Let A ( u ) A(u) A ( u ) is a differentiable vector function of u u u and ∣ ∣ A ( u ) ∣ ∣ = 1 , ||A(u)||=1 , ∣∣ A ( u ) ∣∣ = 1 , let us prove that d A d u \frac{dA}{du} d u d A is perpendicular to A . A . A .
Since ∣ ∣ A ( u ) ∣ ∣ = ( A ( u ) , A ( u ) ) , ||A(u)||=\sqrt{(A(u),A(u))}, ∣∣ A ( u ) ∣∣ = ( A ( u ) , A ( u )) , we conclude that ( A ( u ) , A ( u ) ) = ∣ ∣ A ( u ) ∣ ∣ 2 = 1. (A(u),A(u))=||A(u)||^2=1. ( A ( u ) , A ( u )) = ∣∣ A ( u ) ∣ ∣ 2 = 1.
Therefore,
( d A d u , A ( u ) ) + ( A ( u ) , d A d u ) = 0 , (\frac{dA}{du},A(u))+ (A(u),\frac{dA}{du})=0, ( d u d A , A ( u )) + ( A ( u ) , d u d A ) = 0 , and hence 2 ( d A d u , A ( u ) ) = 0. 2(\frac{dA}{du},A(u))=0. 2 ( d u d A , A ( u )) = 0.
Taking into account that the inner product ( d A d u , A ( u ) ) (\frac{dA}{du},A(u)) ( d u d A , A ( u )) is equal to 0, we conclude that d A d u \frac{dA}{du} d u d A is perpendicular to A . A . A .
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