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Differentiation and dot product

  1. Nov 26, 2005 #1
    Can someone help me with the following question. I've been having trouble with problems of this kind for a while now.

    Q. If the path u(t) (u is a vector) is differentiable at least three times, simplify:

    \frac{d}{{dt}}\left[ {\left( {u' \times u''} \right) \bullet \left( {u' - u} \right)} \right]

    I can't remember the properties of the dot product but I know that the dot product is an inner product. So I will start by using the distributivity of the inner product.

    [tex]\frac{d}{{dt}}\left[ {\left( {u' \times u''} \right) \bullet \left( {u' - u} \right)} \right][/tex]

    = \frac{d}{{dt}}\left[ {\left( {u' \times u''} \right) \bullet u' - \left( {u' \times u''} \right) \bullet u} \right]

    It's not much but it's all I've got at the moment. The cross product will produce a single vector so there probably isn't much that I can do with it. Each of the two expressions inside the square bracket are scalars (scalar functions of t is probably the more accurate description) so I can probably continue as follows.

    = \frac{d}{{dt}}\left[ {\left( {u' \times u''} \right) \bullet u'} \right] - \frac{d}{{dt}}\left[ {\left( {u' \times u''} \right) \bullet u} \right]

    Ok well I'm really stuck here. Does anyone have any suggestions? Thanks.
  2. jcsd
  3. Nov 26, 2005 #2


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    Why not just go ahead and do it?
    I wouldn't even bother to distribute it like you did. Just use the product rule:
    [tex]((u'Xu")\dot(u'- u))'= (u'Xu")'\dot(u'- u)+(u'Xu")\dot(u'- u)'[/tex]
    [tex]= (u"Xu"+ u'X u"')\dot(u'- u)+ (u'Xu")\dot(u"- u')[/tex]
    Last edited by a moderator: Nov 27, 2005
  4. Nov 26, 2005 #3
    I can't see the Latex that you appear to have included in your reply. What I've got is the dot product of a cross product and the difference of two vectors. I don't see how the product rule can be used on the initial expression. Can you please explain further?

    Edit: On second thought. The expression might just be the normal product(like 3 * 5 = 15). In that case I'd be able to this question. However I'd like to know if there is anything that can be done if the product is in fact a dot product rather than just the normal multiplication.
    Last edited: Nov 26, 2005
  5. Nov 27, 2005 #4


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    The usual "product rule" applies to dot product and cross product of vectors also: (u.v)'= u'.v+ u.v', (uxv)'= u'xv+ uxv'.

    Sorry the Latex didn't work. I'm not sure what's wrong.
  6. Nov 28, 2005 #5
    Ok thanks a lot for the help HallsofIvy.
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