MathGroup Archive 2003

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: Re: Derivative of a funtion evaluated at a point in 3D

The question is ill-posed - you have not provided a complete enough 
for anyone to help you. The answer may well be 0, but without "stuff" 
there is no way to know.
If in fact you really want to evaluate D[F,p1,p2,p3,p4] then the answer 
could well be 0 (p1,p2,p3,p4 were most likely taken to be scalars not 
vectors by D).
Consider this trivial example:
F = ({x1, x2, x3} + {y1, y2, y3} + {z1, z2, z3})/3;
D[F, x1, x2, x3] (* mixed partials vanish *)
I suspect what you want is something more like:
{{D[F, x1], D[F, x2], D[F, x3]},{D[F, y1], D[F, y2], D[F, y3]},{D[F, 
z1], D[F, z2], D[F, z3]}}
You should evaluate ?D and follow the links to the help browser entries 
for D to see what this expression (D[F,p1,p2,p3,p4] ) means.

On Nov 10, 2003, at 8:55 PM, mroc wrote:

> mroc_1000 at (mroc) wrote in message 
> news:<boign7$oj1$1 at>...
>> Hello, I am totally new to Mathematica and trying to program a simple
>> FEM-type problem. I am trying to take the partial derivative
>> (symbolically) of an expression that is a function of four points in
>> 3D. All I can think of to do is F[p1,p2,p3,p4] = stuff then
>> D[F,p1,p2,p3,p4] . But I keep getting a zero expression as a result.
>> (where stuff is a nasty combination of these points) Any thoughts?
> Anyone? What am I missing? Is this question too easy or too hard?
Dr Christopher Purcell
Sensors & Actuators Group
DRDC-Atlantic, 9 Grove St., PO Box 1012,
Dartmouth NS B2Y 3Z7 Canada
Tel 902-426-3100 x389
Fax 902-426-9654

  • Prev by Date: Re: orthonormalized eigenvectors
  • Next by Date: Re: Linear integral along a given curve.
  • Previous by thread: Re: Re: Derivative of a funtion evaluated at a point in 3D
  • Next by thread: Re: Derivative of a funtion evaluated at a point in 3D