SF Shawn 0 Posted November 16, 2015 Share Posted November 16, 2015 I want to know about the functions which is path independent and path dependent. Please describe me about this. Link to post Share on other sites

studiot 2240 Posted November 17, 2015 Share Posted November 17, 2015 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? Link to post Share on other sites

SF Shawn 0 Posted November 19, 2015 Author Share Posted November 19, 2015 Yes I want to know about conservative and non conservative fields. Also I want to know about zero and nonzero curl. Link to post Share on other sites

studiot 2240 Posted November 19, 2015 Share Posted November 19, 2015 So how did your query arise? You need to put more in here to get more out. Link to post Share on other sites

SF Shawn 0 Posted November 20, 2015 Author Share Posted November 20, 2015 (edited) 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 November 20, 2015 by SF Shawn Link to post Share on other sites

UT_PQED 0 Posted April 21, 2016 Share Posted April 21, 2016 It is linked to the line only. But the the "integrand" itself may be parametric. Line is the path. Link to post Share on other sites

Country Boy 70 Posted September 22, 2016 Share Posted September 22, 2016 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, [math]\int_p^q f(x,y,z)dx+ g(x,y,z)dy+ h(x,y,z) dy[/math] 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 [math]\frac{\partial F}{\partial x}= f[/math], [math]\frac{\partial F}{\partial y}= g[/math], [math]\frac{\partial F}{\partial z}= h[/math]. But then we would have to have [math]\frac{\partial^2 F}{\partial x\partial y}= f_y= g_x= \frac{\partial F}{\partial y\partial x}[/math], [math]\frac{\partial^2 F}{\partial x\partial z}= f_z= h_x= \frac{\partial F}{\partial z\partial x}[/math], and [math]\frac{\partial^2 F}{\partial y\partial z}= g_z= h_y= \frac{\partial F}{\partial z\partial y}[/math]. (A differential for which that is true is called an exact differential.) As an easy example, if we integrate [math]\int y^2dx+ xdy+ 2dz[/math] on the straight line from (0, 0, 0), we can take as parametric equations x= y= z= t so the integral becomes [math]\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}[/math]. 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 [math]\int_0^1 0 dt= 0[/math]. On (1, 0, 1) to (1, 1, 0) take x= 1, y= t. z= 0. The integral becomes [math]\int_0^1 1 dt= 1[/math]. On (1, 1, 0) to (1, 1, 1) take x= 1, y= 1. z= t. The integral becomes [math]\int_0^1 2dt= 2[/math]. The integral from (0, 0, 0) to (1, 1, 1), along that path is the sum 0+ 1+ 2= 3. Here, of course, [math]\frac{\partial y^2}{\partial y}= 2y\ne 1= \frac{\partial x}{\partial x}[/math] Link to post Share on other sites

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