My Math Forum Need help on an arc length of curve problem

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 September 7th, 2008, 01:03 PM #1 Newbie   Joined: Sep 2008 Posts: 4 Thanks: 0 Need help on an arc length of curve problem The problem says to find the length of the arc of the parabola $y^2= x$ from (0,0) to (1,1). When I work it I end up with crazy (for my math level) integrals that don't seem right for the problem. When working it from $y= \sqrt{x}$ I get $\int_0^1(\sqrt{1 + \frac{1}{4} x^{-1} } )dx$ That makes me do integration by parts, and this problem is from before that section. Working the problem from $y^2= x$ I get $\int_0^1(\sqrt{1 + 4y^2 }) dy$ and that doesn't seem right either. Any help on how to work this problem is greatly appreciated.
 September 7th, 2008, 01:54 PM #2 Guest   Joined: Posts: n/a Thanks: Re: Need help on an arc length of curve problem I think I found a relatively easy way to integrate this. We have $\int\sqrt{1+\frac{1}{4x}}dx$ Now, let $u=4x, \;\ \frac{du}{4}=dx$ Change the limits of integration to 0 to 4. $\frac{1}{4}\int_{0}^{4}\sqrt{\frac{u+1}{u}}du$ Now, let $u=tan^{2}(t), \;\ du=2sec^{2}(t)tan(t)dt$ This gives: $\frac{1}{2}\int_{0}^{tan^{-1}(2)}\sqrt{\frac{sec^{2}(t)}{tan^{2}(t)}}sec^{2}( t)tan(t)dt$ Believe it or not, this whittles down to: $\frac{1}{2}\int_{0}^{tan^{-1}(2)}sec^{3}(t)dt=\frac{1}{4}sec(t)tan(t)+\frac{1 }{4}ln|sec(t)+tan(t)|$ Now, use the limits of integration.
 September 7th, 2008, 02:03 PM #3 Senior Member   Joined: May 2007 Posts: 402 Thanks: 0 Re: Need help on an arc length of curve problem The second one is easy. Just replace $y=\frac{1}{2}\sinh t$. Then you have: $\frac{1}{2}\int ^1 _0 \sqrt{1+\sinh^2t}\cdot \cosh t \,dt= \frac{1}{2}\int ^1 _0 \cosh^2 t \,dt = \frac{1}{8}\int ^1 _0 \left(e^t+e^{-t}\right)^2 \,dt= \frac{1}{8}\int ^1 _0 \left(e^{2t}+e^{-2t}+2\right) \,dt$ You can do this on your own from now...
 September 7th, 2008, 02:04 PM #4 Newbie   Joined: Sep 2008 Posts: 4 Thanks: 0 Re: Need help on an arc length of curve problem Holy cow! That is crazy lol, seems a little above my level. Maybe my problem is getting the actual integral? This is a problem for me to review prior to the test, but we haven't learned most of the stuff you just did. We just started integration by parts, and we haven't gotten into trigonometric integral solutions. I think I may have just came up with the wrong integrals for the problem.
 September 7th, 2008, 02:19 PM #5 Guest   Joined: Posts: n/a Thanks: Re: Need help on an arc length of curve problem No, that's the right integral. Arc length for $y=\sqrt{x}$ from 0 to 1 would be $\int_{0}^{1}\sqrt{1+\frac{1}{4x}}dx$ The general formula is $\int_{a}^{b}\sqrt{1+[\frac{d}{dx}]^{2}}dx$ and that is what you have.
 September 7th, 2008, 02:50 PM #6 Newbie   Joined: Sep 2008 Posts: 4 Thanks: 0 Re: Need help on an arc length of curve problem Oh man, this problem is just nasty then lol, thank you very much for the help
 September 7th, 2008, 04:28 PM #7 Senior Member   Joined: Jul 2008 Posts: 895 Thanks: 0 Re: Need help on an arc length of curve problem Hopefully not intruding on this fine work, but my first thought would be that it is the same length as that for y = x^2. Same trig mess though.
 September 11th, 2008, 09:55 AM #8 Newbie   Joined: Sep 2008 Posts: 4 Thanks: 0 Re: Need help on an arc length of curve problem We went over this "hw" problem in class... The teacher said for us, not having learned trig substitution yet, that it is impossible to solve. Why would a teacher assign a hw problem that is impossible to solve, so we learn how it feels to be demoralized? Thanks for the help though.
September 11th, 2008, 11:43 AM   #9
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Re: Need help on an arc length of curve problem

Quote:
 Originally Posted by Prometheos Why would a teacher assign a hw problem that is impossible to solve, so we learn how it feels to be demoralized?
Good Lord, no! You are surely not the only student in that class, and your teacher is surely not a mind-reader. He/she will begin to recognise capabilities as the year proceeds, but there are always those who rise to a challenge. He/she has to care for all needs, not just yours. If not presented with such challenges now an then, some complain [because they are "bright", of course] that they are not challenged enough. Well, you can please some of the people some of the time .... . The motive might be for you to do some research into future studies. I see absolutely nothing wrong with that since most students seriously studying calculus do so because they are fairly good at it at least, and are ambitious.

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