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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?

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2 Answers 2

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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?

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  • $\begingroup$ er, both of the examples I gave are algebraic. One can clear radicals as necessary by raising to appropriate powers. $\endgroup$ Sep 9, 2010 at 15:31
  • $\begingroup$ 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). $\endgroup$
    – Isaac
    Sep 9, 2010 at 17:15
  • $\begingroup$ @J.M I tried to strikethrough my last comment but couldn't find out how so I deleted. $\endgroup$
    – alext87
    Sep 9, 2010 at 18:47
  • $\begingroup$ 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. $\endgroup$ Sep 9, 2010 at 21:27
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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....

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  • $\begingroup$ That's what the Mathematica code I included in my post does, more or less, except that only three equations are dealt with. $\endgroup$ Sep 9, 2010 at 21:21

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