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Derivative of scalar triple product

  1. Jun 24, 2014 #1
    1. The problem statement, all variables and given/known data

    If u(t) = σ(t) . [σ'(t) x σ''(t)], show that u'(t) = σ(t) . [σ'(t) x σ'''(t)].

    2. Relevant equations

    The rules for differentiating dot products and cross products, respectively, are:

    d/dt f(t) . g(t) = f'(t) . g(t) + f(t) . g'(t)

    d/dt f(t) x g(t) = f'(t) x g(t) + f(t) x g'(t)

    3. The attempt at a solution

    So I applied each individual rule to σ(t) . [σ'(t) x σ''(t)] to get

    σ'(t) . [σ'(t) x σ''(t)] + σ(t) . [σ''(t) x σ''(t) + σ'(t) x σ'''(t)]

    which I can expand into

    σ'(t) . [σ'(t) x σ''(t)] + σ(t) . [σ''(t) x σ''(t)] + σ(t) . [σ'(t) x σ'''(t)]

    meaning σ'(t) . [σ'(t) x σ''(t)] + σ(t) . [σ''(t) x σ''(t)] should equal zero, but I don't see why it would.

    Thanks
     
  2. jcsd
  3. Jun 24, 2014 #2

    pasmith

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    Homework Helper


    Basic facts about the cross product:

    For any vector [itex]\mathbf{a}[/itex], we have [itex]\mathbf{a} \times \mathbf{a} = \mathbf{0}[/itex].

    For any vectors [itex]\mathbf{a}[/itex] and [itex]\mathbf{b}[/itex], the vector [itex]\mathbf{a} \times \mathbf{b}[/itex] is orthogonal to both [itex]\mathbf{a}[/itex] and [itex]\mathbf{b}[/itex], so [tex]
    \mathbf{a} \cdot (\mathbf{a} \times \mathbf{b}) = 0 = \mathbf{b} \cdot (\mathbf{a} \times \mathbf{b}).[/tex]
     
  4. Jun 24, 2014 #3
    The cross product of any vector with itself is zero, so the second term is zero.

    The first term is zero because the cross product σ'(t) x σ''(t) is orthogonal to σ'(t). The dot product of a vector with a vector orthogonal to itself is zero.

    Hope that helps.

    [Edit: Oops...I was posting at the same time as pasmith. What pasmith said.]
     
  5. Jun 24, 2014 #4
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