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 March 5th, 2012, 07:03 AM #1 Newbie   Joined: Feb 2012 Posts: 5 Thanks: 0 path dependent function with a definite path This question is about if I have a path dependent function but a deinite path, then can I take partial derivatives? And are the points that are along path A connected to the points that are along path B. If I consider a non-conservative vector field $\vec{dt}= \frac{\vec{ds}}{V(z)}\$ I must specify a path, in this case the path is always: $p= \frac{sin \Theta}{V(z)} = \frac{dx}{V(z)\sqrt{dx^{2}+dz^{2}}}$ (eq. 1) and integrate along the path to create the field: $t= \int_{0}^{z }\frac{dz}{V(z)\sqrt{1-p^{2}V(z)^{2}}}\$ (eq. 2) and also: $x= \int_{0}^{z }\frac{pV(z)dz}{\sqrt{1-p^{2}V(z)^{2}}}\$ (eq. 3) may I now treat t as path independent? In other words, may I now take the total derivative and evaluate at a constant x like this: $dt= \frac{\partial t}{\partial z} |_{p} dz + \frac{\partial t}{\partial p} |_{z}dp$ (eq. 4) $dx= \frac{\partial x}{\partial z}|_{p} dz + \frac{\partial x}{\partial p}|_{z}dp$ (eq. 5) and then evaluate eq. 5 at a fixed x so that dx = 0 It seems to me that I can't do this because I think that I have defined a relationship between dx and dz in equation 1. Since t is path dependent, p must be in the definition of t so setting dx = 0 violates the constraint that p has set on the relationship between dx and dz. Is what I was thinking correct? Or am I allowed to take the total derivative of X and set dx = 0? Is the point t(x,z) connected to t(x,z+dz)? I am posting this question here and in the differentials section because I am not sure if this is a question about disconnected space or conservative fields.
 March 5th, 2012, 04:27 PM #2 Newbie   Joined: Feb 2012 Posts: 5 Thanks: 0 Re: path dependent function with a definite path I should have mentioned that all paths go through the origin and x and z can only be positive.
March 5th, 2012, 06:30 PM   #3
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Re: path dependent function with a definite path

Quote:
 Originally Posted by aise5668 I should have mentioned that all paths go through the origin and x and z can only be positive.

Is the vector field two dimentional or three dimentional ??

 March 5th, 2012, 06:36 PM #4 Newbie   Joined: Feb 2012 Posts: 5 Thanks: 0 Re: path dependent function with a definite path It is 2 dimensional.

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