Derivative of position [Beginning calculus question.] I saw in a calculus lecture online that for a position vector $\boldsymbol{r}$
$$\left|\frac{d\boldsymbol r}{dt}\right| \neq 
\frac{d\left| \boldsymbol r \right|}{dt}$$
but I don't understand exactly how to parse this. 
It's my understanding that: 


*

*$\frac{d\boldsymbol r}{dt}$ refers to the rate of change in the
position over time (speed?)

*$|\boldsymbol r|$ refers to the magnitude of the position, i.e. the distance (from what to what?)

*$\frac{d\left| \boldsymbol r \right|}{dt}$ refers to the rate of change in distance traveled over time, (a different kind of speed?)


Is there a good way to understand what both of these expressions mean?
 A: We can understand this with one example. Consider angular motion problem where position of a particle is given by $\mathbf{r}=\hat{\imath}\cos(\omega t)+\hat{\jmath}\sin(\omega t)$. Now $\frac{d\mathbf{r}}{dt}=-\hat{\imath}\omega\sin(\omega t)+\hat{\jmath}\omega \cos(\omega t)$, clearly $\lvert \frac{d\mathbf{r}}{dt}\rvert = \omega$ , whereas $\lvert \mathbf{r} \rvert =1$, hence $\frac{d\lvert \mathbf{r}\rvert}{dt}=0$. It means that a particle is at constant position, within unit radius, only angular position is changing, hence derivative of $\lvert \mathbf{r}\rvert$ is zero.
A: The first one is looking at the velocity $\mathbf{v}=\frac{d\mathbf{r}}{dt}$, and taking its norm: it's the value of the speed, i.e. the value of (instantaneous) change in position.
The second is looking at the distance from the point of coordinates $\mathbf{0}$ (the origin), $\lvert \mathbf{r}\lvert$, and taking its derivative: it's the instantaneous change in distance from the origin.
The two indeed need not be equal: 
imagine you are moving very fast, but staying at the same distance from the origin (that is, you're moving very fast on a circle). Then the speed $\left\lvert \frac{d\mathbf{r}}{dt}\right\rvert$ is big (you're moving fast), but $\lvert \mathbf{r}\lvert$ is constant -- so $\frac{d\lvert\mathbf{r}\rvert}{dt} = 0$.
