Differentiation and dot product

[Benny]

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:

<br /> \frac{d}{{dt}}\left[ {\left( {u&#039; \times u&#039;&#039;} \right) \bullet \left( {u&#039; - u} \right)} \right]<br />

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.

\frac{d}{{dt}}\left[ {\left( {u&#039; \times u&#039;&#039;} \right) \bullet \left( {u&#039; - u} \right)} \right]

<br /> = \frac{d}{{dt}}\left[ {\left( {u&#039; \times u&#039;&#039;} \right) \bullet u&#039; - \left( {u&#039; \times u&#039;&#039;} \right) \bullet u} \right]<br />

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.

<br /> = \frac{d}{{dt}}\left[ {\left( {u&#039; \times u&#039;&#039;} \right) \bullet u&#039;} \right] - \frac{d}{{dt}}\left[ {\left( {u&#039; \times u&#039;&#039;} \right) \bullet u} \right]<br />

Ok well I'm really stuck here. Does anyone have any suggestions? Thanks.

 

[HallsofIvy]

Why not just go ahead and do it?
I wouldn't even bother to distribute it like you did. Just use the product rule:
((u&#039;Xu&quot;)\dot(u&#039;- u))&#039;= (u&#039;Xu&quot;)&#039;\dot(u&#039;- u)+(u&#039;Xu&quot;)\dot(u&#039;- u)&#039;
= (u&quot;Xu&quot;+ u&#039;X u&quot;&#039;)\dot(u&#039;- u)+ (u&#039;Xu&quot;)\dot(u&quot;- u&#039;)

 

[Benny]

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.

 

[HallsofIvy]

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.

 

[Benny]

Ok thanks a lot for the help HallsofIvy.