Question

does cartesian equation of a cardiode exist??

Answer 1

i believe it does

Answer 2

yeah i just think its really ugly

Answer 3

can it be obtained from the polar equation...?

Answer 4

(x2 + y2 - 2ax)2 = 4a2(x2 + y2)

Answer 5

how did u obtain it??

Answer 6

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
\[x ^{2}+y ^{2}=\sqrt{x ^{2}+y ^{2}}+y\]

Answer 7

r^2 = x^2 + y^2 is for circle

Answer 8

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

Answer 9

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