Re: Getting only 1 of 3 curves of intersection
- To: mathgroup at smc.vnet.net
- Subject: [mg126312] Re: Getting only 1 of 3 curves of intersection
- From: Bob Hanlon <hanlonr357 at gmail.com>
- Date: Mon, 30 Apr 2012 04:42:40 -0400 (EDT)
- Delivered-to: l-mathgroup@mail-archive0.wolfram.com
- References: <201204290609.CAA10817@smc.vnet.net>
g1 = Plot3D[Sin[x*y], {x, 0, Pi}, {y, 0, Pi},
PlotStyle -> None,
MeshStyle -> Red,
Axes -> True,
AxesLabel -> {"x", "y", "z"}];
g2 = Plot3D[Cos[x*y], {x, 0, Pi}, {y, 0, Pi},
Mesh -> None,
PlotStyle -> {Cyan, Opacity[.8]}];
rSol = Reduce[{z - Sin[x y] == 0, z - Cos[x y] == 0,
0 <= x <= \[Pi], 0 <= y <= \[Pi]}, {y, z}]
(((2*Pi - 2*ArcTan[1 + Sqrt[2]])/Pi <= x <=
Pi && y == (2*Pi - 2*ArcTan[1 + Sqrt[2]])/
x) || (-((2*ArcTan[1 - Sqrt[2]])/Pi) <=
x <= Pi && y == -((2*ArcTan[1 - Sqrt[2]])/
x)) || ((2*Pi - 2*ArcTan[1 - Sqrt[2]])/
Pi <= x <= Pi &&
y == (2*Pi - 2*ArcTan[1 - Sqrt[2]])/x)) &&
z == Sin[x*y]
rSol // FullSimplify
x <= Pi && (((9*Pi)/x == 4*y && 4*x >= 9) ||
(4*x*y == Pi && 4*x >= 1) ||
((5*Pi)/x == 4*y && 4*x >= 5)) &&
z == Sin[x*y]
param = Cases[rSol,
(y == f_) :> {x, f, Sin[x*f]}, Infinity] //
FullSimplify
{{x, (5*Pi)/(4*x), -(1/Sqrt[2])},
{x, Pi/(4*x), 1/Sqrt[2]}, {x, (9*Pi)/(4*x),
1/Sqrt[2]}}
g3 = ParametricPlot3D[param, {x, 0, Pi},
PlotStyle -> {{Magenta, Thickness[.007]}}];
Show[g1, g2, g3,
Background -> LightYellow,
ImageSize -> 500]
Bob Hanlon
On Sun, Apr 29, 2012 at 2:09 AM, Bill <WDWNORWALK at aol.com> wrote:
> Hi:
>
> Consider the following plot:
>
> g1=Plot3D[Sin[x*y],{x,0,Pi},{y,0,Pi},PlotStyle->None,MeshStyle->Red,Axes->True,AxesLabel->{"x","y","z"}];
> g2=Plot3D[Cos[x*y],{x,0,Pi},{y,0,Pi},Mesh->None,PlotStyle->{Cyan,Opacity[.8]}];
> nsSol=NSolve[{z-Sin[x*y],z-Cos[x*y]},{y,z}];//Quiet
> g3=ParametricPlot3D[{x,y,z}/.nsSol[[2]],{x,0,Pi},PlotStyle->{Magenta,Thickness[.007]}];
> Show[g1,g2,g3,Background->LightYellow,ImageSize->500]
>
> Using the above Mathematica 8.0.4 code, I can plot one curve of intersection shown in magenta.
> Within the plotted area, I can see 2 more places where intersection curves should be.
> I've tried using Reduce in place of NSolve, but can't get it to work.
>
> Question: If this can be done, can someone please give me the code?
>
>
> Thanks,
>
> Bill
- References:
- Getting only 1 of 3 curves of intersection
- From: Bill <WDWNORWALK@aol.com>
- Getting only 1 of 3 curves of intersection