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Re: Graphics << Implicit vs ContourPlot

  • To: mathgroup at smc.vnet.net
  • Subject: [mg122788] Re: Graphics << Implicit vs ContourPlot
  • From: John Accardi <johnaccardi at comcast.net>
  • Date: Thu, 10 Nov 2011 06:55:15 -0500 (EST)
  • Delivered-to: l-mathgroup@mail-archive0.wolfram.com
  • References: <201111091124.GAA11084@smc.vnet.net> <4DFC3764-C65E-4C56-B6D2-B72B0035E9E5@gmail.com>

Thank you, Heike,

On 11/9/2011 5:20 AM, Heike Gramberg wrote:
> Setting Axes ->  True, Frame ->  False in your ContourPlot will create a plot with axes instead of a frame.
Great suggestion.
>
> To plot the rotated ellipse it's probably easier to use Circle and Rotate instead of ContourPlot.  For example, this will plot an ellipse with semi-axes 2 and 1 rotated over 45 degrees.
>
> With[{ellipse = Circle[{0, 0}, {2, 1}], angle = 45 Degree},
>   Graphics[{ellipse,
>    {Red, Rotate[{ellipse, Line[{{-3, 0}, {3, 0}}], Line[{{0, -3}, {0, 3}}]}, angle]}},
>      Axes ->  True, PlotRange ->  {{-2.1, 2.1}, {-2.1, 2.1}}]]
This is a great alternative.  Unfortunate I need to stick with 
ContourPlot because the purpose of the lesson is to show students the 
"xy" term in the equation and how it is responsible for the rotation.

By the way, do you know why I need AspectRatio -> 1.25 to have the axes 
at 45 degrees to be perpendicular?  Intuitively I would think 
AspectRatio -> 1 would do this.

Thanks again.
John

>
> Heike
>
> On 9 Nov 2011, at 12:24, John Accardi wrote:
>
>> Goal: Show students a plot of both an ellipse with x axis as ellipse's
>> transverse axis
>> and the same ellipse in an x'y' coordinate plane that is rotated some angle
>> with respect the the original xy coordinate plane.  (All in one plot)
>>
>> I used:
>>
>> <<  Graphics`ImplicitPlot`; ImplicitPlot[{7 x^2 - 6 Sqrt[3] x y +
>>      13 y^2 - 16 == 0, ((x^2)/2^2) + ((y^2)/2^2) == 1, y == x,
>>    y == -x}, {x, -3, 3}, AspectRatio ->  1.25]
>>
>> which works well but I had to hard fix the axis of rotation at 45 degrees
>> and plot it (y == x and y == -x).  I also get an obsolete warning and
>> the suggestion to use the new ContourPlot for this in the future:
>>
>> General::obspkg: "\!\(\"Graphics`ImplicitPlot`\"\) is now obsolete.
>> The legacy version being loaded may conflict with current Mathematica
>> functionality. See the Compatibility Guide for updating information."
>>
>> So I try to accomplish the same graph with ContourPlot:
>>
>> ContourPlot[{7 x^2 - 6 Sqrt[3] x y + 13 y^2 - 16 ==
>>     0, ((x^2)/2^2) + ((y^2)/2^2) == 1, y == x, y == -x, y == 0,
>>    x == 0}, {x, -3, 3}, {y, -3, 3}, AspectRatio ->  1.25]
>>
>> which gets me close but I have lost traditional plotting of the xy axes
>> (no tick marks).  Instead I get ContourPlots boxed style coordinate system.
>>
>> Question: How can I get my old style axes back in the context of
>> ContourPlot?
>>
>> Thank you for any insights.
>



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