find the cartesian equivalent of polar equivalent

[spdrmncoo]

find the cartesian equivalent of polar equation r=1 - cos(theta)
somebody can help me, please?

 

[stapel]

What formulas did they give you that relate x, y, r, and theta?

Thank you.

Eliz.

 

[spdrmncoo]

r = 1 - cos(theta)
x = r cos(theta)
y= r sin (theta)
cos(theta)=x/r --> r= (1-x)/r

 

[galactus]

Don't forget, \(\displaystyle \L\\\sqrt{x^{2}+y^{2}}=r\)

 

[Smily]

is it look right?

r^2 = r - r cos (theta)
x^2 + y^2 = sqrt(x^2 + y^2) - x
y^2 = sqrt(x^2 + y^2) - x- x^4
sqrt(x^2 + y^2) - x- x^2 - y^2 =0