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Weird arc length question

  1. Mar 17, 2005 #1
    I want to find two functions f(n,x) and g(n,x) such that f(n,x)sin(g(n,x)) always has a constant arc length over some interval [a,b]. Where n increases the amplitude but decreases the period.

    Any suggestions?
    Last edited: Mar 17, 2005
  2. jcsd
  3. Mar 17, 2005 #2


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    Well, you've got the trivial choice [tex]f(n,x)=\frac{1}{\sin(g(n,x)}[/tex]

    In the general case, the arc-length will be a function of "n".
    Can you set up a differential equation for f and g which ensures that the arc-length is just the constant function?
  4. Mar 17, 2005 #3
    I did set up a DE. That’s how I know they have to be f(n,x) not just f(n)

    If they were…
    [tex]f(x) = g(n)sin(x*p(n)) [/tex]
    [tex]f’(x) = g(n)p(n)cos(x*p(n)) [/tex]

    So arch length is:

    [tex]\int_{0}^{1} \sqrt{{1+ (g(n)p(n)cos(p(n)) )^2} dx =c[/tex]

    1+ (g(n)p(n)cos(p(n)))^2 – (1+ (g(n)p(n))^2) = 0

    (g(n)p(n)cos(p(n)))^2 = (g(n)p(n))^2)

    cos(p(n)) = 1

    p(n) = 0

    So I know if they were just functions of just n it would be trivial. For functions of x and n (possibly I will need to put a and b in there also) I know how to set up the de, no clue how to solve it.
  5. Mar 17, 2005 #4
    Have you already tried seperation of variables to transform your partial DE into an ordinary DE?

    What I mean is assume g(x,n) = X(x)*N(n).
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