1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    TD

    User Avatar
    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

    TD

    User Avatar
    Homework Helper

    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)}}
    {{2n}}[/tex]

    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]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Arclength TI-89
  1. Ti 83->89 (Replies: 5)

  2. Graphing with ti 89 (Replies: 5)

Loading...