Registered: Apr 2005
Posts: 1

Hi there,
Well i've been using the feature ContourPlot in mathematica recently and i think that i need your help guys.
The problem is that i need to plot the contour lines plus the values of each curve : in another term plot the curve and the values of the isolines as well.
Thx a lot for your response

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Registered: Aug 2003
Posts: 712

In general, for Mathematica help the places to look are first the help browser, then the Mathematica book, and then online resources like the Information Center -

http://library.wolfram.com/infocenter

If you still can't find what you need you can try the MathGroup discussion list, where you can submit a question if there is no relevant thread already -

http://forums.wolfram.com/mathgroup/

This forum is specifically for New Kind of Science subjects, rather than general Mathematica ones.

All that said, to address your actual question, you can probably get what you want with just the optional argument ContourLabels->Automatic. For example -

ContourPlot[x Sin[2 Pi y], {x, -4, 4}, {y, -1, 1}, 
PlotRange -> {-4, 4}, PlotPoints -> 25,
ContourShading -> False,  Contours -> 7, 
ContourLabels -> Automatic] 

Notice, the Contours argument tells it how many lines to draw. If you just give an integer, it draws that many lines, and they are evenly spaced within the specified PlotRange. ContourLabels tells it to label each line with the Z value of that contour.

You can also give a list of z values instead of a integer number of lines, as a value for Contours. E.g.

ContourPlot[x Sin[2 Pi y], {x, -4, 4}, {y, -1, 1}, 
PlotRange -> {-4, 4}, PlotPoints -> 25,
ContourShading -> False,  Contours -> {-3,0,2}, 
ContourLabels -> Automatic] 

That draws only the z = -3, the z = 0, and the z = 2 contours - and labels them.

Notice I also used the PlotPoints argument. With the default number of points, some of the contour lines might look a bit squared off in places. Increasing the number of plot points gives a smoother shape - at some sacrifice in rendering speed.

Incidentally, if you want to see a 3D graphic version of the same, you would use something like -

Plot3D[x Sin[2 Pi y], {x, -5, 5}, {y, -2, 2}, Mesh -> False, PlotPoints -> 50]

I hope this helps.

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Registered: Feb 2006
Posts: 2

I try

ContourPlot[x Sin[2 ð y], {x, -4, 4}, {y, -1, 1}, ContourLabels -> Automatic]

but I get

ContourPlot :: optx :
Unknown option ContourLabels in ContourPlot[x Sin[2 Pi y],{x,-4,4},{y,-,1,1},ContourLabels->Automatic]. More...

Note that

?ContourLabels

gives the answer

Global`ContourLabels

and this means the system recognizes it, isn't it? So what is going on?

I would be very grateful for any help on this.

:) Katerina.

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