Question #277297

If A(u) is a differentiable vector function of u and ||A(u)||=1 , prove that dAdu is perpendicular to A .


1
Expert's answer
2021-12-09T03:34:50-0500

Let A(u)A(u)  is a differentiable vector function of uu  and A(u)=1,||A(u)||=1 , let us prove that dAdu\frac{dA}{du}  is perpendicular to A.A .

Since A(u)=(A(u),A(u)),||A(u)||=\sqrt{(A(u),A(u))}, we conclude that (A(u),A(u))=A(u)2=1.(A(u),A(u))=||A(u)||^2=1.

Therefore,

(dAdu,A(u))+(A(u),dAdu)=0,(\frac{dA}{du},A(u))+ (A(u),\frac{dA}{du})=0, and hence 2(dAdu,A(u))=0.2(\frac{dA}{du},A(u))=0.


Taking into account that the inner product (dAdu,A(u))(\frac{dA}{du},A(u)) is equal to 0, we conclude that dAdu\frac{dA}{du}  is perpendicular to A.A .

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