Cartesian to polar form

By Brad Rodgers (P1930) on Sunday, January 14, 2001 - 05:29 pm :

Is there a general way to convert a Cartesian function to a polar function. I've tried using imaginary units, but this seems to only work when I can get the equation in just the right parametric form. For example, how do I turn

y = a x^{4} + b x^{3} + c x^{2} + m x + n

into a polar function of the form

r = f (θ)

Thanks,

Brad

By Dan Goodman (Dfmg2) on Sunday, January 14, 2001 - 05:38 pm :

Well,

x = r \cos (θ)

and

y = r \sin (θ)

, substituting into the first equation you get:

$r \sin (θ) = a r^{4} \cos^{4} (θ) + b r^{3} \cos^{3} (θ) + c r^{2} \cos^{2} (θ) + m r \cos (θ) + n$ , in other words you have:

$a r^{4} \cos^{4} (θ) + b r^{3} \cos^{3} (θ) + c r^{2} \cos^{2} (θ) + r [m \cos (θ) - \sin (θ)] + n = 0$

Which is a polynomial in $r^{4}$ , this means that you can always solve for $r$ in terms of $θ$ (you wouldn't be able to if you had a degree 5 term in it). However, it won't be easy, if $a = b = 0$ , then you can just use the quadratic formula to express $r$ in terms of $θ$ . I don't think you'll be able to do better than this in general, although for specific functions you might be able to.