I want to plot a curve in R^3 that is defined implicitly by the two constraints f_1(x_1,x_2,x_3)=0 and f_2(x_1,x_2,x_3)=0.
The corresponding problem for Mathematica has been discussed in
http://mathematica.stackexchange.com/questions/5968/
I applied one of the solutions given there to the present problem
f1[x_, y_, z_] := x^2 + y^2 + z^2;
f2[x_, y_, z_] := x + y + z;
ContourPlot3D[{f1[x, y, z] == 1, f2[x, y, z] == 1},
{x, -1, 1}, {y, -1, 1}, {z, -1, 1},
Contours -> {0},
ContourStyle -> Opacity[0],
Mesh -> None,
BoundaryStyle -> {1 -> None, 2 -> None, {1, 2} -> {{Green, Tube[.03]}}}
]
The result is attached. The trick lies in the option BoundaryStyle. I do not know whether an analog solution exists with maple.
Try this:
with(plots):
f_1:=(x_1,x_2,x_3)->x_1^2+x_2^2+x_3^2;
f_2:=(x_1,x_2,x_3)->x_1+x_2+x_3; implicitplot3d([f_1(x_1,x_2,x_3)=1,f_2(x_1,x_2,x_3)=1],x_1=-2..2,x_2=-2..2,x_3=-2..2,grid=[13,13,13],style=patchnogrid,orientation=[103,103]);
Demetris, thank you. So I guess the cutting set of the two surfaces is the curve. Is there also a way to plot only the curve?
I think you have to solve for the points of the spacecurve first, and then use spacecurve, e.g.
>with(plots):
f_1:=(x_1,x_2,x_3)->x_1^2+x_2^2+x_3^2;
f_2:=(x_1,x_2,x_3)->x_1+x_2+x_3;
p:=proc (t)
local x,y,z,a,v;
x:=t;
assign(fsolve(map(v->subs(a=x,v),{f_1(a,y,z)=1,f_2(a,y,z)=0.5}),{y,z}));
[t,y,z];
end;
p(0);p(-0.5);
thispoints := [seq(p(t/20.),t=-10..10)]:
spacecurve(thispoints);
The range of x in this example, for which you can solve the implicit equations, you might estimate using the implicitplot3d command as in the example of Demetris.
Best regards
Herbert
You can also try this, but you have to read about Groebner basis a lot!
with(Groebner);
[f_1(x_1,x_2,x_3)-1,f_2(x_1,x_2,x_3)-1];
Basis(%, plex(x_1,x_2,x_3));solve(%,[x_1,x_2,x_3]):sol:=allvalues(%):%[1]:s1:=%[1]:[subs(x_3=t,rhs(%[1])),subs(x_3=t,rhs(%[2])),t];rt1:=%:p1:=spacecurve(rt1,t=-1/3..1,numpoints=1000):sol:%[2]:s2:=%[1]:[subs(x_3=t,rhs(%[1])),subs(x_3=t,rhs(%[2])),t];rt2:=%:p2:=spacecurve(rt2,t=-1/3..1,numpoints=1000):
display([p1,p2],orientation=[103,103]);
Why did I put the range [-1/3,1]?
The result is attached.
The corresponding problem for Mathematica has been discussed in
http://mathematica.stackexchange.com/questions/5968/
I applied one of the solutions given there to the present problem
f1[x_, y_, z_] := x^2 + y^2 + z^2;
f2[x_, y_, z_] := x + y + z;
ContourPlot3D[{f1[x, y, z] == 1, f2[x, y, z] == 1},
{x, -1, 1}, {y, -1, 1}, {z, -1, 1},
Contours -> {0},
ContourStyle -> Opacity[0],
Mesh -> None,
BoundaryStyle -> {1 -> None, 2 -> None, {1, 2} -> {{Green, Tube[.03]}}}
]
The result is attached. The trick lies in the option BoundaryStyle. I do not know whether an analog solution exists with maple.
Another method to consider is to use numeric ODEs. Regard x,y,z as functions of an independent variable t. Use the equations fro dF/dt and dG/dt = 0 (since F and G are zero), and maybe add (dx/dt)^2 + (dy/dt)^2+(dz/dt)^2=1. Now there are three DEs in three dependent variables. Find a single point on the intersection curve to give initial conditions.
This method is useful in situations where algebraic approaches might fall short e.g. if F and G have transcendental functions that do not lend themselves to algebraic methods. It requires that one find points on all components of the intersection though, so it might not be straightforward to automate.
Daniel, thank you for your comments, but may I ask you a question about my own answer above?
The pdf-file produced is quite large. I believe this is due to the fact that unvisible portions of the plot could not be removed, since Mathematica allows the user to rotate the produced graphics arbitrarily. So I guess the unvisible parts of the plot are still part of the pdf file. Is that true?
If I am interested in on viewpoint only, how could I remove the hidden objects of the from the plot?
Thanks.
(I hope my answering a question raised in the responses is not against RG policy..)
Gunther, I am not certain but I believe there are invisible parts, as you suspect, from the (surface) contours with opacity of 0. Regardless, Mathematica graphics do tend to be large. What I usually do to shrink graphics sizes, e.g. for conversion to pdf, is to use Rasterize on the result. It often makes a huge difference. In fact I just did this yesterday in order to create a pdf of reasonable size so I could upload it to RG.
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