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Crazy Function for Arc Length!

  1. Jan 23, 2008 #1
    Hi! Here's my question on finding arc length. If I've taken the derivative correctly, is there anyway I can simplify it before putting it into the arc length formula?

    1. The problem statement, all variables and given/known data
    Find the arc length where 0[tex]\leq[/tex]x[tex]\leq[/tex]2
    y=(x[tex]^{3}[/tex]/3)+x[tex]^{2}[/tex]+x+1/(4x+4)


    2. Relevant equations
    L=[tex]\int[/tex]ds=[tex]\sqrt{1+(dy/dx)^{2}}[/tex]



    3. The attempt at a solution
    I've only taken the derivative so far:
    (dy/dx)=x[tex]^{2}[/tex]+2x+1-4(4x+4)[tex]^{-2}[/tex]
    =(x+1)[tex]^{2}[/tex]-4(4x+4)[tex]^{-2}[/tex]


    I tried expanding the equation, but that only makes it more complex.

    I know to find the arc length I need to square the derivative and place it in the formula (and possibly using substitution), but I'm just wondering how I can simplify the above equation to make it easier to square and calculate!
     
    Last edited: Jan 23, 2008
  2. jcsd
  3. Jan 24, 2008 #2

    Gib Z

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    Polynomial division!! =]
     
  4. Jan 24, 2008 #3

    Defennder

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    Yeah you should get this if you do the polynomial division correctly:

    [tex]\int_0^2 (x+1)^2 + \frac{1}{4(x+1)^4} dx[/tex].

    To make things easier, denote (x+1) by A or some other letter, then it'll look easier.

    EDIT: Fixed error
     
    Last edited: Jan 24, 2008
  5. Jan 24, 2008 #4

    HallsofIvy

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    I would look at it like this:
    [tex]y'= x^2+ 2x+ 1- \frac{4}{(4x+ 4)^2}= x^2+ 2x+ 1- \frac{1}{4(x+1)^2}[/tex]
    [tex]= (x+ 1)^2- \left(\frac{1}{2(x+1)}\right)^2[/tex]
    so
    [tex](y')^2= (x+1)^4- 1/2+ \left(\frac{1}{2(x+1)}\right)^4[/tex]
    [tex](y')^2+ 1= (x+1)^4+ 1/2+ \left(\frac{1}{2(x+1)}\right)^4[/tex]
    [tex]= \left((x+1)^2+ \left(\frac{1}{2(x+1)}\right)^2\right)^2[/tex]
    so that
    [tex]\sqrt{(y')^2+ 1}= (x+1)^2+ \frac{1}{2(x+2)}[/tex]
     
  6. Jan 24, 2008 #5

    Defennder

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    There's a small error here, the last term should read 1/4(x+1)^2 instead.
     
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