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Re: Solve Question - 2 Non zero values

  • To: mathgroup at smc.vnet.net
  • Subject: [mg112973] Re: Solve Question - 2 Non zero values
  • From: Peter Pein <petsie at dordos.net>
  • Date: Fri, 8 Oct 2010 04:50:49 -0400 (EDT)
  • References: <i7seni$pph$1@smc.vnet.net>

Am Tue, 28 Sep 2010 10:06:42 +0000 (UTC)
schrieb "Lea Rebanks" <lrebanks at netvigator.com>:

> Dear All,
> 
>  
> 
> Enclosed are the workings to Solve for BOTH values of V1 and V2.
> 
> I have already shown below the Solve for just one value either V1 or
> V2 and there is no problem.
> 
> The correct values required for V1 and V2 are shown below when I
> worked them out individually.
> 
> However, when I setup a Solve function to evaluate BOTH V1 and V2 in
> the same Solve function then I run into problems.
> 
>  
> 
> Question 1:-In the instance shown below the result shows V1 and V2 to
> equal' zero'.
> 
> Therefore to avoid this could someone show me how to set Solve, (or
> Reduce, or whatever) up to provide a result greater than zero.(I am
> sure this is easy, but I don' t know how to do it.)
> 
>  
> 
> Secondly, I suspect there is a deeper or more involved mathematical
> method to extract these 2 required values. Any ideas anyone?
> 
> Many thanks for everyones' help & attention.
> 
> Best regards,
> 
> Lea...
> 
...

Hi Lea,

your posting has been a while ago but let's see...

It is almost ever a good idea to simplify expressions before feeding
them into Solve or Reduce.

In[1]:=
M=Simplify[45-ArcTan[Sqrt[(4-Pi)/Pi]]/Degree//FunctionExpand]//Apart
radiusD[x_]=Tan[M*Degree]*x//FunctionExpand//ExpandAll//TrigExpand//Simplify
Out[1]= 45-(180 ArcTan[Sqrt[-1+4/\[Pi]]])/\[Pi] Out[2]=
((-Sqrt[4-\[Pi]]+Sqrt[\[Pi]]) x)/(Sqrt[4-\[Pi]]+Sqrt[\[Pi]])

so we've got a simple radiusD :-)

With the same simplifications and replacement of floating point numbers
by their exact values, the first equation reduces to 18 == V1^2 and the
second equation to V2^2 == 8.

 So we get V1 == 3 Sqrt[2] and V2 == 2 Sqrt[2].

The list of equations reduces to two times the same equation:
Simplify[TrigExpand[ExpandAll[FunctionExpand[
         {(((15*360)*
             Sqrt[radiusD[-(((15*Sqrt[2])*V1)/(Tan[135*Degree] - 
                    Tan[M*Degree]))]^2 + 
                             (((15*Sqrt[2])*V1)/(Tan[135*Degree] - 
                    Tan[M*Degree]))^2])/(360*15))^2*Pi - 
               (((45*360)*
             Sqrt[radiusD[-((((45/2)*Sqrt[2])*V2)/(Tan[135*Degree] - 
                                       Tan[M*Degree]))]^2 + ((((45/2)*
                    Sqrt[2])*V2)/(Tan[135*Degree] - 
                                     Tan[M*Degree]))^2])/(360*45))^2*
        Pi == 0, 
           (((13*360)*
             Sqrt[radiusD[-(((13*Sqrt[2])*V1)/(Tan[135*Degree] - 
                    Tan[M*Degree]))]^2 + 
                             (((13*Sqrt[2])*V1)/(Tan[135*Degree] - 
                    Tan[M*Degree]))^2])/(360*13))^2*Pi - 
               (((39*360)*
             Sqrt[radiusD[-((((39/2)*Sqrt[2])*V2)/(Tan[135*Degree] - 
                                       Tan[M*Degree]))]^2 + ((((39/2)*
                    Sqrt[2])*V2)/(Tan[135*Degree] - 
                                     Tan[M*Degree]))^2])/(360*39))^2*
        Pi == 0}]]]]
gives:
{4*V1^2 == 9*V2^2, 4*V1^2 == 9*V2^2}

Using Reduce[4*V1^2 == 9*V2^2 && V1 > 0 < V2, {V1, V2}] yields V1 > 0
&& V2 == (2*V1)/3, which confirms the relation between V1 and V2 found
above.


Regards,
 Peter


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