How can I convert the polar equation r=1-cos(theta) into its cartesian equivalent? How about r=1+sin(theta) and 2/r=4+cos(theta)?

Question

caozeyuan · Answer

@amistre64  Help me!!!

caozeyuan · Answer

I'm dying! some one save me please!!!!

amistre64 · Answer

some things to keep in mind

x = r cos(t)
y = r  sin(t)

r^2 = x^2+y^2

caozeyuan · Answer

I just got a result. For the first one, it's x^2+y^2=sqrt(x^2-y^2)-y. Is that correct

amistre64 · Answer

the polar plot of r = 1-cos(t) is heart shaped

x^2 + y^2 = r - rcos(t)

x^2 + y^2 = r - x

x^2 - x + y^2 = r

(x-1/2)^2 + y^2 - 1/4 = r

yeah, these things have always been a pain to me too

amistre64 · Answer

but your solution so far is not quite correct yet http://www.wolframalpha.com/input/?i=x%5E2%2By%5E2%3Dsqrt%28x%5E2-y%5E2%29-y

amistre64 · Answer

my getup is pointing in the wrong direction, but is getting closer :) http://www.wolframalpha.com/input/?i=%28x-1%2F2%29%5E2%2By%5E2-1%2F4%3Dsqrt%28x%5E2%2By%5E2%29

amistre64 · Answer

if i change the (x-1/2)  into (-x-1/2) we get a better result

amistre64 · Answer

pfft, i see what i did to begin with,  i should have add x to each side, that where i introduced my error at

caozeyuan · Answer

you see. r=1-cos theta, r^2 = r-rcos theta. x^2+y^2=r-y, x^2+y^2=sqrt(x^2+y^2)-y
why, is my reasoning wrong?

caozeyuan · Answer

@amistre64

amistre64 · Answer

because rcos(t) = x,  not y

amistre64 · Answer

otherwise, your spot on

caozeyuan · Answer

I'sooooo stupid!