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        riddles >> medium >> MEDIUM: CYCLOID
        
(Message started by: srowen on Jul 30^th, 2002, 8:47am)

Title: MEDIUM: CYCLOID
Post by srowen on Jul 30^th, 2002, 8:47am

Is the path it traces an ellipse? I've forgotten too much... but not that it would mean the path's length is 1/2(pi * 2 * pi) = pi^2.

Whether that's right or not I suspect there is a sneakier way to the answer... any thoughts?

Title: Re: MEDIUM: CYCLOID
Post by bartleby on Jul 30^th, 2002, 12:12pm

The answer is: 8.

Title: Re: MEDIUM: CYCLOID
Post by bartleby on Jul 30^th, 2002, 12:13pm

Helpful, eh?

Title: Re: MEDIUM: CYCLOID
Post by bartleby on Jul 30^th, 2002, 12:24pm

This is a fairly stodgy old trigonometry problem, I don't think it deserves to be included in this otherwise clever list of "Ah-HA!" style puzzles.

Here's the math:

If the cycloid of radius a has a cusp at the origin, its equation in Cartesian coordinates is

x = a cos^-1 ( (a-y) / 1 ) +- sqrt( 2ay - y² )

In parametric form,

x = a(t - sin t)
y = a(1 - cos t)

Taking the derivatives:

x' = a(1 - cos t)
y' = a(sin t)

dy/dx = y'/x' = (a sin t) / a(1-cos t) = sin t / (1 - cos t) =
2 sin (t/2) cos (t/2) / (2 sin²(t/2) ) = cot (t/2)

The squares of the derivatives are:

x'² = a²(1 - 2 cos t + cos² t)
y'² = a² sin² t

So the arc length of the cycle is:

L = Sds = S0->2pi sqrt( x'² + y'² ) dt

...which works out to

8

QED.

Title: Re: MEDIUM: CYCLOID
Post by bartleby on Jul 30^th, 2002, 12:26pm

No I'm not walking around with all that in my head!

Here is the web page I copied from:

http://mathworld.wolfram.com/Cycloid.html

Title: Re: MEDIUM: CYCLOID
Post by william wu on Aug 3^rd, 2002, 6:43pm

on 07/30/02 at 12:24:27, bartleby wrote:

This is a fairly stodgy old trigonometry problem, I don't think it deserves to be included in this otherwise clever list of "Ah-HA!" style puzzles.

Yea, I agree. I don't know why I put it up. If I ever get around to databasing all the riddles, I'll delete it.