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Please help me spot my mistake:

I have an equation $$(u(x)^{-2} + 4u'(x)^2)^{\frac{1}{2}} - u'(x)\frac{d}{du'}(u(x)^{-2} + 4u'(x)^2)^{\frac{1}{2}} = k$$ where $k$ is a constant.

I am quite sure that if I take $u(x) = \sqrt{y(x)}$ I would have the brachistochrone equation, hence I am expecting a cycloid equation if I let $u(x) = \sqrt{y(x)}$ in the result, but I don't seem to get it :(

My workings are as follows: $$u(x)^{-2} + 4u'(x)^2- 4u'(x)^2 = k \times (u(x)^{-2} + 4u'(x)^2)^{\frac{1}{2}}$$ $$\implies u(x)^{-4} = k^2 \times (u(x)^{-2} + 4u'(x)^2)$$ $$\implies u'(x)= \frac{1}{2k}\sqrt{u(x)^{-4} - k^2u(x)^{-2}}$$ $$\implies \int \frac{1}{u \sqrt{u^2 - k^2}} du = \int \frac{1}{2k} dx$$ Change variable: let $v = \frac{u}{k}$ $$\implies \int \frac{1}{v \sqrt{v^2 - 1}} dv = \frac{x+a}{2}$$, where $a$ is a constant $$\implies \operatorname{arcsec}(v) = \frac{x+a}{2} $$ $$\implies \operatorname{arcsec}\left(\frac{\sqrt{y}}{k}\right) = \frac{x+a}{2}$$ which does not seem to describe a cycloid...

Help would be very much appreciated! Thanks.

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1 Answer 1

up vote 2 down vote accepted

In the line after

$$ 2k u' = \sqrt{ u^{-4} - k^2 u^{-2}} $$

(the fourth line of your computations) when you divided, you divided wrong. The integrand in the LHS should be

$$ \int \frac{u^2}{\sqrt{1-k^2 u^2}} \mathrm{d}u $$

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Aha! Thanks you very much! My brain is dead... –  distressed Aug 17 '11 at 16:48

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