'Nother Inverse Function Question

*To*: mathgroup at smc.vnet.net*Subject*: [mg126607] 'Nother Inverse Function Question*From*: Bill Freed <billfreed at shaw.ca>*Date*: Fri, 25 May 2012 04:54:46 -0400 (EDT)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com

Thanks for the previous hints on using InverseFunction. Will be helpful for me. I am working on parameterizing regions bounded by 4 curves in the plane or 6 surfaces in 3D. Below is an example for the region bounded by x=2+1.2(2-y)^2, x=-3.2-1/3(y-1/3)^2 y=(-x/3)^3+x/2+2.5, y=(x/4)^3+1 Table[FindRoot[{x == s (2 + 1.2 (2 - y)^2) + (1 - s) (-3.2 - 1/3 (y - 1.3)^2), y == t (-(x/3)^3 + x/2 + 2.5) + (1 - t) ((x/4)^3 + 1)}, {x, 0}, {y, 2}], {s, 0, 1, .1}, {t, 0, 1, .1}]; x = ListInterpolation[x /. %, {{0, 1}, {0, 1}}]; y = ListInterpolation[y /. %%, {{0, 1}, {0, 1}}]; ParametricPlot3D[{x[s, t], y[s, t], 0}, {s, 0, 1}, {t, 0, 1}, ViewPoint -> {0, 0, +Infinity}, Boxed -> False, Axes -> None, PlotPoints -> {25, 11}] The problem with this homemade inverter is frequent error messages concerning accuracy and convergence and also problems in choosing the starting points, here {x,0}, {y,2}. Is there a way of using InverseFunction or other Mathematica command that is more robust? Thanks Bll Freed

**Follow-Ups**:**Re: 'Nother Inverse Function Question***From:*Ingolf Dahl <ingolf.dahl@telia.com>

**Re: 'Nother Inverse Function Question***From:*Bob Hanlon <hanlonr357@gmail.com>