polar equation of circle
- To: mathgroup at smc.vnet.net
- Subject: [mg102958] polar equation of circle
- From: Narasimham <mathma18 at hotmail.com>
- Date: Thu, 3 Sep 2009 05:38:55 -0400 (EDT)
To algebraically find polar equation R[th] for a circle of radius a
and tangent length T from origin lying outside the circle (Conversion
from cartesian form is easier, just tried it nevertheless-- between
two th domain limits / within an interval ), I attempted to use DSolve
without luck. The message says it should be non-algebraic, but
involving several log functions. What should be chosen for #1? Can
somone help to simplify the result? TIA. Narasimham
Eq = R'[th]/R[th] == Sqrt [ (2 a R[th]/(R[th]^2 - T^2) )^2 - 1 ];
DSolve[Eq, R, th] ;
{{R -> Function[{th},
InverseFunction[((-Log[#1^2] +
Log[-2 a^2 - T^2 + #1^2 +
Sqrt[-4 a^2 #1^2 + (-T^2 + #1^2)^2]] +
Log[-2 a^2 #1^2 +
T^2 (T^2 - #1^2 +
Sqrt[-4 a^2 #1^2 + (-T^2 + #1^2)^2])]) Sqrt[-4 a^2 \
#1^2 + (-T^2 + #1^2)^2])/(
2 Sqrt[-T^2 - 2 a #1 + #1^2]
Sqrt[-T^2 + 2 a #1 + #1^2]) &][\[ImaginaryI] th + C[1]]]}}