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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
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