       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.

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)}, {x -> 1/Sqrt}}

There are 3 possible solutions.  However, with x->0, the distance is not the
shortest:

d /. %

{1, Sqrt/2, Sqrt/2}

Therefore, the remaining 2 solutions would be the shortest.  Ie..

{x -> -(1/Sqrt)}, {x -> 1/Sqrt}

--

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, y -> 3/2

>

> but x also has another answer: +1/Sqrt. 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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