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Mathematics Graphics and Visualization Wolfram Language

Plot a function (x,y) in polar coordinates?

Muhammad Abbas

Posted 9 years ago

Hi, I've this function f(x,y) such that (x,y) are the coordinates of an ellipse, that is (x/a)^2 + (y/b)^2 = 1.......so like the function yields an output for all coordinates of the ellipse....I want the plot in polar coordinates, like I want to have all the outputs laid down on a circular plot with degrees shown all around and the values of f(x,y)...i've absolutley no idea how to do this so please help me out from the basics, as i'm new to mathematica...

POSTED BY: Muhammad Abbas

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Marco Thiel

Marco Thiel, University of Aberdeen - Dept. of Physics/Mathematics

Posted 9 years ago

POSTED BY: Marco Thiel

Sander Huisman

Sander Huisman, University of Twente

Posted 9 years ago

Or even, yet simpler: equation=(x/a)^2+(y/b)^2==1; equation=equation/.{a->2,b->3} plot=ContourPlot[Evaluate[equation],{x,-6,6},{y,-6,6},PlotPoints->100,Axes->False,Frame->False]; plot2=PolarPlot[3,{\[Phi],0,2\[Pi]},PlotStyle->None,PlotRange->5,PolarAxes->Automatic,PolarTicks->{"Degrees",Automatic}]; Show[{plot2,plot}]

Or even, yet simpler:

equation=(x/a)^2+(y/b)^2==1;
equation=equation/.{a->2,b->3}

plot=ContourPlot[Evaluate[equation],{x,-6,6},{y,-6,6},PlotPoints->100,Axes->False,Frame->False];
plot2=PolarPlot[3,{\[Phi],0,2\[Pi]},PlotStyle->None,PlotRange->5,PolarAxes->Automatic,PolarTicks->{"Degrees",Automatic}];
Show[{plot2,plot}]

POSTED BY: Sander Huisman

Sander Huisman

Sander Huisman, University of Twente

Posted 9 years ago

of course one does not have to go back and forth polar coordinates if one has the equation in cartesian coordinates: equation=(x/a)^2+(y/b)^2==1; equation=equation/.{a->2,b->3} plot=ContourPlot[Evaluate[equation],{x,-6,6},{y,-6,6},PlotPoints->100]; lines=Cases[Normal[plot],Line[_],\[Infinity]]; max=Max[Cases[lines,(x:{_?NumericQ,_?NumericQ}):>Norm[x],\[Infinity]]]; PolarPlot[1.2max,{\[Phi],0,2\[Pi]},Epilog->lines,PlotStyle->None,PlotRange->5,PolarAxes->Automatic,PolarTicks->{"Degrees",Automatic}] giving the same plot...

of course one does not have to go back and forth polar coordinates if one has the equation in cartesian coordinates:

equation=(x/a)^2+(y/b)^2==1;
equation=equation/.{a->2,b->3}

plot=ContourPlot[Evaluate[equation],{x,-6,6},{y,-6,6},PlotPoints->100];
lines=Cases[Normal[plot],Line[_],\[Infinity]];
max=Max[Cases[lines,(x:{_?NumericQ,_?NumericQ}):>Norm[x],\[Infinity]]];
PolarPlot[1.2max,{\[Phi],0,2\[Pi]},Epilog->lines,PlotStyle->None,PlotRange->5,PolarAxes->Automatic,PolarTicks->{"Degrees",Automatic}]

giving the same plot...

POSTED BY: Sander Huisman

Sander Huisman

Sander Huisman, University of Twente

Posted 9 years ago

great answer, as always! The transformation rules can also be obtained like this: rules = Thread[{x, y} -> CoordinateTransform[{"Polar" -> "Cartesian", 2}, {r, \[Phi]}]] which might be easier to grasp... Another new way of doing the entire thing, including the polar axes is: equation=(x/a)^2+(y/b)^2==1; equation=equation/.{a->2,b->3} rules=Thread[{x,y}->CoordinateTransform[{"Polar"->"Cartesian",2},{r,\[Phi]}]]; equation=equation/.rules; plot=ContourPlot[Evaluate[equation],{r,0,6},{\[Phi],-\[Pi]+10^-8,\[Pi]},PlotPoints->100]; lines=Cases[Normal[plot],Line[x_]:>Line[FromPolarCoordinates[x]],\[Infinity]]; max=Max[Cases[lines,(x:{_?NumericQ,_?NumericQ}):>Norm[x],\[Infinity]]]; PolarPlot[1.2max,{\[Phi],0,2\[Pi]},Epilog->lines,PlotStyle->None,PlotRange->5,PolarAxes->Automatic,PolarTicks->{"Degrees",Automatic}] giving:

great answer, as always!

The transformation rules can also be obtained like this:

rules = Thread[{x, y} -> CoordinateTransform[{"Polar" -> "Cartesian", 2}, {r, \[Phi]}]]

which might be easier to grasp...

Another new way of doing the entire thing, including the polar axes is:

equation=(x/a)^2+(y/b)^2==1;
equation=equation/.{a->2,b->3}

rules=Thread[{x,y}->CoordinateTransform[{"Polar"->"Cartesian",2},{r,\[Phi]}]];
equation=equation/.rules;

plot=ContourPlot[Evaluate[equation],{r,0,6},{\[Phi],-\[Pi]+10^-8,\[Pi]},PlotPoints->100];
lines=Cases[Normal[plot],Line[x_]:>Line[FromPolarCoordinates[x]],\[Infinity]];
max=Max[Cases[lines,(x:{_?NumericQ,_?NumericQ}):>Norm[x],\[Infinity]]];
PolarPlot[1.2max,{\[Phi],0,2\[Pi]},Epilog->lines,PlotStyle->None,PlotRange->5,PolarAxes->Automatic,PolarTicks->{"Degrees",Automatic}]

giving:

enter image description here

POSTED BY: Sander Huisman

Marco Thiel

Marco Thiel, University of Aberdeen - Dept. of Physics/Mathematics

Posted 9 years ago

POSTED BY: Marco Thiel

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