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Create a parametric plot, with only positive axes - stability diagram?

GROUPS:

Hello everyone!

I started using Mathematica today (have only a little bit OF Matlab experience).

I got the following code:

r[x_]= 2x / (x^2+1)^2;
s[x_] = (x^2 (-1+x^2))/(1+x^2)^2;
ParametricPlot[{r[x],-s[x]}, {x,0,10}, PlotStyle -> {Blue}, 
AxesLabel->{Style[r,Large],Style[s,Large]},LabelStyle->Directive[Bold]]

My task seems to be quite easy, but I fail: I want to get this plot, but only with positive s-axis - no negative s-values needed.

Can anybody help me, please?

Thank you so much!

POSTED BY: m s
Answer
2 months ago

Welcome to Wolfram Community! Please make sure you know the rules and how to format your code properly, which you can find here: https://wolfr.am/READ-1ST If you do not format code, it may become corrupted and useless to other members. We edited your posts to make code blocks start on a new paragraph and look framed and colored. Use code button in future:

enter image description here

POSTED BY: Moderation Team
Answer
2 months ago

Look up PlotRange in docs: http://reference.wolfram.com/language/ref/PlotRange.html

Make it a habit to read docs carefully. If you would carefully read ParametricPlot docs you would find examples there that solves your problem.

POSTED BY: Kapio Letto
Answer
2 months ago

I allready looked up PlotRange and tried out a lot, but I was not able to get it working.

I now tried some things and fixed the Problem with

...  ParametricPlot[{r[x],-s[x]}, {x,0,**1**},...

even though I mathematically don't understand what I did... how would ich have to arrange the "PlotRange"?

POSTED BY: m s
Answer
2 months ago

(Click "Reply" to reply to a specific comment to have comments structured). Something like this?

r[x_] = 2 x/(x^2 + 1)^2;
s[x_] = (x^2 (-1 + x^2))/(1 + x^2)^2;
ParametricPlot[{r[x], -s[x]}, {x, 0, 10}, PlotStyle -> {Blue}, 
 FrameLabel -> {Style[r, Large], Style[s, Large]},
 LabelStyle -> Directive[Bold],
 PlotRange -> {Automatic, {0, .13}},
 AspectRatio -> 1,
 PlotTheme -> "Detailed"]

enter image description here

POSTED BY: Kapio Letto
Answer
2 months ago

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