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Arc length problem.

  1. Sep 30, 2012 #1
    Hi folks. I'm working on an arc length problem, and my answer doesn't match the answer that our professor gave, so I was hoping that someone could point out where I'm going wrong here.

    1. The problem statement, all variables and given/known data
    Find the arc length determined by x^2=y^3 from (0,0) to (1,1).


    all of the work is in the attached Word doc. I realize that I'm severely limiting the help that I'll receive by using an attachment, but copying all of that formatted work to bbcode (I assume) formatting isn't exactly trivial, so... *shrug* I'd very much appreciate any help or discussion about this, though.

    Edit: I had to .zip the file because .docx isn't a recognized format here. Sorry again for that.
     

    Attached Files:

  2. jcsd
  3. Sep 30, 2012 #2

    SammyS

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    I assume that unzipping that file gives a .docx Word file back, which many of us will not open.

    Can you at least output your document to a pdf file and post that ?
     
  4. Sep 30, 2012 #3
    sure. uh... here you go.
     

    Attached Files:

  5. Sep 30, 2012 #4
    It appears that you've incorrectly done u substitution here. In this step --

    [tex] s = \int_{0}^{1} u^{1/2} du (u = 1 + 4/9 x^{-2/3}) [/tex]

    -- you've implicitly assumed that the derivative of u is present in the integrand before substitution, but it isn't. When [itex] u = 1 + 4/9 x^{-2/3} [/itex], [itex] du = -8/27 x^{-5/3}dx [/itex], but that term isn't present in the unsubstituted integral. What you'll have to do is solve for dx in terms of u and du and substitute that into the integrand. It might help if you simply define [itex] u = x^{-2/3} [/itex] instead of the entire argument of the square root. You'll also probably have to do another round of u substitution and maybe use some trigonometric functions as well. Don't forget to solve for the new boundaries of integration. This is a pretty nasty integral.
     
  6. Sep 30, 2012 #5
    yea, I was kindof suspisious about that u substitution...
    Thanks. I'll start working at it again.
     
  7. Sep 30, 2012 #6
    humm... ok, I'm not sure that u substitution is going to help here at all.
    If [itex] u = x^{-2/3} [/itex], then [itex] u' = {2/3}x^{-5/3} [/itex], so... that doesn't really help anything.
     
  8. Sep 30, 2012 #7
    Just so people don't have to go hunting through this whole thread:
    The problem is how to integrate

    [itex] s = \int_{0}^{1} √(1+{4/9}x^{-2/3}) dx [/itex]
     
  9. Sep 30, 2012 #8
    You have [itex] du = -2/3 x^{-5/3} dx [/itex] and [itex] u = x^{-2/3} [/itex] which means [itex] x = u^{-3/2} [/itex]. Substitute u in for x in the equation for du and then solve for dx in terms of u and du. Then you'll have an integral that you can start to work with.

    If this is homework, I'd suggest going to a TA or your professor and getting help walking through the integral. This will be a rather long integral, and I can't think of any ways to simplify it off the top of my head. It'd be better to have help in person, since leading you through every step on the forums is going to get arduous.
     
  10. Sep 30, 2012 #9
    Yea... my professor is apparently sick, so I was hoping to get some help from my fellow students. Only one has even tried, which brought me here.

    This helps though, so thanks. I need to go eat something, and I'll come back and look at it again.
     
  11. Sep 30, 2012 #10
    ok, let's see... looking at these fractional exponents is making my eyes swim, but I'm working at it.

    I don't see how [itex] x = u^{-3/2} [/itex] helps. I staired at it long enough (a few seconds, you know?) to see what you're getting at there, but I'm looking at the integral and I don't see how it fits back in there. I went back and did a couple of substitution examples too, just to make sure that I have that concept down. I don't know, if there's something there substitution wise (with that u), I'm just not quite seeing it.

    One thing that I'm thinking about right now is simplifying out the square root by converting it to an exponent and distributing it, so that the integral looks like:
    [itex]s=\int_{0}^{1} 1+{2/3}x^{-1/3} dx [/itex] (I'm fairly certain, at least... that took a little work). However, that basically leaves me in the same place with a slightly different set of products.... if [itex]f(x)=x^{-1/3}[/itex], then [itex]f'(x)={-1/3}x^{-4/3}[/itex], so [itex]u=x^{-1/3}[/itex] and [itex]du={-1/3}x^{-4/3}dx[/itex]... same place, basically.
     
  12. Sep 30, 2012 #11
    With [itex] x = u^{-3/2} [/itex] (and remembering [itex]du = -2/3 x^{-5/3} dx[/itex]), then we can substitute [itex] u^{-3/2} [/itex] for x into [itex]du = -2/3 x^{-5/3} dx[/itex], obtaining [itex] du = -2/3 u^{5/2} dx[/itex]. We can solve for an expression of dx in terms of u and du, and plug that expression in for dx into the integral. We then obtain an expression like [itex] \int_{0}^{1} \sqrt{1+4/9u}*(-3/2 u^{-5/2})du [/itex]. Do you see my point now?

    In any case, I've found an easier method (so you can essentially ignore everything we've done up to this point): try taking the integral with respect to y. Solve for dx/dy -- the argument of the square root turns into a simple u substitution this way.
     
  13. Sep 30, 2012 #12
    Of course! *facepalm*
    OK, I'll work on that.

    Anyway,
    ...yea, sorta. I'm getting a little burnt out (been at this all day, and this isn't the only problem that I've been working), so I think that I'll set this aside for the night and approach it fresh in the morning. However, just from looking at it, I'm missing how: "obtaining [itex] du = -2/3 u^{5/2} dx[/itex]" is happening, for some reason. It's probably obvious, but right now it doesn't seem... correct (not that I'm doubting you, you understand).

    I (very much) appreciate all of the help so far, though!
     
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