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?
Derivative of cross-product of two vectors
2 Answers
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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0But using the product rule, if I understand correctly, would require me to differentiate like this, right?: $\vec{u'(t)}\times\vec{v(t)} + \vec{u(t)}\times\vec{v'(t)}$ – 2012-05-25
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1@Dylan: That's the RHS side, yes. The LHS is *already* 'finding the cross-product before differentiating'... – 2012-05-25
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0My 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? – 2012-05-25
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0@Dylan I understood your question backwards. Yes, you can find the cross product first, and then differentiate it. – 2012-05-25
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0@Dylan: The reason we have this rule in the first place is not that it makes explicit computation more efficient necessarily, but rather because we can, if need be, use the properties of the cross product and tangent vectors to expedite derivations involving them without having to write out sums and components (especially when we already know something about $u$ and $v$...). – 2012-05-25
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2Whether 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. – 2012-05-25
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1@Ayman Thanks. Using the first method seems to involve a lot more differentiation, whereas the latter method seems much simpler to me. – 2012-05-25
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0@Robert Okay. So it just depends on the question. I will keep that in mind. – 2012-05-25
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)