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Re: Help with Solve

  • To: mathgroup at smc.vnet.net
  • Subject: [mg112385] Re: Help with Solve
  • From: ADL <alberto.dilullo at tiscali.it>
  • Date: Sun, 12 Sep 2010 03:11:31 -0400 (EDT)
  • References: <i6a5i6$kne$1@smc.vnet.net> <i6fj5e$p7r$1@smc.vnet.net>

What probably is puzzling you (and also puzzled me at first) is to
miss that your equations imply a relationship among the apparently
independent parameters (n, m, v, r, g).

In fact, since Cos[x]^2 + Sin[x]^2 == 1,  it must be ((m v^2)/(n r))^2+
((m g)/n)^2==1.

Mathematica, tends to explicitate this last relationship, thus making
the output cumbersome.

If you look at:

eq1 = Sin[x] == a;
eq2 = Cos[x] == b;
cons = {a > 0, b > 0, x \[Element] Reals};
sys = Join[{eq1, eq2}, cons];
sol = Reduce[sys, x]

and then to the resulting expression of Tan[x]:

FullSimplify[Tan[x /. ToRules@Last[sol]], Most[sol]]

this may become clearer.

What would be desirable is perhaps a less "mixed-up" solution form.

ADL


On 11 Set, 11:46, Eduardo  Cavazos <wayo.cava... at gmail.com> wrote:
> On Sep 9, 3:23 am, Eduardo  Cavazos <wayo.cava... at gmail.com> wrote:
>
> > Here's a couple of equations:
>
> > eq1 = n*Sin[x] == (m*v^2)/r;
> > eq2 = n*Cos[x] == m*g;
>
> > The goal is to solve for 'x'.
>
> > I can do this in a roundabout way via:
>
> >     Solve[eq1 /. Solve[eq2, n], x]
>
> I should point out that the answer that I get from using Solve this
> way is:
>
>     {{x -> ArcTan[v^2/(g r)]}}
>
> which is form the answer I'm looking for. :-) Again, I'm just
> wondering if Mathematica can do the work for me, instead of having to
> manually eliminate 'n' in a separate step.
>
> Sjoerd and Bob recommended using Reduce. Thanks for the tip! I tried
> to use Reduce but the results it produced were not in the above form.
>
> Ed



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