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Thread: parametrization

  1. #1
    Member akhayoon's Avatar
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    parametrization

    how would you go about parameterizing

    $\displaystyle x^{2/3}+y^{2/3}=1$??
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  2. #2
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    Hello, akhayoon!

    Parametize: .$\displaystyle x^{\frac{2}{3}}+y^{\frac{2}{3}}\:=\:1$
    . . . . $\displaystyle \begin{array}{ccc}x &=&\cos^3\!\theta \\ y &=&\sin^3\!\theta \end{array}$

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  3. #3
    Member akhayoon's Avatar
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    so how do I know when to you cos and sin when parameterizing? is there a general rule of some sort?
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  4. #4
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    Hello, akhayoon!

    so how do I know when to use cos and sin when parameterizing?
    is there a general rule of some sort?
    Good questions . . . I wish knew the answers.


    From experience, when I see $\displaystyle x^n + y^n$ (both to the same power),
    . . I suspect that I can hammer it into the form: .$\displaystyle \sin^2\!\theta + \cos^2\!\theta \:=\:1$


    Example: .$\displaystyle x^3 + y^3 \:=\:8$

    Here is my rather primitive approach . . .

    Divide by 8: .$\displaystyle \left(\frac{x}{2}\right)^3 + \left(\frac{y}{2}\right)^3 \:=\:1$

    We want this to equal: .$\displaystyle \sin^2\!\theta + \cos^2\!\theta \:=\:1$

    So we want: .$\displaystyle \begin{array}{ccc}\left(\dfrac{x}{2}\right)^3 &=&\cos^2\theta \\ \\ [-4mm] \left(\dfrac{y}{2}\right)^3 &=&\sin\theta \end{array}$


    Solve for $\displaystyle x\text{ and }y$

    . . $\displaystyle \left(\frac{x}{2}\right)^3 \:=\:\cos^2\theta \quad\Rightarrow\quad \frac{x}{2} \:=\:(\cos\theta)^{\frac{2}{3}}\quad\Rightarrow\qu ad \boxed{x \:=\:2(\cos\theta)^{\frac{2}{3}}}$

    . . $\displaystyle \left(\frac{y}{2}\right)^3 \:=\:\sin^2\theta \quad\Rightarrow\quad \frac{y}{2} \:=\:(\sin\theta)^{\frac{2}{3}} \quad\Rightarrow\quad \boxed{y \:=\:2(\sin\theta)^{\frac{2}{3}}} $


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    If we have the form: .$\displaystyle x^n - y^n \:=\:C$

    . . force it into the form: .$\displaystyle \sec^2\!\theta - \tan^2\!\theta \:=\:1$

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