Re: Solve and Reduce

• To: mathgroup at smc.vnet.net
• Subject: [mg52143] Re: [mg52128] Solve and Reduce
• From: Bob Hanlon <hanlonr at cox.net>
• Date: Fri, 12 Nov 2004 02:13:57 -0500 (EST)
• Sender: owner-wri-mathgroup at wolfram.com

```eqn = 2500*c^2-25*c^3+3500*c*q-
320*c^2*q-1104*c*q^2-1152*q^3\[Equal]0;

Needs["Graphics`"];

ImplicitPlot[eqn, {c,0,100},{q,0,5},
AspectRatio->.6,ImageSize->{288,178},
Frame->True, Axes->False,
FrameLabel->{"c","q\n"}];

With[{b=(5*(-109199+1497*Sqrt[5489]))/2744},
DisplayTogether[
Plot[Root[-2500*c^2+25*c^3-3500*c*#1+
320*c^2*#1+1104*c*#1^2+1152*#1^3&,1],
{c,b,100}, PlotStyle->Blue],
Plot[Root[-2500*c^2+25*c^3-3500*c*#1+
320*c^2*#1+1104*c*#1^2+1152*#1^3&,3],
{c,0,b}, PlotStyle->Red],
PlotRange->{0,5},
Frame->True, Axes->False,
FrameLabel->{"c","q\n"}]];

If c is defined before the equation is solved, then the roots are sorted and the
positive result is last

Table[Chop[(q/.
Solve[eqn /. c->{.1,1.,5.,10.}[[n]], q])],
{n,4}]

Clear[qr];
qr[c_/;0<c<100]:=
q/.Solve[eqn,q][[3]];

Plot[qr[c],{c,0,100},
PlotRange->{0,5},
Frame->True, Axes->False,
FrameLabel->{"c","q\n"}];

If the equation is solved for q before c is defined then your desired result is
the second one

Chop[(q/.
Solve[eqn, q]) /. c->{.1,1.,5.,10.}]

Clear[qr];
qr[c_/;0<c<100]=
q/.Solve[eqn,q][[2]];

Plot[qr[c],{c,0,100},
PlotRange->{0,5},
Frame->True, Axes->False,
FrameLabel->{"c","q\n"}];

Bob Hanlon

>
> From: "Carol Ting" <tingyife at msu.edu>
To: mathgroup at smc.vnet.net
> Date: 2004/11/11 Thu AM 04:53:03 EST
> To: mathgroup at smc.vnet.net
> Subject: [mg52143] [mg52128] Solve and Reduce
>
>
> Hello list,
>
> I want to find q as a function of c, q(c), given the following
> equation:
>
> 2500*c^2 - 25*c^3 + 3500*c*q - 320*c^2*q - 1104*c*q^2 - 1152*q^3
== 0
>
> However, each of the following three methods gives different results.
> I check the Mathematica Book but still cannot figure out why there are
> such differences.  Could someone please explain this to me?  Thanks a
> lot!
>
> (1) Use "Reduce"
>
> In[5]:=
> q1[c_] = Reduce[{2500*c^2 - 25*c^3 +3500*c*q - 320*c^2*q
-1104*c*q^2
> -1152*q^3 == 0, c > 0,q > 0}, q]
>
> Out[5]=
> 0<c<=(5*(-109199 + 1497*Sqrt[5489]))/2744] &&q == Root[-2500*c^2 +
> 25*c^3 - 3500*c*#1 + 320*c^2*#1 + 1104*c*#1^2 + 1152*#1^3 & ,3] ||
> (5*(-109199 + 1497*Sqrt[5489]))/2744 < c < 100 && q ==Root[
-2500*c^2 +
> 25*c^3 -3500*c*#1 + 320*c^2*#1 + 1104*c*#1^2 + 1152*#1^3 & , 1]
>
> In[6]:=
> Plot[Root[-2500*c^2 + 25*c^3 - 3500*c*#1 + 320*c^2*#1 + 1104*c*#1^2
+
> 1152*#1^3 & ,1], {c, 0, 100}]
> Plot[Root[-2500*c^2 + 25*c^3 - 3500*c*#1 + 320*c^2*#1 + 1104*c*#1^2
+
> 1152*#1^3 & ,3], {c, 0, 100}]
>
> Out[6]=
> Graphics[]
>
> Out[7]=
> Graphics[]
>
>
> (2) Use "Solve" and Immediate assignment
>
> In[32]:=
> qdroot1[c_] = q /. Solve[2500*c^2 - 25*c^3 + 3500*c*q - 320*c^2*q -
> 1104*c*q^2 - 1152*q^3 == 0,q][[1]]
> qdroot3[c_] = q /. Solve[2500*c^2 - 25*c^3 + 3500*c*q - 320*c^2*q -
> 1104*c*q^2 - 1152*q^3 == 0,q][[3]]
>
> In[34]:=
> Plot[qiroot1[c], {c, 0, 100}]
>
> Out[34]=
> Graphics[]
>
> In[35]:=
> Plot[qiroot3[c], {c, 0, 100}]
>
> Out[35]=
> Graphics[]
>
> (3) Use "Solve" and delayed assignment
>
> In[28]:=
> qdroot1[c_] := q /. Solve[2500*c^2 - 25*c^3 + 3500*c*q - 320*c^2*q -
> 1104*c*q^2 - 1152*q^3 == 0,q][[1]]
> qdroot3[c_] := q /. Solve[2500*c^2 - 25*c^3 + 3500*c*q - 320*c^2*q -
> 1104*c*q^2 - 1152*q^3 == 0,q][[3]]
>
> In[30]:=
> Plot[qdroot1[c], {c, 0, 100}]
>
> Out[30]=
> Graphics[]
>
> In[31]:=
> Plot[qdroot3[c], {c, 0, 100}]
>
> Out[31]=
> Graphics[]
>
>
> Carol
>
>

```

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