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Curve C traced by a particle, did i draw these vectors right?

  1. Oct 8, 2005 #1
    Ello again. The problem states:
    The figure shows the curve C traced by a particle with position vector r(t) at time t.

    (a) Draw a vector that represents the average velocity of the particle over the time interval [3,3.2]
    (b) Write an expression for the velocity v(3)
    (c) Write an expression for the unit tangent vect0or T(3) and draw it.


    http://img208.imageshack.us/img208/83/lastscan7jy.jpg

    I drew the vector, and in b, i think thats what they want but i'm not sure, and with C, i drew the Tagent vector t, but i'm not sure about how to write an epxression for unit vecotr T. THanks.
     
  2. jcsd
  3. Oct 8, 2005 #2

    HallsofIvy

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    I see a graph that has a single mark on the x-axis that appears to be labled "1". But the problem asked for the average velocity over t= 3 to 3.2?

    The problem asked you to write an expression for v(3) and you have
    v(3)= r'(3). That's a formula certainly. Are you not given any more information about r(t)? Where did you get the graph?

    Your formula for the unit tangent vector at t= 3 makes no sense to me at all. You have two formulas:
    [tex]k= \frac{|T'(t)|}{|r'(t)|}[/tex]
    which doesn't seem to me to have anything to do with the question and
    [tex]|T'(t)|= C|r'(t)|[/tex]
    which doesn't make sense for several reasons- first |T'(t)| is NOT a vector and secondly that you haven't told us what "C" is. Did you mean the "k" from the first equation? If so it still doesn't make sense! You need to know |T'(t)| to find k so you can't use k to find |T'(t)|. At any rate, since you were asked to find T, why are you using T'?
     
  4. Oct 8, 2005 #3

    mezarashi

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    The instantaneous velocity or the derivative of the position vector is naturally tangent to the path of the object. The unit vector would simply by the instantaneous velocity divided by its own magnitude.
     
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