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Math Help - Writing an equation to catenary curve.

  1. #1
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    Writing an equation to catenary curve.

    A catenary curve is an idealised hanging chain or cable. The curve is the graph of a hyperbolic cosine function and is similar in appearance to a parabola. The equation of a catenary can be expressed in two ways: y = c cosh (x/c) or y = 2/a (e^bx + e^-bx)

    To find the equation of this catenary curve two points much be chosen, and their coordinates must be substituted into the equation to find the values of a and b.
    The points which are used to find the equation are (0,2) and (5,16)
    Substituting (0,2) into the equation:

    y = 2/a (e^bx + e^-bx)
    2 = 2/a (e^b0 + e^-b0)
    2 = 2/a (e^0 + e^-0)
    2 = 2/a (1 + 1)
    2 = 2/a * 2
    2 = a

    The value of a is 2.
    Therefore the equation becomes:

    y = e^bx + e^-bx

    By substituting (5,16) into the equation you can find the value of b:
    y = ebx + e-bx
    16 = e5b + e-5b
    loge16 = loge e5b + logee-5b
    2.773 = 5b -5b

    I have to get an equation so is my 'a' value correct? And how do I get my b value because I have them cancelling out. Any help on where I've gone wrong or what I do?
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  2. #2
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    Check the equation of catenary in Wikipedia.
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  3. #3
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    Quote Originally Posted by eriiin View Post
    ...
    16 = e5b + e-5b
    loge16 = loge e5b + logee-5b <---
    ...

    Where did you learn this step?

    16=e^{5b} + e^{-5b}~\implies~16 e^{5b} = \left(e^{5b}\right)^2+1

    This is a quadratic in e^{5b}. Use substitution to solve the equation.
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  4. #4
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    log(x+ y)\ne log(x)+ log(y)
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