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Topic: MEDIUM: CYCLOID (Read 3059 times) |
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S. Owen
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MEDIUM: CYCLOID
« on: Jul 30th, 2002, 8:47am » |
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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?
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bartleby
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The answer is: 8.
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bartleby
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Helpful, eh?
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bartleby
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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.
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william wu
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Re: MEDIUM: CYCLOID
« Reply #5 on: Aug 3rd, 2002, 6:43pm » |
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on Jul 30th, 2002, 12:24pm, 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.
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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.
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