Having trouble figuring out the total time it takes for something following the

Question

Having trouble figuring out the total time it takes for something following the path of a brachistochrone. Starting at some point along and falling due to gravity to the lowest point. Equation I have is (a^.5/g^.5)*(1-cos(t))^.5/(cos(t_o)-cos(t))^.5 a and g are constants and t_o is the initial angle at which the particle is at in relation to the brachistochrone. Can not get this function to integrate on Mathematica or Calculator; no consistent answers. Mathematica returns part of the integrate as an imaginary number and I don't know why.

Explanation / Answer

t = Integral ( ds/v)

where ds = (dx^2 + dy^2) = (1 + y' ^2) dx

and v = (2gy)

t = Integral ( (1 + y' ^2)/(2gy) )dx

We know Brachistochrone curve is actually a cycloid, given as

x = (1/2)*k^2*( - sin )

y = (1/2)*k^2*(1- cos )

Also, using Beltrami Identity, it can be derived that

1/( (1 + y' ^2) * (2gy) ) = C

Where C and k are positive constants related as C^2*k^2 = 1/2g

Using these results, we get

t = Integral ( (2C/(1-cos )) (1/2)*k^2*(1 - cos )d )

t = C*k^2* [(ending point) - (starting point)]

The constant k can be found by using that the cycloid curve passes through the starting and ending points

( that is put values of x and y in the parametric equations and solve for k and )

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