What is the Cartesian form of #r^2+r = 2theta-sec^2theta+4csctheta #?

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))/y#

Explanation:

From the given:
#r^2+r=2theta-sec^2 theta+4*csc theta#
We make use of the equivalent of
#r=sqrt(x^2+y^2)#
#theta=arctan (y/x)#

Direct substitution
#r^2+r=2theta-sec^2 theta+4*csc theta#
#(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#

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))/y#
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.

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