Deriviate proof d/dt[r (v a)]= r(v a)

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SUMMARY

The discussion focuses on proving the derivative of the expression \(\frac{d}{dt}[r (v \times a)] = r (v \times \frac{d}{dt}a)\), where \(d\), \(v\), and \(a\) represent position, velocity, and acceleration, respectively. Participants emphasize the application of the chain rule and the properties of vector cross products, specifically that the cross product of a vector with itself is zero. The key takeaway is that understanding vector calculus and the behavior of derivatives in relation to cross products is essential for solving this proof.

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  • Vector calculus fundamentals
  • Understanding of derivatives and the chain rule
  • Knowledge of vector cross products
  • Familiarity with physical concepts of position, velocity, and acceleration
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agentnerdo
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Hey guys, I really do not even know how to get this question started.

Homework Statement



\frac{d}{dt}[r (v x a)] = r (v x a)

the last a is supposed to have a period on top

as such it is \frac{d}{dt}[r (v x a)] = r (v x \frac{d}{dt}a)

d, v, and a are position, velocity, and acceleration

the last a, after the = sign is d/dt of a, as mentioned above






The Attempt at a Solution



I do not even know how to start...I tried doing different degrees of deriviatives of both left and right sides but, it seems like I need to get ride of one unit of /s (time).
 
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that was an epic fail on trying to use brackets...


derivative of [r (v times a)] = r ( v times a*)

a* is the derivative of a.

question asks to prove that both sides are equal.
 
Simply apply the chain rule and remember that the cross product of a vector with itself is 0, and that \vec a \cdot (\vec a \times \vec b)=0 (Convince yourself this is true, because a x b is perpendicular to a, and the dot product of a vector with another vector to which it is orthogonal, is 0)
 

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