How do you convert #r = 1 - sin (theta) # into cartesian form?

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Answer 1 · 2016-07-07T12:33:03

# x^2 + y^2 = sqrt(x^2 + y^2) - y #

you have

#x = r cos theta#
#y = r sin theta#

and from that

#r = sqrt(x^2 + y^2)#

so we have

# sqrt(x^2 + y^2) = 1- y/r #

# sqrt(x^2 + y^2) = 1- y/sqrt(x^2 + y^2) #

so time by the radical and you have

# x^2 + y^2 = sqrt(x^2 + y^2) - y #