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Finding flow lines (vector calc problem)

  1. Mar 21, 2009 #1
    1. The problem statement, all variables and given/known data

    F = (x^2 / y) i + y j + k

    a) Use parametric equations to determine the equation for the flow line for the function F which passes thru the point (1,1,0)
    b) Show that the flow line also passes thru the point (e,e,1)


    2. Relevant equations

    F = F1 i + F2 j + F3 k
    dx/F1 = dy/F2 = dz/F3

    3. The attempt at a solution

    I didn't think parametric equations were actually needed here, but I think we're supposed to use them...

    Somebody told me
    x'[t] = x[t]2/y[t], y'[t] = y[t], z'[t] = 1 with the initial conditons that x[0] = 1, y[0] = 1, and z[0] = 0.
    The solution is: x[t] = et, y[t] = et, z[t] = t
    Setting t = 1 proves part b of the problem.

    But I get
    dx/(x^2 / y) = dt
    dy/y = dt
    dz = dt

    Thus,
    -y/x = t + c1 => x = -y / (t + c1)
    ln|y| = t + c2 => y = c3e^t
    z = t + c4

    where c1, c2, c3, c4 are constants
    What am I doing wrong? Also, why is t = 0?
     
  2. jcsd
  3. Mar 21, 2009 #2

    tiny-tim

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    Hi ! :smile:

    You can't integrate dx/(x^2 / y) = dt to get -y/x = t + c1 …

    you have to eliminate y first, by expressing it as a function of t (or of x, I suppose).

    And you're given (1,1,0), so if you're using t = z as a parameter, then obviously (1,1,0) corresponds to t = 0 :smile:
     
  4. Mar 21, 2009 #3
    Thanks but I'm still not getting x = e^t, y = e^t. How do you eliminate y?
     
    Last edited: Mar 22, 2009
  5. Mar 22, 2009 #4
    You can't just take the antiderivative of x and treat y as a constant? (Thus getting -y/x)? When I try eliminating y I get

    dx / (x^2 / y) = dy/y
    y*dx/x^2 = dy/y
    dx/x^2 = dy/y^2
    -1/x = -1/y => x = (1/y + c5)^-1

    If y = c3e^t as calculated earlier,
    x = [(1/c3e^t)+c5]^-1 = c6e^t
    Is it ok to eliminate y in this manner (sort of after the fact since I still take dx/(x^2/y)? How else would you do it?
     
    Last edited: Mar 22, 2009
  6. Mar 22, 2009 #5

    tiny-tim

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    Hi jaejoon89! :smile:

    (try using the X2 and X2 tags just above the Reply box :wink:)
    No! :rofl:

    yes, that's fine, except it would have been a lot simpler to leave dt as it was, and just substitute y = c3et, so as to give:
    dx /x2 = c3e-t dt :wink:
     
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