# Fastest curve from A to B under gravity!

1. Jun 6, 2009

### Unit

Here is a situation I made up that I'm working on. I would greatly appreciate any corrections to my mistakes in logic, or suggestions on where to go from here.

I know that the fastest way from point A to B solely under the influence of gravity is a cycloid curve (http://en.wikipedia.org/wiki/Brachistochrone" [Broken]), and I know it's called the brachistochrone problem, but I'm interested in finding the equations of motion for many different kinds of functions which define a curved path from A to B.

Imagine a rectangular region bounded by (0, 0), up to (0, h) and out across to (l, h) and down to (l, 0). Let's take any continuous function that is differentiable over $$0 \leq x \leq l$$, and that starts at (0, h) and goes all the way to (l, 0), but not going out beyond the edges of the rectangular region. If we put a particle of whatever mass at (0, h), we want gravity to pull it down.

Here is a picture I drew:
http://img192.imageshack.us/img192/6776/fastestcurveproblem.png [Broken]

Gravity is always straight down (dark blue line); it can be broken into two components (in light blue). Its acceleration down a ramp is

$$\frac{\mathrm{d^2} x}{\mathrm{d}t^2} = g \sin{\theta}$$.

The angle varies with the slope of the line tangent to the curved function (dark red). The two components of the tangent line (light red) are the rise and the run.

If the function defining the curve is $$y$$, then

$$\tan{\theta} = \texttt{slope of y}$$

$$\tan{\theta} = \frac{\mathrm{d}y}{\mathrm{d}x}$$,

at any given x position.

$$\theta = \arctan{\frac{\mathrm{d}y}{\mathrm{d}x}}$$

So, now that there is an expression for the angle, go back to our acceleration:

$$\frac{\mathrm{d^2} x}{\mathrm{d}t^2} = g \sin{(\arctan{(\frac{\mathrm{d}y}{\mathrm{d}x})})}$$.

There is this cool identity:
$$\arctan{x} = \arcsin{\frac{{x}}{\sqrt{1 + {x}^2}}$$

So the expression for acceleration becomes:

$$\frac{\mathrm{d^2} x}{\mathrm{d}t^2} = - g \frac {(\frac{\mathrm{d}y}{\mathrm{d}x}) } { \sqrt{1 + (\frac{\mathrm{d}y}{\mathrm{d}x})^2 } }$$

It is interesting how the denominator is the integrand of the arc length formula. Does anyone know why?

The g is negative because the tangent of the angle is taken backwards (angle opening up to the left, sorry for bad terminology), because the general slope of the curved path is negative, so that angle will be 2nd-quadrant, so yeah, I made it negative and it makes sense. Also, because the curve of the ramp is down, the derivative will give you a negative slope, but really, it's a positive acceleration, in my case.

Just so that I knew I wasn't going crazy, I used a parabola to check things out.

$$y = \frac{h}{l^2} (x-l)^2$$

$$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2h}{l^2}(x - l)$$

So putting it into the formula and simplifying (somewhat):

$$\frac{\mathrm{d^2} x}{\mathrm{d}t^2} = \frac{-2gh(x - l)}{ \sqrt{(l^2)^2 + (2h(x-l))^2} }$$

This turns out to be a mess but it looks pretty accurate, and I've tried it with numbers (h = 3 and l = 5). Anyways, what do I do from here? I want to solve the differential equation for acceleration, but I don't really know how to do differential equations whatsoever so please help

My goal is to then find a general expression for the time taken for the particle to travel down the curved ramp. That could probably be done by finding a function for x in terms of t. And, I eventually want to use a general equation to model any function that agrees with the original conditions.

Thanks!

Last edited by a moderator: May 4, 2017