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Anitderivative of a position

  1. May 29, 2009 #1
    Well sitting in math class today learning about antiderivatives we talked about the following

    Accelerations antiderivative is velocity

    and,

    velocities antiderivative is position.

    So I asked my teacher what the antiderivative of a position would be then, and he did not know.

    So if someone could tell me what he antiderivative of a position would represent that would be a massive releif to my curiosity.

    Thanks.
     
  2. jcsd
  3. May 29, 2009 #2

    Cyosis

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    Integrating the position function with respect to time would give you a quantity with units m s (length*time). Do you know of any physical quantity with those units? I don't.
     
  4. May 29, 2009 #3
    well i have read this in line integrals that to integrate a continuous function f(x,y,z) which is here position function of an object over curve c integratef(g(t),h(t),k(t))lv(t)ldt where v(t)=dr/dt and coordinates x,y,z are functions of time i guess it has something to do this with this type of problem
     
  5. May 29, 2009 #4

    Cyosis

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    Line integrals are something different. Seeing as the examples he listed were [itex]\int a(t)dt=v(t), \int v(t)dt=x(t)[/itex] it stands to reason he meant [itex]\int x(t)dt=X(t)[/itex]. This is of course not a line integral.
     
  6. May 29, 2009 #5
    if position is integrated with respect to time it must be position function and its coordinates can be function of time also.Is it not so?
     
  7. May 29, 2009 #6

    Cyosis

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    I am not quite sure what you mean with that, could you elaborate and perhaps give a concrete example?
     
  8. May 29, 2009 #7
    suppose position function is f(x,y,z) of a particle where x,y,z are functions of time t i.e x=h(t) y=g(t) and z=k(t) then it can be integrated i think so
     
  9. May 29, 2009 #8

    Cyosis

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    Of course it can. The question is whether or not there is a useful application for the integral of the position. I cannot think of one, that doesn't mean there isn't one of course. But your example just extends it to three dimensions which doesn't really change the physical interpretation.
     
  10. May 29, 2009 #9

    HallsofIvy

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    Yes, no one is disputing that the position function can be integrated. What is in dispute is the idea that its integral must have some specific physical meaning. As Cyosis said in the first response to your post, if the position function is integrated with respect to time, then the result must have units of "length times time" or "meters times seconds". There is not regularly defined physics quantity that even has those units!
     
  11. May 29, 2009 #10
    Units:
    Acceleration has the units of length /time^2. Taking the antiderivative of the acceleration causes you to multiply the acceleration units by time. So you end up with length/time which are the units of velocity. Taking the antiderivative of velocity, multiply by time again, to get units of length, which is ...um..length or distance. If you take the antiderivative of distance, you have to multiply by seconds again to get units of
    length*time. Can you think of any physical measurement that you can do that gets you units of length*time?

    You can certainly take antiderivatives of a function all day long if you want. Nothing is preventing you. The question though, is does that answer have any physical meaning? I could take cats and divide by the color red. Does that have any physical meaning? (I hope not :)
     
  12. May 29, 2009 #11
    Bah...beaten to the punch.
     
  13. Jun 12, 2009 #12
    thanks a lot for the responses everyone, so the antiderivative of a position would give the units length*time is what i gathered. so something like m*s, which is kind of interesting weather it has practical physical use or not haha. Ill have to run that by my teacher and hopefully get a bonus mark or two :) .
     
  14. Jun 12, 2009 #13

    HallsofIvy

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    Not a slap on the ears? A very patient teacher!:rofl:
     
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