I have put in the formula for arc length and am fairly sure I have the derivatives figured correctly. The problem is, I'm unable to put the equation into a form that is integratable.

http://img267.imageshack.us/img267/4885/10350nc3.jpg

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- May 22nd 2008, 09:52 PMCrazyLondArc length from parametric
I have put in the formula for arc length and am fairly sure I have the derivatives figured correctly. The problem is, I'm unable to put the equation into a form that is integratable.

http://img267.imageshack.us/img267/4885/10350nc3.jpg - May 22nd 2008, 10:04 PMo_O
You forgot to square dx/dt and dy/dt ;)

- May 22nd 2008, 11:18 PMCrazyLond
Thanks for pointing that out. Anyway, I seem to have made another mess of it. I've been doing this calculus for hours now and can't seem to see straight anymore so I'm going to take a break. This is what I have so far but it is not pretty. Maybe trig substitution isn't the way to go but I can't see how else to do it.

http://img389.imageshack.us/img389/2200/10350aw9.jpg - May 22nd 2008, 11:29 PMCrazyLond
It looks like I may have gotten a sign wrong. Now I'm left with something that looks quite simple, I might be on the right track.

http://img144.imageshack.us/img144/4656/10350ev0.jpg - Jun 20th 2008, 09:03 AMMathstud28
$\displaystyle dx=\frac{1}{\sqrt{1-t^2}}$

$\displaystyle dy=\frac{-t}{1-t^2}$

Arc length then is

$\displaystyle \int_0^{\frac{1}{2}}\sqrt{\frac{1}{1-t^2}+\frac{t^2}{(1-t^2)^2}}dt$

Seeing that our quantity inside our squareroot is equal to

$\displaystyle \frac{(1-t^2)}{(1-t^2)^2}+\frac{t^2}{(1-t^2)^2}=\frac{1}{(1-t^2)^2}$

So we have

$\displaystyle \int\frac{dt}{1-t^2}$

if you look all the values within our limits of integration give a positive result, so we can forgoe our absolute value. So we have

$\displaystyle \int_1^{\frac{1}{2}}\frac{dt}{1-t^2}={\rm{arctanh}}(t)\bigg|_{0}^{\frac{1}{2}}=\fr ac{\ln(3)}{2}$