What is the Cartesian form of r^2+theta = -tan^2theta+4cottheta r2+θ=−tan2θ+4cotθ?

1 Answer
Cesareo R.
May 16, 2016

y^3 -4 x^3 + x^4 y + x^2 y^3 + x^2 y arctan(y/x)y3−4x3+x4y+x2y3+x2yarctan(yx)

Explanation:

From the pass equations
x = r cos(theta)x=rcos(θ)
y = r sin(theta)y=rsin(θ)
We find out that
y/x=sin(theta)/cos(theta) = tan(theta)->theta = arctan(y/x)yx=sin(θ)cos(θ)=tan(θ)→θ=arctan(yx)
cot(theta) = 1/tan(theta) = x/ycot(θ)=1tan(θ)=xy
r = x/cos(theta)r=xcos(θ)

Related questions
See all questions in Converting Between Systems
Impact of this question
1232 views around the world
You can reuse this answer
Creative Commons License