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March 24, 2017

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Catenary; y= a/2(e^bx +e^-bx)
goes through the origin (0,0) and has point (-9,18) & (9,18)
solve for a and b.

  • maths - ,

    sub in (0,0)

    0 = a/2(e^0 + e^)
    0 = a/2(2)
    a = 0

    something is not right here, a catenary of the above form does not pass through (0,0)
    Only if a = 0, but then the equation would collapse to
    y = 0

  • maths - ,

    What about if we move the turning point (0,0) to (9,0) and the other points are (0,18) and (18,18, what then?

  • maths - ,

    If we use the above points,I think
    y= a/2(e^bx +e^-bx)
    18 = a/2 (e^b0 + e^-b0)
    18 = a/2 (e^0 + e^0)
    18 = a/2 (1 + 1)
    18 = a/2 (2)
    18 = a

    To find b;
    then substitute in y= a/2(e^bx +e^-bx)
    at point (9,0)
    0= 18/2(e^b9 +e^-9b)
    0= 9(e^9b +e^-9b)
    0 = 9e^9b + 9e^-9b
    0 = 9e^9b + 9/e^9b (Xe^9b)
    0 = 9(e^9b)^2 + 9

    let e^9b=x
    ie 0 = 9x^2 + 9
    9x^2 = -9
    x^2 = -1
    x= (-1)^1/2
    Still it doesn't work out
    or do I differentiate d(y)/d for a gradient of 0/ Still doesn't work.

    What am I missing here?

  • maths - ,

    Using these two points (0,2) & (8,7) can y find a & b using ghe formula y=aĆ·2(e^bx+e^-bx)

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