       Re: multiple derivatives?

• To: mathgroup at smc.vnet.net
• Subject: [mg45436] Re: multiple derivatives?
• From: Roland Franzius <roland.franzius at uos.de>
• Date: Wed, 7 Jan 2004 17:31:00 -0500 (EST)
• Organization: Universitaet Hannover
• References: <200312271000.FAA02128@smc.vnet.net> <bt0ujs$hf8$1@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

Michael Alfaro wrote:
> Hi all,
>
> I have been struggling with the following problem and would be grateful
> for any advice on how to proceed.  The rather ugly function below (uKT)
> describes a four-bar mechanism found in the jaws of some fishes and
> relates lower jaw opening to motion expressed at another jaw bone (uKT
> is a measure of mechanical advantage).
>
> (uDiagonal[input_, fixed_, angleA_] := Sqrt[input^2 + fixed^2 -
> 2*input*fixed*Cos[angleA*(Pi/180)]]; )*
>     (uAngYZ[fixed_, input_, coupler_, output_, angleA_] :=
>       ArcCos[(coupler^2 + output^2 - uDiagonal[input, fixed,
> angleA]^2)/(2*coupler*output)]*(180/Pi); )*
>     (uAngW[fixed_, input_, coupler_, output_, angleA_] :=
> ArcCos[(uDiagonal[input, fixed, angleA]^2 + fixed^2 - input^2)/
>          (2*uDiagonal[input, fixed, angleA]*fixed)]*(180/Pi); )*
>     (uAngX[fixed_, input_, coupler_, output_, angleA_] :=
> ArcCos[(coupler^2 + uDiagonal[input, fixed, angleA]^2 - output^2)/
>          (2*coupler*uDiagonal[input, fixed, angleA])]*(180/Pi); )*
>     (uAngY[f_, i_, c_, o_, A_] := 90 - uAngW[f, i, c, o, A] - uAngX[f,
> i, c, o, A]; )*
>     (uAngZ[f_, i_, c_, o_, A_] := uAngYZ[f, i, c, o, A] - uAngY[f, i, c,
> o, A]);
> uKT[f_, i_, c_, o_, A_] := (uAngZ[f, i, c, o, A + 30] - uAngZ[f, i, c,
> o, A])/30
> uKT[1, 0.409923448, 0.420293528, 0.558332627, 30]
>
> I am interested in how uKT changes as a of function to i, c, and o, and
> have been calculating partial derivatives of each of these variables
> for a number of fish species.  Here is an example of the partial
> derivative of i evaluated for one species :
>
> Derivative[0, 1, 0, 0, 0][uKT][1, 0.7, 0.9, 0.74, 30]
>
> However, what I would really like to know is, given a unit small amount
> of change that can be distributed over all three variables in any way,
>   how much does uKT change.  I initially thought that I could get at
> that by taking the multiple derivative of all three variables.
>
> Derivative[0, 1, 1, 1, 0][uKT][1, 0.7, 0.9, 0.74, 30]
>
> However, I now have doubts that the multiple partial derivative
> actually tells me this.  Does anyone have any insight into this
> problem?  Thanks!
>
> Michael
>

Really no Mathematica user knows? Could be the effect of using
Mathematica to much instead of mathematics.

The linear approximation of a realfunction f of n variables

X[n_,x_String]:=Table[ToExpression[x<>ToString[k]],{k,1,n}]

at a fixed point of approximation

Xa={1.1, 2.2, ... , n.n}

one gets by linear combining the linear approximations in the directions
of the n variables

f@@(X[3, "x"] + X[3, "h"]) ->
f@@X[3, "x"] + Dot[X[3, "h"], (D[f @@ X[3, "x"], #] &) /@ X[3, "x"]]

evaluates

f[h1 + x1, h2 + x2, h3 + x3] ->
f[x1, x2, x3] +
h3*Derivative[0, 0, 1][f][x1, x2, x3] +
h2*Derivative[0, 1, 0][f][x1, x2, x3] +
h1*Derivative[1, 0, 0][f][x1, x2, x3]

This is the famous formula of multidimensional calculus

f[\vec x + \vec h] =
f[x] + sum_k h_k \partial/\partial x_k  f[x]  + o(h^2)

--

Roland Franzius



• Prev by Date: RE: multiple list plot and frameticks
• Next by Date: RE: Re: Compile
• Previous by thread: multiple derivatives?
• Next by thread: Mathematica exports curves in pieces to Illustrator