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