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MathGroup Archive 2005

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Re: Integrate is driving me crazy, please help!

  • To: mathgroup at smc.vnet.net
  • Subject: [mg56205] Re: [mg56184] Integrate is driving me crazy, please help!
  • From: DrBob <drbob at bigfoot.com>
  • Date: Tue, 19 Apr 2005 04:55:03 -0400 (EDT)
  • References: <200504180708.DAA20024@smc.vnet.net>
  • Reply-to: drbob at bigfoot.com
  • Sender: owner-wri-mathgroup at wolfram.com

Integrate[hillb*((f0 + hilla)/
      (2*Pi*f*a*Cos[ArcSin[x/a]] + hillb)) -
    hilla, {x, -a, a}]

(-4*a^2*f*hilla*Pi*
     Sqrt[hillb^2 - 4*a^2*f^2*
        Pi^2] + a*f0*hillb*Pi*
     Sqrt[hillb^2 - 4*a^2*f^2*
        Pi^2] + a*hilla*hillb*Pi*
     Sqrt[hillb^2 - 4*a^2*f^2*
        Pi^2] + 2*a*(f0 + hilla)*
     hillb^2*ArcTan[(2*a*f*Pi)/
       Sqrt[hillb^2 - 4*a^2*f^2*
          Pi^2]] - I*Sqrt[a^2]*
     (f0 + hilla)*hillb^2*
     Log[-((I*Sqrt[a^2]*hillb)/
        (2*a*Sqrt[hillb^2 -
           4*a^2*f^2*Pi^2]))] +
    I*Sqrt[a^2]*f0*hillb^2*
     Log[(I*Sqrt[a^2]*hillb)/
       (2*a*Sqrt[hillb^2 -
          4*a^2*f^2*Pi^2])] +
    I*Sqrt[a^2]*hilla*hillb^2*
     Log[(I*Sqrt[a^2]*hillb)/
       (2*a*Sqrt[hillb^2 -
          4*a^2*f^2*Pi^2])])/
   (2*a*f*Pi*Sqrt[hillb^2 -
      4*a^2*f^2*Pi^2])

% /. {hilla -> 3, hillb -> 50,
    f0 -> 8, a -> 1, f -> 1}

(1/(2*Pi*Sqrt[2500 - 4*Pi^2]))*
   (538*Pi*Sqrt[2500 - 4*Pi^2] +
    55000*ArcTan[(2*Pi)/
       Sqrt[2500 - 4*Pi^2]] -
    27500*I*Log[-((25*I)/
        Sqrt[2500 - 4*Pi^2])] +
    27500*I*Log[(25*I)/
       Sqrt[2500 - 4*Pi^2]])

N[%]
14.037200672826035 + 0.*I

Bobby

On Mon, 18 Apr 2005 03:08:44 -0400 (EDT), Jim Martin <jim.martin at utah.edu> wrote:

> Hello Mathematica Experts:
>
> I am a biomechanist and work mostly in the area of muscle contraction. I
> do a lot of numerical computations using excel, but right now I need an
> analytical solution that represents force as a function of position
> integrated over a shortening amplitude. I downloaded a trial version of
> Mathematica and have tried to obtain a solution for this:
>
> Integrate[(hillb*((f0 + hilla)/(2*pi*f*a*Cos(ArcSin(x/a)) + hillb))) -
> hilla, {x, -a, a}]
>
> Mathematica returns this:
> (-4 a ArcSin Cos f hilla pi + (f0 + hilla) hillb (-Log[hillb - 2 a
>      ArcSin Cos f pi] + Log[hillb + 2 a ArcSin Cos f pi]))/(4 a ArcSin
> Cos f pi)
>
> I know the line wrap makes this hard to read so please feel free to
> email me and I can send you the output as a picture.
>
> In a sample data set, hilla=3, hillb=50, f0=8, a=1, f=1
>
> I can numerically integrate this function and obtain a value for that
> sample data set of 14.04. When I put those sample values into the
> solution that Mathematica produces, I get 10.01.
>
> Can any of you please give a hand here? I must be making some simple
> Mathematica-beginner error but I just can't see it.
>
> In Mathematica, Log is Log to base e, right (LN in excel)? Did I use
> variables that have intrinsic functions in Mathematica? Maybe I am
> misunderstanding the output with regard to implicit parentheses etc.
> Any help appreciated!
>
> Thanks,
>
> Jim
>
>
>
>
>



-- 
DrBob at bigfoot.com


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