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 x4 +b x3 +c x2+m x+n
into a polar function of the form
r=f(q)
Thanks,
Brad
By Dan Goodman (Dfmg2) on Sunday,
January 14, 2001 - 05:38 pm :
Well, x=rcos(q) and y=rsin(q), substituting
into the first equation you get:
rsin(q)=a r4cos4(q)+b r3 cos3(q)+c r2cos2(q) +m rcos(q)+n, in other words you have:
a r4cos4(q)+b r3cos3(q)+c r2cos2(q)+r[mcos(q) -sin(q)]+n=0
Which is a polynomial in r4, this means that you can always solve for r
in terms of q (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 q. I don't think you'll be able to
do better than this in general, although for specific functions you might be
able to.