Answer to Question #276691 in Calculus for Nawrin Dewan

Question

If B(u)

 is a differentiable vector function of u

 and ||B(u)||=1

, prove that du

 is perpendicular to B

Answer 1

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

Taking into account that $||B(u)||=\sqrt{(B(u),B(u))},$

we conclude that $(B(u),B(u))=||B(u)||^2=1.$

It follows that

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

Since the inner product $(\frac{dB}{du},B(u))$ is equal to 0, we conclude that $\frac{dB}{du}$  is perpendicular to $B .$