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Thread: Calculation of y^n.

  1. #1
    May 2011

    Calculation of y^n.

    I'm doing some exercises and I came across this one that starts requesting an implicit differentiation of the second derivative:


    I've done a few of these implicit differentiation by now but what's confusing me about this one is that the first step of the solutions manual transforms the equation into:


    Later of course the right side of the equation is set to 0 and x' and y' are calculated. This first step above though has me wondering how the first equation which is a product of x^2a^2 and y^2b^2 gets changed to a division of each respectively looks strange to me. I'm wondering if anyone may have a better understanding of how those might be equivalent?

    Thanks in advance...
    Last edited by sepoto; Nov 11th 2013 at 07:31 PM.
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  2. #2
    MHF Contributor
    Nov 2010

    Re: Calculation of y^n.

    It looks like the solution's manual was suggesting that so long as a and b are nonzero, this is the equation for an ellipse. If a\neq 0 then \dfrac{1}{a}\neq 0. So, the solutions manual really should have written \dfrac{x^2}{a_1^2} + \dfrac{y^2}{b_1^2}=1 where a_1 = \dfrac{1}{a} and b_1 = \dfrac{1}{b}. In other words, they are a different a and b.
    Thanks from sepoto
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