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Cable length

  1. Oct 26, 2006 #1
    A cable hangs between two poles of equal height and 37 feet apart.
    At a point on the ground directly under the cable and
    x feet from the point on the ground halfway between the poles
    the height of the cable in feet is
    h(x)=10 +(0.4)( x^{1.5})
    The cable weighs 15.1 pounds per linear foot.
    Find the weight of the cable.


    so i find h'(x) i square it and add one to find the length of the curve
    i.e
    (1+ h'(x)^2 ) ^(1/2)

    my question is is that right? and what are the limits of integration??


    blumfeld0
     
  2. jcsd
  3. Oct 26, 2006 #2

    HallsofIvy

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    Well, more correctly "i find h'(x) i square it and add one" , take the square root and integrate "to find the length of the curve" :approve:

    Since your variable, x, is the "from the point on the ground halfway between the poles", x= 0 there. At one pole x= -37/2 and at the other x= 37/2. Integrate with respect to x from -37/2 to 37/2.
     
  4. Oct 26, 2006 #3
    ok so i integrate (1 + h'(x)^2 )^(1/2) with repect to x from -37/2 to 37/2
    i use mathematica and i get 39.25+ 24.9 i

    where "i" is imaginary
    why do i get an imaginary answer??


    also what does the 15.1 pounds per linear foot have to do with it?

    thanks
     
  5. Oct 26, 2006 #4

    HallsofIvy

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    Ah! Misinterpretation on my part. Since x is a distance it is always positive. The height "h(x)=10 +(0.4)( x^{1.5})" isn't even defined for x negative. Integrate from 0 to 37/2 to find the length of 1/2 and then double.
     
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