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Given an implicit equation such as $x^2+y^2=1$ , I know it corresponds to the parametric equations $ \begin{cases} x=\cos t\\ y=\sin t \end{cases}$.

But I don't know how to get from the implicit equations to the parametric one, only the other way around.

I'm interested in a general algorithm in solving for the parametric equations given the implicit equation. If such algorithm doesn't exist I want to know how to approach this problem given different types of implicit equations:

  • Polynomial
  • Trigonometric
  • Exponential
  • Other types and/or a combination of any of the above

Maybe some of them can be transformed and some others can't (How can I tell?).

To illustrate such an algorithm, please show me how to get parametric equations for $y\sin x+x\cos y=1$.

Thank you!

Sine of the Time
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aku jack
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    If you find such algorithm, invite a round of beers for the Field medal. – ajotatxe Dec 11 '22 at 22:52
  • Then I guess this question shows how little do I know about this stuff @ajotatxe. – aku jack Dec 11 '22 at 22:53
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    @akujack There are a few tricks that work on some curves. For instance, for $x^2+y^2=1$, let $y=tx$, then $x^2+t^2x^2=1$ hence $x=\dfrac{1}{\sqrt{1+t^2}}$ and $y=\dfrac{t}{\sqrt{1+t^2}}$ is a parameterization. – Jean-Claude Arbaut Dec 11 '22 at 22:55
  • @Jean-ClaudeArbaut, Thank you for the trick, but that parametrization works only for half a unit circle since it doesn't yield any negative values for x. – aku jack Dec 11 '22 at 23:02
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    Indeed, it's often the case with a parameterization. Of course here you get the other half with $x=-\dfrac{1}{\sqrt{1+t^2}}$, $y=-\dfrac{t}{\sqrt{1+t^2}}$ (when solving for $t$ I picked only the + sign above, you get another part with the - sign). The same applies to $y=\sqrt{1-x^2}$, that you can write $x=t, y=\sqrt{1-t^2}$. – Jean-Claude Arbaut Dec 11 '22 at 23:03
  • Only a little example: $e^x+\sin (y^{1.8})=7\pi$. Now tell me the algorithm.... – ajotatxe Dec 11 '22 at 23:06
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    This one is easy: solve for $x$ as a function of $y$, and let $y=t$. It gets complicated when you can't solve easily for one of the variables. – Jean-Claude Arbaut Dec 11 '22 at 23:07
  • Parametrization isn't unique. You could just as well have $(x, y) = (\sin t, \cos t)$. Or $(t, \pm \sqrt{1-t^2})$. – Dan Dec 15 '22 at 19:43
  • Solving for $x$ is fairly similar, so for $a y+\cos(y) x=0$ use this inverse – Тyma Gaidash Jan 16 '23 at 13:26

1 Answers1

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The parametric equation $\begin{cases}x=\cos t\\ y=\sin t\end{cases}$ you wrote, is actually the entire solution set of the implicit equation $x^2+y^2=1$. In this sense, if we cannot solve the implicit equation in closed-form using standard mathematical functions with respect to any variable, then we cannot obtain the parametric equation either.

Let's consider the variable $y$, as a fixed number. We want to see the solution of the equation $y\sin x+x\cos y=1$, with respect to $x:$

$$\begin{align}&\sin x=\frac {1-x\cos y}{y}\\ \implies &\sin x=-\frac {\cos y}{y}x+\frac 1y\end{align}$$

Then, note that the equation $\sin x=kx,~k≠0$ (except $x=0$) and $\sin x=ax+b$, where $a,b≠0$ are types of transcendental equations that do not have a general closed-form solution.

Therefore, the implicit equation $y\sin x+x\cos y=1$ cannot be expressed by means of closed - form parametric equations, in general.

However, the equations we noted as transcendental, do not have only general closed-form solutions. For instance, the specific transcendental equation $\sin x=\frac 3\pi x$ have closed-form solutions. Indeed, $x_1=0,~x_2=\frac \pi 6,~x_3=-\frac \pi 6$ are only possible solutions.

To summarize, the implicit equation $y\sin x+x\cos y=1$, cannot be expressed with parametric equations, even through any special functions defined in mathematics.

lone student
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