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Parametric Equation Speed

  1. Feb 12, 2009 #1
    1. The problem statement, all variables and given/known data
    Can someone please tell me how to get the average speed of a particle moving along a path represented by parametric equations? Is it [itex]\frac{1}{b-a}\int_{a}^{b}\sqrt{\frac{dx }{d t}^2 + \frac{d y}{d t}^2}[/itex]

    Isn't this the arc length formula?
     
  2. jcsd
  3. Feb 12, 2009 #2
    This is the arc length formula. The average value formula is Favg=(1/b-a)INT[f(x)dx]. It seems you combined two formulas.
     
  4. Feb 12, 2009 #3
    But if I wanted the speed of a particle moving with a parametric graph, woldn't everything under the radical be my speed function?
     
  5. Feb 12, 2009 #4
    Actually, you may be right. I think that might actually work.
     
  6. Feb 13, 2009 #5

    Dick

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    No, no, no. The average speed is displacement over time. It has nothing to do with arc length. It's sqrt((x(b)-x(a))^2+(y(b)-y(a))^2)/(b-a) where a is the intiial time and b is the final time. Right?
     
  7. Feb 13, 2009 #6
    Couldn't you also do the average value of the absolute value of the velocity graph?
     
  8. Feb 13, 2009 #7

    Dick

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    Yes, you could. In which case that would be correct. Distance travelled/time could also be considered an average speed. I was only thinking of the displacement/time definition.
     
  9. Feb 13, 2009 #8
    Alirght, thank you for the help.

    Also, is there any way to determine if a particle traveling on a parametric path is increasing in speed? I know I can determine if the x and y are accelerating, but I can I determine if the particle itself is increasing?

    What if it was accelearating in the x direction but decelerating in the y? Would the particle's speed be increasing or decreasing?
     
  10. Feb 13, 2009 #9

    Dick

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    The 'speed' is sqrt((dx/dt)^2+(dy/dt)^2), isn't it? Just look at whether that quantity is increasing or decreasing.
     
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