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Work done by Force Field

  1. May 30, 2011 #1
    Find the work done by the force field F= 3yi - x^2j
    in moving a particle along the curve y=x^(1/2) from (1,1) to (4,2)


    so i parameterize and get x=1+3t, y= 1+t

    so

    3(1+t)-(1+t)^2

    then take the derivative of x and y and multiple to ge
    9(1+t)-(1+t)^2

    and i integrate that from 0-1
    and i get 13/2


    but the answer should be 39/5

    so my question is where am i going wrong
     
  2. jcsd
  3. May 30, 2011 #2

    hunt_mat

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    The work done is given by:
    [tex]
    W=\int_{\gamma}\mathbf{F}\cdot d\mathbf{r}
    [/tex]
    where [itex]d\mathbf{r}=dx\mathbf{i}+dy\mathbf{j}[/itex]
     
  4. May 30, 2011 #3
    yes it is
     
  5. May 30, 2011 #4

    hunt_mat

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    if y=x^1/2, then dy=...

    Also on the curve you know that y=x^1/2, and so....
     
  6. May 30, 2011 #5
    dy=1/2x^-(1/2)
     
  7. May 30, 2011 #6

    hunt_mat

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    the correct answer is [itex]dy=1/2x^-(1/2)dx[/itex], insert this into F.dr to find that F.dr=...
     
  8. May 30, 2011 #7
    would i have to parameterize the x?
     
  9. May 30, 2011 #8

    hunt_mat

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    No, just work out F.dr in terms of x and dx.
     
  10. May 30, 2011 #9
    I still dont see how that is going to work
     
  11. May 30, 2011 #10

    HallsofIvy

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    This is wrong. If y= 1+ t, then t= y- 1 so x= 1+ 3(y- 1)= 3y- 2. That is linear- it is the straight line through (1, 1) and (4, 2).

    Use [itex]x= t^2[/itex], [itex]y= t[/itex], with t from 1 to 2 instead.
    Now F= 3ti- t^4 j, dx= 2tdt, dy= dt so
    [tex]\int_{t= 1}^2 (3t)(2t)dt- (t^4)dt= \int_{t=1}^2 (6t^2- t^4)dt[/tex]

     
  12. May 30, 2011 #11
    Thanks Halls, that made sense. I have one more question how do you know to to parameterize it the way you did? is it because y=x^1/2 so choose x=t^2 ad y=t
     
  13. May 30, 2011 #12

    HallsofIvy

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    Yes. Of course you could always use [itex]x= t[/itex], [itex]y= t^{1/2}[/itex] with t from 1 to 4. There are many different ways to parameterize any curve.
     
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