Proving the Vector Identity: a dot d(a)/dt = ||a|| x ||da/dt||

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SUMMARY

The vector identity stating that the dot product of a vector a with its time derivative d(a)/dt equals the product of the magnitudes of a and da/dt is confirmed through the application of the product rule for derivatives. The equation can be expressed as (a)·(d(a)/dt) = ||a|| × ||da/dt||. The discussion highlights the misconception that the angle between the vector and its derivative is always zero, clarifying that this identity holds under specific conditions.

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Angry Citizen
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Homework Statement



Prove the following vector identity:

Any vector a dotted with its time derivative is equal to the vector's scalar magnitude times the vector's derivative's scalar magnitude.

Homework Equations



(a)dot(d(a)/dt)=||a|| x ||da/dt||

The Attempt at a Solution



I tried separating into components, but then it hit me. This identity doesn't make any sense. If a-dot represents the time derivative, then it also represents the velocity tangent to the curve formed by a, which is obviously not always going to be in the same direction. The identity presumes that the angle formed between the time derivative and the vector is zero. How can this be accurate?
 
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Angry Citizen said:

Homework Statement



Prove the following vector identity:

Any vector a dotted with its time derivative is equal to the vector's scalar magnitude times the vector's derivative's scalar magnitude.

Homework Equations



(a)dot(d(a)/dt)=||a|| x ||da/dt||

The Attempt at a Solution



I tried separating into components, but then it hit me. This identity doesn't make any sense. If a-dot represents the time derivative, then it also represents the velocity tangent to the curve formed by a, which is obviously not always going to be in the same direction. The identity presumes that the angle formed between the time derivative and the vector is zero. How can this be accurate?
Use the product rule for the derivative of a scalar product.

[itex]\displaystyle \frac{d}{dt}\left(\textbf{b}\cdot\textbf{c}\right)=\frac{d\textbf{b}}{dt}\cdot\textbf{c} +\textbf{b}\cdot\frac{d\textbf{c}}{dt}[/itex]
 

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