In the VectorCalculus package, use %Divergence to apply the unevaluated divergence operator. The value command applied to this expression will cause the divergence to be evaluated. The downside to this approach is that the return of %Divergence is precisely that.

Alternatively, work within the Physics:-Vectors package. The inertization operator is the same, namely, %, but the display is the Nabla (Del) dot the expression. Again, the value command causes evaluation.

The Physics:-Vectors package has one additional advantage. It supports what are called "unprojected" vectors, that is, vectors whose components don't have to be declared. (A "projected" vector is one for which components are explicitly given.) Hence, a certain amount of symbolic vector calculus can be implemented in this package, something available nowhere else in Maple.

RJL Maplesoft

F:=evalc(Re(1/(1-x-I*y)));

f:=exp(F);

F^2 < f, as shown by plot3d(f-F^2,x=-1..1,y=-sqrt(1-x^2)..sqrt(1-x^2),view=-1..3);

Since f and F^2 are positive, their integrals are likewise ordered.

q:=int(F^2,y=-sqrt(1-x^2)..sqrt(1-x^2));

int(q,x=-1..1) => Maple returns the infinity symbol.

The integral of a smaller quantity diverges, so the original integral, which is bigger, diverges.

I wonder if this is the "by hand" approach originally taken?

RJL Maplesoft

See the ApproximateInt tutor in the Multivariate Calculus package. Access it from the Tools/Tutors menu.

At the bottom of the tutor you will also see the command upon which the tutor is based. You can copy/paste that to obtain similar results without the tutor.

If your need is for code that accomplishes these tasks from first principles, then you will have to learn a bit more about programming in Maple.

RJL Maplesoft

I believe I would use the tools in the VectorCalculus packages, probably the Student version, which is more lenient with respect to declaring coordinate systems.

To create a vector that knows where it's tail should be, use the RootedVector command with argument "root" that takes the point at which the tail is to reside. Then, use the PlotVector command whose argument can be a list of vectors, RootedVectors, etc.

I find the use of the primitives, either the arrow command in the plottools package, or the arrow command in the plots package, tedious, requiring too many parameters to get the vectors to look right and to be in the right places.

RJL Maplesoft

The VectorCalculus packages in Maple do not treat tensors. Use either the Physics package or the DifferentialGeometry package to deal in any way with tensors.

Moreover, the M in your calculation is not a tensor in Maple. It is seen in Maple as a matrix. Mathematically, one can blur the lines between structures but in a computer algebra system, a structure is what it is. And M is just a matrix as far as Maple knows.

RJL Maplesoft

Solve works find in Maple 18.02. I suspect that you had previously assigned something that caused the error messages. Try a restart.

RJL Maplesoft

It appears that Maple uses variation of parameters to obtain a particular solution. This method has the propensity to provide a particular solution that contains solutions of the homogeneous equation.

I am not aware of any simple way to filter out the homogeneous solutions from such a particular solution.

Since the ode is linear, constant-coefficient, and the right-hand side yields to the method of undetermined coefficients for finding a particular solution, that might be the best way to obtain the "simpler" particular solution.

RJL Maplesoft

If I understand the question correctly, simply add the option "axes=boxed" to the plot command.

RJL Maplesoft

Assuming that you really have a circular cylinder ( what appears in your message isn't the equation for such), and taking a=1 and b=1/2, the following code will generate the graph I think you want.

p1:=plots:-implicitplot3d(x^2+y^2=1,x=-1..1,y=-1..1,z=0..Pi,style=surface,transparency=.1):
p2:=plots:-spacecurve([cos(t),sin(t),t/2],t=0..2*Pi,color=black,thickness=3):
plots:-display(p1,p2,scaling=constrained); 

RJL Maplesoft

Maple's top-level diff command will not take a derivative with respect to a function. In the Physics package, diff has been modified to do this task. The simplest way to access this modified version is to replace diff with Physics:-diff.

RJL Maplesoft

Look at the recorded solution from the Maple web site. It's at the end of this link:

http://www.maplesoft.com/teachingconcepts/detail.aspx?cid=3

RJL Maplesoft

The difficulty with the graph is the singularity at the origin. Here's a trick I learned from one of the Maplesoft programmers, and in my Red Book of Maple Magic it's dated December 19, 2012.

U:=min(3.5, 1/sqrt(x^2+y^2));

plots:-contourplot(U,x=-1..1,y=-1..1,grid=[70,70]);

The limit on heights imposed by the min command keeps contourplot from distorting the contours. The differing views shown in the post probably come from having evaluation points fall on the singularity or not.

RJL Maplesoft

An unevaluated integral constructed with the Int command can be sent to the built-in numeric integrator (an adaptive procedure) by applying the evalf command to the integral. For example:

q:=Int(x,x=0..1);

evalf(q)

Not sure how important it is to your work that the integral actually be evaluated by the Trapezoid rule.

RJL Maplesoft

In the Tools menu, select Tutors, then Calculus-Single Variable, then Newton's Method.

That tutor will probably give  you what you need. Notice the command at the bottom of the tutor. You can copy and paste that command and use it to fine-tune. You will h ave to load the Student Calculus 1 package to make that command work. Use the Tools menu again, but this time select Load Package and then select the Student Calculus 1 package.

RJL Maplesoft

The feasible region is a triangle in the first octant, vertices are (3,0,0), (0,3,0), (0,0,3);

The function value at each of the three vertices is approximately 18.4.

Now search on each edge of the triangle. The function has a minimum along each edge.

Now search for an "interior" max by eliminating z. Look for a max in f(x,y,3-x-y). Apply plot3d to see that the highest point is in the plane x=3. Therefore, look for a max in f(3,y,-y). This max is approximately 19.8 (by fsolve and the usual techniques of the calculus).

RJL Maplesoft