Question

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

Answer 1

Given that $\vec{A}(U)$ is vector function.

Given that $\|\vec{A}(u)\|=1$

 $\begin{aligned} &\Rightarrow(\|\vec{A}(u)\|)^{2}=(1)^{2} \\ &\Rightarrow \vec{A}(u) \cdot \vec{A}(u)=1 \\ &\Rightarrow\left[\begin{array}{c} \because \cdot \vec{B} \|^{2}=\vec{B} \cdot \vec{B} \\ \text { for any vector } \vec{B} \end{array}\right] \\ &\Rightarrow \frac{d}{d u}(\vec{A}(u) \cdot \vec{A}(u))=\frac{d}{d u}(1) \\ &\Rightarrow \vec{A}(u) \cdot \frac{d}{d u}(\vec{A}(u))+\frac{d}{d u}(\vec{A}(u)) \cdot \vec{A}(u)=0 \\ &\because \frac{d}{d u}(\operatorname{constan} t)=0 \\ &\frac{d}{d u}(\vec{B} \cdot \vec{C})=\vec{B} \cdot \frac{d}{d u}(\vec{C})+\frac{d}{d u}(\vec{B}) \cdot \vec{C} \\ &\operatorname{here} \vec{B}=\vec{C}=\vec{A}(u) \\ &\Rightarrow \vec{A} \cdot \frac{d \vec{A}}{d u}+\frac{d \vec{A}}{d u} \cdot \vec{A}=0 \end{aligned}$

$\begin{aligned} &\Rightarrow \vec{A} \cdot \frac{d \vec{A}}{d u}+\vec{A} \cdot \frac{d \vec{A}}{d u}=0 \\ &\because \vec{B} \cdot \vec{C}=\vec{C} \cdot \vec{B} \\ &\text { here } \left.\vec{B}=\frac{d \vec{A}}{d u}, \quad C=\vec{A}\right] \\ &\Rightarrow \quad 2\left(\vec{A} \cdot \frac{d \vec{A}}{d u}\right)=0 \\ &\Rightarrow \quad \vec{A} \cdot \frac{d \vec{A}}{d u}=0 \\ &\Rightarrow \frac{d \vec{A}}{d u} \text { is perpendicular to } \vec{A} \text {. } \\ &\text { ar } \vec{B} \cdot \vec{C}=0 \\ &\Rightarrow \vec{C} \text { is perpendicular to } \vec{B}] \\ &\text { so } \frac{d \vec{A}}{d u} \text { is perpendicular to } \vec{A} \text {. } \\ &\text { Hence proved. } \end{aligned}$