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Math Help - Hyperbolic Cosine/Derivative question

  1. #1
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    Hyperbolic Cosine/Derivative question

    I don't even understand what this problem is asking, much less how to solve it. Any help would be appreciated!



    Problem: The cable between two towers of an overhead utility cable hangs in the shape of the curve

    y = \frac{T}{w} cosh(\frac{wx}{T})

    where T is the tension in the cable at its lowest point and w is the weight of the cable per unit length. Suppose the cable stretches between the points  x = \frac{-T}{w} and x = \frac{T}{w}.

    a.) Sketch a graph of the cable.
    b.) Find an expression for the sag in the cable (the distance between the highest point on the cable and the lowest point).
    c.) Show, by computing both sides, that f\prime\prime = \frac{w}{T}\sqrt{1 + (y\prime)^2}
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by lysserloo View Post
    I don't even understand what this problem is asking, much less how to solve it. Any help would be appreciated!



    Problem: The cable between two towers of an overhead utility cable hangs in the shape of the curve

    y = \frac{T}{w} cosh(\frac{wx}{T})

    where T is the tension in the cable at its lowest point and w is the weight of the cable per unit length. Suppose the cable stretches between the points  x = \frac{-T}{w} and x = \frac{T}{w}.

    a.) Sketch a graph of the cable.
    b.) Find an expression for the sag in the cable (the distance between the highest point on the cable and the lowest point).
    c.) Show, by computing both sides, that f\prime\prime = \frac{w}{T}\sqrt{1 + (y\prime)^2}
    b) as \cosh is an even function the lowest point is at x=0, and the highest points are at either end, so the sag is:

    s=\frac{T}{w} \cosh(1)-\frac{T}{w} \cosh(0)=\frac{T}{e}(e-1)=T(1-e^{-1})

    CB
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