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Derivatives Hyper functions

  1. Sep 11, 2005 #1
    Could anyone give a hand in solving this problem?

    In shallow ocean water, the frequency of the wave action (number of wave crests passing a given point per second) is given by f= [(g/(2PiL))tanh(2Pih/L)]^1/2, where g is the acceleration due to gravity, L is the distance between wave crests, and h is the depth of the water. Get an equation for the rate of change of f with respect to h assuming g and L are constants.

    Well this is what I've gotten so far...I'm not sure if it is right or not...

    df/dh=1/2((g/2PiL) tanh(2Pih/L))^-1/2 * ((g/L^2)sech^2(2Pih/L))
    =((g/L^2)sech^2(2Pih/L))/2√((g/2PiL)tanh(2Pih/L))

    well seems that the final answer is

    (gPi/2L^3)^1/2 csch^1/2(2Pih/L)sech^3/2(2Pih/L)

    So, I have no idea how this answer was derived, looking at it seems that they integrated it (sech^3/2)... is that so?

    It would be of great help you anyone could give a hand on this one....


    Thank you so much... guys...
     
  2. jcsd
  3. Sep 12, 2005 #2

    FredGarvin

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    Science Advisor

    I am very rusty on my derivatives. But I will say that it doesn't make any sense to integrate something while looking for the derivative of a function. It appears to be a nasty little hyperbolic function...You may want to repost this in the homework or math sections for a bit more help.
     
  4. Sep 12, 2005 #3

    Astronuc

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    Staff: Mentor

    cunhasb - you seem to be on the right track. Make sure constants are correct, and remember

    tanh x = sinh x/ cosh x = sech x/csch x, csch x = 1/sinh x, sech x = 1/cosh x

    --------------------------------------

    let f = [itex](\frac{g}{2\pi L} tanh (\frac{2\pi h}{L})^{1/2}[/itex]

    or f = [itex] (a g)^{1/2}[/itex], so

    df/dh = f ' = 1/2 a1/2 g-1/2 g '

    now if g = tanh (bh) , then g ' = b sech2 (bh) where b = [itex]\frac{2\pi}{L}[/itex]

    so

    f ' = 1/2 [itex] (\frac{g}{2\pi L})^{1/2} (tanh (\frac{2\pi}{L}))^{-1/2}\,sech^2 (\frac{2\pi h}{L})\,\,\frac{2\pi}{L} [/itex]
     
  5. Sep 23, 2005 #4
    thank you guys... Sorry not to thank you before... but I was away for a few days....

    Thank you so much...
     
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