Re: Total Derivative and Output Display Question

*To*: mathgroup at smc.vnet.net*Subject*: [mg65142] Re: Total Derivative and Output Display Question*From*: Paul Abbott <paul at physics.uwa.edu.au>*Date*: Wed, 15 Mar 2006 06:30:31 -0500 (EST)*Organization*: The University of Western Australia*Sender*: owner-wri-mathgroup at wolfram.com

Matt wrote: > 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). Here is another approach: D[z /. z -> z[x, y] /. {x -> x[r, s], y -> y[r, s]} /. {r -> r[t], s -> s[t]}, t] % /. {Derivative[1, 0][f_][__] :> HoldForm[D[f, r]], Derivative[0, 1][f_][__] :> HoldForm[D[f, s]], Derivative[1][f_][_] :> HoldForm[D[f, t]]} > 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. What you are after is an _implicit_ method of declaring the variable dependency of a function; no such functionality is built-in. > 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. There are no differential operators in your final expression, only derivatives of functions -- so the order of terms does not matter here. Also, functionally the original output is perfectly suitable for further calculations, so the "preferred" simpler typeset form is really for aesthetics. Enter x-1 into Mathematica. StandardForm re-orders this to -1 + x. If you prefer x-1 convert to TraditionalForm. Perhaps the discussion in "Structure and Interpretation of Classical Mechanics" by Gerald Jay Sussman and Jack Wisdom, available online at http://mitpress.mit.edu/SICM/ is of interest. In particular, have a read of the Preface at http://mitpress.mit.edu/SICM/book-Z-H-5.html Cheers, Paul _______________________________________________________________________ Paul Abbott Phone: 61 8 6488 2734 School of Physics, M013 Fax: +61 8 6488 1014 The University of Western Australia (CRICOS Provider No 00126G) AUSTRALIA http://physics.uwa.edu.au/~paul

**Re: 3D-plot over a triangle**

**Finding a shorter form for coplanar point in 3D space**

**Re: Total Derivative and Output Display Question**

**Plot resolution**