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Topic: Parabolic Perimeter (Read 2862 times) |
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ThudnBlunder
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Let P be a point on the parabola y = x2 other than the origin. The parabola and the normal at P enclose a bounded region.
Find the exact coordinates of P which minimize the perimeter of the bounded region.
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pex
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Re: Parabolic Perimeter
« Reply #1 on: Jul 16th, 2008, 9:18am » |
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Hm. I seem to get an x coordinate
sqrt(7)/3 * cos( arccos( sqrt(7)/14 ) / 3) - 1/6,
which doesn't seem to simplify. Not exactly what I expected...
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SMQ
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Re: Parabolic Perimeter
« Reply #2 on: Jul 16th, 2008, 11:12am » |
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I get the x coordinate as the positive root of 8x3 + 4x2 - 4x - 1. Is {-1 +   [(7/2)(1 + -27)] + [(7/2)(1 - -27)]} / 6 "exact" enough for an answer?
General approach:
Given a positive x-coordinate x, the x-coordinate of the other corner is -x - 1/(2x). The length of the line segment is thus
(4x2 + 1)3/2 / (4x2),
the length of the parabola segment is
 -(2x^2 + 1)/(2x)x (4t2 + 1) dt
= 0x (t2 + 1) dt + 0(2x^2 + 1)/(2x) (t2 + 1) dt,
and obviously the length of the entire perimeter P is just the sum of the two. To find the minimum value we take the derivative with respect to x. The fundamental theorem of calculus lets us take the derivative without ever evaluating the integrals, so we find
dP/dx = (2x2 - 1)(4x2 + 1) / (2x3) + (4x2 + 1) + (2x2 - 1)(x2 + 1)(4x2 + 1) / (2x3)
= {2x3 + (2x2 - 1)[1 + (x2 + 1)]}(4x2 + 1) / 2x3.
Setting this equal to zero gives:
2x3 + (2x2 - 1)[1 + (x2 + 1)] = 0 (since (4x2 + 1) can not be 0 for positive x)
 (x2 + 1) = 2x3/(1 - 2x2) - 1
x2 + 1 = 4x6 / (1 - 2x2)2 - 4x3 / (1 - 2x2) + 1
(1 - 2x2)2 = 4x4 - 4x(1 - 2x2)
8x3 + 4x2 - 4x - 1 = 0
and a quick graph of the cubic shows only one positive root, which must then be the answer we're looking for.
--SMQ
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| « Last Edit: Jul 16th, 2008, 11:46am by SMQ » |
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Eigenray
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Re: Parabolic Perimeter
« Reply #3 on: Jul 16th, 2008, 11:14am » |
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It does simplify: Let f(x) be the minimal polynomial of that number. Now work out f( (t+1/t)/2 ).  (Confession: I only found this using Plouffe.)
So, is there a reason for this, or is it just a coincidence (as in, we just end up with some random cubic)?
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pex
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Re: Parabolic Perimeter
« Reply #4 on: Jul 16th, 2008, 11:38am » |
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SMQ's and my answer are the same, and, using Eigenray's hint, equal to cos( 2 pi / 7 ) .
To be quite honest, I don't see why this would not be "just a coincidence"...
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Immanuel_Bonfils
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Re: Parabolic Perimeter
« Reply #5 on: Jul 16th, 2008, 11:43am » |
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Did I see an -27 somewhere?
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SMQ
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Re: Parabolic Perimeter
« Reply #6 on: Jul 16th, 2008, 11:52am » |
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on Jul 16th, 2008, 11:43am, Immanuel_Bonfils wrote:| Did I see an -27 somewhere? |
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Yep, you did indeed. That's what happens when you try to express the solution to trig equations without using trig. 
--SMQ
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| « Last Edit: Jul 16th, 2008, 11:52am by SMQ » |
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ThudnBlunder
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Re: Parabolic Perimeter
« Reply #7 on: Jul 17th, 2008, 11:16am » |
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Sorry for posting a boring coordinate geometry puzzle. I didn't have anything better at the time.
on Jul 16th, 2008, 11:12am, SMQ wrote:I get the x coordinate as the positive root of 8x3 + 4x2 - 4x - 1. Is {-1 + [(7/2)(1 + -27)] + [(7/2)(1 - -27)]} / 6 "exact" enough for an answer?
--SMQ |
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Nice work, as usual. But I should have stipulated no radicals.
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Immanuel_Bonfils
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Re: Parabolic Perimeter
« Reply #8 on: Jul 18th, 2008, 12:09pm » |
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Eigenray or/and pex. Could you, please, explain (please, more explicit) how to get to so a compact answer?
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| « Last Edit: Jul 18th, 2008, 12:13pm by Immanuel_Bonfils » |
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Eigenray
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Re: Parabolic Perimeter
« Reply #9 on: Jul 18th, 2008, 12:54pm » |
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As SMQ shows, the x-coordinate satisfies 8x3 + 4x2 - 4x - 1 = 0. If we substitute x=(t+1/t)/2, this turns into
(t7-1)/(t-1) = t6 + t5 + t4 + t3 + t2 + t + 1 = 0,
so t = e2pi i k/7 is a primitive 7th root of unity, and x = (t+1/t)/2 = cos(2pi k/7). Since x > 0, this means x = cos(2pi/7). Of course, this derivation only works if we assume that x is the cosine of something nice.
pex, I'm curious, how did you get your answer?
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Immanuel_Bonfils
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Re: Parabolic Perimeter
« Reply #10 on: Jul 18th, 2008, 2:12pm » |
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Nice. Thanks!!!
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ThudnBlunder
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Re: Parabolic Perimeter
« Reply #11 on: Jul 18th, 2008, 3:43pm » |
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on Jul 16th, 2008, 11:43am, Immanuel_Bonfils wrote:| Did I see an -27 somewhere? |
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It is casus irreducibilis.
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SMQ
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Re: Parabolic Perimeter
« Reply #12 on: Jul 18th, 2008, 4:24pm » |
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on Jul 18th, 2008, 12:54pm, Eigenray wrote:| pex, I'm curious, how did you get your answer? |
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I won't speak for pex, of course, but I obtained something very similar by trying to evaluate the real part of the complex cube root using polar forms and trig. I didn't pursue that line to its conclusion so I don't know if I would have arrived at pex's formula or not.
--SMQ
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Immanuel_Bonfils
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Re: Parabolic Perimeter
« Reply #13 on: Jul 19th, 2008, 6:15pm » |
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Thanks people (SMQ and ThudanBlunder). I already knew about the sum of the complex conjugateds given real root in the third degree equation; but that was the point: I was wandering of the others two real roots... then I notice that the equation only holds for positive x.
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Immanuel_Bonfils
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Re: Parabolic Perimeter
« Reply #14 on: Jul 20th, 2008, 9:33am » |
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Eigenray, how did you browse Plouffe, to get such a hint?
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pex
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Re: Parabolic Perimeter
« Reply #16 on: Jul 21st, 2008, 3:36am » |
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on Jul 18th, 2008, 12:54pm, Eigenray wrote:| pex, I'm curious, how did you get your answer? |
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I made pretty much the same derivations SMQ did, and tried to work out the cube roots by hand.
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