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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?

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2 Answers 2

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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.

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  • $\begingroup$ But using the product rule, if I understand correctly, would require me to differentiate like this, right?: $\vec{u'(t)}\times\vec{v(t)} + \vec{u(t)}\times\vec{v'(t)}$ $\endgroup$
    – dtg
    May 25, 2012 at 20:57
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    $\begingroup$ @Dylan: That's the RHS side, yes. The LHS is already 'finding the cross-product before differentiating'... $\endgroup$
    – anon
    May 25, 2012 at 20:58
  • $\begingroup$ 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? $\endgroup$
    – dtg
    May 25, 2012 at 20:58
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    $\begingroup$ 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. $\endgroup$ May 25, 2012 at 21:00
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    $\begingroup$ @Ayman Thanks. Using the first method seems to involve a lot more differentiation, whereas the latter method seems much simpler to me. $\endgroup$
    – dtg
    May 25, 2012 at 21:01
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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)\\ & \quad + \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)\\ & \quad + \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 $\delta t\to 0$, which gives

$$ \frac{d}{dt}\left( \vec{u}\times\vec{v} \right) = \frac{d\vec{u}}{dt}\times\vec{v} + \vec{u}\times\frac{d\vec{v}}{dt} $$

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)

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