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Re: Is this a problem in mathematica?
*To*: mathgroup at smc.vnet.net
*Subject*: [mg74639] Re: Is this a problem in mathematica?
*From*: "Dana DeLouis" <dana.del at gmail.com>
*Date*: Thu, 29 Mar 2007 02:33:08 -0500 (EST)
> Minimize[minDist, y == 1 + x^2, {x, y}]
> but x also has another answer: +1/Sqrt[2].
Here's an alternative using Derivatives.
Plot[x^2 + 1, {x, -2, 2},
AspectRatio -> Automatic,
GridLines -> Automatic];
pt = {0, 2};
{x, 1 + x^2} - pt
{x, -1 + x^2}
Distance Equation: I used Simplify to remove Abs[ ]
d = Simplify[Norm[%], x >= 0]
Sqrt[1 - x^2 + x^4]
Take the Derivative...
D[d, x]
(-2*x + 4*x^3)/(2*Sqrt[1 - x^2 + x^4])
Solve[% == 0, x]
{{x -> 0}, {x -> -(1/Sqrt[2])}, {x -> 1/Sqrt[2]}}
There are 3 possible solutions. However, with x->0, the distance is not the
shortest:
d /. %
{1, Sqrt[3]/2, Sqrt[3]/2}
Therefore, the remaining 2 solutions would be the shortest. Ie..
{x -> -(1/Sqrt[2])}, {x -> 1/Sqrt[2]}
--
HTH :>)
Dana DeLouis
Windows XP & Mathematica 5.2
"traz" <t_raz at yahoo.com> wrote in message news:eud3g1$l0t$1 at smc.vnet.net...
> Let's say I wanna solve this problem:
>
> Determine point(s) on y = x^2 +1 that are
> closest to (0,2).
>
> So in mathematica:
>
> minDist = (x - 0)^2 + (y - 2)^2;
> Minimize[minDist, y == 1 + x^2, {x, y}]
>
> Output will give you: x -> -1/Sqrt[2], y -> 3/2
>
> but x also has another answer: +1/Sqrt[2]. Is this a problem in
mathematica or can my code be changed to output the other value of x for the
minimum distance?
>
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