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# Prove that for a vector Ā d/dt (A^2) =2Ā.dĀ/dt Hence, show that if a vector has a constant magnitude then

it is always perpendicular to its time derivative.

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- Lv 71 decade agoFavorite Answer
I am afraid that I don't understand your notation. However,

Let l denote the vector's length. The components of the vector (x(t), y(t), z(t) ...) then satisfy x^2 + y^2 + z^2 + ... = l^2. The derivative of this vector is (x', y', z'...). And implicitly differentiating the expression above yields 2xx' + 2yy' + 2zz' + ... = 0, wheras the dot-product of the vector and its derivative is half of this, or xx' + yy' + zz' + ... Hence it is also zero - and the vector is perpendicular to its derivative.

I think somewhere in here is the answer to your question...

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