Re: Partial Differentiation of Implicit Functions
- To: mathgroup at smc.vnet.net
- Subject: [mg83187] Re: Partial Differentiation of Implicit Functions
- From: Jens-Peer Kuska <kuska at informatik.uni-leipzig.de>
- Date: Wed, 14 Nov 2007 04:48:19 -0500 (EST)
- Organization: Uni Leipzig
- References: <fhc3pe$510$1@smc.vnet.net>
- Reply-to: kuska at informatik.uni-leipzig.de
Hi,
E1 = -2 + x^2 + x z + y^2 + z^2;
eqn2 = (0 == x'[z] /. Solve[D[E1 /. x -> x[z], z] == 0, x'[z]][[1]]) /.
x[z] -> x
Eliminate[{E1 == 0, eqn2}, z]
??
Regards
Jens
Jay wrote:
> Hi,
>
> I have some equations of the form:
> AA x^2 + BB y^2 + CC z^2 + DD x y +EE x z + FF y z + GG x + HH y + II z +
> JJ== 0
>
> I want to solve for e.g. partial dy/dz = 0 and then combine the result with
> the original equation to get a new implicit equation.
>
> E.g.
>
> E1 := -2 + x^2 + x z + y^2 + z^2
>
> (where I require E1 == 0)
>
> Manually performing the differentiation gives:
>
> partial dx/dz (y constant) = (x+2 z)/(2x + z) == 0
>
> I can then go back into mathematica and do
>
> Eliminate[{E1 == 0, 2*z + x == 0}, {z}]
>
> giving:
>
> 4 x y + 4 y^2 == 8 - 5 x^2
>
> This is what I want but how do I do everything in Mathematica? I expected to
> be able to do something like:
>
> Solve[E1==0,D[x,y]]
>
> but it doesn't seem to work (says "0" is not a valid variable)
>
> Thanks,
>
> Jay.
>
>
>
>
>
>
>