Comparing Brachistochrone curve with a Hypocycloid curve

Question

I want to compare the time that it takes to slide a particle in a frictionless hypocycloid curve, so time would be given by the arclength divided by the velocity

So I need first compute the arclength of the hypocycloid curve, but in general the arclength is given by

And by conservation of energy, velocity is given by

Substituting in the integral results

Solving the indefinite integral results in

So now I would just substitute the function y corresponding to the hypocycloid curve

Is my reasoning right?

Then finally to compare times I would just make a graph of the time functions corresponding to the brachistochrone and the hypocycloid

Answer 1

Your reasoning is almost perfect. Just one small, trivial problem that can be easily fixed. You assert that from conservation of energy $v = \sqrt{2gy}$. However, what conservation of energy really states is tha the initial potential energy equals the kinetic energy plus the potential energy at a state. Hence, $\frac{1}{2}mv^2 + mgy = mgy_0$, where $y_0$ is the initial hieght. From this, we obtain that $v = \sqrt{2g(y_0 - y)}$, where $y_0$ is a constant which depends on where you set your object to start rolling. Once $y_0$ is determined, you have your formula for velocity. From there, you solve the integral to obtain a formula similar to the one you have. The best thing to do from here is to either do as you stated and substitute the function for $y$ in terms of $x$, or to write $y'$ in terms of $y$ and substitute. If you the first way, you get a function of $t$ in terms of $x$. If you do the second way, you get a function of $t$ in terms of $y$. Both ways are okay, but the first way is by far the easiest way to go. Which ever way you do, when you are done you could either graph or do a functional analysis.