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The principle is simple: if diff(f(x,y), x) = 0 then f(x,y) is constant with respect to x, which means that f(x,y) depends only on y, i.e. f(x,y) = g(y).

If you want to see this in Maple:

pdsolve( diff(f(x,y), x) = 0, f(x,y) );

         f(x, y) = _F1(y)

Actually, the fact is true only locally: in a non-convex domain, f(x,y) could depend on x (!). Maths is difficult...

So, you want a partial fraction form. Unfortunately Maple does it only for rational expressions and a substitution is needed.

S:=sqrt(N__s):
subs(x=S, convert(subs(S=x,S^2=x^2, ex1), parfrac));

 

 

 

 

The standard plot is over a rectangle (i.e. the xy projection is a rectangle).

plot3d(4*x^2+9*y^2, x=-5..5, y=-4..4);

 

You probably want an ellipse, obtained by cutting the paraboloid by an horizontal plane.

solve(4*x^2+9*y^2<=100,[x,y])

[[x <= 5, -5 <= x, y <= (2/3)*(-x^2+25)^(1/2), -(2/3)*(-x^2+25)^(1/2) <= y]]

(1)

plot3d(4*x^2+9*y^2, x=-5..5, y=-(2*sqrt(-x^2 + 25))/3..(2*sqrt(-x^2 + 25))/3)

 

 

A simpler method is to use the view option

plot3d(4*x^2+9*y^2, x=-5..5, y=-5..5, view=0..100);

 

The loci are two circles centered at B and C.

(The geometry package cannot be used generally for such problems; they must be solved directly.)


 

restart;

A triangle ABC with fixed B and C vertices is considered in the plane, A being variable so that AB+AC remains constant and equal to a given length L.
We call P, T, T1  the points of contact of the excircle in the angle B with the sides BC, AB and AC respectively.
Show that P is fixed and is a vertex of the ellipse described by point A.

What are the locus of T and T1? How to animate the drawing when A move ? Thank you.

 

The equation of the ellipse is x^2/p^2 + y^2/q&2 = 1 (0<q<p);   f is the focal length
a,b,c are the sides of the triangle ABC.

 

c:=p+u*f/p:
b:=p-u*f/p:
a:=2*f:

A:=[u,v]:
B:=[-f,0]:
C:=[f,0]:

PC:=(a+b+c)/2-a:

PO=f+PC;

PO = p

(1)

# ==> P is a vertex

P:=[p,0]:

PB:=p+f:

T:=B+(A-B) *~ (PB/c):

[x,y]-simplify(T):

elim:=eliminate([%[], u^2/p^2+v^2/q^2-1, f^2=p^2-q^2], [u,v]);

[{u = p*(f^2-p*x)/(f^2-f*p+f*x-p^2), v = y*(f^2-p^2)/(f^2-f*p+f*x-p^2)}, {f^2-p^2+q^2, f^2*y^2+2*f*p*q^2-2*f*q^2*x+p^2*q^2-p^2*y^2-q^2*x^2}]

(2)

eqT:=simplify(elim[2][2], [f^2=p^2-q^2]);

q^2*(2*f*p-2*f*x+p^2-x^2-y^2)

(3)

Student:-Precalculus:-CompleteSquare(eqT,{x,y});

-q^2*y^2-q^2*(x+f)^2+q^2*(2*f*p+p^2)+f^2*q^2

(4)

 ==>  The locus of T is a circle having the center at B.

 

T1:=C+(A-C) *~ (PC/b):

[x,y]-simplify(T1):

elim1:=eliminate([%[], u^2/p^2+v^2/q^2-1, f^2=p^2-q^2], [u,v]);

[{u = p*(f^2-p*x)/(f^2+f*p-f*x-p^2), v = y*(f^2-p^2)/(f^2+f*p-f*x-p^2)}, {f^2-p^2+q^2, f^2*y^2-2*f*p*q^2+2*f*q^2*x+p^2*q^2-p^2*y^2-q^2*x^2}]

(5)

eqT1:=simplify(elim1[2][2], [f^2=p^2-q^2]);

-q^2*(2*f*p-2*f*x-p^2+x^2+y^2)

(6)

Student:-Precalculus:-CompleteSquare(eqT1,{x,y});

-q^2*y^2-q^2*(x-f)^2-q^2*(2*f*p-p^2)+f^2*q^2

(7)

==> The locus of T1 is a circle having the center at C.

##### Plot ####

p:=5;q:=3;f:=4;
v:=solve(u^2/p^2+vv^2/q^2-1,vv)[1]:

5

 

3

 

4

(8)

pp:=proc(u_)
global u; uses plots;
u:=u_;
display(
  plot([A,B,C,A]),
  plot([T,T1,P], style=point, color=blue, symbolsize=20),
  implicitplot(x^2/p^2+y^2/q^2-1, x=-10..10, y=-10..10, color=pink),
  implicitplot(eqT,  x=-20..10, y=-1..10, color=green),
  implicitplot(eqT1, x=-20..10, y=-1..10, color=blue),
  scaling=constrained, view=[-15..6, 0..10]
) end proc:

Explore(pp(u_), u_=-p*1.0 .. p)

 

 


 

 

 

Download geom-loci-ellipse-circles.mw

You have some inaccurate formulations, starting with the definition of a defective eigenvalue (see the cited wiki article).
Anyway, the defective eigenvectors can be obtained directly from the Jordan form of the matrix (they are the columns of Q):

A :=Matrix([[1,-2],[2,5]]):
J,Q:=LinearAlgebra:-JordanForm(A, output=['J','Q']);

         

v1:=Q[..,1];
v2:=Q[..,2];


(A-3).v1,  (A-3).v2,  (A-3)^2 . v2;

restart;
HB1 := HeunB(5/2, -5^(1/4)*(2*M - R)/sqrt((2*M - R)*(4*M - R)), (17*sqrt(5))/10, (3*5^(3/4)*(2*M - R))/(2*sqrt((2*M - R)*(4*M - R))), 5^(1/4)*sqrt((2*M - R)*(4*M - R))*r^2/(2*(2*M - R)*R^2)):
A:=op(5, HB1):
answer := series(eval(series(eval(HB1, A=z), z), z=A), r);

Your difficulties are related to maths rather than Maple.

C is a curve represented geometrically as a segment with endpoints (1,0), (2,2).
So, you have to find a parametric representation for C [for a computation by hand].

You are asked for the path integral (not line integral -- Maple terminology).  The computation is straightforward using VectorCalculus.

VectorCalculus:-PathInt( 2*x + y^2, [x,y] = Line(<1,0>, <2,2>) );

        (13*sqrt(5))/3

You do not need Maple for this. Denote by {0, 1, a} the set, where 0 and 1 are the min and max elements.
Then there is a unique lattice (0 < a < 1), whose graph is

    0
    |
    a
    |
    1

Of course, if the max and min elements are not distinguished, there are 3! = 6 (isomorphic) lattices. 

radsimp does a symbolic simplification, which means (among other things) that Maple is allowed to choose for sqrt(x^2) any of the values x or -x. Actually this command is deprecated, see the help page.

radsimp( b-a - sqrt((b-a)^2) ); 
    2 b - 2 a

Note that it is not correct to speak about a right or wrong result here beacuse the sign of b-a is not known.
Thr preferred approach is to use assuming:

simplify( b-a - sqrt((b-a)^2) ) assuming b>a;
                               0
simplify( b-a - sqrt((b-a)^2) ) assuming b<a;
                           2 b - 2 a
 

 

 

h := x -> 2/f(x):
eval( D(h)(-1),  [f(-1)=4, D(f)(-1)=2] );

                    - 1/4

 

# Denote  u = tan(B/2),  v=tan(C/2)

formulae:= [sin(B) = 2*u/(1+u^2), sin(C) = 2*v/(1+v^2),
            cos(B) = (1-u^2)/(1+u^2), cos(C) = (1-v^2)/(1+v^2)];

[sin(B) = 2*u/(u^2+1), sin(C) = 2*v/(v^2+1), cos(B) = (-u^2+1)/(u^2+1), cos(C) = (-v^2+1)/(v^2+1)]

(1)

# b + c = L (constant)

L = a*(sin(B)+sin(C))/sin(B+C);  # law of sines

L = a*(sin(B)+sin(C))/sin(B+C)

(2)

expand(%)

L = a*sin(B)/(sin(B)*cos(C)+cos(B)*sin(C))+a*sin(C)/(sin(B)*cos(C)+cos(B)*sin(C))

(3)

eval(%, formulae);

L = 2*a*u/((2*u*(-v^2+1)/((u^2+1)*(v^2+1))+2*(-u^2+1)*v/((u^2+1)*(v^2+1)))*(u^2+1))+2*a*v/((2*u*(-v^2+1)/((u^2+1)*(v^2+1))+2*(-u^2+1)*v/((u^2+1)*(v^2+1)))*(v^2+1))

(4)

simplify(%);

L = -a*(u*v+1)/(u*v-1)

(5)

isolate(%,u*v);

v*u = (L-a)/(L+a)

(6)

# So, u*v = const

 

 

If you just need the gradient, it can be obtained directly.

restart;
f:= (x,y,z)->exp(x*y^3*z^2):
Gf:=[D[1](f), D[2](f), D[3](f)]:  # list
Gf(1, -1, 2);

         [-4*exp(-4), 12*exp(-4), -4*exp(-4)]

Vector(%);   # ==> Vector 

Your problem is equivalent to the generalized eigenvalue problem A = lambda * A^+
(A^+ is the transpose)
This problem is solved very efficiently by Maple. Try:

with(LinearAlgebra):
A:=<2,3;4,5>;
Eigenvectors((A^(-1))^+ . A);
Eigenvectors(A , A^+);
A:=RandomMatrix(18, generator=-5..5, datatype=float[8]);
Eigenvectors(A , A^+);
Digits:=200;
A:=RandomMatrix(18, generator=-5..5, datatype=float);
Eigenvectors(A , A^+);

 

IsEulerPath:=proc(G::list(set), excludecirc:=true)
  local e,  a, a1, ok;
  for a1 in G[1] do  # a1 = first vertex
    ok:=true;
    a:=(G[1] minus {a1})[];
    for e in G[2..] do
      if   a = e[1] then a := e[2]
      elif a = e[2] then a := e[1]
      else ok:=false; break  fi;
    od;
    if ok then break fi 
  od;  
  if not ok then return false fi;
  if excludecirc and (a = a1) then return false fi;
  true
end proc:

G:=[{1,2}, {1,3}, {1,4}, {3,4}$2, {2,4}]:
select(IsEulerPath, combinat:-permute(G)); nops(%)
      [[{1, 2}, {2, 4}, {1, 4}, {1, 3}, {3, 4}, {3, 4}],
        [{1, 2}, {2, 4}, {3, 4}, {1, 3}, {1, 4}, {3, 4}],
        [{1, 2}, {2, 4}, {3, 4}, {3, 4}, {1, 4}, {1, 3}],
        [{1, 3}, {1, 2}, {2, 4}, {3, 4}, {3, 4}, {1, 4}],
        [{1, 3}, {1, 4}, {3, 4}, {3, 4}, {2, 4}, {1, 2}],
        [{1, 3}, {3, 4}, {1, 4}, {1, 2}, {2, 4}, {3, 4}],
        [{1, 3}, {3, 4}, {2, 4}, {1, 2}, {1, 4}, {3, 4}],
        [{1, 4}, {3, 4}, {1, 3}, {1, 2}, {2, 4}, {3, 4}],
        [{1, 4}, {3, 4}, {3, 4}, {2, 4}, {1, 2}, {1, 3}],
        [{1, 4}, {2, 4}, {1, 2}, {1, 3}, {3, 4}, {3, 4}],
        [{3, 4}, {1, 4}, {1, 2}, {2, 4}, {3, 4}, {1, 3}],
        [{3, 4}, {1, 4}, {1, 3}, {3, 4}, {2, 4}, {1, 2}],
        [{3, 4}, {3, 4}, {1, 3}, {1, 2}, {2, 4}, {1, 4}],
        [{3, 4}, {3, 4}, {1, 3}, {1, 4}, {2, 4}, {1, 2}],
        [{3, 4}, {2, 4}, {1, 2}, {1, 3}, {3, 4}, {1, 4}],
        [{3, 4}, {2, 4}, {1, 2}, {1, 4}, {3, 4}, {1, 3}]]

                               16
 
Edit: a bug corrected

The equation is solvable [in terms of radicals] only for some values of k.

For example, for k=2,
galois(6*w^5 - 15*w^4 + 10*w^3 - 2, w)
returns the group "5T5"  which is not solvable.

But for k=1, solve finds the roots.

In https://en.wikipedia.org/wiki/Quintic_function  there are other methods to check the solvability.

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