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