RE: Plotting an implicit function

• To: mathgroup at smc.vnet.net
• Subject: [mg50584] RE: [mg50559] Plotting an implicit function
• From: "David Park" <djmp at earthlink.net>
• Date: Sat, 11 Sep 2004 06:44:35 -0400 (EDT)
• Sender: owner-wri-mathgroup at wolfram.com

```David,

Use r_ and M_ in the definition, and I prefer a definition as a function and
not an equation so we can test evaluate or Plot the function for various
cases.

f[r_, M_] := 1 + (-1 - 0.03*M^2 + 1.2*M)*r -
((0.05*E^(r*M))/r - 0.05/r + E^(r*M)*(1 - 0.05*M))

The following seems to give the plot that you are looking for. I fixed the
aspect ratio to get a better representation and also extended the M domain
to obtain a better picture of the function.

Needs["Graphics`ImplicitPlot`"]

ImplicitPlot[f[r, M] == 0, {M, 0, 30}, {r, 0.0001, 0.03},
AspectRatio -> 1,
Frame -> True,
FrameLabel -> {M, r}];

David Park

From: David Friskin [mailto:david.friskin at upe.ac.za]
To: mathgroup at smc.vnet.net

I have the following implicit function:
f[r, M]=1 + (-1 - 0.03*M^2 +
1.2*M)*r - (0.05*E^(r*M)/r - 0.05/r + E^(r*M)*(1 - 0.05*M))==0

where r is a function of M i.e. r(M).

I would like to plot r(M) vs M, for {M,0,20}.

I tried using
ImplicitPlot[{f[r,M] == 0},{M,0,20}], but Solve is having problems
finding the roots to my function.

Could someone help me with the code to do this, perhaps using FindRoot
somehow?

Thanks
David

P.S. r > 0

```

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