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Proper acceleration question

  1. Aug 27, 2009 #1
    I have been trying to learn the proper acceleration math , and dont understand d[tex]^{2}[/tex]x , or d[tex]^{2}[/tex]t as it appears , where d seems to have the normal meaning of delta.

    Is this simply equivalent to dx[tex]^{2}[/tex] or dt[tex]^{2}[/tex] ????

    Thanks
     
  2. jcsd
  3. Aug 27, 2009 #2

    tiny-tim

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    Hi Austin0! :smile:

    (try using the X2 tag just above the Reply box :wink:)
    No.

    d is an operator (a sort of function), and d2 means that you perform the operation twice.

    If you perform it both times with respect to the same variable, say y, you put dy2 on the bottom, so it's d2/dy2 (x), or d2x/dy2.

    If you perform it once with respect to y and once with respect to z, you put dxdy on the bottom, so it's d2/dydz (x), or d2x/dydz. :smile:

    If you want to see it in ∆s …

    define ∆+x to be the increase in x if you increase y by ∆y, and ∆-x to be the decrease in x if you decrease y by ∆y.

    Then dx/dy = lim ∆+x/∆y = lim ∆-x/∆y

    and d2x/dy2 = lim ∆(dx/dy))/∆y

    = lim (∆+x - ∆-x)/(∆y)2

    (and, by comparison, (dx/dy)2 = dx/dy = lim (∆x)/(∆y)2, which is not the same :wink:)

    For example, if x = yn,

    then ∆+x = nyn-1∆y + (1/2)n(n-1)yn-2(∆y)2 + …,

    and ∆-x = nyn-1∆y - (1/2)n(n-1)yn-2(∆y)2 + …,

    so dx/dy = lim ∆+x/∆y = lim ∆-x/∆y = nyn-1,

    and d2x/dy2 = lim (∆+x - ∆-x)/(∆y)2 = n(n-1)yn-2 :smile:
     
    Last edited: Aug 27, 2009
  4. Aug 27, 2009 #3
    Hi tiny-tim You have been very helpful. I didn't have the correct understanding of d at all.
    I thought it was simply the total change in whatever it was applied to ,irrespective of whatever function it was used in. I.e. Simply the total interval. The distance represented by dx as the difference between two specific coordinate x's.
    I think your explication has made it clear but I guess time will tell.
    Thanks for your responce and friendly greeting :smile:
     
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