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Delta V/ delta t vs. dv/dt

  1. Sep 17, 2013 #1
    I know delta means change , but I don't what the difference between Δv/Δt vs dv/dt is ?

    I am at the noob end of calculus so trying to grasp how to interpret things like dv/dt or what what dv would mean if it were standing alone.

    TIA
     
  2. jcsd
  3. Sep 18, 2013 #2

    arildno

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    1. When we use "delta" rather than "d", we mean that, say "delta v"/"delta t" is the average change in "v" over the non-zero time interval "delta t".
    2. Now, dv/dt is what we call the instantaneous rate of change in "v", that is the limit of "delta v"/"delta t" as we let the time interval "delta t" shrink to zero.
     
  4. Sep 18, 2013 #3

    HallsofIvy

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    The "delta t shrinking to 0" is the limit process:
    [tex]\dfrac{dx}{dt}= \lim_{\Delta t \to 0}\dfrac{\Delta x}{\Delta t}[/tex]

    Calculus texts typically spend a good deal of time on the "limit" concept.
     
  5. Sep 28, 2013 #4
    Suppose that at time ##t_0## a car starts at a point ##x_0## and travels to point x arriving at time t. Then the average velocity of the car over its trip is ## \Delta x/ \Delta t = \frac {x-x_0}{t-t_0}## -- that is the change in distance divided by the change in time.

    This tells you nothing about the velocity at any given moment. The car likely started at 0, gained speed up to a certain point, stopped for a light, got lost and had to go back a block, etc. So if you want to know how fast the car was traveling, and in what direction, at any given moment, ## \Delta x/ \Delta t ## tells you almost nothing.

    But if you look at the average velocity of a small time period, that is closer to the velocity at any given time ##t_0## in the period. Make the time period smaller yet, and you are closer yet to the velocity at ##t_0##.

    The genius of calculus was to see that you can let that time interval go to 0 and ## \Delta x/ \Delta t## may "converge" to a simple number -- say 30 mph. That is dx/dt -- the instantaneous velocity.
     
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