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, 10:52 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, 11:04 PMo_O
You forgot to square dx/dt and dy/dt ;)

- May 23rd 2008, 12:18 AMCrazyLond
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 23rd 2008, 12:29 AMCrazyLond
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 - June 20th 2008, 10:03 AMMathstud28