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Derivative of vector functions

  1. Feb 9, 2007 #1
    Okay, so I am having trouble with this concept (i actually didnt think i was until i realized i was getting all the wrong answers)...
    As an example, if you want to take the derivative of:

    d/dt [ r(t) X r'(t)]

    I just used the general rule of d/dt [r1(t) X r2(t)] = r1(t)Xr2'(t) + r1'(t)Xr2(t)

    I ***thought**** I was doing the problem correctly but apparently not.

    I just did the following by using the generic equation above:
    d/dt [ r(t) X r'(t)] = r(t)Xr''(t) + r'(t)Xr'(t)

    What is wrong with what I did? At the back of my book it says the answer is r(t) Xr''(t) which is the first part of my answer (but doesnt include the second portion). Can someone explain where i'm wrong? I'd really appreciate it.
  2. jcsd
  3. Feb 9, 2007 #2

    D H

    Staff: Mentor

    For any vector [itex]\vec a[/itex], what is [itex] \vec a \times \vec a[/itex]?
  4. Feb 9, 2007 #3
    the cross product?? the determinant of the components?

    Sorry i'm really shaky on vector calculus...
  5. Feb 9, 2007 #4

    D H

    Staff: Mentor

    There is no vector calculus here. I asked what is [itex] \vec a \times \vec a[/itex]. Hint: The answer is the same for all vectors.

    Step back: What is the magnitude of the cross product of two vectors [itex] \vec a[/itex] and [itex] \vec b [/itex]? What does that become when you set [itex] \vec b = \vec a[/itex]?
  6. Feb 9, 2007 #5
    a^2 ? so how does that help solve the problem?
  7. Feb 9, 2007 #6
    okay um nevermind i mean 0!
  8. Feb 9, 2007 #7

    D H

    Staff: Mentor

    Excellent. So now, what is

    [tex]\vec r(t) \times \vec r\;^{\prime\prime}(t) + \vec r\;^\prime(t) \times \vec r\;^\prime(t)[/tex] ?
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