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Math Help - Second derivative of the equation of an ellipse

  1. #1
    Junior Member Hardwork's Avatar
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    Second derivative of the equation of an ellipse

    I've \frac{x^2}{a^2}+\frac{y^2}{b^2} = 1. I'm asked to find \frac{d^2y}{dx^2}. I've implicitly differentiated it twice, but I can't get the required answer. What I did:

    y = \frac{x^2}{a^2}+\frac{y^2}{b^2} = 1 \Rightarrow \frac{2x}{a^2}+\frac{2y}{b^2}\frac{dy}{dx} = 0 \Rightarrow \frac{2}{a^2}+\frac{2}{b^2}\frac{dy}{dx}\frac{dy}{  dx}+\frac{2y}{b^2}\frac{d^2y}{dx^2} = \frac{2}{a^2}+\frac{d^2y}{dx^2}\left(\frac{2}{b^2}  +\frac{2y}{b^2}\right) = 0

    From this I got \frac{d^2y}{dx^2} = \frac{-b^2}{a^2(y+1)}, whereas I was expected to get something quite different. So what has possibly gone wrong?
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  2. #2
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    Quote Originally Posted by Hardwork View Post
    I've \frac{x^2}{a^2}+\frac{y^2}{b^2} = 1. I'm asked to find \frac{d^2y}{dx^2}. I've implicitly differentiated it twice, but I can't get the required answer. What I did:

    y = \frac{x^2}{a^2}+\frac{y^2}{b^2} = 1 \Rightarrow \frac{2x}{a^2}+\frac{2y}{b^2}\frac{dy}{dx} = 0 \Rightarrow \frac{2}{a^2}+\frac{2}{b^2}\frac{dy}{dx}\frac{dy}{  dx}+\frac{2y}{b^2}\frac{d^2y}{dx^2} = \frac{2}{a^2}+\frac{d^2y}{dx^2}\left(\frac{2}{b^2}  +\frac{2y}{b^2}\right) = 0

    From this I got \frac{d^2y}{dx^2} = \frac{-b^2}{a^2(y+1)}, whereas I was expected to get something quite different. So what has possibly gone wrong?
    \displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2} = 1

    b^2x^2 + a^2y^2 = a^2b^2

    \displaystyle 2b^2 x + 2a^2 y \cdot \frac{dy}{dx} = 0

    \displaystyle \frac{dy}{dx} = -\frac{b^2}{a^2} \cdot \frac{x}{y}

    \displaystyle \frac{d^2y}{dx^2} =  -\frac{b^2}{a^2} \cdot \frac{y - x \cdot \frac{dy}{dx}}{y^2}

    \displaystyle \frac{d^2y}{dx^2} =  -\frac{b^2}{a^2} \cdot \frac{y + x \cdot \frac{b^2}{a^2} \cdot \frac{x}{y}}{y^2}

    multiply numerator and denominator by \displaystyle \frac{y}{b^2} ...

    \displaystyle \frac{d^2y}{dx^2} =  -\frac{b^2}{a^2} \cdot \frac{\frac{y^2}{b^2} + \frac{x^2}{a^2}}{\frac{y^3}{b^2}}

    \displaystyle \frac{d^2y}{dx^2} =  -\frac{b^4}{a^2y^3}
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  3. #3
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    I would say

     \displaystyle y = \sqrt{b^2\left(1-\frac{x^2}{a^2}\right)}

    then differentiate explicitly twice.
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  4. #4
    Junior Member Hardwork's Avatar
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    Thanks guys. @ Pickslides, that's a wise suggestion, but it's from an implicit differentiation exercise.
    Quote Originally Posted by skeeter View Post
    \displaystyle \frac{d^2y}{dx^2} =  -\frac{b^4}{a^2y^3}
    That's indeed the expected answer, thanks. But I can't manage to find anything that's wrong with the one I got either.
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  5. #5
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    Quote Originally Posted by Hardwork View Post
    y = \frac{x^2}{a^2}+\frac{y^2}{b^2} = 1 \Rightarrow \frac{2x}{a^2}+\frac{2y}{b^2}\frac{dy}{dx} = 0 \Rightarrow \frac{2}{a^2}+\frac{2}{b^2}\frac{dy}{dx}\frac{dy}{  dx}+\frac{2y}{b^2}\frac{d^2y}{dx^2} = \frac{2}{a^2}+\frac{d^2y}{dx^2}\left(\frac{2}{b^2}  +\frac{2y}{b^2}\right) = 0

    From this I got \frac{d^2y}{dx^2} = \frac{-b^2}{a^2(y+1)}, whereas I was expected to get something quite different. So what has possibly gone wrong?
    3rd step to 4th ...

    \frac{dy}{dx} \cdot \frac{dy}{dx} \ne \frac{d^2y}{dx^2}
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  6. #6
    Junior Member Hardwork's Avatar
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    Quote Originally Posted by skeeter View Post
    3rd step to 4th ...

    \frac{dy}{dx} \cdot \frac{dy}{dx} \ne \frac{d^2y}{dx^2}
    Oh, my, my! I can't thank you enough, Skeeter. That assumption was what frustrated me!
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