theta - tan theta # to Cartesian form?

Question

1766 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-18T17:32:06

#sqrt(x^2+y^2)=-(y/x+2tan^-1(y/x))#

#"Given,"r=-2theta-tantheta#

#x=rcostheta, y=rsintheta#

#tantheta=y/x#

#r=sqrt(x^2+y^2), theta=tan^-1(y/x)#

Thus,

#r=-2theta-tantheta " becomes, "#
# sqrt(x^2+y^2)=-2tan^-1(y/x)-y/x#

Rearranging

#sqrt(x^2+y^2)=-(y/x+2tan^-1(y/x))#