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Parametrization of implicit curve

  1. Sep 26, 2015 #1
    1. The problem statement, all variables and given/known data
    [itex]y^2 + 3x - x^3 = C, C\in\mathbb{R}\setminus\{0\}[/itex]

    2. Relevant equations


    3. 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.
     
  2. jcsd
  3. Sep 26, 2015 #2

    verty

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    Homework Helper

    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.
     
  4. Sep 26, 2015 #3
    Hmm, from what you say I gather there are a number of criteria for a curve to be "parametrizable".
     
  5. Sep 26, 2015 #4

    Ray Vickson

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    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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