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        riddles >> medium >> Parabolic Perimeter
        
(Message started by: ThudanBlunder on Jul 15^th, 2008, 5:31pm)

Title: Parabolic Perimeter
Post by ThudanBlunder on Jul 15^th, 2008, 5:31pm

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.

Title: Re: Parabolic Perimeter
Post by pex on Jul 16^th, 2008, 9:18am

Hm. I seem to get an x coordinate
[hide]sqrt(7)/3 * cos( arccos( sqrt(7)/14 ) / 3) - 1/6[/hide],
which doesn't seem to simplify. Not exactly what I expected...

Title: Re: Parabolic Perimeter
Post by SMQ on Jul 16^th, 2008, 11:12am

I get the x coordinate as [hide]the positive root of 8x³ + 4x² - 4x - 1[/hide]. Is [hide]{-1 + http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sup3.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif[(⁷/₂)(1 + http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif-27)] + http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sup3.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif[(⁷/₂)(1 - http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif-27)]} / 6[/hide] "exact" enough for an answer?

General approach:

[hide]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

_{http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif}_{-(2x^2 + 1)/(2x)}^x http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif(4t² + 1) dt

= _{http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif}₀^x http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif(t² + 1) dt + _{http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif}₀^{(2x^2 + 1)/(2x)} http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif(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)http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif(4x² + 1) / (2x³) + http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif(4x² + 1) + (2x² - 1)http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif(x² + 1)http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif(4x² + 1) / (2x³)
= {2x³ + (2x² - 1)[1 + http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif(x² + 1)]}http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif(4x² + 1) / 2x³.

Setting this equal to zero gives:

2x³ + (2x² - 1)[1 + http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif(x² + 1)] = 0 (since http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif(4x² + 1) can not be 0 for positive x)

http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gif http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif(x² + 1) = 2x³/(1 - 2x²) - 1

http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gif x² + 1 = 4x⁶ / (1 - 2x²)² - 4x³ / (1 - 2x²) + 1

http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gif (1 - 2x²)² = 4x⁴ - 4x(1 - 2x²)

http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gif 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.
[/hide]
--SMQ

Title: Re: Parabolic Perimeter
Post by Eigenray on Jul 16^th, 2008, 11:14am

It does simplify: [hide]Let f(x) be the minimal polynomial of that number. Now work out f( (t+1/t)/2 ).[/hide] :o (Confession: I only found this using [hide]Plouffe[/hide].)

So, is there a reason for this, or is it just a coincidence (as in, [hide]we just end up with some random cubic[/hide])?

Title: Re: Parabolic Perimeter
Post by pex on Jul 16^th, 2008, 11:38am

SMQ's and my answer are the same, and, using Eigenray's hint, equal to [hide] cos( 2 pi / 7 ) [/hide].

To be quite honest, I don't see why this would not be "just a coincidence"...

Title: Re: Parabolic Perimeter
Post by Immanuel_Bonfils on Jul 16^th, 2008, 11:43am

Did I see an http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif-27 somewhere?

Title: Re: Parabolic Perimeter
Post by SMQ on Jul 16^th, 2008, 11:52am

on 07/16/08 at 11:43:35, Immanuel_Bonfils wrote:

Did I see an http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif-27 somewhere?

Yep, you did indeed. That's what happens when you try to express the solution (http://mathworld.wolfram.com/TrigonometryAnglesPi7.html) to trig equations without using trig (http://mathworld.wolfram.com/CubicFormula.html). ;)

--SMQ

Title: Re: Parabolic Perimeter
Post by ThudanBlunder on Jul 17^th, 2008, 11:16am

Sorry for posting a boring coordinate geometry puzzle. I didn't have anything better at the time.

on 07/16/08 at 11:12:16, SMQ wrote:

Nice work, as usual. But I should have stipulated no radicals.

Title: Re: Parabolic Perimeter
Post by Immanuel_Bonfils on Jul 18^th, 2008, 12:09pm

Eigenray or/and pex. Could you, please, explain (please, more explicit) how to get to so a compact answer?

Title: Re: Parabolic Perimeter
Post by Eigenray on Jul 18^th, 2008, 12:54pm

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 = e^{2pi 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?

Title: Re: Parabolic Perimeter
Post by Immanuel_Bonfils on Jul 18^th, 2008, 2:12pm

Nice. Thanks!!!

Title: Re: Parabolic Perimeter
Post by ThudanBlunder on Jul 18^th, 2008, 3:43pm

on 07/16/08 at 11:43:35, Immanuel_Bonfils wrote:

Did I see an http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif-27 somewhere?

It is casus irreducibilis (http://en.wikipedia.org/wiki/Casus_irreducibilis).

Title: Re: Parabolic Perimeter
Post by SMQ on Jul 18^th, 2008, 4:24pm

on 07/18/08 at 12:54:10, Eigenray wrote:

pex, I'm curious, how did you get your answer?

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

Title: Re: Parabolic Perimeter
Post by Immanuel_Bonfils on Jul 19^th, 2008, 6:15pm

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.

Title: Re: Parabolic Perimeter
Post by Immanuel_Bonfils on Jul 20^th, 2008, 9:33am

Eigenray, how did you browse Plouffe, to get such a hint?

Title: Re: Parabolic Perimeter
Post by Eigenray on Jul 20^th, 2008, 2:27pm

Like [link=http://bootes.math.uqam.ca/cgi-bin/ipcgi/lookup.pl?Submit=GO+&lookup_type=simple&number=0.62348980]this[/link]. If you bookmark
[link=http://bootes.math.uqam.ca/cgi-bin/ipcgi/lookup.pl?Submit=GO+&lookup_type=simple&number=%s]http://bootes.math.uqam.ca/cgi-bin/ipcgi/lookup.pl?Submit=GO+&lookup_type=simple&number=%s[/link] in Firefox with keyword 'plouffe' you just need to type 'plouffe 0.62348980'.

Title: Re: Parabolic Perimeter
Post by pex on Jul 21^st, 2008, 3:36am

on 07/18/08 at 12:54:10, Eigenray wrote:

pex, I'm curious, how did you get your answer?

I made pretty much the same derivations SMQ did, and tried to work out the cube roots by hand.