Kitonum

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17 years, 136 days

MaplePrimes Activity


These are replies submitted by Kitonum

@acer  You wrote  "And how about requesting their spacing evenly by arclength rather than by value of the independent parameter? "

See the post  https://www.mapleprimes.com/posts/213452-Uniform-Point-Plot  with a procedure for this. 

@Carl Love  Sorry, but it seems to me that for a beginner, your method will be difficult to understand.

Another way for the same:

plots:-display(plot(sin(x), x= -Pi..Pi), plot(sin(x),x= -Pi..Pi, style=point, symbol= box, numpoints= 16, adaptive= false));


We can write very short for more symbols:

plot(sin(x), x= -Pi..Pi, style=pointline, symbol= box, numpoints= 30, adaptive= false);

                  

@ecterrab  OP wants to plot the equation  |z+1|*|z-1|=1  (he mistakenly calls it a function), not independent graphs of the left and right sides of this equation.

@abdulganiy  Use the plots:-textplot  command.

@JAMET You have to do it yourself. It's quite easy if you first study the solutions I've shown.

@AHSAN  See my updated answer.

@janhardo I think it is useful to know the physical meaning of this curvilinear integral: if  2*x+y^2  is the linear density at the point  (x,y) of the segment, then the value of the integral is the mass of this segment. 

@Preben Alsholm  Thank you for the improvement.

@ActiveUser  See the corrected answer.

You have already built it. cos(Pi/4)  is a constant. And what is the question?

@Scot Gould You're right. Using  add  instead of  sum  avoids many problems. 

@sunit  You are not required to apply the command to every equation. Build a system and apply the command to this system as in the example:

convert({1.0*x+0.3*y=1,2*x-1.0*y=3}, rational);

                              

@nguyenhuyenag 

This is also written in the help.

Solution for your example:

restart;
bl:=sqrt(2)-1>0:
evalb(evalf(bl));

                               true

@Carl Love  Thanks. The  seq  is my favorite command in Maple. Owners of older versions of Maple (< Maple 13 or 14) , who do not know about  zip ,  can simply write this:

A:=[1,2,3]:
B:=[7,8,9]:
[seq(f(A[i],B[i]),i=1..nops(A))];


The  seq  command can completely replace map, zip, and ~  (which of course make the code a little shorter).

Here's another animation in which the ellipse of a given shape (defined by the parameter  k ) rotates around a given triangle:


 

restart;
Eq:=(x-x0)^2/a^2+(y-y0)^2/b^2=1:
b:=k*a: k:=2/3:
Eq1:=subs([x=x*cos(alpha)+y*sin(alpha),y=-x*sin(alpha)+y*cos(alpha)],Eq);
T:=[[0,0],[2,3],[6,0]]:
R:=[seq([x,y]=~t,t=T)];
Sys:=map(p->eval(Eq1,p),R);
Sol:=solve(Sys,{a,x0,y0}, explicit)[1];
eval(Sol,alpha=Pi/4);
F:=t->plots:-implicitplot(eval(Eq1,[eval(Sol,alpha=t)[],alpha=t]), x=-3..10, y=-10..10, gridrefine=3);
plots:-animate(F,[t], t=0..Pi, frames=90, background=plots:-display(plottools:-polygon([[0,0],[6,0],[2,3]],color="LightBlue", thickness=2)), scaling=constrained);

(x*cos(alpha)+y*sin(alpha)-x0)^2/a^2+(9/4)*(-x*sin(alpha)+y*cos(alpha)-y0)^2/a^2 = 1

 

[[x = 0, y = 0], [x = 2, y = 3], [x = 6, y = 0]]

 

[x0^2/a^2+(9/4)*y0^2/a^2 = 1, (2*cos(alpha)+3*sin(alpha)-x0)^2/a^2+(9/4)*(-2*sin(alpha)+3*cos(alpha)-y0)^2/a^2 = 1, (6*cos(alpha)-x0)^2/a^2+(9/4)*(-6*sin(alpha)-y0)^2/a^2 = 1]

 

{a = (1/72)*(31625*sin(alpha)^6-25500*cos(alpha)*sin(alpha)^5-11850*sin(alpha)^4-5700*cos(alpha)*sin(alpha)^3+40605*sin(alpha)^2+11760*cos(alpha)*sin(alpha)+56260)^(1/2), x0 = (5/4)*sin(alpha)^2*cos(alpha)+3*cos(alpha)-(85/24)*sin(alpha)^3+(49/24)*sin(alpha), y0 = -(85/54)*sin(alpha)^2*cos(alpha)+(49/54)*cos(alpha)-(5/9)*sin(alpha)^3-(22/9)*sin(alpha)}

 

{a = (1/576)*630565^(1/2)*8^(1/2), x0 = (187/96)*2^(1/2), y0 = -(281/216)*2^(1/2)}

 

proc (t) options operator, arrow; plots:-implicitplot(eval(Eq1, [(eval(Sol, alpha = t))[], alpha = t]), x = -3 .. 10, y = -10 .. 10, gridrefine = 3) end proc

 

 

 


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