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Line Integral

  1. Nov 7, 2008 #1
    1. The problem statement, all variables and given/known data
    Evaluate the line integral, where C is the given curve:

    [tex]\int_C\,y\,ds \, \,\,\,\,C:\,x=t^2\,\,\,\,y=t[/tex]

    for t between 0 and 2 (out of curiousity, how do you make the "greater than or equal 2" in LateX)

    Okay, in the book the line integral is defined as [tex]\int_C F\cdot dr=\int_a^bF(r(t))\cdot r'(t)\,dt[/tex]


    Okay now what is F in this case? What is "ds"? what happened? Why is the problem format so different from the definition?

    Can someone help me out conceptually on this one?


    Thanks!!
    Casey
     
  2. jcsd
  3. Nov 7, 2008 #2

    gabbagabbahey

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    Hi Casey,

    [itex]\geq[/itex] is just \geq :smile:

    In this format, [itex]s[/itex] is the arc length of the path that is described by your parametric equations for [itex]x(t)[/itex] and [itex]y(t)[/itex] and so [itex]ds[/itex] represents an infinitesimal change in arc length....look at your notes; how is arc length defined?....how do you express it as a function of x and y?
     
  4. Nov 7, 2008 #3

    HallsofIvy

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    \ge

    The first form is in terms of parametric equations. The second is in terms of vector functions. One can always be converted to the other.
    "ds" is normally the differential of arclength. In this case, with x= t2, y= t, [itex]ds= \sqrt{(dx/dt)^2+ (dy/dt)^2}dt= \sqrt{4t^2+ t^2}dt= t\sqrt{5}dt[/itex] and y is, of course, t. The integral is
    [tex]\int_0^2 t (t\sqrt{5}dt)= \sqrt{5}\int_0^2 t^2 dt[/tex]
     
  5. Nov 7, 2008 #4
    What I am really confused about is which represents my Field and which is my r ?
     
  6. Nov 7, 2008 #5

    gabbagabbahey

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    hmmmm... isn't [tex]\frac{dy}{dt}=1[/tex]?
     
  7. Nov 7, 2008 #6

    gabbagabbahey

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    Remember that the speed along a curve, is equal to the rate of change of the arc lentgh :

    [tex]|\vec{r'}(t)|=|\frac{d \vec{r}(t)}{dt}|=\frac{ds}{dt}[/tex]

    And that the velocity along the curve is just speed*tangent vector

    [tex]\Rightarrow \vec{r'}(t)dt=\hat{T} \left(\frac{ds}{dt} \right) dt= \hat{T}ds[/tex]

    So when you compare that to the form of the path integral in your book, you see that the field in this case could be:

    [tex]\vec{F}(\vec{r})=y\hat{T}[/tex]

    ...Anyways, this is unimportant for this problem: the integral has already been parameterized for you, so all you need to do is express [itex]y[/itex] and [itex]ds[/itex] in terms of [itex]t[/itex] and integrate.

    Do you follow?
     
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