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Complicated Arc length problem

  1. Jan 14, 2014 #1
    1. The problem statement, all variables and given/known data
    The length L or the curve given by
    [tex]\frac{3y^{4}}{2}+\frac{1}{48y^{2}}-5[/tex]
    from y=1 to y=2

    2. Relevant equations



    3. The attempt at a solution
    Setting up the formula is easy. First I found the derivative of f(y) which is:
    [tex]f'(y)=6y^{3}-\frac{1}{24y^{3}}[/tex]
    Then I plugged all the given data into the formula
    [tex]L=\int_{1}^{2}\sqrt{1+(6y^{3}-\frac{1}{24y^{3}})^{2}}~dy[/tex]
    I expanded out the binomial and simplified to
    [tex]L=\int_{1}^{2}(36y^{6}+\frac{1}{48y^{6}}+\frac{1}{2})^{1/2}~dy[/tex]
    It is from here I get stuck. It looks to me like I need to use a u-substitution. I checked out the step-by-step solution from wolfram alpha's integral calculator and it uses the following
    [tex]u=\frac{1}{x^{6}}~\therefore~du=-\frac{6}{x^{7}}dy[/tex]
    I could go on and list each next step (another substitution is used down the line), but that would do me little good. I do not understand how the value [itex]\frac{1}{x^{6}}[/itex] for ##u## was chosen.
     
  2. jcsd
  3. Jan 14, 2014 #2

    LCKurtz

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    That number is wrong. Fix it and you should get a perfect square under the square root sign.
     
  4. Jan 14, 2014 #3

    SteamKing

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    It's not clear how you went from:

    [itex]{1+(6y^{3}-\frac{1}{24y^{3}})^{2}}[/itex] to

    [itex](36y^{6}+\frac{1}{48y^{6}}+\frac{1}{2})[/itex]

    There seems to be some algebra which doesn't look quite right when expanding the squared expression.
    [itex](a + b)^{2} = a^{2} + 2ab + b^{2}[/itex]
     
  5. Jan 14, 2014 #4

    Ray Vickson

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    When I do the simplification (using Maple) I get
    [tex] L = \int_1^2 \sqrt{ 36y^6+\frac{1}{576 y^6}+\frac{1}{2}} \, dy [/tex]
    (because ##24^2 = 576##). It turns out that
    [tex] \sqrt{ 36y^6+\frac{1}{576 y^6}+\frac{1}{2}} [/tex]
    has the nice form of
    [tex] a + b y^3 + \frac{c}{y^3} [/tex]
    for some constants ##a,b,c## (again thanks to Maple). Just find these constants and you are essentially done.
     
  6. Jan 15, 2014 #5
    Now I feel retarded. That's what I get for going at coursework for 8 hours straight while tired. I clearly have 576, not 48 written the line above on my scratch paper.
     
  7. Jan 15, 2014 #6
    I fear I am going to have to slap myself again, but with this corrected equation:
    [tex] L = \int_1^2 \sqrt{ 36y^6+\frac{1}{576 y^6}+\frac{1}{2}} \, dy [/tex]
    I am still having difficulty despite this advice:
    [tex] a + b y^3 + \frac{c}{y^3} [/tex]
    What exactly is it that I am missing?
     
  8. Jan 15, 2014 #7

    Ray Vickson

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    Have you tried solving
    [tex] \left(a + b y^3 + \frac{c}{y^3}\right)^2 \equiv \frac{1}{2}+36 y^6 + \frac{1}{576 y^6}[/tex]
    for ##a, b,c##?
     
  9. Jan 15, 2014 #8

    LCKurtz

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    Look at what you got when you squared that expression before you simplified it by adding the 1.


    Now look at what changed after you added the 1. It's very similar... (It might help you recognize it if you put the 1/2 in the middle of your expression.)
     
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