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Significance of parabola in a line integral?

  1. May 7, 2012 #1
    If $${\tmmathbf{r}}$$ and $${\tmmathbf{s}}$$ are piecewise smooth paths, which have the same graph, then they are said to be equivalent paths.

    They either trace out a set of points in the same direction, or in the opposite direction.

    If they trace out a curve $${C}$$ in the same direction

    $${\int_C \tmmathbf{F} \cdot \tmmathbf{r}' (t) d t =}$$ $${{\int_{C_{}} \tmmathbf{F} \cdot \tmmathbf{s}' (t) d t}}$$

    If they trace out a curve $${C}$$ in the opposite direction

    $${\int_C \tmmathbf{F} \cdot \tmmathbf{r}' (t) d t =}$$$${{- \int_{C_{}} \tmmathbf{F} \cdot \tmmathbf{s}' (t) d t}}$$

    Evaluate the path integral $${\int_C \tmmathbf{F} \cdot \tmmathbf{r}' (t) d t}$$ where

    $${\tmmathbf{F}=\tmmathbf{i}}$$

    and

    $${C}$$ is the parabola $${y = x^2}$$ traced out by $${\tmmathbf{r}}$$ so that the path goes from (0,0) to (1,1).

    What's the significance of mentioning that C is a parabola y=x^2? I mean can't we do without it?
     
  2. jcsd
  3. May 7, 2012 #2


    You should "preview post" before "submit post": something's wrong with your writing.

    DonAntonio
     
  4. May 7, 2012 #3
    I agree. So here I've rewritten it:
    Evaluate the path integral
    ∫F.dr (restricted to path C)

    where F=i
    and C is the parabola y=x^2 traced out by r.

    so that the path goes from (0,0) to (1,1).

    What's the significance of mentioning that C is a parabola y=x^2? I mean if we forget that C is the parabola y=x^2 and just take y=x^2 can't we do the question.
     
  5. May 7, 2012 #4


    I'm not sure I get your point: are asking why didn't they just wrote "..and C is y=x^2..", instead of "...and C is the parabola y = x^2..."??

    Well, who cares? As it happens, y = x^2 is a parabola, so they just point out his trivial fact.

    DonAntonio
     
  6. May 7, 2012 #5
    Thanks for clearing that part of the question.

    Now the curve is y=x^2. So what would r be? The same as the path or would it be ti +tj since we're going from (0,0) to (1,1).

    And now the last part (promise) I'm confused about.

    When we do ∫ on curve C of F.dr we only are going to do as follows:

    Define the terms of F in terms of r. (i.e. Suppose r=10ti+5tj and F= (2y+1)i +10xyj; F would be (10t+1)i+500t^2j.)

    Take dot product of F and dr ( dr=in this case would be 10dti+5dtj;)

    Integrate the result of the above dot product.

    That's all there is to it right. And that is this integrating over a curve?
     
  7. May 7, 2012 #6

    *** You have to fo from (0,0) to (1,1) along [itex]\,\,y=x^2\,\,[/itex] , or in other words: if you write parametrically

    this parabola, you get [itex]\,\,\{(t, t^2)\,;\,\,t\in\mathbb{R}\}\,\,[/itex] . Well, what is the range of values

    t has to obtain for the above parametric curve to go from (0,0) to (1,1)?




    Yes, that's all...and the important thing here is "integrating over (or better, along) the given curve, not ANY curve...

    DonAntonio
     
  8. May 8, 2012 #7
    Okay I get why the curve part is necessary. So in your e.g.( {(t,t^2 );t∈R} ) r would be ti +t^2j.

    And then using this value of r we would do the whole procedure of finding the integral of the dot product of F and dr. And this will integrating along the given curve since the r we are using is along the given curve. Right?
     
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