Re: Crossing of 3D functions

*To*: mathgroup at smc.vnet.net*Subject*: [mg57203] Re: Crossing of 3D functions*From*: Maxim <ab_def at prontomail.com>*Date*: Fri, 20 May 2005 04:44:21 -0400 (EDT)*References*: <d5sks6$o2t$1@smc.vnet.net> <d69uuf$94p$1@smc.vnet.net> <d6hfac$d8l$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

On Thu, 19 May 2005 07:30:20 +0000 (UTC), Maxim <ab_def at prontomail.com> wrote: > On Mon, 16 May 2005 11:07:59 +0000 (UTC), Paawel <fonfastik at interia.pl> > wrote: > >> Hello again >> >> The point of this question is to obtain the function of crossing of >> these supersurfaces. Once I have this fcrossing function I want to >> find a local minimum (1<r<2, 90<f<180). So I want to find a local >> minimum, and plot this function. >> >> Regards >> > > I assume that you want to minimize another function (func below) along > the > curve F1 == F2. Then you can either solve for f explicitly and use > FindMinimum or search for a constrained minimum with NMinimize: > > In[1]:= > {{r1, r2}, {f1, f2}} = {{1, 2}, {90, 180}}; > > F1[r_, f_] = 3.92 + 3.758/E^(6.152*(-1.184 + r)) - (6.288336759430112*(1 > - > 0.00008280000000000001*(180 - f)^2))/E^(3.076*(-1.184 + r)); > > F2[r_, f_] = 0.479 + 0.479/E^(9.986*(-1.375 + r)) - (0.958*(1 - > 0.00107*(133 - f)^2))/E^(4.993*(-1.375 + r)); > > func[r_, f_] = 50000*(r - 1.2)^2 + (f - 135)^2; > > Lsol = Solve[F1[r, f] == F2[r, f], f]; > > FindMinimum[func[r, f] /. Lsol[[2]], {r, (r1 + r2)/2, r1, r2}] > > Out[6]= > {176.96819, {r -> 1.1620375}} > > In[7]:= > NMinimize[{func[r, f], F1[r, f] == F2[r, f]}, > {{r, r1, r2}, {f, f1, f2}}] > > Out[7]= > {176.96819, {f -> 145.24258, r -> 1.1620374}} > > Also you can use FindRoot, which can handle equations that don't have > closed-form solutions and is faster than NMinimize: > > In[8]:= > R[f_?NumericQ] := > r /. FindRoot[F1[r, f] == F2[r, f], {r, (r1 + r2)/2, r1, r2}] > > Plot[func[R[f], f], {f, f1, f2}] > > FindMinimum[func[R[f], f], {f, (f1 + f2)/2, f1, f2}] > > Out[10]= > {176.96819, {f -> 145.24259}} > > This works because there is only one root for each value of f. > > Maxim Rytin > m.r at inbox.ru > One additional comment: NMinimize[foo, {{x, x1, x2}}] is not quite the same as NMinimize[{foo, x1 <= x <= x2}, x] in that in the first case x1 and x2 just determine the starting region and do not act as constrains. Maxim Rytin m.r at inbox.ru