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Line integral calculations

  1. Jun 20, 2013 #1
    Can someone tell me where my calculations are going wrong.

    I am integrating over C2: (Note Line integral over C1 and C3 are zero.)

    NOTE: The vector function f(x,y,z) that I am integrating over C2 is highlighted in red in the paint doc.

    The equation that I am using is: ∫[f (dot) unit tangent]ds

    Equation of C2: y = 1 - x
    Parameterization of C2: y = 1 - s and x = s

    Vector function for C2 is: r(s) = (s,1-s)
    d(r(s))/ds = (1,-1)
    Unit d(r(s))/ds = (1,-1)/√2

    f(x,y,z) = (y,-x) = (1-s,-s)


    f (dot) unit tangent = (1-s,-s) dot (1,-1)/√2 = 1/√2

    Integrating line integral from 0 to √2 I get: 1/√2(s) from 0 to √2 = 1.

    The answer is -1.
     

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  3. Jun 20, 2013 #2

    Dick

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    1) pay some attention to which direction C2 goes in, r(s)=(s,1-s) goes in the wrong direction for s in [0,1]. 2) You are just integrating F(r).dr. dr doesn't have to be a unit vector and your parameter s isn't in [0,sqrt(2)], it's in [0,1]. You are making two mistakes that are cancelling and letting you get close to the correct answer.
     
  4. Jun 21, 2013 #3
    I see where the interval s in [0,1] comes from. It comes from the fact that x = s.
    From my equation r(s)=(s,1-s) I would have to integrate from 1 to 0.
    However I don't understand why I wouldn't integrate the dot product of f and the unit tangent to the curve. I used the equation ∫[f (dot) unit tangent ]ds
     
  5. Jun 21, 2013 #4

    HallsofIvy

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    "Unit tangent to the curve". The path in this problem is a straight line. Any "tangent" vector just points along the line itself. Even with a true curve, the unit tangent is not really necessary. The length will be taken care of in the differential.
     
  6. Jun 21, 2013 #5
    But the reason why we dot the vector field with the unit tangent to the curve is so we can obtain only the component of the vector field in the direction of the curve or parallel to the unit tangent.

    and of course you have to dot it with the unit tangent: Look at the equation it is....
    f (dot) ds, where f is the vector field we wish to integrate along the curve and ds is the vector in the direction of the curve.

    The vector ds is equal to (unit tangent)ds where unit tangent is the directional vector in the direction of motion along the curve and ds is the magnitude of the directional vector.

    What you guys are saying is we should use (tangent vector)ds whish is equal to (unit tangent)(ds)2
     
    Last edited: Jun 21, 2013
  7. Jun 21, 2013 #6

    Dick

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    Just use (tangent vector)ds. Not (unit tangent vector)ds.
     
  8. Jun 21, 2013 #7

    haruspex

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    Your argument would be correct if you had defined s to be the distance along C2. But you didn't, you defined it as a parameter going from 0 to 1.
     
  9. Jun 23, 2013 #8
    The only logical explanation I can think of is by parameterizing the equation for the curve then creating a vector function for that curve (which can be thought of as the position vector for the curve) the direction of the curve would some how automatically be given by the tangent vector which is equal to the derivative of the position vector I defined above. It does this because the original position vector that I defined above isn't just related to the curve but it is the actual definition of the curve in vector form just as the equation for the curve defines that curve. Therefore the tangent is related the same way and therefore automatically gives the proper direction without having to calculate the unit tangent.

    That is my only explanation.

    I have a definition that I would like someone to verify.
    If my above explanation is correct then the following should be correct.

    given the function G = g(x,y,z) that defines the curve C and a vector field F = f(x,y,z) the following line integral
    ∫(F(x,y,z))dot(unit tangent(x,y,z))ds is equivalent to the following integral
    ∫(F(x(s),y(s),z(s))dot(tangent(x(s),y(s),z(s)) where tangent(x(s),y(s),z(s)) = dG/ds and x = x(s),y=y(s), and z = z(s) are the parameterizations of the curve C that create the position vector function g(x(s),y(s),z(s)) to C.
     
  10. Jun 26, 2013 #9

    Dick

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    The definition of the line integral is independent of parametrization. You will only get a unit vector for the derivative of the curve if you chose arc length parametrization. You don't have to. That's the whole story.
     
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