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Determing the equation of a streamline

  1. Sep 12, 2015 #1
    1. The problem statement, all variables and given/known data
    A fluid has velocity components of u=[30/(2x+1)]m/s and v=(2ty)m/s,where x and y are in meters and t is in seconds. Determine the equation of the streamline that passes through point (2 m, 6 m) at time t = 2 s. Write the equation in the form y={y(x)}m, where x is in meters.

    2. Relevant equations
    dy/dx = v/u

    3. The attempt at a solution
    So when I separate my variables, I get udy=vdx. That means my equations are (30/(2x+1))dy = (2ty)dx. Is it really just a matter of integrating such "simple" equations? Or do I need to do some other method/algebra to get the (2x+1) on the dx side and the (2ty) on the dy side? Then all I have to do is integrate and find my constants, correct?
     
  2. jcsd
  3. Sep 14, 2015 #2

    rude man

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    Separation of variables would be dy/v = dx/u. I don't see why you can't do as you wrote: separate variables, solve the ODE including the "initial conditions". If you want we can compare answers if you post yours first.
     
  4. Sep 14, 2015 #3
    So:

    dy/v = dy/2ty
    dx/u = (1/30)(2x+1)dx

    I get x2 + x + C1 and ln(y)/2t + C2

    How do I solve for those constants to get y=f(x) form using the given boundaries?
     
  5. Sep 14, 2015 #4

    rude man

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    What happened to "30"? Then, put an equal sign somewhere & I'm a buyer.
    Since it's a 1st order ODE you know there is only one boundary condition and one constant to be determined by the boundary condition. You can obviously combine your C1 and C2 into one constant. Have another shot at it, you're getting warmer.
     
  6. Sep 14, 2015 #5
    I divided the thirty into the dx equation.

    So it's legal to set these equal to each other, put in the conditions, solve for the constant (which, now I understand why no one was putting in both C1 and C2), then rearrange to have y on one side and x on the other?
     
  7. Sep 14, 2015 #6

    rude man

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    But it's missing in "I get x2 + x + C1 and ln(y)/2t + C2 ". Put it back in!
    Your equation is dy/v = dx/u. This has all y on the left and all x on the right. Integrate both sides, then solve for y explicitly. Try to show your work or I can't tell if & where you slipped.
     
  8. Sep 14, 2015 #7
    Right right.

    (1/30) x2 + x + C = ln(y)/2t

    I plug in my initial x,y,t points:

    0.2 + C/30 = 0.44794

    Solving for C yields C = 7.4382

    So now (and I'm going to just drop my thirty in the denominator):

    (x2 + x + 7.4382)/30 = ln(y)/2t
    (2tx2 + 2tx + 14.8764t)/30 = ln(y)
    e(2tx2/30 + 2tx/30 + 14.8764t/30) = y
     
  9. Sep 15, 2015 #8

    rude man

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    When you solved for the constant of integration, did you let t=2?
    Do that and you got it! My numbers agree if you let t=2 and compute the numerical value of e(14.8764t/30).
     
  10. Sep 16, 2015 #9
    I think I did, though I might have accidentally did 0 instead . . .
     
  11. Sep 16, 2015 #10

    rude man

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    Your last entry was e(2tx2/30 + 2tx/30 + 14.8764t/30) = y. Needed to get rid of "t".
     
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