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Basic Line-Integral: Just trying to know what is being asked

  1. Oct 16, 2016 #1
    Hello. I'm new to physics, and the problem I have seems so basic, mathematically speaking. I'm just failing to grasp exactly what is being asked. If I can find that, I believe I can find the answer. Here it is:

    1. The problem statement, all variables and given/known data

    Let A = x2ˆx + y2ˆy + z2ˆz

    Consider the parabolic path y2 = x between the points (0, 0) and (2, √2).

    By integrating over x, compute the line integral ∫(A ⋅ ds)

    2. Relevant equations
    • ds = (dx/dt)
    • dy/dx = (1/2)x

    3. The attempt at a solution
    Ok, so we're given a function, y = √(x), and asked to compute a line-integral "over x" under this curve. My questions at this point are:
    • Does "over x" mean with respect to x? (This may just be a problem of semantics.)
    • But, what is ds? Above is my guess at what it should be. Is it just a unit of change along the function y?
    I would very much appreciate any help you may be able to provide.
     
  2. jcsd
  3. Oct 17, 2016 #2

    haruspex

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    You are not integrating a function y(x). The thing to be integrated is the vector function A. This is a function of the position vector (x, y, z). When integrating such a beast, a path or area or volume has to be specified to integrate it over. In this case, you are given the path (x, √x) from (0,0) to (2, √2).
    ds is a vector element along that line, (dx, dy). It is not dx/dt.
    What does not make sense is that A is specified as a 3D vector, but everything else is as though we are working in 2D.
     
  4. Oct 22, 2016 #3
    Thank you, haruspex.

    I was able to find a solution to this problem.

    Like you said, the goal is to integrate a vector function (A = x2ˆx + y2ˆy + z2ˆz) dotted with the derivative of the position vector (ds), over a specified path [i.e. from (0, 0) to (2, √2)]. Even though the vector function A is given in three dimensions, we can just ignore the z-dimension, because the function we're integrating over has only two dimensions.

    But what is ds? It's just the derivative of s, which is a position vector, or a vector whose coordinates point to any position along our function (y2 = x...or...y = √x). Because it's a vector, it can be represented as s = (some explanation of the behavior of x)ˆx + (some explanation of the behavior of y)ˆy. Again, the z-dimension can be ignored. It is important to get both of those relationships in terms of x. The x-coordinate is obvious (why?) and the manner in which the y-coordinate behaves is easily explained by the given function. So, we have:

    s = xˆx + √xˆy

    To find ds, which is an infinitesimal change along the function's path, we just find the derivative of s in terms of x, as below. Please forgive the crude notation.

    ds = dxˆx + (½x)ˆy → ds = dxˆx + (½x)ˆy * dx​

    Once we know all that, we just "pull the lever" and do the calculations we're asked to. First we take the dot product of those two vectors (A and ds), then integrate the result, and evaluate it in terms of our given x- and y-coordinates.

    (Important side-note: How does A relate to the given function? Of that, I'm not sure, and it's really necessary information to explain all of what's happening.)
     
  5. Oct 22, 2016 #4

    haruspex

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    I do not understand the question. ##x^2\hat x+y^2\hat y+z^2\hat z ## is just some arbtrary vector function of position. A is the name the question setter has given it.
     
  6. Oct 22, 2016 #5
    Right; but I'm not sure how A relates to the function, ##y = \sqrt x##. It seems that the function gives us our path, but then what is A used for?
     
  7. Oct 22, 2016 #6

    haruspex

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    It is just the function to be integrated. It need not have any physical significance, and there need be no relationship between A and the path function.
    Here's a scalar example which has physical meaning: suppose the coefficient of kinetic friction of a level floor varies from place to place and is described by f(x,y). The integral of Mgf along some path p(x,y)=0 gives the work done in dragging an object weight Mg along that path.
     
  8. Oct 22, 2016 #7
    Ah! I see! I just find it counter-intuitive that A need not have any relationship to ##y = \sqrt x##. I now understand that my original question should have been how the integral we're asked to evaluate, ##\int (A \cdot ds)##, is relevant to the function ##y##. They just seemed like separate ideas, but I now see that ##y## tells us both about the bounds of this integral and about ds.

    Thank you very much!
     
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