Arclength TI-89

1. Jan 6, 2006

Trepidation

$$\int^{6}_{0} \sqrt{1-n^2x^2}dx=\pi+e$$

I need to solve this for n. I believe there should only be one possible function of the form $$y=x^n$$ that gives an arclength of $$\pi+e$$ 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!

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

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:)

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

Anyway, again, thank you.

Last edited: Jan 6, 2006
2. Jan 6, 2006

TD

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. Jan 6, 2006

Trepidation

Thank you, TD... Mm... Why not, and how would you get a numerical approximation?

4. Jan 6, 2006

TD

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

Integration yields

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

So what you want to solve for n is

$$3\sqrt {1 - 36n^2 } + \frac{{\arcsin \left( {6n} \right)}} {{2n}} = e + \pi$$

Last edited: Jan 6, 2006
5. Sep 25, 2007

camilus

This is an interesting problem. I found that n is approx. 1000/16201, but I havent found an elegant solution for n.

6. Sep 25, 2007

camilus

I refined my solution to:

$$n \approx {10000000 \over 162011025}$$