Deriving dA/dt = \omega x A: Proof for Constant Vector A and Vector \omega

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The discussion centers on proving the equation dA/dt = ω x A, where A is a constant vector and ω is another vector representing angular velocity. Participants clarify that A must be constant in magnitude, leading to the conclusion that the derivative of A can be expressed using the cross product with ω. The hint provided emphasizes differentiating the squared magnitude of A, which simplifies the proof process. Ultimately, the proof is confirmed to be straightforward once the correct approach is understood.

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nna
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I hope somebody can help me.. this is the problem i have to proof that if A is a constant vector then I can write its derivate as dA/dt = \omega x A.. where \omega is a vector, and the "x" is the cross product
 
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Welcome to PF!

Hi nna! Welcome to PF! :smile:

(have an omega: ω :wink:)
nna said:
I hope somebody can help me.. this is the problem i have to proof that if A is a constant vector then I can write its derivate as dA/dt = \omega x A.. where \omega is a vector, and the "x" is the cross product

(You mean "if A is constant in magnitude".)

Hint: if the magnitude is constant, then so is the magnitude squared, which is A.A. :wink:
 
Sorry :( but I don't understand how that helps...
 
Differentiate it …

what do you get? :smile:
 
ok ok thank you so much! it really helps... jaja it was very easy sorry
 

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