What is the Cartesian form of r^2+r = 2theta-sec^2theta+4csctheta r2+r=2θ−sec2θ+4cscθ?

1 Answer
Leland Adriano Alejandro
Mar 25, 2016

Cartesian form:
x^2+y^2+sqrt(x^2+y^2)=2*arctan(y/x)-((x^2+y^2)/x^2)+(4*sqrt(x^2+y^2))/yx2+y2+√x2+y2=2⋅arctan(yx)−(x2+y2x2)+4⋅√x2+y2y

Explanation:

From the given:
r^2+r=2theta-sec^2 theta+4*csc thetar2+r=2θ−sec2θ+4⋅cscθ
We make use of the equivalent of
r=sqrt(x^2+y^2)r=√x2+y2
theta=arctan (y/x)θ=arctan(yx)

Direct substitution
r^2+r=2theta-sec^2 theta+4*csc thetar2+r=2θ−sec2θ+4⋅cscθ
(sqrt(x^2+y^2))^2+sqrt(x^2+y^2)=2*arctan(y/x)-(sqrt(x^2+y^2)/x)^2+(4*sqrt(x^2+y^2))/y(√x2+y2)2+√x2+y2=2⋅arctan(yx)−(√x2+y2x)2+4⋅√x2+y2y

The graph of x^2+y^2+sqrt(x^2+y^2)=2*arctan(y/x)-((x^2+y^2)/x^2)+(4*sqrt(x^2+y^2))/yx2+y2+√x2+y2=2⋅arctan(yx)−(x2+y2x2)+4⋅√x2+y2y
graph{x^2+y^2+sqrt(x^2+y^2)=2arctan(y/x)-((x^2+y^2)/x^2)+(4sqrt(x^2+y^2))/y [-20, 20, -10, 10]}

God bless....I hope the explanation is useful.

Related questions
See all questions in Converting Between Systems
Impact of this question
1351 views around the world
You can reuse this answer
Creative Commons License