Stuck on proof Proving cross product derivative

[mr_coffee]

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 that's what i can't figure out! any help would be great! here is what I have:
http://img135.imageshack.us/img135/5540/opopo3ej.jpg

 

[amcavoy]

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 that's what i can't figure out! any help would be great! here is what I have:
http://img135.imageshack.us/img135/5540/opopo3ej.jpg

One way (although not the nicest) is to write out the components and take the derivative.

u(t)=\left<x(t),y(t),z(t)\right>

v(t)=\left<a(t),b(t),c(t)\right>

...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.

 

[George Jones]

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

 

[nrqed]

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

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

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

\epsilon_{ijk} u_j^{'} v_k + \epsilon_{ijk} u_j v_k^{'}

so that

\bigl( \vec u \times \vec v \bigr)^{'} = \vec u^{'} \times \vec v + \vec u \times \vec v^{'}

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

 

[Gellmann]

Try this way:

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}

= \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}

= \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}

=\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}

=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}

= u'(t)\cdot v(t) + u(t)\cdot v'(t)



Regards
Roman

 

[mr_coffee]

thaniks for the replies everyone, Roman, is that the proof for the dot product? Or did u mean to type cross prodcut?

 

[Gellmann]

its the proof for the dot and for the cross product. just change "dot" into "cross"

 

[mr_coffee]

awesome thanks man!

 

[mr_coffee]

wait, how is the dot prodcut the same as the cross product?

 

[Gellmann]

of course is the dot product not the same as the cross product, but the proof is the same

 

[mr_coffee]

Ahh i c, it made sense once I looked back on it! Thanks Roman!