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Graphs of f(t) and g (t)

  1. Sep 19, 2007 #1
    1. The problem statement, all variables and given/known data
    [​IMG]

    the answers given are:

    none of these
    The particle traces out a square
    The particle moves from ( -1, 1 ) to ( 1, 0 ) to ( 0, -1 ) to ( -1, 0 ) to ( 0, -1 )
    The particle moves from ( -1, 1 ) to ( 1, -1 ) to ( -1, 1 ) to ( 1, -1 ) to ( -1, 1 )

    could anyone help me out a little with this because im not sure where to start? im not looking for an answer to the question...just any explanation...thanks
     
    Last edited: Sep 19, 2007
  2. jcsd
  3. Sep 19, 2007 #2

    G01

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    What point does the particle start at, and what point is it at at 1 second?
     
  4. Sep 19, 2007 #3
    i think thats what i dont understand...how to look at that?
     
  5. Sep 19, 2007 #4
    Are you familiar with forier transforms? You could find the equations of the graphs and plug the parametric equation into a calculator.
     
  6. Sep 19, 2007 #5
    The particle's position and any given time is going to be the coordinate (f(t), g(t))

    So at t = 0 the particle will be at the point (f(0), g(0))

    Can you find what f(0) is? Can you find what g(0) is? You have the graphs of them...
     
  7. Sep 20, 2007 #6

    HallsofIvy

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    As has been said, x= f(t), y= g(t) so when t= 0, the particle is at (f(0), g(0)). When t= 1, the particle is at (f(1), g(1)). Get the x and y values of those points from the graphs given and plot them on an xy-graph. What do you think the particle does between t=0 and t=1?
     
  8. Sep 20, 2007 #7
    "( 1, 0 ) "

    Try to find t when x = 1 and y =0.

    And it doesn't matter if you increase t value by 1 or any other integer
     
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