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Topic: arc length sequence (Read 408 times) |
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jarls
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arc length sequence
« on: Jun 12th, 2009, 1:18pm » |
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I've come up with a problem I am having trouble solving.
There exists some function 'f'. a series of points is placed on the function such that the arc length between any two points is some constant. What sequence at which the function 'f' is evaluated will be a solution for the function value of the nth point. That is, if a series of points is placed placed on the x-axis such that every point on the function will have a corresponding point on the x-axis, just beneath it, what sequence (function of 'n') will designate the distance from the origin of the nth point on the x-axis?
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towr
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Re: arc length sequence
« Reply #1 on: Jun 12th, 2009, 2:34pm » |
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I think you want the inverse of
 x_nx_0  (1+f '(x)2) dx
(which should give the arclength from point 0 to point n on the curve, given the x-value of the 0th and nth points)
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| « Last Edit: Jun 12th, 2009, 2:41pm by towr » |
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jarls
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Re: arc length sequence
« Reply #2 on: Jun 12th, 2009, 3:13pm » |
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on Jun 12th, 2009, 2:34pm, towr wrote:I think you want the inverse of
x_nx_0 (1+f '(x)2) dx
(which should give the arclength from point 0 to point n on the curve, given the x-value of the 0th and nth points)
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This is what I'm having trouble with. Let's say that the arc length function is integral(g(x)).
If i had the correct sequence, s(n), and plugged it in for x, the value of integral(g(s(n))) should be the constant arc length interval, 'c', multiplied by 'n'.
integral(g(s(n)))=cn
differentiate both sides,
g(s(n))=c
so inverse of g(x) is needed.
If f(x) is x^2 then, sqrt(1+[f'(x)]^2) --> sqrt(1+(2x)^2) and this becomes x when 1) the whole thing is squared, 2) 1 is subtracted, 3) the square root is taken and 4) the result is divided by two.
Therfore the inverse (g-inverse(x)) is sqrt(x^2-1)/2.
At this point we have g(s(n))=c
It should then be so that
g-inverse(g(s(n)))=g-inverse(c)=s(n)
This can't be though since before this is done neither g-inverse nor the right-hand side has anything to do with 'n'.
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| « Last Edit: Jun 12th, 2009, 3:14pm by jarls » |
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Obob
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Re: arc length sequence
« Reply #3 on: Jun 12th, 2009, 4:39pm » |
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This is basically just elaborating on what towr said. I'm not 100% clear on the question, but I think this is what you want.
Let g(x) = integral from x_0 to x of (sqrt(1+f'(t)^2) dt, where (x_0, f(x_0)) is the first point that has been placed. So g(x_0) = 0, while if (x_n, f(x_n)) is the (n+1)st point then g(x_n) = nc, where c is the constant arc length. Let h be the inverse of g. Then h(0) = x_0, h(c) = x_1, h(2c) = x_2, etc., and h(nc) = x_n. So the sequence s(n) that you want is s(n) = h(nc).
Of course, except in some very simple or contrived cases, the function h will not be expressible in closed form.
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| « Last Edit: Jun 12th, 2009, 4:41pm by Obob » |
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towr
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Re: arc length sequence
« Reply #4 on: Jun 13th, 2009, 5:25am » |
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on Jun 12th, 2009, 3:13pm, jarls wrote:This is what I'm having trouble with. Let's say that the arc length function is integral(g(x)).
If i had the correct sequence, s(n), and plugged it in for x, the value of integral(g(s(n))) should be the constant arc length interval, 'c', multiplied by 'n'.
integral(g(s(n)))=cn |
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That's not right, though. You have to plug in s(n), after taking the integral; not before.
For f(x) = x2, if you filled in x=s(n) in the integral you'd get
(2*s(n)*sqrt(1 + 4*s(n)^2) + arcsinh(2*s(n)))/4 = c*n
Differentiating both sides to n, would give
[ (2/sqrt(1 + 4*s(n)^2) + (8*s(n)^2)/sqrt(1 + 4*s(n)^2) + 2*sqrt(1 + 4*s(n)^2))/4 ] * s'(n) = c
Which isn't at all helpful; especially not bringing in that s'(n) (due to the chain rule).
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jarls
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Re: arc length sequence
« Reply #5 on: Jun 13th, 2009, 8:43am » |
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on Jun 13th, 2009, 5:25am, towr wrote:
That's not right, though. You have to plug in s(n), after taking the integral; not before.
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yes that makes more sense.
on Jun 13th, 2009, 5:25am, towr wrote:
For f(x) = x2, if you filled in x=s(n) in the integral you'd get
(2*s(n)*sqrt(1 + 4*s(n)^2) + arcsinh(2*s(n)))/4 = c*n
Differentiating both sides to n, would give
[ (2/sqrt(1 + 4*s(n)^2) + (8*s(n)^2)/sqrt(1 + 4*s(n)^2) + 2*sqrt(1 + 4*s(n)^2))/4 ] * s'(n) = c
Which isn't at all helpful; especially not bringing in that s'(n) (due to the chain rule).
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So, you've approached it correctly and still run into the same trouble.
Do you have any idea how to better approach the problem?
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Obob
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Re: arc length sequence
« Reply #6 on: Jun 13th, 2009, 9:16am » |
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He hasn't "run into the same trouble." The point is that going about it the way you were won't help you find the solution. There isn't a better solution to the problem than he and I discussed earlier; you won't find a way of expressing the solution in any way other than the inverse function of an integral. Given a particular f, you might be able to evaluate the integral and take the inverse function. But for almost all f, this will be impossible.
Wildly differentiating things for no reason usually isn't such a hot idea.
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towr
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Re: arc length sequence
« Reply #7 on: Jun 13th, 2009, 2:57pm » |
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Let's take an example with a function f that does work out relatively nicely.
Take f(x) = 2/3 (x-1)3/2, then f'(x) = (x-1) and so we have (1 + f'(x)2) = x. Therefor our integral (if we start with point 0 at x=1) becomes, s(n)1 x dx = 2/3 s(n)3/2 - 2/3 and this must equal c*n
And so we end up with: s(n) = (3/2*c*n+1)2/3
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