@Art Kalb Sorry that I didn't spot that. The whole purpose of the inner subsindets is to protect those alphas (by freezing them).

This works:

thaw(
     subsindets[2](
          subsindets[flat](z2, {identical(alpha), identical(alpha)^anything}, freeze),
          `^`, 1, op
     )
);

To be a bit more general, we can freeze any RootOf. Just replace identical(alpha) with specfunc(RootOf):

thaw(
     subsindets[2](
          subsindets[flat](z2, {specfunc(RootOf), specfunc(RootOf)^anything}, freeze),
          `^`, 1, op
     )
);

@billyp245 What do you mean by both derivatives? The expression D2(a,b) can only be one derivative. If you had two derivatives, say D2__a and D2__b, then you could solve for them being 0 with

solve({D2__a, D2__b}(a,b), {a,b});

If you have another question, please upload a worksheet.

For what it's worth, continuation doesn't help with this problem. I was able to successfully apply continuation (I multiplied the right side of the first boundary condition by the continuation parameter) only to get the much-more-difficult-to-deal-with error Newton iteration is not converging.

I need to know the modulus, i.e., the degree-8 irreducible polynomial, that you used to create GF(2^8).

Here's Kitonum's procedure with my improvements.

Download ContoursWithLabels.mw

@Kitonum Okay, I made it an Answer, and I'll make it a Comment to your post.

I did just what I said: changed x to x(t) and y to y(t). Then you just press ENTER and you should get the plot.

Don't repeat your Questions. It's very annoying to the moderators.

You got an Answer in about 35 minutes. I think that that's a pretty good response rate.

@WernerP There's no need for implicitplot3d or high values of numpoints. Your code can be simplified to

A:= plot3d(5, theta= -Pi..Pi, phi= arccos(4/5)..arccos(3/5), coords= spherical):
B:= plot3d(5, theta= -Pi..Pi, phi= 0..Pi/2, coords= spherical, transparency= .6):
C1:= plot3d([3, theta, z], theta= -Pi..Pi, z= 0..4, coords= cylindrical):
C2:= plot3d([4, theta, z], theta= -Pi..Pi, z= 0..3, coords= cylindrical):
plots:-display(
     {A, B, C1, C2}, scaling= constrained, labels= [X,Y,Z],
     axes= normal, style= patchnogrid, color= yellow
);

@epostma Thanks! Unfortunately, I think that it'll require that kernel work because according to the Maple Programming Guide, section 9.6 "Overloading Built-in Routines," select, etc., aren't on that short list of overloadables. This is surprising because I'd expect to be able to use select on any container object.

To get the plot in the lower hemisphere also, as well as the plot of the sphere itself, you can use

f:= (r,phi)-> sqrt(25-r^2):
P1:= plot3d(
     [[r*cos(phi), r*sin(phi), f(r,phi)], [r*cos(phi), r*sin(phi), -f(r,phi)]],
     r= 3..4, phi= 0..2*Pi, labels= ['x','y','z'],
     color= black, style= patchnogrid, glossiness= 0
):
P2:= plot3d(5, phi= -Pi/2..Pi/2, theta= 0..2*Pi, coords= spherical):
plots:-display([P2,P1], axes= box, scaling= constrained);

@farazhedayati Unfortunately Kitonum's procedure relies on the variables in the expression being x and y. Your variables are x and a. So, call it like this:

ContoursWithLabels(subs(a= y, PP), -3..3, -3..3);

Of course, you may want to change the ranges.

I'll post a correction to the procedure in a little while, if someone else doesn't beat me to it.

@a_simsim Get rid of the line return z. The fsolve itself should be the return value. The fsolve doesn't assign the value of z; it just returns a numeric value for z.

@acer To appreciate the difference in timing, you'd need to do a bunch of short messages. I didn't do it, but it's obvious that there'd be some savings. Every little bit helps.

@acer Sure it can be done that way:

BySubs:= module()
local
     b, k,
     All:= seq(Bits:-Split(k, bits= 8), k= 0..254, 2),
     Subs:= {seq(b= [b[1..2][], 1 -~ b[3..7][], b[8]], b= All)},
     ModuleApply:= block-> subs(Subs, block)
;
     All:= 'All'  #garbage collection
end module: