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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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