How can I prove derivative of cross vectors

[esradw]

Hello,

I am new here and I am very happy to find a webpage like this.
It has been 6 years sice I graduated from Physics and I haven't dealt with derivatives of vectors for a long time. Recently I have been accepted for the master degree and trying to built my memory again.

My homework quetion is that prove the product rule for two dependent vectors is

d/dt ( r X u ) = r X du/dt + dr/dt X u

I desperatly need help , I just can't remember. thanks

 

[Cyrus]

You will want to start distributing.

Major Clue:

u (t) = < f_1(t),f_2(t),f_3(t)>

v (t) = < g_1(t),g_2(t),g_3(t)>

This should be in any standard math text as a formal proof already done for you. Or available online.

 

[esradw]

Thank you very much, it was a little but powerfull hint for me that helped me solve the problem :)

 

[dextercioby]

Assuming the basis vectors are constant in time, then writing

\frac{d}{dt}\left(\epsilon_{ijk}r_{i}u_{j}\vec{e}_{k}\right)

and expanding using the Leibniz rule will lead you to the result...

Daniel.

 

[Cyrus]

Daniel, this is way over I think both my and the origional posters heads, but I think for being a grad student, it would serve him well to know your method as well. I have never even heard about this notation for cross product before. Can you work it out and show us the anwser to it, (But try to keep it as simple as possible, so I can at least try to understand what's going on.) Thanks,

 

[Tide]

You could just go back to the calculus definition of derivative

\frac {df}{dt} = \lim_{t \rightarrow 0} \frac {f(t + \Delta t) - f(t)}{\Delta t}

and apply it to the vector product.

 

[dextercioby]

Using the constant Levi-Civita (pseudo)tensor, one can write the vector/cross product the way i did in my previous post. Then applying the rules of differentiation and rewriting a term such

\epsilon_{ijk}\frac{du_{i}}{dt}v_{j}\vec{e}_{k}

as \frac{d\vec{u}}{dt}\times\vec{v}

will easily lead to the result.

Daniel.

 

[George Jones]

Daniel's way is probably the shortest, but be careful, he's using the summation convention, which you may not have seen. Repeated indices are summed over.

I started a post last night that used the same idea as in Tide's post, but did not finish. Here are most of the details. The same trick that is used for proving the product rule in ordinary calculus can be used - adding zero.

<br /> \begin{equation*}<br /> \begin{split}<br /> \frac{d}{dt} \left(\vec{r} \times \vec{u} \right) \left(t\right) &amp;= \lim_{h \rightarrow 0} \left[ \frac{\vec{r} \left( t + h \right) \times \vec{u} \left( t + h \right) - \vec{r} \left(t\right) \times \vec{u} \left(t\right)}{h} \right]\\<br /> &amp;= \lim_{h \rightarrow 0} \left[ \frac{\vec{r} \left( t + h \right) \times \vec{u} \left( t + h \right) - \vec{r} \left( t + h \right) \times \vec{u} \left( t \right) + \vec{r} \left( t + h \right) \times \vec{u} \left( t \right) - \vec{r} \left(t\right) \times \vec{u} \left(t\right)}{h} \right]\\<br /> &amp;= \lim_{h \rightarrow 0} \left[ \frac{\vec{r} \left( t + h \right) \times \vec{u} \left( t + h \right) - \vec{r} \left( t + h \right) \times \vec{u} \left( t \right)}{h} \right] + \lim_{h \rightarrow 0} \left[ \frac{\vec{r} \left( t + h \right) \times \vec{u} \left( t \right) - \vec{r} \left(t\right) \times \vec{u} \left(t\right)}{h} \right]\\<br /> &amp;= \lim_{h \rightarrow 0} \left[ \frac{\vec{r} \left( t + h \right) \times \left\{ \vec{u} \left( t + h \right) - \vec{u} \left( t \right) \right\}}{h} \right] + \lim_{h \rightarrow 0} \left[ \frac{\vec{r} \left( t + h \right) - \vec{r} \left(t\right)}{h} \right] \times \vec{u} \left(t\right)\\<br /> \end{split}<br /> \end{equation*}<br />

Regards,
George

 

[Cyrus]

I still have no clue about what daniel is talking about, :( (Im studying engineering, we don't go into that depth of math he's doing) I have not dealt with tensor math before. I have done linear algerba, but I doubt the two are the same.

 

[Tide]

Using the constant Levi-Civita (pseudo)tensor, one can write the vector/cross product the way i did in my previous post. Then applying the rules of differentiation and rewriting a term such
\epsilon_{ijk}\frac{du_{i}}{dt}v_{j}\vec{e}_{k}
as \frac{d\vec{u}}{dt}\times\vec{v}
will easily lead to the result.
Daniel.

Just remember to alternate signs between even and odd permutations of indices.

 

[leright]

oops...double post.

 

[leright]

You could just go back to the calculus definition of derivative
\frac {df}{dt} = \lim_{t \rightarrow 0} \frac {f(t + \Delta t) - f(t)}{\Delta t}
and apply it to the vector product.

This seems like the most straight forward, nuts and bolts method to me and I have a feeling this is the approach your professor expects you to take.