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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.

    [tex]u(t)=\left<x(t),y(t),z(t)\right>[/tex]

    [tex]v(t)=\left<a(t),b(t),c(t)\right>[/tex]

    ...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.

    Regards,
    George
     
  5. Oct 1, 2005 #4

    nrqed

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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]

    QED.

    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.


    Patrick
     
  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]


    :wink:
    Regards
    Roman
     
  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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