1
$\begingroup$

enter image description here

Can someone please help me to draw the attached figure in Mathematica?

$\endgroup$
1
  • 1
    $\begingroup$ Maybe the title should be modified to specify that the figures have arrows as this is the most difficult part I think (and the one that is maybe most useful to the community). Perhaps: " drawing two figures with directed lines specified by arrows" or "drawing two figures with arrows that flow along curves". $\endgroup$ Aug 5 at 0:02

2 Answers 2

2
$\begingroup$

Try e.g.:

cu1 = {Arrow[Line@Table[{x, 1/(x - 1) + 1}, {x, 1.1, 10, .1}]]} ;
cu2 = cu1 /. p : {_, _} -> ReflectionTransform[{1, 0}][p];
cu3 = cu1 /. p : {_, _} -> ReflectionTransform[{0, 1}][p];
cu4 = cu2 /. p : {_, _} -> ReflectionTransform[{0, 1}][p];
Graphics[{Arrowheads[{-.05, -.05}], cu1, cu2, cu3, cu4, 
  Arrowheads -> {-0.1, 0.1}, Thickness[0.01], 
  Arrow[{{0, 11}, {0, -11}}]}]
Graphics[{Arrowheads[{.05, .05}], cu1, cu2, cu3, cu4, 
  Arrowheads -> {0.1, -0.1}, Thickness[0.01], 
  Arrow[{{0, 10}, {0, -10}}]}]

enter image description here

$\endgroup$
0
$\begingroup$

Edits: Show[plot1,plot2]->Plot[{func1,func2}], Row->GraphicsRow

You might find better answers from people here that know how to use Graphics better than me but a semi-manual approach is possible with drawing tools in Mathematica.

Step 1:

Plot[{1/x,-1/x}, {x, -1, 1}, Axes -> False, PlotStyle -> Black]

Step 2:

Use line and arrows in drawing tools by right-clicking the plot and then clicking on drawing tools.

output:

plot

Step 3:

Do similar manipulations for the other plot

Step 4:

copy each plot (right click copy graphics), stack them side by side and use :

GraphicsRow[{plot1,plot2}]

I personally used Inkscape when I wanted to modify plots.

$\endgroup$
2
  • 2
    $\begingroup$ Your first step can be simplified to Plot[{1/x, -1/x}, {x, -1, 1}, Axes -> False, PlotStyle -> Black] $\endgroup$
    – Bob Hanlon
    Aug 4 at 22:14
  • $\begingroup$ @BobHanlon Thank you. I will edit the code $\endgroup$ Aug 4 at 22:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.