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Delta Squared?

  1. Jan 26, 2010 #1
    How does one deal with a delta squared ? For example, volume is a function of del Diameter and del Length.

    If i let v = d^2(pi)L/4, and then call the d and L del. d and del L, when i separate for d i get del d^2

    So i'm wondering if there are any special rules for squaring a "del"
     
  2. jcsd
  3. Jan 26, 2010 #2

    tiny-tim

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    Hi Zerius! :smile:

    (have a del: ∆ and try using the X2 tag just above the Reply box :wink:)
    You mean as in π(∆d)2(∆L)/4 ?

    (∆d)2 is just the square of ∆d.

    What is the context? Is this part of some larger problem? :confused:
     
  4. Jan 26, 2010 #3

    HallsofIvy

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    All you have done here is call ordinary numbers "delta d" and "delta L". There is nothing at all different about that.

    If you are talking about differentials, then with [itex]V= \pi r^2 L[/itex], [itex]dV= 2\pi r L dr+ \pi r^2 dL[/itex]. There is only one "dr".

    By the way "del" is a common way of refering to the [itex]\nabla[/itex] operator so it is not a good idea to shorten "delta" to "del".
     
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