wu :: forums - Print Page


    
      
        wu :: forums
        (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi)
      

        riddles >> general problem-solving / chatting / whatever >> arc length sequence
        
(Message started by: jarls on Jun 12^th, 2009, 1:18pm)

Title: arc length sequence
Post by jarls on Jun 12^th, 2009, 1:18pm

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?

Title: Re: arc length sequence
Post by towr on Jun 12^th, 2009, 2:34pm

I think you want the inverse of
http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif^x_n_{x_0} http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif(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)

Title: Re: arc length sequence
Post by jarls on Jun 12^th, 2009, 3:13pm

on 06/12/09 at 14:34:09, towr 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

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'.

Title: Re: arc length sequence
Post by Obob on Jun 12^th, 2009, 4:39pm

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.

Title: Re: arc length sequence
Post by towr on Jun 13^th, 2009, 5:25am

on 06/12/09 at 15:13:58, jarls wrote:

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).

Title: Re: arc length sequence
Post by jarls on Jun 13^th, 2009, 8:43am

on 06/13/09 at 05:25:51, towr wrote:

That's not right, though. You have to plug in s(n), after taking the integral; not before.

yes that makes more sense.

on 06/13/09 at 05:25:51, 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).

So, you've approached it correctly and still run into the same trouble.

Do you have any idea how to better approach the problem?

Title: Re: arc length sequence
Post by Obob on Jun 13^th, 2009, 9:16am

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.

Title: Re: arc length sequence
Post by towr on Jun 13^th, 2009, 2:57pm

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) = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif(x-1) and so we have http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif(1 + f'(x)²) = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gifx. Therefor our integral (if we start with point 0 at x=1) becomes, http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gif^s(n)₁ http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gifx 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