Maple 15 Questions and Posts

These are Posts and Questions associated with the product, Maple 15

Spiral (equidistant) around the curve.  In this case, a spiral around the spiral.
So without any sense. 
If we re-save the animation with the program Easy GIF Animator, its size is reduced by about 10 times, and sometimes much more.

Imitation coloring both sides of the polygon in 3d.  We  build a new polygon in parallel with our polygon on a very short distance t. (We need any three points on the polygon plane, do not lie on a straight line.) This place in the program is highlighted in blue.

Paint the polygons are in different colors.

In a post of April 15, 2013 by Kitonum, the procedure named Picture accepts a list of polygon segments, creates a plot of these as a 2D polygon's boundaries and fills the polygon with a color.

The code below attempts to modify Picture to produce a 3D filled polygon in a plane parallel to the xy plane.

When invoked by the code below the procedure, the filling color conforms to the straight line boundaries but overflows the curved, parabolic boundary. How can this be corrected?

Picture:=proc(L, C, N::posint:=100, Boundary::list:=[linestyle=1])

 local i, var, var1, var2,e, e1, e2,e3, P, h ;

 global Q,Border;

 for i to nops(L) do    

#` set P`[i] = list of points for each segment.    

#` for a segment defined as a list of points, P[i] = the segment's definition`

#` for a curve definition, approximate it with a list of [x,y] points of its function evaluated at N even intervals in its

# range`  

  if type(L[i],listlist(algebraic))  then P[i]:=op(L[i]);   else  

  #` for curve def'n, set var = def'n and h= `(variable range)/(2)

  var:=lhs(L[i,2]);  var1:=lhs(rhs(L[i,2]));  var2:= rhs(rhs(L[i,2])); h:=(var2-var1)/(N);

  #` for function def'n, set e=function`

 if type(L[i,1], algebraic) then  e:=L[i,1];

  #` for polar function r=f(t) create N values of the [cos*r,sin*r] i.e. the equivalent [x,y] values for r valued at N even

  # divisions of its range`  

 if nops(L[i])=3 then P[i]:=seq(subs(var=var1+h*i,[e*cos(var), e*sin(var)]), i=0..N);  else

    #` for non-polar function y=f(x) create N values of [x,y] for x values at N even divisions of its range`  

 P[i]:=seq([var1+h*i, subs(var=var1+h*i,e)], i=0..N)  fi;  else

 #` for parametric function [f`(t),g(t)] create N values of [f(t),g(t)] for t values at N even divisions of its range.

     e1:=L[i,1,1];  e2:=L[i,1,2];

#` P`[i]:=seq(subs(var=var1+i*h,[e1, e2]), i=0..N):

 P[i]:=seq([subs(var=var1+i*h,e1), subs(var=var1+i*h,e2),0], i=0..N) fi; fi; od;  #`  MODIFIED FOR 3 D `[f(t), g(t), 0] 

  Q:=[seq(P[i], i=1..nops(L))];

 Border:=plottools[curve]([op(Q), Q[1]],  op(Boundary));

     #` the shaded figure is a polygon whose vertices are Q, whose interior color is C`  

 #` return a list of the polygon and its border`

   [plottools[polygon](Q, C),  Border];

 end proc: 

L := [[[0, 0, 0], [0, 1, 0]], [[x, x^2+1, 0], x = 0 .. 2], [[2, 5, 0], [2, 2, 0]], [[x, x, 0], x = 2 .. 0]]:

plots[display](Picture(L, color = yellow), axes = normal, scaling = constrained)

I seperate the variables in Real and Imigneray parts,  as follows:

------------------------- Defining the nature of the variables used ----------------------

t0:=0.0:tN:=30.0: M1:=5000;:th:=evalf((tN-t0)/M1):
ini1:=u(0)=Re(y(0)), v(0)=Im(z(0)),w(0)=x(0);
u(0) = 1, v(0) = 0, w(0) = -1
dsys1 :=diff(w(t),t)=-(N1+M*cos(2*omega*t))*w(t)-1+2*u(t)*cos(2*omega*t)+2*v(t)*sin(2*omega*t), diff(u(t),t)=-N1*u(t)+Delta*v(t)-2*M+(2*M*u(t)-N1-w(t))*cos(2*omega*t)-2*M*v(t)*sin(2*omega*t), diff(v(t),t)=-N1*v(t)-Delta*u(t)-2*M+(2*M*u(t)-N1-w(t))*sin(2*omega*t)+2*M*v(t)*cos(2*omega*t):
dsol1 :=dsolve({dsys1,ini1},var,numeric, output=listprocedure, abserr=1e-9, relerr=1e-8,range=0..1,maxfun=5000):
Warning, cannot evaluate the solution further right of .46544244e-3, maxfun limit exceeded (see ?dsolve,maxfun for details)
t1:=array(0..M1,[]): u1:=array(0..M1,[]): v1:=array(0..M1,[]): w1:=array(0..M1,[]): pt1:=array(0..M1,[]):pt2:=array(0..M1,[]):pt3:=array(0..M1,[]): 
for i from 0 to M1 do t1[i]:=evalf(th*i):u1[i]:=evalf(dsolu(t1[i]));v1[i]:=evalf(dsolv(t1[i])):w1[i]:=evalf(dsolw(t1[i])):pt1[i]:=[t1[i],u1[i]]:pt2[i]:=[t1[i],v1[i]]:pt3[i]:=[t1[i],w1[i]]:od:
Error, (in dsolu) cannot evaluate the solution further right of 0.46544244e-3, maxfun limit exceeded (see ?dsolve,maxfun for details)

plot(mytab3,t=0..5,tickmarks=[6, 6],axes=boxed);

but I got an error

I have downloaded the zip file for CalcP7, unzipped it, and can access its commands in a worksheet after issuing the command with(CalcP7), but "No Matches Found" displays when entering the command ?CalcP7. The download included a file named "aplication" (one "p") of type HDB, but Maple15 can't seem to access its contents.

Are CalcP7's help pages displayable? If so, what is necessary to access them?

I have had no trouble downloading the user package DirectSearch and accessing both its commands and its help pages.

I tried to load my document containing some notes, but then I got the message "There were problems during the loading process, Your worksheet may become incomplete", and as the message said my worksheet were incomplete. Is there a way to restore the document? I have tried following this and added the line it suggested:

But it didn't work.

I have attached the file.

plots[implicitplot3d](max(-x+y+z, x-y+z, x+y-z) = 1.0, x = 0 .. 1, y = 0 .. 1, z = 0 .. 1);

The help page for max does not explain or show an example of max(sequence of expressions)= a constant.

Please let me know if this link correctly accesses my worksheet. If not, I will copy its contents into this question.

Which ODE in the worksheet, if any, provides the correct answer?


f := proc (x) local t; if not type(evalf(x), 'numeric') then ('procname')(x) else evalf(Int(exp(-(1/10)*t^2), t = 0 .. x)) end if end proc

solA := dsolve({diff(y(x), x) = y(x)+f(x), y(0) = 0}, numeric, known = f)


[x = 1., y(x) = HFloat(0.7081492947996167)]


f2 := evalf(Int(exp(-(1/10)*t^2), t = 0 .. 1)); f(1)





solB := dsolve({diff(y(x), x) = y(x)+f2, y(0) = 0}, numeric, output = listprocedure)


[x(1) = 1., (y(x))(1) = HFloat(1.6626837619970016)]


YinSolB := subs(solB, y(x))

YinSolBeval := solve(YinSolB(a) = .7081, a); solB(YinSolBeval)



[x(.5491485953) = .5491485953, (y(x))(.5491485953) = HFloat(0.7081000000284681)]





  I would need to get access to the values of euler angles defining the orientation of a 3D plot.

I generate the 3D plot with a given orientation, then the user is allowed to rotate it with mouse orbit. How can I get the values of the new euler angles to use them in further computations?

In the plot/structure I cannot find any data concerned with orientation...


Thank you.



Is there a help page which explains why braces provide the partial text evaluation in this code?

RopeLen := 30;RopeAddLen := .5;

plots[textplot]([1, 1, typeset("%1", (({(1/2)*RopeLen}+{(1/2)*RopeAddLen})^2-{(1/2)*RopeLen}^2)^(1/2))])

eq1 := z = y*log(x): eq2 := z = y+x*log(x):

DispIntersecting := implicitplot3d([eq1, eq2], x = 0 .. 10, y = -30 .. 30, z = -40 .. 40, color = [blue, green]):

solve({eq1, eq2}, [x, y, z]);


DispIntersection := spacecurve([x, y, z], x = 0.1e-2 .. 10, color = red, view = [0 .. 10, -30 .. 30, -40 .. 40]):

display(DispIntersecting, DispIntersection, axes = boxed, scaling = constrained);

Hello guys and gals!

I'm not strong enough with maple to get what the result I want.

It seems that it's because I'm asking for two lenths, and not a lenth and an angle, but I have no Idea how to tackle it diferently.

If you know a trick, please share it!


Here's an image:

And here's the maple file attached (I think)


Happy new year!

has anybody experienced issues with Maple connectiong to Solidworks?  There doen't seem to be much info ion trouble shooting a failed "OpenConnection" call.





If I have this equation

and I need to use different Z, it's like

how can I solve it

Many Thanks


Hello I am trying to plot this differential equation,  have not had any success can any one help me.

plot(q(t)= Ce^-1/2 + sin2t +cos4t)

not sure what i am doing wrong and i just used the plot command


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