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PolarConversiion

John Lutz
Posted 12 years ago

In Mathematica, how can I convert an equation written in Polar coordinates to Cartesian coordinates with x and y values?
POSTED BY: John Lutz
12 Replies
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John Lutz
John Lutz
Posted 12 years ago

David, I implemented your myTransformedField function. It is great. Years ago I bought a software package called Scientific Notebook by MacKichan Software Inc. It is a very user friendly front end to the old Maple version 5.5. It is no longer being upgraded. I wish someone would look at it to see the features and concept. Maybe something like it could be done for Mathematica.

Thanks again.
POSTED BY: John Lutz
David Keith
David Keith
Posted 12 years ago

Thanks, Eric -- I edited it to correct the error.
POSTED BY: David Keith
John Lutz
John Lutz
Posted 12 years ago

Thanks David. That worked! I finally have the conversion.
POSTED BY: John Lutz
David Keith
David Keith
Posted 12 years ago
POSTED BY: David Keith
POSTED BY: Eric Johnstone
John Lutz
John Lutz
Posted 12 years ago
POSTED BY: John Lutz

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POSTED BY: Simon Cadrin

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POSTED BY: Simon Cadrin
David Keith
David Keith
Posted 12 years ago

The built-in functions I'm aware of (CoordinateTransform (above) and FromPolarCoordinates) operate on vectors, not equations. Perhaps you could say more about the application.
POSTED BY: David Keith
Alexey Popkov
Alexey Popkov
Posted 12 years ago

Perhaps CoordinateTransform?

CoordinateTransform["Polar" -> "Cartesian", {r, \[Theta]}]
(* => {r Cos[\[Theta]], r Sin[\[Theta]]} *)
POSTED BY: Alexey Popkov
John Lutz
John Lutz
Posted 12 years ago

I can do it on paper. I worded my question poorly. What I am looking for is a Mathematica command that will do this. Some commands will convert to polar but I haven't found one that converts to Cartesian. Thanks
POSTED BY: John Lutz
David Keith
David Keith
Posted 12 years ago

You could use the usual substitution. I use ArcTan[y,x] rather than ArcTan[y/x] because it accounts for the quadrant.

* Below edited to correct error *

In[1]:= fromPolarRule = {r -> Sqrt[x^2 + y^2], \[Theta] -> 
    ArcTan[x,y]};

In[2]:= eq = r Cos[\[Theta]] == r Sin[\[Theta]];

In[3]:= eqxy = eq /. fromPolarRule

Out[3]= x == y
POSTED BY: David Keith
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