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

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Answer 1 · 2018-07-08T16:34:03

#( x^2 + y^2 )^1.5( x^2 - 4 sqrt( x^2 + y^2 ))+ x^4 = 0#

Using the conversion formula

# ( x, y ) = r ( cos theta, sin theta ), r = sqrt ( x^2 + y^2 ) >= 0#,

# r = - sin^2theta + 4 csc^2theta in [0, oo ), theta ne kpi#,

#k = 0, +-1, +-2, +-3, ...# becomes

#x^2( x^2 + y^2 )^1.5 = - x^4 + 4 ( x^2 + y^2 )^2# giving

#( x^2 + y^2 )^1.5( x^2 - 4 sqrt( x^2 + y^2 ))+ x^4 = 0#

Graph of # r = - sin^2theta + csc^2theta#:
graph{(y-(x^4 + ( x^2 + y^2 )^1.5( x^2 -4 sqrt( x^2 + y^2 ))))=0[-40 40 -20 20]}

Observe the inner loop bracing the pole, from blow.
graph{(y-(x^4 + ( x^2 + y^2 )^1.5( x^2 -4 sqrt( x^2 + y^2 ))))=0[-2 2 -1.5 0.5]}