Differentiation of position vector wrt time

Question

I want to differentiate $\frac{\mathbf{r}}{r}$ wrt t (time). I know the following; $$\frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t} = \mathbf{\dot{r}}$$ $$\frac{\mathrm{d}{r}}{\mathrm{d}t} = {\dot{r}}$$ and so $$\frac{\mathrm{d}\frac{1}{r}}{\mathrm{d}t} = -\frac{1}{r^2}\dot{r}$$ and putting this together gives; $$\frac{\mathrm{d}\frac{\mathbf{r}}{r}}{\mathrm{d}t} = \frac{\mathbf{\dot{r}}}{r} - \frac{\dot{r}\mathbf{r}}{r^2}$$

Is this correct as I would have instinctively thought that since $\frac{\mathbf{r}}{r} = \mathbf{\hat{{r}}}$, the correct answer would be $\mathbf{\hat{\dot{r}}}$

Answer 1

It's correct. You obtain $$\frac{\mathrm{d}\frac{\mathbf{r}}{r}}{\mathrm{d}t} = \frac{\mathbf{\dot{r}}}{r} - \frac{\dot{r}\mathbf{r}}{r^2}=\frac{\mathbf{\dot{r}}r-\dot{r}\mathbf{r}}{r^2}$$ as direct differentiating in ratio shows.

Answer 2

You are correct. First, $\hat r(t)=\frac{\vec r(t)}{|\vec r(t)|}$ and hence

$$\begin{align} \dot{\hat r}(t)&\equiv \frac{d\hat r(t)}{dt}\\\\ &=\frac{d}{dt}\left(\frac{\vec r(t)}{|\vec r(t)|}\right)\\\\ &=\frac{1}{|\vec r(t)|}\frac{d\vec r(t)}{dt}-\frac{\vec r(t)}{|\vec r(t)|^2}\frac{d|\vec r(t)|}{dt}\\\\ &=\frac{\dot{\vec r}(t)}{|\vec r(t)|}-\frac{\vec r(t)}{|\vec r(t)|^2}\,\dot{|\vec r(t)|} \end{align}$$