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Help Setting Up Work Over a Path Problem

  1. Jul 11, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the work done by force F from (0,0,0) to (1,1,1).

    F=3yi+2xj+4zk

    The path C_3 U C_4 consisting of the line segment from (0,0,0) to (1,1,0) followed by the segment (1,1,0) to (1,1,1)

    2. Relevant equations

    None

    3. The attempt at a solution

    I am trying to find r(t) for both line segments.

    For r(t)_1 I tried (1-0)i+(1-0)k+(0-0)j. I got that from x_2-x_1, etc. This worked out for the first r(t) netting me ti+tj

    I then tried this for the second line segment and ended with tk only. The correct r(t) is i+j+tk

    So clearly I am solving the r(t) part incorrectly. The book doesn't go over how to get the r(t) - that is where I need the help. I am thinking it might have something to do with the original (0,0,0) to (1,1,1) but I'm not sure.

    How does one obtain both r(t)'s when only points are given?
     
  2. jcsd
  3. Jul 11, 2010 #2
    Hey, don't worry, your method is basically right:

    so your r(t)_1 is completely right and your r(t)_2 makes sense but has a small error: you were thinking of a vector only, correct? When you use the x_2 - x_1, etc... you're determining the direction of the path, that sounds logical, because you're only using the differences (ending point minus start point), but take in account that r(t) is not only a vector pointing in the right direction (your direction is correct!), but is a path that has a beginning and end point! Make a drawing of the x,y,z-axes, the points (0,0,0), (1,1,0) and (1,1,1) and then the arrows pointing from (0,0,0) to (1,1,0) and from (1,1,0) to (1,1,1), you'll see where you went wrong: tk is the path from the origin to (0,0,1), but that's not the path you're taking: it has to go through the point (1,1,0). So how do you "fix" the tk correctly? r(t) = tk + constant, with the constant determining where the path is (the direction is already fixed, of course), and as r(0) = (1,1,0), we get the right answer r(t) = i + j + tk
     
  4. Jul 12, 2010 #3
    Got it. That makes sense. I'll try a few more after work and see if I've got it down.

    Thanks for the help!
     
  5. Jul 12, 2010 #4
  6. Jul 12, 2010 #5
    Turns out, I don't get it. :confused:

    I have another problem where I am supposed to evaluate [tex]\int_{C}xydx+(x+y)dy[/tex] along the curve y=x^2 from (-1,0) to (2,4).

    If I start subtracting the two points I get 3i + 3j. The correct answer for r(t) is ti + t^2j.

    I see in the solutions manual they set x=t and therefore get y=t^2...I understand that - although not sure exactly why they used x=t, I am assuming just for ease of solving sake.

    So anyway, how did they arrive at xi +yj for r(t)?
     
  7. Jul 12, 2010 #6

    vela

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    That's the definition of r(t).
     
  8. Jul 13, 2010 #7
    That's where I am having trouble. We just started using r(t) with no explanation of what it is and I can't see anywhere that explains it in the book.

    It almost seems like r(t)= i+j+k and then points are used to find the values. But that's only from the problem sets I've done.

    I don't really understand what r(t) is so I can't conceptually get it or look it up anywhere outside of my book.
     
  9. Jul 13, 2010 #8

    vela

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    It's just a vector from the origin to a point on the path, so r(t)=x(t)i + y(t)j + z(t)k. As t varies, the tip of the vector moves along the path.

    For example, in your first problem, you had
    For C3, you could use r(t) = (1,1,0)t = t i + t j, so as t varied from 0 to 1, the tip would follow the line segment joining (0,0,0) and (1,1,0). For C4, you could use r(t) = (1,1,0)+(0,0,1)t = 1 i + 1 j + t k so as t goes from 0 to 1, r would go from (1,1,0) to (1,1,1) along the line segment connecting them.

    In this latest problem, they told you how x and y are related. If you figure out some expression for x, you can just square it to find y. The simplest thing you can do is just set x to the parameter t. Note that this isn't the only parameterization you could use. If you wanted to make things more difficult for yourself, you could, for instance, use x=t2, then y=t4. As long as (x(t),y(t)) follows the curve, you're fine.

    By the way, are you sure the problem says from (-1,0) to (2,4) and not (0,0) to (2,4)? The curve y=x2 doesn't pass through (-1,0).
     
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