more clearly exactly what you want, The code which you posted executes without error and produces the plot which you asked for - ie 'V_TEV-VP' as a function of 'E' and 'T'.

Apparently this does not ptoduce the plot which you want?? So exactly what do you want to plot? Please don't use terms such as abscissa and ordinate which only really apply to 2D plots, just specify what is the x-variable, what is the y-variable and what is the z-variable in your desired plot

For future reference, please use the big green up-arrow in the Mapleprimes toolbar to upload you actual worksheet - it make life easier for responders.

The code which you posted seems a bit "repetitive" and generally "disorganised". The code below (and in the attachment) is in a more "logical" order and (IMHO) much easier to read. It produces the same plot (see below)

  restart;
#
# Some "parameters". NB the parameter 'a' is
# never used - should it be?
#
  h:= 4: m:= 0.035: g:= h*m:            
  k:= 0.15: a:= 0.25: F_ANR:= 4*m:
  r:= 0.1:
#
# Define a couple of "utility" functions
#
  theFunc:= z-> piecewise
                ( z < 9984, 0,
                  z < 14926, (1008.7*(z-9984)/10000+1400)*(z-9984)/10000,
                  z < 58596, (206.43*(z-14926)/10000+2397)*(z-14926)/10000+938.24,
                  z < 277826, 0.42*z-9267.53,
                  0.45*z-17602.28
                ):
  GewSt:= z-> piecewise( z < 24500, 0,
                         (z-24500)*g
                       ):
#
# Computation of V_TEV
#
  K:= 0.6*E*(1-k-g)*(1+r*(1-k-g))^T:
  V_TEV:= K/0.6*(1 - 0.6*theFunc(K)/K):

#
# Computation of VP
#
  Er:= E+(E-theFunc(E))*r:
  V_P:= ( E-theFunc(E)+GewSt(E)-min( theFunc(E),
                                     GewSt(E),
                                     E*F_ANR
                                   )
        )
        *
        (1+r*(1-theFunc(Er)/E ))^T:

#
# difference between V_TEV and VP
#
  VT:= V_TEV-V_P:
  plot3d( VT,
          E = 0 .. 500000,
          T = 1 .. 15,
          axes = boxed,
          style=surface,
          color=red,
          labels=["E", "T", "VT"],
          labelfont=[times, bold, 12]
        );

which produces

Attachment finCalc.mw

The solution for one particular innitial condition is shown in the attached

odeSol.mw

all you have is a lot of "simple" syntax errors. I can only suggest that you read the 'help' for the commands in the geometry package ot carefully.

For your specific exampl,e the syntax-corrected code s shown in the attached.

Download basicGeom.mw

the execution group I have added in the attached

odeplots.mw

using the geometry() package is shown in the attached,

Download circtri.mw

it is a bad idea to make "unnecessary" assignments. You can access the value to which the variable names in sol[] have been equated by using eval( varName, sol), as in the code below

  restart;
  sol := { c[1, 1] = 18.00000000, c[1, 2] = -0., u[1, 1] = -.9000000000,
           u[1, 2] = 0., x[1, 1] = 3.600000000, x[1, 2] = -0. }:
#
# You can access the values to which the parameters
# in sol have been equated, by using
#
#    eval(varName, sol)
#
# as in the following
#
  eval( c[1,1], sol);
  eval( c[1,2], sol);
  eval( u[1,1], sol);
  eval( u[1,1], sol);
  eval( x[1,1], sol);
  eval( x[1,2], sol);

                          18.00000000

                              -0.

                         -0.9000000000

                         -0.9000000000

                          3.600000000

                              -0.

Of course if tou really, reallly want to convert these to assignments, then just apply the assign command to sol[], in an elementwise fashion, as in the code below

  restart;
  sol := { c[1, 1] = 18.00000000, c[1, 2] = -0., u[1, 1] = -.9000000000,
           u[1, 2] = 0., x[1, 1] = 3.600000000, x[1, 2] = -0. }:
#
# If you really, really want to make *assignments* then just
# apply the assign() command elementwise to sol
#
  assign~(sol):
  c[1,1];
  c[1,2];
  u[1,1];
  u[1,1];
  x[1,1];
  x[1,2];

more of an illustration.

The attached worksheet defines a curve in 3D, as it happens, a conical helix, (you do not supply the curve you are interested in, so I feel free to use anything I want), together with tangents at various points. As a visual check the curve and the tangents are plotted. The tangent at an arbitray point is also calculated. So please expalin clearly for your curve (which you should specify exactly), what do you find difficult?

Download 3Dtangents.mw

the plot of the analytic solution is the function y(t) versus the function x(t), I assume you want the same thing from the numeric solution. The method for this is shown in the attached.

Inline display - broken AGAIN

ytVsxt.mw

that the directory exists and is wrtable, using FileTools commands - for example

  restart;
  with(FileTools):
#
# Change the following directory name to something
# appropriate for wuor machine
#
  dirName:="C:/Users/TomLeslie/Desktop":
#
# Check that the directory exists
#
  Exists(dirName);
#
# Check that it is a directory
#
  IsDirectory(dirName);
#
# Check that the directory is writable
#
  IsWritable( "C:/Users/TomLeslie/Desktop" );
#
# Check whether the file you are interested in
# exists already
#
  fileName:="tuples.mw":
  Exists( cat(dirName, "/", fileName));
  fileName:="doesNotExist.mw":
  Exists( cat(dirName, "/", fileName));

                              true

                              true

                              true

                              true

                             false

tuple10:=proc(searchstop,diff1,diff2,diff3,diff4,diff5,diff6);

has only seven parameters. However in both of the examples (one in the worksheet, and one in the text of your question), you supply eleven arguments. It is not obvous (to me at least) what you want (if anything) to do with the last four arguments. As written, your procedure iwill just ignore them.

I have assumed that you always want to pass an argument for 'searchstop', and at least two more arguments. As  written the attached will accept any number of arguments (>2) and print the results in the way you wish

Download tuples.mw

just because

1. Alternatives are good
2. I happpen to think that one should probably use Maple's geometry() package for ptoblems in - well "geometry"

Download doTri.mw

Since yyou don't actually supply the file you are interested in ( ie "Indifferenzkurve.csv") I can't produce the code to do this bu the appoach would be

1. Import the whole file
2. Split the resulting matrix into a list of columns using the the LinearAlgebra:Column() function
3. Use a select() command to pick out only those columns whose first entry corrresponds to one of "E","T", and "S_P and S_TEV"
4. combine these selected columns into a new matrix

This is actually pretty trivial - and if you had supplied the file "Indifferenzkurve.csv", probably someone here would have supplied you with the code to do it by now!

the same calculation - but in two conceptually different ways. Why should this matter? Well with floating-point arithmetic, sometimes it does!

When you use unapply(), it will evaluats its arguments before converting to a function definition. So the way you have defined the function f(), it will generate an expression contianing 10000 terms, each of which contains the passed argument 'r'. On the other hand, since you have placed unevaluaion quotes arounnd  the add() function in the definition of of the function g(), this will generate one single term to which the add() will be applied.

On a subsequent call to the function f(), the passed argument will be inserted into each of the 10000 terms in the function definition and the sum will be computed. On the other hand, a subsequent call to the function g() will insert the passed argument into a single term,  and the resulting sum will be computed. Essentially two different ways of doing the "same" (floating-point) calculation. Looks like exactly the same thing - doesn't it!! However if you do the (logically) same calcultion two different ways in floating point arithmetic, it is perfectly possible to get two different answers, depending on phenomena such as "rounding" or "catastrophic cancellation".

To resolve your problem, in the attached, I increased the setting of "Digits" to 50, so that all floating point calculations will be carried out to 50 digits - and the two results now agree. Isn't floating-point arithmetic wonderful!

Download digFloat.mw

1. You have syntax issues - your entries t = time, rage = 0 .. 1, should be time=t, range=0..4. The end points of the range, must correspond with the values you have defined in the boundary conditions (which are x=0 and x=4).
2. Maple's numeric PDE solvers cannot handle boundary conditions with derivatives which are parallel to the boundary, in your case a derivative with respect to 'x' at the value t=0. and you have D[1](u)(x, 0) = 0. Derivatives must be normal to the boundary - so, for example, D[2](u)(x, 0) = 0 would be acceptable - although it is not the condition that you want!

Thus the code

PDE := diff(u(x, t), x, x) + diff(u(x, t), t, t)/0.5^2 - 2*diff(u(x, t), x, t)/0.5^2 = -0.0013*u(x, t) + 8.6510*u(x, t)^3;
IBC := {u(0, t) = 0.01, u(4, t) = 0, u(x, 0) = 0, D[2](u)(x, 0) = 0};
pds := pdsolve(PDE, IBC, numeric, time=t, range = 0 .. 4);

will execute - although the derivative boundary condition is not what you want

was a straightforward translation from 'linalg()' to 'LinearAlgebra()'. I didn''t pay much attention to the calculation you were attempting but it looks like you want to rotate/translate an arbitrary parabola, to one which has its vertex at the origin, and its axis paralle with the x-axis.

There are easier ways to do this! See the attached, which unfortunately won't display inline, but it produces the following output and plots

GeometryDetail(["name of the object",        pa],
               ["form of the object",        parabola2d],
               ["vertex",                    [-17/40, 17/20]],
               ["focus",                     [-1/40, 1/20]],
               ["directrix",                 -1/5*x*sqrt(5) + 2/5*sqrt(5)*y - 33/40*sqrt(5) = 0],
               ["equation of the parabola",   4*x^2 + 4*x*y + y^2 - 8*x + 16*y - 17 = 0])

GeometryDetail(["name of the object",       pc],
               ["form of the object",       parabola2d],
               ["vertex",                   [0, 0]],
               ["focus",                    [2/5*sqrt(5), 0]],
               ["directrix",                -2/5*sqrt(5) - x = 0],
               ["equation of the parabola", -8*x*sqrt(5) + 5*y^2 = 0])

parabolae.mw