Answer to Question #277297 in Calculus for Ridhwan

Question

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

Answer 1

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

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

Therefore,

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

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