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MathGroup Archive 2006

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Total Derivative and Output Display Question

  • To: mathgroup at smc.vnet.net
  • Subject: [mg64993] Total Derivative and Output Display Question
  • From: "Matt" <anonmous69 at netscape.net>
  • Date: Sat, 11 Mar 2006 05:15:28 -0500 (EST)
  • Sender: owner-wri-mathgroup at wolfram.com

Hello Mathgroup,
   I have been experimenting with the Dt[] function in Mathematica.
The reason for this is that as I review or work with new topics in
math, I try to see how the same results achieved by hand can or will be
realized in Mathematica.  I recently needed to review the concept of
the total derivative, and as is my usual practice, to ensure I
understood what I was doing, I came up with an example that would be a
challenge, that is, a challenge for me.
  Here's what I did on paper (the actual example was two levels deeper
than this, but this should suffice for the sake of argument):

z = f(x,y);
x = g(r,s);
y = h(r,s);
r = a(t,u);
s = b(t,u);

What is the total (partial) derivative of z with respect to 't'?

On paper, I worked this out to (all derivatives should be considered as
partial):

dz/dt = dz/dx(dx/dr dr/dt + dx/ds ds/dt) + dz/dy(dy/dr dr/dt + dy/ds
ds/dt).

As you can see, there are no f's, g's, h's, a's, or b's, because we all
understand that when we state something like z = f(x, y), we are
stating that z is the dependent variable and x and y are the
independent variables and 'f' is just part of the notation indicating
that z is a function of x and y.

I then wanted to see how I could achieve this same result in
Mathematica in a general way.  I tackled the same problem, but I left r
and s as functions of only t.  I won't attempt to get the results from
my Mathematica attempt in this message.  Suffice it to say that I went
through many iterations in order to achieve my less than satisfactory
results.

Here is the Mathematica code (converted to InputForm):

Clear[a, b, f, g, h, r, resultOne, resultTwo, s, t, x, y, z, ζ, ξ]
r = a[t];
s = b[t];
x = g[r, s];
y = h[r, s];
z = f[x, y];
resultOne = Dt[z, t]
resultOne /. {Replacement Rules}

Unfortunately, converting my replacement rules to InputForm didn't work
very well.  I will post what I have done as a notebook.  Here is the
direct link (please replace 'notvalid' with 'anonmous':

http://home.comcast.net/~notvalid/mathematica/TotalDerivativeQuestion.nb

As those of you who download it will see, there are numerous problems
with my approach.  The first is that it appears overly burdensome to
convert from Mathematica's default output form, to the more
'traditional' mathematical form.  Secondly, even after I convert, I end
up with the differential operators being in the wrong order, and I'm
not really sure about how to avoid that.

Thanks for any and all help,

Matt


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