Implicit arc length differentiation and perimeter computation?
- To: mathgroup at smc.vnet.net
- Subject: [mg33213] Implicit arc length differentiation and perimeter computation?
- From: jflanigan at netzero.net (jose flanigan)
- Date: Sat, 9 Mar 2002 03:19:35 -0500 (EST)
- Sender: owner-wri-mathgroup at wolfram.com
Hi, Suppose I have the following equation: f(x,y) = const. Now, as I understand it, this implicitly defines y as a function of x, y(x). Suppose that I am able to solve for this explicitly. Next, to compute the perimeter defined by this equation: ArcLength = integrate[ sqrt(1+y'(x)^2) dx] Now, we are up to my question: If I derive a unit vector tangent to the curve defined by f(x,y)=const. called v such that <v,del f(x,y)> = 0 at every point on the contour, how can I use this to compute the same perimeter using polar coordinates. Note that del is the usual gradient vector. I am thinking of something like ArcLength = integrate[F,dphi] where F=<v,del some_other_function> which would evaluate to the correct perimeter. The question is how can I get at this mysterious some_other_function Any help appreciated. Thanks in advance. By the way, I have tried a number of things like some_other_function = ArcTan[x,y].