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Parametric curves applications

  1. Oct 5, 2007 #1
    Q: A particle is following the path C: f(t)=(2cos(t), 2sin(t), t), t>=0, and flies off on the tangent line at time t=3pi/2. Find the position of the particle at time t=5pi/2.


    Equation of the tangent line:
    l(s)=(0,-2,3pi/2) + s(2,0,1)

    s=0, (0,-2,3pi/2)
    So s=5pi/2 - 3pi/2 = pi gives position at t=5pi/2

    l(pi)=(0,-2,3pi/2) + pi (2,0,1)
    = (2pi, -2, 5pi/2) [answer]

    I don't understand the red part. How come s=pi gives position at t=5pi/2 ? What is the relation between s and t? Are they realted linearly?

    It would be nice if someone can explain this part. Thanks!
  2. jcsd
  3. Oct 5, 2007 #2


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    [tex]s = t - \frac{3\pi}{2}[/tex]

    It just represents the starting time when the particle leaves the curve.
  4. Oct 5, 2007 #3
    But I don't get WHY...
    Why not, say, s = t^2 -3pi/2 or s= t - 2* 3pi/2?
  5. Oct 5, 2007 #4


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    It's simple variable substitution. Here's an analogy. Say you have the equation [itex]y = m(x - x_0) + y_0[/itex]. For convenience, I can let [itex]s = x - x_0[/itex], which means [itex]y = ms + y_0[/itex]. Same thing is happening here.

    Though I haven't seen parametric equations written in this form before :yuck:, I think the equation of the tangent line could have been written (w/o variable substitution) as:

    l(t)=(0,-2,3pi/2) + (t-3pi/2)(2,0,1)

    letting s = t-3pi/2 results in:

    l(s)=(0,-2,3pi/2) + s(2,0,1)

    When t = 5pi/2, s = 5pi/2-3pi/2 = pi
    Last edited: Oct 5, 2007
  6. Oct 6, 2007 #5


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    If it is 3:00 now, how long will be until it is 5:00?

    Would you even consider saying s = (5)2 -3 or s= 5 - 2(3)?
  7. Oct 8, 2007 #6
    Why would the position on the tangent line be moved by (2,0,1) per second?
    Say, l(s)=(0,-2,3pi/2) + s(8,0,4), I changed the direction vector, but this would still represent the same line, how can I figure out how s and t are related now?

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