Calculating the time of a Brachistochrone

Question

I derived the general equation of a Brachistochrone, which is a cycloid.

$y=A(1-\cos\theta)$

$x=A(\theta-\sin\theta)$

I'm now trying to calculate the time needed to go from the origin to a point $(x,y)$.

From previous analysis I found that the time is equal to

$T=\sqrt{\frac{1}{2g}}\int_0^x\sqrt{\frac{1+(y')^2}{y}}dx$

I'm struggling to solve this. This is what I found so far

$\frac{dx}{d\theta}=A(1-\cos\theta)$

$\frac{dy}{d\theta}=A\sin\theta$

Applying the chain rule: $\frac{dy}{dx}=\frac{dy}{d\theta}\frac{d\theta}{dx}=\frac{A\sin\theta}{A(1-\cos\theta)}=\frac{\sin\theta}{1-\cos\theta}$

Therefore, $(\frac{dy}{dx})^2=(\frac{\sin\theta}{1-\cos\theta})^2$

Therefore,

$T=\sqrt{\frac{1}{g}}\int_0^x\sqrt{\frac{1-\cos^2\theta}{(1-\cos\theta)^2}}d\theta$

Is this right? How do I go on from here? I have the coordinates of two points (and therefore I could derive the equation of the brachistochrone curve between them) and I would like to find the time taken to fall from the initial to the final point along the brachistochrone under acceleration g. Thank you!

Answer 1

Changing variables from $x$ to $\theta$ via $x=A(\theta - \sin\theta)$, the integral that represents the time to go from $(x,y)=(0,0)$ to $(x_0,y_0)$ becomes $$ T=\sqrt{\frac 1{2g}}\int_{\theta=0}^{\theta_0}\sqrt{\frac{1+(y')^2}y}\frac{dx}{d\theta}\,d\theta $$ where $\theta_0$ is the value of $\theta$ that corresponds to $x_0$. You've calculated $y'=\sin\theta/(1-\cos\theta)$ and $dx/d\theta=A(1-\cos\theta)$ correctly, now plug everything in. The integral will simplify to $$ T=\sqrt{\frac A{2g}}\int_{\theta=0}^{\theta_0}\sqrt{\frac{(1-\cos\theta)^2+\sin^2\theta}{1-\cos\theta}}\,d\theta=\sqrt{\frac Ag}\int_{\theta=0}^{\theta_0}\sqrt{\frac{1-\cos\theta}{1-\cos\theta}}\,d\theta=\sqrt{\frac Ag}\theta_0. $$

Answer 2

From $ x = A(\theta - \sin\theta) $ and $ y = A(1 - \cos\theta) $ we have: $$ v = \frac{ds}{d\theta} = \sqrt{\left(\frac{dx}{d\theta}\right)^2 + \left(\frac{dy}{d\theta}\right)^2} = A\sqrt{(1 - \cos\theta)^2 + (\sin\theta)^2} = A\sqrt{(2(1 - \cos\theta)} = \sqrt{2Ay}$$

By the law of conservation of energy we get: $$ \frac{ds}{dt} = v = \sqrt{2gy} . $$ Therefore: $$ \frac{d\theta}{dt} = \frac{\frac{ds}{dt}}{\frac{ds}{d\theta}} = \frac{\sqrt{2gy}} {\sqrt{2Ay}} = \sqrt{\frac{g}{A}} \quad \text{and} \quad T = \int_0^\theta\sqrt{\frac{A}{g}} \mathrm{d}t = \sqrt{\frac{A}{g}}·\theta $$