Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

In finding the derivative of the cross product of two vectors $\frac{d}{dt}[\vec{u(t)}\times \vec{v(t)}]$, is it possible to find the cross-product of the two vectors first before differentiating?

share|improve this question
add comment

2 Answers 2

You can evaluate this expression in two ways:

  • You can find the cross product first, and then differentiate it.
  • Or you can use the product rule, which works just fine with the cross product:

$$ \frac{d}{dt}(\mathbf{u} \times \mathbf{v}) = \frac{d\mathbf{u}}{dt} \times \mathbf{v} + \mathbf{u} \times \frac{d\mathbf{v}}{dt} $$

Picking a method depends on the problem at hand. For example, the product rule is used to derive Frenet Serret formulas.

share|improve this answer
    
But using the product rule, if I understand correctly, would require me to differentiate like this, right?: $\vec{u'(t)}\times\vec{v(t)} + \vec{u(t)}\times\vec{v'(t)}$ –  Dylan May 25 '12 at 20:57
1  
@Dylan: That's the RHS side, yes. The LHS is already 'finding the cross-product before differentiating'... –  anon May 25 '12 at 20:58
    
My question was, is it similar to the case with differentiating dot-product, can't you just take the determinate form of the inside before differentiating? –  Dylan May 25 '12 at 20:58
2  
Whether it's easier or not will depend on the particular problem. For example, you might know something about $u'$ and $v'$ that makes the problem easy. –  Robert Israel May 25 '12 at 21:00
1  
@Ayman Thanks. Using the first method seems to involve a lot more differentiation, whereas the latter method seems much simpler to me. –  Dylan May 25 '12 at 21:01
show 3 more comments

Working from the first principles: \begin{aligned}\vec{u}\left(t+\delta t\right)\times\vec{v}\left(t+\delta t\right)-\vec{u}\left(t\right)\times\vec{v}\left(t\right) & =\vec{u}\left(t+\delta t\right)\times\vec{v}\left(t+\delta t\right)-\vec{u}\left(t\right)\times\vec{v}\left(t+\delta t\right)+\\ & =\vec{u}\left(t\right)\times\vec{v}\left(t+\delta t\right)-\vec{u}\left(t\right)\times\vec{v}\left(t\right)=\\ & =\left[\vec{u}\left(t+\delta t\right)-\vec{u}\left(t\right)\right]\times\vec{v}\left(t+\delta t\right)+\\ & =\vec{u}\left(t\right)\times\left[\vec{v}\left(t+\delta t\right)-\vec{v}\left(t\right)\right] \end{aligned} Now divide by $\delta t$ and take limit as $\delta t\to 0$

On the other hand $$\frac{d}{dt}\left|\begin{array}{ccc} i & j & k\\ v_{x} & v_{y} & v_{z}\\ u_{x} & u_{y} & u_{z} \end{array}\right|=\left|\begin{array}{ccc} i & j & k\\ \frac{dv_{x}}{dt} & \frac{dv_{y}}{dt} & \frac{dv_{z}}{dt}\\ u_{x} & u_{y} & u_{z} \end{array}\right|+\left|\begin{array}{ccc} i & j & k\\ v_{x} & v_{y} & v_{z}\\ \frac{du_{x}}{dt} & \frac{du_{y}}{dt} & \frac{du_{z}}{dt} \end{array}\right|$$ Using the rule of differentiation of a determinant. One useful application of it is in the proof of Abel's identity (which before Wikipedia was known to me as Ostrogradski-Liouville formula)

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.