Derivative of magnitude of position vector

Question

I have encountered this derivative in the context of orbital physics but I cannot understand the steps taken to achieve the last line of the equation.

$$\frac{dr}{dt}=\frac{d}{dt}\sqrt{\vec r\cdot\vec r}$$

$$=\frac{\vec r}{r}\cdot\frac{d\vec r}{dt}$$ Where $r=|\vec r|$

My attempt:$$\frac{d}{dt}\sqrt{\vec r\cdot\vec r}=\frac{d}{dt}\frac{\vec r \cdot \vec r}{r}$$ $$=\frac{r((\frac{d\vec r}{dt}\cdot \vec r)+(\vec r\cdot \frac{d\vec r}{dt}))-\frac{dr}{dt}(\vec r \cdot \vec r)}{r^2}$$ $$=\frac{r((\frac{d\vec r}{dt}\cdot \vec r)+(\vec r\cdot \frac{d\vec r}{dt}))}{r^2}-\frac{dr}{dt}$$ $$=\frac{((\frac{d\vec r}{dt}\cdot \vec r)+(\vec r\cdot \frac{d\vec r}{dt}))}{r}-\frac{dr}{dt}$$

I have tried writing $\sqrt{\vec r\cdot\vec r}$ as $\frac{\vec r \cdot \vec r}{r}$ and then using the quotient rule, but I get an erroneous result. I believe I am mistreating the dot product in my working or something else that is not apparent to me.

I would like to know steps taken to achieve this result. Thanks

Answer 1

continuing Your attempt, last line:

$$ \begin{aligned} \frac{dr}{dt}&=2\frac{\frac{d\vec{r}}{dt}\cdot\vec{r}}{r}-\frac{dr}{dt}\\ 2\frac{dr}{dt}&=2\frac{\frac{d\vec{r}}{dt}\cdot\vec{r}}{r}\\ \frac{dr}{dt}&=\frac{\frac{d\vec{r}}{dt}\cdot\vec{r}}{r} \end{aligned} $$

Answer 2

You can make the following substitution $$\vec r\cdot\vec r=u,$$ and perform differentiation to get the required result. $$\frac{dr}{dt}= \frac{d}{dt} \sqrt u $$ $$=\frac{1}{2\sqrt u}\frac{du}{dt}$$ $$=\frac{1}{2r}\frac{d}{dt}(\vec r\cdot\vec r)$$ $$=\frac{1}{2r}(2\vec r\cdot\frac{d\vec r}{dt})$$ $$=\frac{\vec r}{r}\cdot\frac{d\vec r}{dt}$$