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Calculus of variations. See attached file for full problem description.

#### Solution Preview

The derivation is in the attached pdf file.

I chose a different constant of convenience because I think it makes things simpler:
my b and the C of your texts are related as b = 1/2C^2
Also, I took negative y below zero, unlike the reversed direction of y in your texts.
These details do not affect the essence, and I think my choice is better than that suggested in the text.

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Here is the plain TEX source

centerline{bf Brachistochrone }

The motion starts from \$(x,y)=(0,0)\$ and the mass moves down to positive x.
The potential energy transformed into kinetic energy is \$-mgy\$ and the kinetic energy is \$mv^2/2\$,
therefore the speed is
\$\$
v=sqrt{-2gy}.
eqno(1)
\$\$
The time spent on travelling infinitesimal displacement is
\$\$
dt = {sqrt{dx^2+dy^2}over v} = {dxsqrt{1+y'^2}over sqrt{-2gy}},
eqno(2)
\$\$
where \$'\$ means the derivative with respect to x (\$y'=dy/dx\$).
From equation (2) we find
\$\$
f(y.y',x) = {1over sqrt{2g}} sqrt{1+y'^2over -y} ...

#### Solution Summary

The famous Brachistochrone Variation problem is worked out in an attached PDF.

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