Re: Convert function from polar to Cartesian
- To: mathgroup at smc.vnet.net
- Subject: [mg112793] Re: Convert function from polar to Cartesian
- From: Bob Hanlon <hanlonr at cox.net>
- Date: Thu, 30 Sep 2010 04:54:06 -0400 (EDT)
- Reply-to: hanlonr at cox.net
$Version
"7.0 for Mac OS X x86 (64-bit) (February 19, 2009)"
fPolar[r_, a_] = r^2 Cos[2 a];
fCart[x_, y_] = fPolar[Sqrt[x^2 + y^2], ArcTan[x, y]];
fCart[x, y] // FullSimplify
(x - y) (x + y)
Bob Hanlon
---- Sam Takoy <sam.takoy at yahoo.com> wrote:
=============
Hi,
I've stumbled onto an unexpected problem. I have a function in polar
coordinate and I need to convert it to Cartesian coordinates:
fPolar[r_, a_] = r^2 Cos[2 a];
fCart[x_, y_] = fPolar[Sqrt[x^2 + y^2], ArcTan[x, y]];
fCart[x, y]
I get: (x^2 + y^2) Cos[2 ArcTan[x, y]].
But how to get rid of the ArcTan[x, y]? I've tried this:
fCart[x, y] //
TrigExpand /. {Cos[ArcTan[x, y]] -> x/Sqrt[x^2 + y^2],
Sin[ArcTan[x, y]] -> y/Sqrt[x^2 + y^2]}
but it does not do the substitution I need it to do. So what's the
proper way to get rid of the ArcTan[x, y] and for the simple example
above, to obtain x^2 - y^2?
Many thanks in advance,
Sam