r:=25:

Sol:=dsolve({diff(x(t),t)=-10*x(t)+10*y(t), diff(y(t),t)=r*x(t)-y(t)-x(t)*z(t), diff(z(t),t)=(-8/3)*z(t)+x(t)*y(t), x(0)=-1, y(0)=-1, z(0)=1}, numeric):

plots[odeplot](Sol,[x(t), y(t), z(t)], 0..30, axes=normal, numpoints=10000);

It turns out a very interesting picture:

See  plots[matrixplot]  command.

f := (x, y) -> ((2*x^2+y^2*(x^2*(-2+sqrt(1/x^4+4/(x^2*y^2)+4/y^4+4))-1))/(4*x^2-2*y^2+4))^(1/2) ;

g := unapply(diff(f(x, y), x), x,y);

plot([g(.1, y), g(1, y), g(3, y)], y = 0 .. 10);

A:=Matrix([[2,2,8,5], [6,3,4,9], [5,5,7,4], [2,1,3,2]]);

interface(rtablesize=100):

B:=LinearAlgebra[DiagonalMatrix]([A, -A, A]);

Instead of L in fsolve command after comma write L=0..infinity

You have a rather complicated transcendental equation. Solve command does not solve such equations. Replace solve command by fsolve command.

A:=plot(x, x=0..1):

B:=plot(x^2, x=0..1):

C:=plot(x^5, x=0..1):

plots[display](array([A, B, C]), scaling=constrained, thickness=2);

The function

x->2*a/Pi*arcsin(sin(2*Pi*x/b))

gives a triangular signal between -a and a with the period b .

An example:

a:=1: b:=4:

plot(2*a/Pi*arcsin(sin(2*Pi*x/b)), x=0..8, thickness=2, scaling=constrained);

Dr:=[[1,5],[2,6],[3,4],[4,5],[5,7],[6,8],[7,4],[8,6],[9,5],[10,7]]:

Y:=proc(i) Dr[i,2] end; ty:=proc(i) Dr[i,1] end;

P[X](omega)=(1/2)*{(Sum('Y'(j)*cos(omega*('ty'(j)-1/(2*omega*tan((Sum(sin(2*omega*'ty'(j)),j=1..N[0]))/(Sum(cos(2*omega*'ty'(j)),j=1..N[0]))))))^2,j=1..N[0]))/(Sum(cos(omega*('ty'(j)-1/(2*omega*tan((Sum(sin(2*omega*'ty'(j)),j = 1..N[0]))/(Sum(cos(2*omega*'ty'(j)),j=1..N[0]))))))^2,j= 1..N[0]))+(Sum('Y'(j)*sin(omega*('ty'(j)-1/(2*omega*tan((Sum(sin(2*omega*'ty'(j)),j=1..N[0]))/(Sum(cos(2*omega*'ty'(j)),j=1..N[0]))))))^2,j=1..N[0]))/(Sum(sin(omega*('ty'(j)-1/(2*omega*tan((Sum(sin(2*omega*'ty'(j)),j=1..N[0]))/(Sum(cos(2*omega*'ty'(j)),j =1 .. N[0]))))))^2,j=1..N[0]))};

Walk1 := proc(n)

local pick, i,j,edge,step,L,a, A, B, C, E;   

pick := rand(0..1);    (i,j) := (0,0);  L:=[[0,0]];  edge := n/2;   

for step do a:=pick();       

if a=0 then i := i + 2*pick()-1; L:=[op(L), [i,j]]; fi;       

if a=1 then j := j + 2*pick()-1; L:=[op(L),[i,j]]; fi;       

if edge < abs(i) or edge < abs(j) then  print(step); break;       

end if; end do;

print(L);

A:=plot([seq([t, i, t=-n/2..n/2],i=-n/2..n/2), seq([i, t, t=-n/2..n/2],i=-n/2..n/2)], color=black);

B:=seq(plottools[disk](L[i],0.1, color=red), i=1..nops(L));

E:=seq(plottools[curve]([seq(L[k],k=1..i)], color=blue, thickness=5), i=1..nops(L));

C:=seq(plots[display](op([A, B[i], E[i]]))$10, i=1..nops(L));

print(plots[display](seq(C[i],i=1..10*nops(L)),insequence=true,scaling=constrained,view=[-n/2-1..n/2+1, -n/2-1..n/2+1]));

end proc:

The procedure returns the number of steps, a list of nodes traversed, and the animation of the random walk.

An example:

Walk1(4);

It is quite simply:

L1:=seq(2*Pi*i/20,i=0..20):

L:=op(evalf([L1]));

K:=op(evalf([seq(-5*sin(L1[i]),i=1..21)]));

For example as follows:

F:=Student[VectorCalculus][Tangent](t -> <t, t^2, t^3>, t = 2):

sort(Student[VectorCalculus][DotProduct]( <F(1)[1]-F(0)[1], F(1)[2]-F(0)[2], F(1)[3]-F(0)[3]>, <x-F(0)[1], y-F(0)[2], z-F(0)[3]> ))=0;

    x + 4y + 12z - 114 = 0

You have a complicated non-linear system of 4 equations with 4 unknowns and three parameters. Maple is usually not solve these systems symbolically. This system can be solved numerically, but it must be given the values ​​of all parameters, such as:

restart;

eq1:=A2*t2*exp(zt*t2*om)*sin(t2*wd)+A3*t3*sin(t3*wd)*exp(zt*t3*om) = 0;

eq2:=A2*t2*exp(zt*t2*om)*cos(t2*wd)+A3*t3*cos(t3*wd)*exp(zt*t3*om) = 0;

eq3:=              A2*exp(zt*t2*om)*sin(t2*wd)+A3*sin(t3*wd)*exp(zt*t3*om) = 0;

eq4:=1-A2-A3+A2*exp(zt*t2*om)*cos(t2*wd)+A3*cos(t3*wd)*exp(zt*t3*om) = 0;

zt:=1/2: wd:=2: om:=3:

s1:=fsolve({eq1,eq2,eq3,eq4},{t2=0..infinity, t3=0..infinity, A2=0..infinity,A3=0..infinity});

Write so:

A:= Matrix(2,2,[-5,2,-3,4]):

B[1] := [1, 1];

A[op(B[1])];

plot3d(x^2+y^2, x = -1 .. 1, y = -sqrt(1-x^2) .. sqrt(1-x^2), style=surfacecontour, contours = 4);