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Another Line Integral Problem

  1. Apr 29, 2005 #1
    Greetings All Again,

    I wanted to thank you for the reply on my other problem, it was indeed very helpful and this is a very strange problem. So here goes :yuck: :

    Compute the work done by the force field F(x,y,z) = <4y,2xz,3y> acting on an object as it moves along the helix defined parametrically by
    x=2 cos t, y=2 sin t, z=3t, from the point (2,0,0) to the point (-2,0,3(pi)).

  2. jcsd
  3. Apr 29, 2005 #2
    this is just

    [tex]\int_C 4y \ dx + 2xy \ dy + 3y \ dz[/tex]

    where C is the curve you defined above. You know the parameterization. Just solve for the values of [itex]t[/itex] that give you the endpoints (to get the appropriate bounds), and sub in the parameterized x, y, z, dx, dy, dz.
  4. Apr 30, 2005 #3
    With this problem, assuming i worked it correctly, i got the values for the parameticize. I am confused about what values to plug into x,y, and z. either what i just solved or the other values for x,y, and z. x = 2cost, y = 2sint, and z = 3t.

    r(t)=(1-t)<2,0,0> + t<-2,0,3(pi)>


    I hope this makes sense.
  5. Apr 30, 2005 #4
    i have absolutely no idea what you're doing there. The parameterization is given in the question, x(t) = 2cos t, y(t) = 2sin t, z(t) = 3t. All you need to do is find the bounds on t (using the endpoints given) and take the integral.
  6. Apr 30, 2005 #5
    physicsmajor, what you've done on this probllem, with yoru parametricization, is turned your path from a helix into a straight line. you should be computing along the helix.

    now, if your vector field is conservative, you can do that, the straight line approach, which makes things much easier, but, i'd imagine it's not conservative or you wouldn't be told to integrate along the helix.
  7. Apr 30, 2005 #6
    thanks trance, i actually figured that out a few hours after i made that post. i guess i am having trouble finding the limits to integrate from. I believe they are from 0 < t < 3(pi), but i am not sure.
  8. Apr 30, 2005 #7
    0<t is correct, but at the second endpoint you need [itex]z=3t=3\pi[/itex], and does [itex]t=3\pi[/itex] solve that?
  9. Apr 30, 2005 #8
    no it does not, so its from 0<t<(pi)
  10. Apr 30, 2005 #9
    sounds good to me.
  11. Apr 30, 2005 #10
    you the man data!!!!!!!!
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