       Re: Solve can solve it with some help

• To: mathgroup at smc.vnet.net
• Subject: [mg115396] Re: Solve can solve it with some help
• From: "Sjoerd C. de Vries" <sjoerd.c.devries at gmail.com>
• Date: Mon, 10 Jan 2011 02:36:24 -0500 (EST)
• References: <igbnk8\$hj9\$1@smc.vnet.net>

```Hi Eduardo,

You may ask Solve to eliminate vf in the process of solving for Theta:

s = {0 == 1/2*m*vf^2 + 0 - m*g*(R - R*Cos[\[Theta]]), m*vf^2/R ==
m*g*Cos[\[Theta]]};

Solve[s, \[Theta], vf]

During evaluation of In:= Solve::bdomv: Warning: vf is not a valid
domain specification. Mathematica is assuming it is a variable to
eliminate. >>

During evaluation of In:= Solve::ifun: Inverse functions are being
used by Solve, so some solutions may not be found; use Reduce for
complete solution information. >>

Out= {{\[Theta] -> -ArcCos[2/3]}, {\[Theta] -> ArcCos[2/3]}}

Cheers -- Sjoerd

On Jan 9, 8:20 am, Eduardo Cavazos <wayo.cava... at gmail.com> wrote:
> Hello,
>
> Solve doesn't come up with anything for these two equations:
>
> {
>   0 == 1/2*m*vf^2 + 0 - m*g*(R - R*Cos[\[Theta]]),
>
>   m*vf^2/R == m*g*Cos[\[Theta]]
>   };
> Solve[%, \[Theta]]
>
> If you manually solve one of them for vf, Solve can take care of the
> rest:
>
> 0 == 1/2*m*vf^2 + 0 - m*g*(R - R*Cos[\[Theta]]);
> % /. Solve[m*vf^2/R == m*g*Cos[\[Theta]], vf][];
> Solve[%, \[Theta]]
>
> {{\[Theta] -> -ArcCos[2/3]}, {\[Theta] -> ArcCos[2/3]}}
>
> Of course, Reduce can handle the original set. However, Solve is nice
> due to the brevity of output. (Side question: is there a way to extract
> an equation from the results of Reduce based on variable name? Sometimes
> the results from Reduce can be so verbose, it'd be nice to say "extract
> equation for theta".)
>
> My main question: is there a way to get Solve to solve the original set
> of two equations without taking the manual approach?
>
> Ed

```

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