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Parametrization of implicit curve

  • Thread starter nuuskur
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  • #1
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Homework Statement


[itex]y^2 + 3x - x^3 = C, C\in\mathbb{R}\setminus\{0\}[/itex]

Homework Equations




The Attempt at a Solution


Keeping in mind that ##\cos ^2\alpha + \sin ^2\alpha = 1##
I would go about it
[tex]\left (\frac{y}{\sqrt{C}}\right )^2 + \left (\frac{\sqrt{3x-x^3}}{\sqrt{C}}\right )^2 = 1 [/tex]
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.
 

Answers and Replies

  • #2
verty
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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.
 
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  • #3
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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.
Hmm, from what you say I gather there are a number of criteria for a curve to be "parametrizable".
 
  • #4
Ray Vickson
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Homework Statement


[itex]y^2 + 3x - x^3 = C, C\in\mathbb{R}\setminus\{0\}[/itex]

Homework Equations




The Attempt at a Solution



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

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.
 

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