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Work done by F for a particle moving along C

  1. Nov 15, 2006 #1
    Here's the question:
    If C is the curve given by http://ada.math.uga.edu/webwork2_files/tmp/equations/01/126d71403184689027c86bb27202771.png [Broken],[/URL] http://ada.math.uga.edu/webwork2_files/tmp/equations/56/f3b3a7b9c3e5d804f53afad7a07bc71.png [Broken] and F is the radial vector field http://ada.math.uga.edu/webwork2_files/tmp/equations/28/71b024e85380a1ae00268ee0af38721.png [Broken],[/URL] compute the work done by F on a particle moving along C.

    This is what I've done so far...
    F = r because F is only x, y, z
    dr/dt = 2cos(t)i+5sin(2t)j+6cos(t)sin^2(t)k

    Then F dot dr/dt =

    Now cos(pi/2) = 0 and sin(pi) = 0 - from 5sin(2t) - so when plugging in pi/2 the answer should be zero for all.
    sin(0) = 0 and cos(0) = 1, so when evaluated at 0 the only answer should be 2 (2cos(0)*1) so by my account the answer should be -2.

    Apparently this isn't the right answer, if someone could tell me what I missed it'd be awesome.

    Thanks a lot.
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Nov 15, 2006 #2
    F dot dr is the incremental work along the path... you need to integrate this over the entire path, not just plug in a value!

    In this case, W = Int[0, pi/2] F dot dr/dt dt.

    Also be careful... you lost a 2 in 2cos(t)(1 + 2sin(t))!
  4. Nov 15, 2006 #3
    oh wow, that's a great point.

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