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Product rule for vector derivative

  1. Jun 29, 2015 #1
    Say I have a position vector

    p = e(t) p(t)

    Where, in 2D, e(t) = (e1(t), e2(t)) and p(t) = (p1(t), p2(t))T

    And if I conveniently point the FIRST base vector of the frame at the particle, I can use: p(t) = (r1(t), 0)T

    I want the velocity, so I take

    v = d(e(t))/dt p(t) + e(t) d(p(t))/dt

    And from there... blah blah.. I can take the rate of change of the frame, etc... but that is not my concern.

    My concern is that I KNOW the product rule for functions: I can prove it and use the rule and it is for functions.
    But here, I am using it here NOT for a product of two functions but for a product of a base frame and functions.

    So, I can ask my question two ways and I hope someone can answer it both ways.

    First, what is it about the frame e(t) that can enable me to treat it like function and blithely apply the product rule.
    OR
    Second, what is it about the product rule for functions that can enable me to apply it to base vectors so expeditiously?

    In other words: does the base frame have the nature of a function, OR how does product rule for functions extend to the product of things that are not functions?
     
  2. jcsd
  3. Jun 29, 2015 #2

    BiGyElLoWhAt

    User Avatar
    Gold Member

    I'm not sure that I see your concern. e is a unit vector? Is it dynamic or constant? If it's dynamic, I would refrain from expressing it as (r1(t),0), as the second compenent will not always be zero. Also, why is the product e*p not a function?
     
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