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Determining Motion from a Derivative

  1. Jan 7, 2015 #1
    1. The problem statement, all variables and given/known data
    Given position function r(t) and r'(t) = c X r(t), where c is some constant vector, describe the path of the particle. In other words, describe r(t).

    2. Relevant equations
    //

    3. The attempt at a solution


    a) r'(t) points in the direction of motion. If we can understand how r'(t) changes direction, we can understand how r(t) moves.
    b) r'(t) is the cross product of c and r(t). Therefore, the function is perpendicular to both c and r(t).
    c) If r'(t) is perpendicular to c, r'(t) lies on a plane that c is normal to.
    d) ---- No other leads
     
  2. jcsd
  3. Jan 7, 2015 #2

    RUber

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    Consider iterating a few times with Newton's method to see what happens.
    For starters, assume that C = Identity and evaluate the cross product.
     
  4. Jan 10, 2015 #3
    what does C = identity mean, sorry? a constant function?

    if c is given by <C1,C2,C3> and I cross that by r(t) <R1,R2,R3>, I get a cross product.:
    <C2(R3) - C3(R2), C1(R3) - C3(R1), C1(R2) - C2(R1)>

    I then multiply that cross product by time interval t. This gives the position function's next approximate position.

    $<t[C2(R3) - C3(R2)], t[C1(R3) - C3(R1)], t[C1(R2) - C2(R1)]t>

    Subsequent iterations become very messy. Are you sure about this? No conceptual way of understanding the problem?
     
  5. Jan 10, 2015 #4

    haruspex

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    Dot both sides with r(t) and solve. What does this tell you about |r(t)|?
     
  6. Jan 10, 2015 #5

    vela

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    You might find it helpful to choose a coordinate system to make the analysis easier. Orient the coordinate system so that ##\vec{c}## points along the z-axis. That'll give you a much simpler expression for ##\vec{c}\times\vec{r}##.
     
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