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For $x^2+y^2=a^2$ show that $y''=-(a^2/y^3)$

I got that

$y^2=a^2-x^2$

$y'=-x/y$

$y''=(-1-y'^2)/y$

But then I get stuck.

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    $\begingroup$ Here's a MathJax tutorial :) $\endgroup$
    – Shaun
    Commented Sep 29, 2017 at 12:07
  • $\begingroup$ Thank you, I was looking for that :) $\endgroup$
    – user341207
    Commented Sep 29, 2017 at 13:26

2 Answers 2

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Implicit differentiation gives $$ 2x+2yy'=0 \tag{*} $$ Differentiate again (after removing the common factor $2$): $$ 1+(y')^2+yy''=0 \tag{**} $$ Now (*) implies $y'=-x/y$, so you can substitute in (**): $$ 1+\frac{x^2}{y^2}+yy''=0 $$ Isolate $y''$ and go on:

$$y''=-\frac{y^2+x^2}{y^3}$$

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I would prefer to use parametric equation for these type of problems

$x=a\cos \alpha$

$y=a\sin \alpha$

$\frac{dy}{d\alpha}=a \cos \alpha$

$\frac{dx}{d\alpha}=-a \sin \alpha$

$\frac{dy}{dx}=- \cot \alpha$

$\frac{{d}^2y}{d{x}^2}=-\frac{d}{d\alpha} \cot \alpha \cdot\frac{d\alpha}{dx}$

$\frac{{d}^2y}{d{x}^2}=\frac{ (\operatorname{cosec} \alpha)^2} {-a \sin \alpha}$

$\frac{{d}^2y}{d{x}^2}=\frac{ 1} {-a (\sin \alpha)^3}$

$\frac{{d}^2y}{d{x}^2}=\frac{-a^2} {y^3}$

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