How do you write the cartesian equation for #r=1+3sinx#?

1 Answer
sankarankalyanam
Jul 9, 2018

As below.

Explanation:

#r = 1 + 3 sin theta#

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

#sqrt(x^2 + y^2) = 1 + 3 * y/sqrt(x^2 + y^2)#

#x^2 + y^2 = sqrt(x^2 + y^2) + 3y#

![http://www3.wolframalpha.com/Calculate/MSP/MSP72311c46dfa9cfg334b300002adeff9987e0ge2d?MSPStoreType=image/gif&s=41&w=300.&h=280.&cdf=RangeControl](useruploads.socratic.org)

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