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Homework Help: Stuck on proof! Proving cross product derivative!

  1. Oct 1, 2005 #1
    Hello everyone, i'm stuck on trying to prove the cross product rule for derivatives. I Have to add the right terms and its suppose to be easy but thats what i can't figure out! any help would be great! here is what I have:
    http://img135.imageshack.us/img135/5540/opopo3ej.jpg [Broken]
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Oct 1, 2005 #2
    One way (although not the nicest) is to write out the components and take the derivative.



    ...now take the cross product of these, and take the derivative. Once you're there, you can rearrange and get it to look like what you want.

    Anyone know of a better way to do this?

    Edit: Actually, use the properties of the cross product in your last equation (in the picture). Something can be said about the addition and subtraction you have.
    Last edited by a moderator: May 2, 2017
  4. Oct 1, 2005 #3

    George Jones

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    Hint: change the signs of the middle 2 terms in your bottom line, and then take a common factor of v(t + h) out of the first 2 terms and a common factor of u(t) out of the last 2 terms.

  5. Oct 1, 2005 #4


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    Well, you may write that the "ith" component of the cross product is

    [itex] \bigl( \vec u \times \vec v \bigr)_i = \epsilon_{ijk} u_j v_k [/itex]

    then you can use the usual rule for the derivative of a product of functions, so that the derivative of that is simply

    [itex] \epsilon_{ijk} u_j^{'} v_k + \epsilon_{ijk} u_j v_k^{'} [/itex]

    so that

    [itex] \bigl( \vec u \times \vec v \bigr)^{'} = \vec u^{'} \times \vec v + \vec u \times \vec v^{'} [/itex]


    If you are not allowed to use the fact that teh derivative of fg is f' g + fg', then you can just prove this as usual but applied on the expressions with the levi-civita tensor. That way, the proof is no more difficult than with a usual product of functions.

  6. Oct 1, 2005 #5
    Try this way:

    [tex]y=f(t)=u(t)\cdot v(t) \hspace{2cm}\rightarrow \frac{dy}{dt}=u'(t)v(t) + v'(t)u(t) =\lim_{h\to 0} \frac{u(t+h)\cdot v(t+h) - u(t)\cdot v(t)}{h}[/tex]

    [tex]= \lim_{h\to 0} \frac{u(t+h)\cdot v(t+h) - u(t)\cdot v(t)}{h} \hspace{1.5cm}\pm\frac{u(t+h)\cdot v(t)}{h}[/tex]

    [tex]= \lim_{h\to 0} \frac{u(t+h)\cdot v(t+h) - u(t)\cdot v(t) + u(t+h)\cdot v(t) - u(t+h)\cdot v(t)}{h} [/tex]

    [tex]=\lim_{h\to 0} \frac{u(t+h)\cdot\Bigl(v(t+h)-v(t)\Bigr) + v(t)\cdot\Bigl(u(t+h) - u(t)\Bigr)}{h} [/tex]

    [tex]=v(t)\cdot\lim_{h\to 0} \frac{u(t+h)-u(t)}{h} + \lim_{h\to 0} u(t+h)\cdot \lim_{h\to 0} \frac{v(t+h)-v(t)}{h}[/tex]

    [tex]= u'(t)\cdot v(t) + u(t)\cdot v'(t)[/tex]

  7. Oct 2, 2005 #6
    thaniks for the replies everyone, Roman, is that the proof for the dot product? Or did u mean to type cross prodcut?
  8. Oct 2, 2005 #7
    its the proof for the dot and for the cross product. just change "dot" into "cross"
  9. Oct 2, 2005 #8
    awesome thanks man!
  10. Oct 2, 2005 #9
    wait, how is the dot prodcut the same as the cross product?
  11. Oct 2, 2005 #10
    of course is the dot product not the same as the cross product, but the proof is the same
  12. Oct 2, 2005 #11
    Ahh i c, it made sense once I looked back on it! Thanks Roman!
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