Arclength TI-89

  • #1

Main Question or Discussion Point

[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:

Answers and Replies

  • #2
TD
Homework Helper
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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.
 
  • #3
Thank you, TD... Mm... Why not, and how would you get a numerical approximation?
 
  • #4
TD
Homework Helper
1,022
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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)}}
{{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]
 
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  • #5
146
0
This is an interesting problem. I found that n is approx. 1000/16201, but I havent found an elegant solution for n.
 
  • #6
146
0
I refined my solution to:

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

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