Jump to content


Photo
- - - - -

Calculus


  • Please log in to reply
6 replies to this topic

#1 SF Shawn

SF Shawn

    Lepton

  • New Members
  • 4 posts

Posted 16 November 2015 - 02:24 PM

I want to know about the functions which is path independent and path dependent. Please describe me about this.


  • 0

#2 studiot

studiot

    Genius

  • Senior Members
  • 6,484 posts
  • LocationSomerset, England

Posted 17 November 2015 - 08:45 AM

What is this in relation to?

 

Are you asking about conservative and non conservative fields or forces?

 

Gravity is conservative, friction is non conservative.

 

Or are you asking about zero v nonzero curl?


  • 0

#3 SF Shawn

SF Shawn

    Lepton

  • New Members
  • 4 posts

Posted 19 November 2015 - 07:12 PM

Yes I want to know about conservative and non conservative fields. Also I want to know about zero and nonzero curl.


  • 0

#4 studiot

studiot

    Genius

  • Senior Members
  • 6,484 posts
  • LocationSomerset, England

Posted 19 November 2015 - 09:01 PM

So how did your query arise?

 

You need to put more in here to get more out.


  • 0

#5 SF Shawn

SF Shawn

    Lepton

  • New Members
  • 4 posts

Posted 20 November 2015 - 05:05 PM

My previous question posted by mistake. I considered that there is a relation between them. Actually I just wanted to know about the line integral. Is every line integral independent of path?


Edited by SF Shawn, 20 November 2015 - 05:13 PM.

  • 0

#6 UT_PQED

UT_PQED

    Lepton

  • New Members
  • 8 posts

Posted 21 April 2016 - 12:47 PM

It is linked to the line only. But the the "integrand" itself may be parametric. Line is the path. :unsure:


  • 0

#7 HallsofIvy

HallsofIvy

    Baryon

  • Senior Members
  • 299 posts

Posted 22 September 2016 - 08:08 PM



My previous question posted by mistake. I considered that there is a relation between them. Actually I just wanted to know about the line integral. Is every line integral independent of path?

 

I am puzzled by this question.  Any text that introduces path integrals will tell you the conditions under which this is true.  If a path integral, \int_p^q f(x,y,z)dx+ g(x,y,z)dy+ h(x,y,z) dy  is independent of the path, then we could define a function F(x,y,z) by "F(x,y,z) is the integral from (0, 0, 0) to (x, y, z)".  But then it would follow that \frac{\partial F}{\partial x}= f, \frac{\partial F}{\partial y}= g, \frac{\partial F}{\partial z}= h.  But then we would have to have \frac{\partial^2 F}{\partial x\partial y}= f_y= g_x= \frac{\partial F}{\partial y\partial x}, \frac{\partial^2 F}{\partial x\partial z}= f_z= h_x= \frac{\partial F}{\partial z\partial x}, and \frac{\partial^2 F}{\partial y\partial z}= g_z= h_y= \frac{\partial F}{\partial z\partial y}. (A differential for which that is true is called an exact differential.)

 

As an easy example, if we integrate \int y^2dx+  xdy+ 2dz on the straight line from (0, 0, 0), we can take as parametric equations x= y= z= t so the integral becomes \int_0^1 t^2dt+ tdt+ 2dt= \int_0^1 (t^2+ t+ 2)dt= \left[\frac{1}{3}t^3+ \frac{1}{2}t^2+ 2t\right]_0^1= \frac{1}{3}+ \frac{1}{2}+ 2= \frac{17}{6}.

 

But if we integrate that same integrand from (0, 0, 0) to (1, 1, 1) by taking the path (0, 0, 0) to (1, 0, 0), then to (1, 1, 0), then to (1, 1, 1) we get:

  On (0, 0, 0) to (1, 0, 0) take x= t, y= 0, z= 0.  The integral becomes \int_0^1 0 dt= 0.

  On (1, 0, 1) to (1, 1, 0) take x= 1, y= t. z= 0.  The integral becomes \int_0^1 1 dt= 1.

  On (1, 1, 0) to (1, 1, 1) take x= 1, y= 1. z= t.  The integral becomes \int_0^1 2dt= 2.

The integral from (0, 0, 0) to (1, 1, 1), along that path is the sum 0+ 1+ 2= 3.

 

  Here, of course, \frac{\partial y^2}{\partial y}= 2y\ne 1=  \frac{\partial x}{\partial x}


  • 0




0 user(s) are reading this topic

0 members, 0 guests, 0 anonymous users