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Line Integrals and Finding Parametric Equations

  1. Mar 2, 2014 #1

    dwn

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    I am having a difficult time finding the parametric equations x = x(t) and y = y(t) for line integrals. I know how to find them when dealing with circles, but when it comes to finding them for anything else, I don't see the method...it all seems very random. I did fine with finding the parametric equations earlier in the semester, but this seems very different.

    For example: ∫ ydx + x^2dy (0,-1) to (4,-1) to (4,3)

    How do you determine the parametric equations for this problem?
     
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  3. Mar 2, 2014 #2

    CAF123

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    You first have to specify the path with which you want to integrate your function over. The given points is suggestive of a path consisting of two straight lines, one from (0,-1) to (4,-1) and an another from (4,-1) to (4,3).

    So, in this case, you have to parametrise a straight line since this is the path you are integrating over.
     
  4. Mar 2, 2014 #3

    dwn

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    Does that mean you can choose any linear function(t)? x = t y = 2t? Are there infinitely many possibilities?

    Which interval is integrated?

    thanks
     
  5. Mar 2, 2014 #4

    PeroK

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    You don't always need to parameterise a straight line. If you are integrating wrt x from (0, -1) to (4, -1), then y = -1 and x goes from 0 to 4. There's no need to introduce an additional variable. Also, dy = 0 on a horizontal line.
     
  6. Mar 2, 2014 #5
    No, the parametric equations have to correspond to the two points. There are infinitely many possibilities, but they all only differ in the interval of t.

    One fairly simple way of determining an equation is to first find a function in terms of x and y -- for two points, you can find the slope and then use point-slope form to find the equation. Then let x equal t, and then solve for y in terms of t. The interval of t in this case would be the interval in x.

    Another, more creative way of finding them is to first start with your starting points -- for the first interval, [itex](0, -1)[/itex] -- then think about how you get to your end point. To go from 0 to 4, you need an increase of 4, and you don't need to change -1 at all. If you think about t being a step, this means x has to go from 0 to 4 in t steps. So maybe 4 in 1 step, 2 in 2 steps, 3 in 4/3 steps, etc. This is where the infinitely many paths comes in. The simplest is usually 1 step -- that is, [itex]0 \leq t \leq 1[/itex]. This would give you the parametric equations:

    [itex]x(t) = 0 + 4t, y(t) = -1, 0\leq t \leq 1[/itex]. You can check this by plugging in 0 and 1 for t and seeing that you get the points corresponding to your endpoints.
     
  7. Mar 2, 2014 #6

    PeroK

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    What is the purpose of that parameterisation? Why not use x as the parameter?
     
  8. Mar 2, 2014 #7
    Yes, I know using t in this case is a bit overkill, but it's important to know how to parameterize things since you don't always have just one variable changing.
     
  9. Mar 2, 2014 #8

    PeroK

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    Okay, so you've got two ways to do this. Maybe try it both ways and make sure you get the same result:

    a) Use an additional parameter t (going from 0 to 1) in each case.

    b) Use x and y as your parameters for the first and second parts of your curve.
     
  10. Mar 2, 2014 #9

    dwn

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    What about the point (4,3)? Where does this come into play?

    Is it just common practice to setup the interval 0 to 1 for t, when you parameterize in terms of t?


    Thanks for all the help, I'm taking this class online and this has me stumped.
     
  11. Mar 2, 2014 #10
    The other point is the endpoint of the second line -- you have two paths, so you'll have to do an integral for each one. And yes, it's just convention to have an interval of 0 to 1 since this is usually the easiest interval to set up, and also easiest to evaluate as bounds on your integral. You can use any interval you want, so long as the interval correctly describes the path in question.
     
  12. Mar 2, 2014 #11

    dwn

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    How I understand it based on your responses...is this correct?

    Code (Text):
    ∫ ydx + x^2dy      

    y = -1   x = 4t
    dy = 0   dx = 4
     0 to 1  interval for t
    ∫ -1(4)dt + 4t^2(0)dt

    -4t |[SUP]1[/SUP]  = -4
    Δx = 4-0, therefore x = 0 +4t → x = 4t

    Why is it -1 and not Δy = -1 - (-1) = 0 ?
     
    Last edited: Mar 2, 2014
  13. Mar 3, 2014 #12

    CAF123

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    It is correct, yes, so now do the same again for the other straight line segment comprising your path. Note that in some cases, the conventional parametrisation for t in the interval [0,1] will not allow you to actually compute the integral. Here though, it is ok.

    I am not quite sure I understand the question. On segment 1, Δy = 0 and only on segment 2, which is treated separately, does y change.
     
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