Differential Model

**Naples** · March 30th 2011, 08:59 PM

A classical problem in the calculus of variations is to find the shape of a curve such that a bead, under the influence of gravity, will slide from point A(0,0) to point B (x1,y1) in the least time. It can be shown that a nonlinear differential for the shape y(x) of the path is $y[1 + (dy/dx)^2] = k$ , where k is a constant. First solve for dx in terms of y and dy and then use the substitution $y = ksin^2(theta)$ to obtain a parametric form of the solution. The curve turns out to be a cycloid.

Not entirely sure what to do with this problem, mainly b/c I don't really understand what the problem is asking, which includes what a cycloid is...or how you find the parametric solution for one.

Not sure if this is right, but it's what I've done so far:
$(dy/dx)^2 = k/y - 1$
$dy/dx = (k/y - 1)^1^/^2$
$dx = dy/[(k/y - 1)^1^/^2]$

Thanks in advance.

**Ackbeet** · March 31st 2011, 01:20 AM

Looking good so far. I'm assuming that the problem defines y positive downward, so that you want the positive square root for the derivative. If y is positive up, then you'd need to take the negative square root, since the bead obviously has to be going down under the influence of gravity. So you have

$dx=\dfrac{dy}{\sqrt{k/y-1}}=dy\sqrt{\dfrac{y}{k-y}}.$

The suggested substitution is $y=k\sin^{2}(\theta),$ with

$dy=2k\sin(\theta)\cos(\theta)\,d\theta.$ We obtain the new DE

$dx=2k\sin(\theta)\cos(\theta)\,d\theta\sqrt{\dfrac {k\sin^{2}(\theta)}{k-k\sin^{2}(\theta)}}.$

Can you continue from here?

Here is a link to information about cycloids.

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