Question

Let $\mathbf{r}\left(t\right)$ be a differentiable vector function. Prove that role="math" localid="1649602115972" $\frac{d}{dt}\left(\Vert \mathbf{r}\Vert \right)=\frac{1}{\Vert \mathbf{r}\Vert }\mathbf{r}·{\mathbf{r}}^r·$ (Hint: role="math" localid="1649602160237" $\left.\Vert \mathbf{r}{\Vert }^2=\mathbf{r}·\mathbf{r}\right)$

Short answer

$$\begin{gathered} \frac{d}{dt}\left(\Vert r{\Vert }^2\right)=\frac{d}{dt}(r·r) \\ \Rightarrow 2\Vert r\Vert \frac{d}{dt}\left(\Vert r\Vert \right)=r·\frac{d}{dt}\left(r\right)+r\frac{d}{dt}\left(r\right) \\ =\frac{1}{\Vert r\Vert }\left(r·r^{\text{'}}\right) \end{gathered}$$Ans:

Step-by-step solution

Step 1. GIven information: 

$r\left(t\right)$is a differentiable vector function.

The objective is to prove that localid="1649602256109" $\frac{d}{dt}\left(\Vert r\Vert \right)=\frac{1}{\Vert r\Vert }r·r^{\text{'}}$

To prove this, we use$\Vert r{\Vert }^2=r·r$

Consider $\Vert r{\Vert }^2=r·r$

Step 2. Proving:

Differentiating both sides, with respect to 't'

$$\begin{gathered} \frac{d}{dt}\left(\Vert r{\Vert }^2\right)=\frac{d}{dt}(r·r) \\ \Rightarrow 2\Vert r\Vert \frac{d}{dt}\left(\Vert r\Vert \right)=r·\frac{d}{dt}\left(r\right)+r\frac{d}{dt}\left(r\right)\text{formula for}\frac{d}{dt}\left(wv\right) \\ 2\Vert r\Vert \frac{d}{dt}\left(\Vert r\Vert \right)=r·r^{\text{'}}+r·r^{\text{'}}\frac{d}{dt}\left(r\right)=r^{\text{'}} \\ \Rightarrow 2\Vert r\Vert \frac{d}{dt}(\Vert r\Vert )=2r·r^{\text{'}}add \\ \Rightarrow \Vert r\Vert \frac{d}{dt}(\Vert r\Vert )=r·r·cancel2on\mathrm{both}\mathrm{sides} \\ \Rightarrow \frac{d}{dt}(\Vert r\Vert )=\frac{1}{\Vert r\Vert }\left(r·r^{\text{'}}\right)\mathrm{Divide}\mathrm{both}\mathrm{sides}with\Vert \mathrm{r}\Vert \\ Thus\frac{d}{dt}(\Vert r\Vert )=\frac{1}{\Vert r\Vert }r·r^{\text{'}} \\ \mathrm{Note}:\mathrm{if}\mathrm{r}\left(t\right)isa\mathrm{differentiable}\mathrm{vector}\mathrm{function}\mathrm{such}that\Vert r\left(t\right)\Vert is\mathrm{constant}\mathrm{then}r\left(t\right)·r^{\text{'}}\left(t\right)=0 \end{gathered}$$