Question

Let $\mathbf{r}\left(t\right)$ be a differentiable vector function such that $\mathbf{r}\left(t\right)·{\mathbf{r}}^{\text{'}}\left(t\right)=0$ for every value of $\mathbf{t}$. Prove that $\Vert \mathbf{r}\left(t\right)\Vert$ is a constant.

Short answer

Ans:

$$\begin{gathered} \frac{d}{dt}\left(\Vert r{\Vert }^2\right)=\frac{d}{dt}(r·r) \\ =r\frac{d}{dt}\left(r\right)+r\frac{d}{dt}\left(r\right)=0 \end{gathered}$$

Step-by-step solution

Step 1. GIven information: 

Consider a differentiable vector function $r\left(t\right)$ such that $r\left(t\right)·r^{\text{'}}\left(t\right)=0$ For every value of $\mathbf{t}$.

Step 2. Proving:

Consider

$$\begin{gathered} \frac{d}{dt}\left(\Vert r{\Vert }^2\right) \\ =\frac{d}{dt}(r·r)sicne\Vert r{\Vert }^2=r·r \\ =r\frac{d}{dt}\left(r\right)+r\frac{d}{dt}\left(r\right)\mathrm{Derivativeforproductof}2\mathrm{functions} \end{gathered}$$

$$\begin{gathered} \begin{array}{ll}=r·r^{\text{'}}+r·r^{\text{'}}& \frac{dr}{dr}=r^{\text{'}}\\ =2r·r^{\text{'}}& \text{add}\\ =2\left(0\right)& \text{since}r·r^{\text{'}}=0\\ =0& \text{Multiply}\end{array} \\ \frac{d}{dt}\left(\Vert r{\Vert }^2\right)=0\text{. That means}\Vert r\Vert \text{is a constant, since the derivative of a constant is zero.} \end{gathered}$$

Thus $\Vert r\left(t\right)\Vert$is a constant if $r\left(t\right)·r^{\text{'}}\left(t\right)=0$

Note: The converse is always true. That is, if $r\left(t\right)$is a differentiable vector function such that $\Vert r\left(t\right)\Vert$is a constant then $r\left(t\right)·r^{\text{'}}\left(t\right)=0$.