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

Answer 1

Quinn Anderson

#(x^2+y^2)^(3/2)y^2=-y^4-(x^2+y^2)^3#

The relation between polar coordinates #(r,theta)# and Cartesian coordinates #(x,y)# is

#x=rcostheta# and #y=rsintheta# i.e. #r^2=x^2+y^2# and #y/x=tantheta#

Hence #r=-sin^2theta-r^2csc^2theta# can be written as

#r=-y^2/r^2-r^2xxr^2/y^2=y^2/r^2-r^4/y^2# and multiplying each term by #r^2y^2#, this can be written as

#r^3y^2=-y^4-r^6#

or #(x^2+y^2)^(3/2)y^2=-y^4-(x^2+y^2)^3#

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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