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Path integrals and parameterization

  1. Feb 21, 2013 #1
    1. The problem statement, all variables and given/known data
    Evaluate ∫ F ds over the curve C for:
    a) F = (x, -y) and r(t) = (cos t, sin t), 0 ≤ t ≤ 2∏
    b) F = (yz, xz, xy) where the curve C consists of straight-line segments joining (1, 0, 0) to (0, 1, 0) to (0, 0, 1)

    2. Relevant equations




    3. The attempt at a solution
    a) I first found the norm:
    r'(t) = (-sin t, cos t)
    ||r'(t)|| = 1
    Now my question is, how do I set up the integral? I know that F = (cos t, -sin t), and I have found the norm, but I'm lost about how toI go about setting up the integral.

    b) Let C1 be the vector from (1, 0, 0) to (0, 1, 0) and let C2 be the vector from (0, 1, 0) to (0, 0, 1)
    I parametrize C1: r(t) = (1-t, t, 0)
    C2: r(t) = (0, 1-t, t)
    Again, I can find the norm easily once I have the parametrization, but how do I set up the integral?
     
  2. jcsd
  3. Feb 21, 2013 #2
    Don't you have a formula for this? It should be ##\displaystyle \int_r \mathbf{F} \ ds = \int_{a}^b \mathbf{F}(r(t))|r'(t)| \ dt##
     
  4. Feb 22, 2013 #3

    HallsofIvy

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    Science Advisor

    Yes, the norm for the first problem was the constant 1 because the path is the unit circle. But there is no need to find the norm at all.

    With [itex]\vec{F}= f(x,y)\vec{i}+ g(x,y)\vec{j}[/itex] and [itex]d\vec{s}= dx\vec{i}+ dy\vec{j}[/itex], [itex]\vec{F}\cdot d\vec{s}= f(x,y)dx+ g(x,y)dy[/itex]

    And with parameterization x= x(t), y= y(t), that becomes simply
    [tex]\int \vec{F}\cdot d\vec{s}= \int f(x,y)(dx/dt)dt+ g(x,y)(dy/dt)dt[/tex].

    In problem (b), [itex]\vec{F}= x\vec{i}- y\vec{j}= cos(t)\vec{j}+ sin(t)\vec{k}[/itex] and [itex]d\vec{s}= sin(t)dt\vec{i}- sin(t)dt\vec{j}[/itex].

    For (b), again, the norm is not relevant. Do this as three separate integrals:
    (I) from (1, 0, 0) to (0, 1, 0): x= 1- t, y= t, z= 0. [itex]d\vec{s}= -dt\vec{i}+ dt\vec{j}[/itex].
    (II) from (0, 1, 0) to (0, 0, 1): x= 0, y= 1- t, z= t. [itex]d\vec{s}= -dt\vec{j}+ dt\vec{k}[/itex].
    (III) from (0, 0, 1) to (1, 0, 0): x= t, y= 0, z= 1- t. [itex]d\vec{s}= dt\vec{i}- dt\vec{k}[/itex]
     
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