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Find the length of the curve

  1. Oct 28, 2004 #1
    Can't seem to finish this problem:

    Find the length of the curve:

    [tex]x = 3y^{4/3}-((3/32)y^{2/3})[/tex]

    And -343 <= y <= 125.

    I used the formula:

    [tex]\int_{a}^{b}\ (1 + (x')^{2})^{1/2}dx[/tex]
    [tex]\int_{-343}^{125}\ (1+(4y^{1/3}-(1/16)y^{-1/2})^{2})^{1/2}dx[/tex]

    But how do you find this integral? Is this even the right integral to use? Thanks.
     
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  3. Oct 28, 2004 #2

    Tide

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    First, you want [itex]\left(y'\right)^2[/itex] under ther radical. Then use the relationship between x and y to find y in terms of x from which you can find y'(x) and you should be able to complete the integration.
     
  4. Oct 28, 2004 #3
    But isn't it just easier to do it the way I did (since it is given that y is between two values and not x)?
     
  5. Oct 28, 2004 #4

    Zurtex

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    Have you tried expanding it out?
     
  6. Oct 28, 2004 #5

    T@P

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    can i suggest a calculator?
     
  7. Oct 28, 2004 #6
    What do you mean by expanding it out?

    And the calculator will not give me an answer since 0 is not in the domain of y^(-1/3)...
     
  8. Oct 28, 2004 #7

    Zurtex

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    When I said expand out I mean like this:

    [tex]\left(1+\left(4y^{\frac{1}{3}}-\frac{1}{16}y^{-\frac{1}{2}}\right)^{2}\right)^{\frac{1}{2}} = \left(1 + 16y^{\frac{2}{3}} - \frac{1}{2}y^{-\frac{1}{6}} + \frac{1}{128}y^{-1}\right)^{\frac{1}{2}}[/tex]

    Anyway I messed about with this and found it wasn't intergratable in terms of elementary functions sorry. Which is a bit of a shame as both your limits are cubes so their were some nice substitutions I tried 1st. I tried rearranging it as [itex]y=f(x)[/itex] and found that [itex]1 + y'^2[/itex] wasn't intergratable in terms of elementary functions as well.

    Any way it occurred to me while I was coming up with a numerical solution that the function is not continuous at y=0 therefore the integral doesn't exist at y=0
     
  9. Oct 28, 2004 #8
    So then how do I go about coming up with a numerical solution to it? Am I going to have to use limits?
     
  10. Oct 29, 2004 #9

    Zurtex

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    There is no numerical solution, your function isn't continous for [itex]y \leq 0[/tex]

    Have you tried getting to plot it out? You'll notice generally for y<0 your function isn't even real.
     
  11. Oct 29, 2004 #10

    HallsofIvy

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    It has been suggested repeatedly that you "multiply it out". Did you even try that?

    The whole point is this: [itex]\frac{dx}{dy}= 4y^{\frac{1}{3}}-\frac{1}{16}y^{\frac{-1}{3}}[/itex]. so [itex]\(\frac{dx}{dy}\)^2= 16y^{\frac{2}{3}}- 1/2+ \frac{1}{256}y^{\frac{-2}{3}}[/itex].

    Now add 1 to that and the only change is that the "-1/2" in the middle becomes "+1/2". What's the square root of that?
     
  12. Oct 30, 2004 #11
    I can't figure it out. What is the square root of it? Do you try to form a perfect square with it all or what?
     
  13. Oct 30, 2004 #12
    Oh wait I think I know what you mean now. You just complete the square or whatever and then cancel out the square with the square root. So now I can do the integral. Problem is that I get a negative answer which doesn't make sense for the length of an arc. :\

    Guess I'm still doing it wrong. Here's exactly what I did:

    [tex]\int_{-343}^{125}\ (1+16y^{\frac{2}{3}}- 1/2+ \frac{1}{256}y^{\frac{-2}{3}})^{1/2}dx[/tex]

    [tex]\int_{-343}^{125}\ (16y^{\frac{2}{3}}+ 1/2+ \frac{1}{256}y^{\frac{-2}{3}})^{1/2}dx[/tex]

    [tex]\int_{-343}^{125}\ ((4y^{1/3}+(1/16)y^{-1/2})^{2})^{1/2}dx[/tex]

    [tex]\int_{-343}^{125}\ 4y^{1/3}+(1/16)y^{-1/2}dx[/tex]

    And then from there I got a negative answer and it's incorrect according to this program I'm using. :\
     
  14. Oct 30, 2004 #13

    HallsofIvy

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    Well, there is one obvious mistake:

    That "[itex]y^{-\frac{1}{2}}[/itex]" should be "[itex]y^{-\frac{1}{3}}[/itex]
     
    Last edited: Oct 30, 2004
  15. Oct 30, 2004 #14
    Yeah that was a typo. I actually used y^1/3 there. My final answer ended up being -5330.25 which makes no sense. :\
     
  16. Oct 30, 2004 #15
    How can an arclength be negative? Is that even possible?
     
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