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Arclength TI-89

  1. Jan 6, 2006 #1
    [tex]\int^{6}_{0} \sqrt{1-n^2x^2}dx=\pi+e[/tex]

    I need to solve this for n. I believe there should only be one possible function of the form [tex]y=x^n[/tex] that gives an arclength of [tex]\pi+e[/tex] over the interval x=0 to x=6, and wish to find the value of n that such a function must have.
    Does anyone know how to do this? I haven't the slightest idea, as I only know as much calculus as I've managed to teach myself over the past few months... Thank you!

    Ah, additionally, I'm assuming (as I, regrettably, read somewhere) that

    [tex]\int^{b}_{a} \sqrt{1-[f'(x)]^2}dx[/tex]

    is equal to arclength (actually, I didn't just accept it completely--I lack the mathematics to evaluate whether or not it actually is such a formula, but my TI-89 is capable of calculating for whatever values I plug in so... They have thus far matched up perfectly with the values produced by the method I came up with myself:)

    [tex]\lim_{x \rightarrow 0}\sum^{\frac{m}{x}-1}_{n=0}\sqrt{x^2+(f(x(n+1))-f(nx))^2}[/tex]

    Anyway, again, thank you.
    Last edited: Jan 6, 2006
  2. jcsd
  3. Jan 6, 2006 #2


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    Homework Helper

    Well the integral is well-known, you'll get an arcsin and square root part so evaluating the integral isn't really a problem. I doubt though that you'll be able to solve for n analytically, after that - unless a numerical approximation would satisfy you.
  4. Jan 6, 2006 #3
    Thank you, TD... Mm... Why not, and how would you get a numerical approximation?
  5. Jan 6, 2006 #4


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    The easy way, of course, would be relying on a computer program. I tried it with Mathematica.

    Integration yields

    [tex]\int\limits_0^6 {\sqrt {1 - n^2 x^2 } dx} = 3\sqrt {1 - 36n^2 } + \frac{{\arcsin \left( {6n} \right)}}

    So what you want to solve for n is

    [tex]3\sqrt {1 - 36n^2 } + \frac{{\arcsin \left( {6n} \right)}}
    {{2n}} = e + \pi[/tex]
    Last edited: Jan 6, 2006
  6. Sep 25, 2007 #5
    This is an interesting problem. I found that n is approx. 1000/16201, but I havent found an elegant solution for n.
  7. Sep 25, 2007 #6
    I refined my solution to:

    [tex]n \approx {10000000 \over 162011025}[/tex]
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