Calculating the Integral for the Brachistochrone Problem

Question

I am trying to prove that the shortest possible time of descent in the Brachistochrone problem is:

$$ T = \sqrt\frac{c_1}{2g} \theta_1 $$

I have the following 2 parametric equations:

$$ x = \frac{c_1}{2} (\theta - \sin\theta) \qquad y = \frac{c_1}{2} (1 - \cos\theta) $$

Solving $$ \frac{dx}{d\theta} = \frac{c_1}{2} (1-\cos\theta) \qquad \frac{dy}{d\theta} = \frac{c_1}{2} (\sin\theta) $$ I know that $$ T = \int_0^{x_1} \frac{\sqrt{1 + (y')^2}}{\sqrt{2gy}} dx \tag{1} $$

Now solve $$y' = \frac{dy}{dx} = \frac{dy / d\theta}{ dx / d\theta } = \frac{\sin\theta}{1-\cos\theta} $$

Therefore $$ (y')^2 = \frac{\sin^2\theta}{(1-\cos\theta)^2} = \frac{1+\cos\theta}{1-\cos\theta} $$

Now plugging everything back into \ref{1} and putting the integral in terms of $\theta$: $$T = \int_0^{\theta_1} \frac{\sqrt{1 + \frac{1+\cos\theta}{1-\cos\theta}}}{\sqrt{2g\frac{c_1}{2} (1 - \cos\theta)}} \frac{c_1}{2} (1-\cos\theta) d\theta$$

I tried solving this integral but I am having a hard time with it. I would appreciate some guidance on how to tackle this integral.

Answer 1

I did not check your calculations, but there are a few things which make the problem simpler $$\sqrt{1+\frac{1+\cos (\theta)}{1-\cos (\theta)}}=\csc \left(\frac{\theta}{2}\right)$$ Simplifying,we have $$I=\frac 12 \sqrt{\frac{c_1}{g}}\int \csc \left(\frac{\theta}{2}\right)\sqrt{1-\cos (\theta)} \,d\theta=\frac 12 \sqrt{\frac{c_1}{g}}\int \csc \left(\frac{\theta}{2}\right) \sqrt{2 \sin ^2\left(\frac{\theta}{2}\right) }\,d\theta$$ $$I=\sqrt{\frac{c}{2 g}} \int d\theta=\sqrt{\frac{c}{2 g}}\, \theta$$

Answer 2

We can work with dimensionless variable, $\large X = {2x \over c_1}\;Y = {2y \over c_1}$

$X = θ - \sin(θ)$
$Y = 1 - \cos(θ)$

${dX \over dθ} = 1 - \cos(θ) = Y$
${dY \over dθ} = \sin(θ) = ±\sqrt{1-\cos^2(θ)} = ±\sqrt{1-(1-Y)^2} = ±\sqrt{2Y-Y^2}$

$\displaystyle \left({dX \over dθ}\right)^2 + \left({dY \over dθ}\right)^2 = Y^2 + (2Y-Y^2) = 2Y$

$\displaystyle T = \int_0^{X_1} \sqrt{1+\left({dY \over dX}\right)^2 \over c_1\;g\;Y} \left( {c_1\over2} dX\right) = \sqrt{c_1 \over 4g} \int_0^{θ_1} \sqrt{2Y\over Y} dθ = \sqrt{c_1 \over 2g}\; θ_1 $