does cartesian equation of a cardiode exist??
i believe it does
yeah i just think its really ugly
can it be obtained from the polar equation...?
(x2 + y2 - 2ax)2 = 4a2(x2 + y2)
how did u obtain it??
the one i know, follow and see if it makes sense r = 1+sin t multiply each side by r r^2=r+rsint r^2 also = x^2+y^2 y=rsint \[x ^{2}+y ^{2}=\sqrt{x ^{2}+y ^{2}}+y\]
r^2 = x^2 + y^2 is for circle
yes it is but that fact still holds true in polar because: x = rcost in polar y= rsint in polar x^2 = r^2cos^2t y^2 = r^2sin^2t add x^2 +y^2= r^2cos^2t + r^2sin^2t and factor out the r^2 so x^2 + y^2 = r^2 (cos^2t + sin^2t) and the identity say cos squared plus sin squared = 1 thus x^2 +y^2 = r^2
also x^2+y^2=√(x^2+y^2)+y which is the equation i had given above for the cartiode rearranged is y= x^2+y^2 -√(x^2+y^2) and that is not the equation for a circle. remember while cartesian is rectangular polar is circular so converting polar to cartesian will give equations that often times resemble circles
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