Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Usually, applying the conversion formulae

$r^2=x^2+y^2$

$\cos\;\theta=\frac{x}{r}$

$\sin\;\theta=\frac{y}{r}$

to transform an equation in polar coordinates to an implicit Cartesian equation is quite straightforward for an equation of the form

$r=f(\cos(n\theta),\sin(n\theta))$

with $n$ an integer, thanks to multiple angle formulae. Polar equations of the form

$r=f\left(\cos\left(\frac{p}{q}\theta\right),\sin\left(\frac{r}{s}\theta\right)\right)$

where $p$, $q$, $r$ and $s$ are integers, and $\frac{p}{q}$ and $\frac{r}{s}$ are in lowest terms, are more difficult to handle, but the case where $q$, $s$ are powers of 2 is slightly eased by the existence of identities like $\tan\left(\frac{\theta}{2}\right)=\frac{\sin\;\theta}{1+\cos\;\theta}$.

The really difficult ones to handle are the cases like Cayley's sextic:

$r=4a\cos^3\left(\frac{\theta}{3}\right)$

and in general, the cases with fractions whose denominators are not powers of 2. In particular for Cayley's sextic, I can't seem to find a quick way to exploit the appropriate multiple angle identities.

One "cheat" I have seen is to instead represent the polar equation as a pair of parametric equations, and then make the substitution $\theta=3\arctan\;t$ so that everything is represented algebraically. The problem is that apparently it takes some insight to recognize how to remove the new parameter $t$ easily. For even tougher cases like

$r=\cos\left(\frac{2\theta}{3}\right)-\sin\left(\frac{3\theta}{5}\right)$

the appropriate substitution after transforming to parametric equations is $\theta=\mathrm{LCM}(3,5)\arctan\;t$, but the expressions, though algebraic, look even more nightmare-ish to manipulate.

Mathematica has no trouble figuring out the implicit Cartesian equation, through a judicious use of Gröbner bases:

Factor /@ GroebnerBasis[Append[Thread[{x, y} == TrigExpand[{4a Cos[t/3]^3 Cos[t], 4a Cos[t/3]^3 Sin[t]}]],
Cos[t/3]^2 + Sin[t/3]^2 == 1], {x, y}, {Cos[t/3], Sin[t/3]}, MonomialOrder -> EliminationOrder]

but I'm pretty sure the Cartesian equations have already been determined way before Buchberger was born.

How would one determine the Cartesian equation of such a curve using only classical techniques?

Apropos to this question, is there a quick way to determine the degree of an algebraic curve represented in polar coordinates without having to do a conversion?

share|improve this question

2 Answers 2

up vote 3 down vote accepted

We could do something simple, since we have: \begin{equation} \cos\left(\frac{p}{q}\theta\right)=\frac{1}{2}\left(e^{i\frac{p}{q}\theta}+e^{-i\frac{p}{q}\theta}\right)=\frac{1}{2}\left(\left(\frac{x}{r}+i\frac{y}{r}\right)^{\frac{p}{q}}+\left(\frac{x}{r}-i\frac{y}{r}\right)^{\frac{p}{q}}\right) \end{equation} That is \begin{equation} \cos\left(\frac{p}{q}\theta\right)=\frac{1}{2}\left(\frac{1}{x^2+y^2}\right)^{\frac{p}{2q}}\left(\left(x+iy\right)^{\frac{p}{q}}+\left(x-iy\right)^{\frac{p}{q}}\right) \end{equation} and similarly we have \begin{equation} \sin\left(\frac{p}{q}\theta\right)=\frac{1}{2i}\left(\frac{1}{x^2+y^2}\right)^{\frac{p}{2q}}\left(\left(x+iy\right)^{\frac{p}{q}}-\left(x-iy\right)^{\frac{p}{q}}\right) \end{equation} Now we can use these to express your equations in terms of $x$ and $y$. Is this any use?

share|improve this answer
    
er, both of the examples I gave are algebraic. One can clear radicals as necessary by raising to appropriate powers. –  J. M. Sep 9 '10 at 15:31
    
Using this answer, I was able to take $r=4a\cos^3(\frac{\theta}{3})$ and turn it into $(2(x^2+y^2)-2ax)^3=54a^2(x^2+y^2)^2$, which appears to have the same graph (using Mathematica). –  Isaac Sep 9 '10 at 17:15
    
@J.M I tried to strikethrough my last comment but couldn't find out how so I deleted. –  alext87 Sep 9 '10 at 18:47
    
Isaac: FullSimplify[(2(x^2 + y^2) - 2a x)^3 == 54a^2 (x^2 + y^2)^2 /. {x -> 4a Cos[t/3]^3 Cos[t], y -> 4a Cos[t/3]^3 Sin[t]}] confirms it. :D alex: Okay, you have my upvote. –  J. M. Sep 9 '10 at 21:27

Here's a systematic method, but not one recommended for paper-and-pen calculation.

Let $n$ be a common denominator for all rationals $r$ such that $\sin r\theta$ or $\cos r\theta$ arises in the formula. Let $u=\cos(\theta/n)$ and $v=\sin(\theta/n)$. There are polynomials $\phi_n$ and $\psi_n$ such that $\cos\theta=\phi_n(u,v)$ and $\sin\theta=\psi_n(u,v)$. The curve has a polar equation which can be written as $f(r,u,v)=0$. Now consider the system of equations: $$f(s^2,u,v)=0,\ \ \ x=s^2\phi_n(u,v),\ \ \ y=s^2\psi_n(u,v),\ \ \ u^2+v^2=1.$$ Yes, four equations in five variables. Using some systematic elimination procedure, for instance Groebner bases, eliminate the three variables $s$, $u$ and $v$ to get (one hopes) one equation in $x$ and $y$. It certainly helps to have some computer package to do this.

As an example, for $r=4a\cos^3(\theta/3)$, the system of equations becomes $$s^2=4au^3,\ \ \ x=s^2(u^3-3uv^2),\ \ \ y=s^2(3u^2v-v^3),\ \ \ u^2+v^2=1.$$ Well, I'm not going to do the elimination, but $s^2$ disappears readily enough....

share|improve this answer
    
That's what the Mathematica code I included in my post does, more or less, except that only three equations are dealt with. –  J. M. Sep 9 '10 at 21:21

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.