# Parametrization of implicit curve

1. Sep 26, 2015

### nuuskur

1. The problem statement, all variables and given/known data
$y^2 + 3x - x^3 = C, C\in\mathbb{R}\setminus\{0\}$

2. Relevant equations

3. The attempt at a solution
Keeping in mind that $\cos ^2\alpha + \sin ^2\alpha = 1$
$$\left (\frac{y}{\sqrt{C}}\right )^2 + \left (\frac{\sqrt{3x-x^3}}{\sqrt{C}}\right )^2 = 1$$
would then?
for some interval for $t$
$\sin t = \frac{y}{\sqrt{C}}$
$\cos t = \frac{\sqrt{3x-x^3}}{\sqrt{C}}$
this doesn't look so good, can't get anywhere with it

I can't remember this part at all and every piece of parametrization I have been able to google is either for explicit curves or obvious identity parametrizations, in other words, not very helpful.

2. Sep 26, 2015

### verty

Obs. 1: This curve is symmetrical about the x-axis.
Obs. 2: $3x - x^3$ slopes positively at the origin. It also has the general shape, down-up-down.

Let $z = y^2$, $3x - x^3 = C - z$.

Obs. 3: $C - z$ is the $y$ intercept, $z$ moves the graph down. The question being asked is, what are the x-intercepts when the y intercept is C-z.

Essentially, we are placing the x-axis at $y = C$, mirroring the part below the axis to above the axis, and applying a square-root transformation. When $y = 4$, $z = 16$, etc. It all gets squashed.

So looking at that, I don't see how you could hope to parametrize it.

3. Sep 26, 2015

### nuuskur

Hmm, from what you say I gather there are a number of criteria for a curve to be "parametrizable".

4. Sep 26, 2015

### Ray Vickson

Before attempting to parametrize the curve (if possible), try to understand it first. For example, if you take $C = 1$ you have two curves
$$y = \sqrt{ x^3 - 3 x + 1} \;\;\; \text{(Curve C1)}$$
and
$$y = - \sqrt{ x^3 - 3 x + 1} \;\;\; \text{(Curve C2)}$$

If $x_1, x_2, x_3$ are the real roots of $f(x) = x^3 - 3 x + 1$, we have $x_1 = -1.879385242$, $x_2 = 0.3472963553$ and $x_3 = 1.532088886$. We have $f(x) > 0$ on $(x_1,x_2)$ and on $(x_3 \infty)$, so the values of $y$ are real on those two intervals. Thus, there are really two pairs of (C1,C2) curves, one pair for $x_1 \leq x \leq x_2$ and another pair for $x \geq x_3$. All the curves satisfy $y = 0$ at $x = x_1, x_2, x_3$. Somehow, any parametrization you cook up would need to recognize all those issues.