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I Position vs. time graph and the derivative

  1. Feb 20, 2017 #1
    So lets assume an object moves along a straight line relative to some fixed origin. Clearly we can study this motion with the help of a position vs. time graph which shows how the position varies as time goes on. Now, as far as I understand, the slope of this graph at any time t gives the instantaneous velocity of the object, because it measures the change in position over time.

    Now, the thing that I don't quite understand is,how would such a graph look like if the object were to move in two or three dimensions? Assuming it now moves in space, it could still be moving along a straight line, but what if it started moving in some other direction, how do the new velocities look like on a position vs. time graph? There seems to be a gap in my understanding at this point and I can't quite connect how the slope of a position vs. time graph would represent the change in velocity.
    I hope I stated my concerns clearly and I'm looking forward to any sharing of thoughts on this matter.
  2. jcsd
  3. Feb 20, 2017 #2


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    Position in space is a vector quantity in the sense that it classically has three components, ##\vec{x} = (x,y,z)##. Velocity is the same way in that
    [tex]\vec{V} = (u,v,w) = \left( \dfrac{dx}{dt},\dfrac{dy}{dt},\dfrac{dz}{dt} \right).[/tex]
    In essence, you need three plots.

    I'd also caution you against thinking of derivatives purely as the slopes of graphs. While it's true that a derivative represents the slope of a graph of one of the variables against the other, it's better to think of them as the rate of change of one variable with respect to another. For instance, ##u = dx/dt## is the rate of change of ##x## with respect to ##t##.
  4. Feb 20, 2017 #3


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    You simply take components of velocity along the two or three axes and write separate kinematic equations for them. To find the resultant velocity, simply add the components (vectorially). Look up projectile motion for instance.
  5. Feb 20, 2017 #4
    Thank you for your answer, this totally makes sense because considering each component separately I can understand it in the same manner as in one dimension.
    Yes you are right on the understanding of the derivative, I was just considering the geometric interpretation of it :)
  6. Feb 20, 2017 #5
    Yes, thank you very much, it all makes more sense now :)
  7. Feb 20, 2017 #6


    Staff: Mentor

    Consider an object moving in two dimensions in regular circular motion. If you "plot" that using something like a 3D printer with time as the third dimension then you would wind up with a helix.

    In principle you can do the same thing with 3D motion, but 4D printers and 4D paper are hard to find!
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