How the Derivative of Distance is Different from Speed

Question

If the velocity vector is given by $\frac{d\overrightarrow{{r}}}{dt}$ and physically, the speed observed is just the magnitude of the velocity vector at any time $t$, then what is the physical interpretation of the derivative of the scalar distance with respect to time (i.e. the $\frac{d|\overrightarrow{{r}}|}{dt}$)? Expressing the displacement $\overrightarrow{r}$ in rectangular coordinates, I verified that $\frac{d|\overrightarrow{{r}}|}{dt}$ which is the derivative of the scalar distance is not identical to $|\frac{d\overrightarrow{{r}}}{dt}|$ which is the "speed" of the moving body. So, again, what is the physical or geometric interpretation of $\frac{d|\overrightarrow{{r}}|}{dt}$?

Answer 1

It is the component of the velocity parallel to the displacement. It can be verified by differentiating $\sqrt{{\Delta x}^2+{\Delta y}^2+{\Delta z}^2}$ with respect to time $t$ and comparing the result with the dot product $\frac{d\overrightarrow{r}}{dt} \cdot \frac{\overrightarrow{r}}{|\overrightarrow{r}|}$ which gives an identical result. In rectangular coordinates, $\frac{d|\overrightarrow{{r}}|}{dt}=\frac{ds}{dt}$ represents the rate at which the straight-line distance of a particle from a source point $(x_0,y_0,z_0)$ (for instance, the origin $(0,0,0)$) changes with time whereas in cylindrical coordinates, $\frac{ds}{dt}$ is itself the speed $v$ of the particle.

Answer 2

The difference between a position $\vec r$ and a distance $|\vec r|$ is that the distance is taken with respect to an implicit reference point, the origin of the coordinates. As this is an arbitrary point in space, it cannot have a physical meaning unless there is something special related to it (for instance if this is the center of mass of the system, or a point where a non-moving object is located). The velocity $\frac{\mathrm d\vec r}{\mathrm dt}$ is completely independent from the location of the origin while the derivative of the distance, $\frac{\mathrm d\,|\vec r|}{\mathrm dt}$ is not.

In polar coordinates, $\vec r=r\,\hat u_r(\theta)$, where $\hat u_r(\theta)$ is the unit radial vector and $\theta$ is the direction. Cartesian coordinates are given by $x=r\cos\theta$ and $y=r\sin\theta$. We have $|\vec r|=r$. Derivating $\vec r$ with respect to time, we get $$\frac{\mathrm d\,\vec r}{\mathrm dt}=\left(\frac{\mathrm dr}{\mathrm dt}\right)\hat u_r+r\left(\frac{\mathrm d\theta}{\mathrm dt}\right)\hat u_\theta.$$

This formula proves that @Paul's answer is right if he defines displacement as the vector for an origin to the object. It also shows that the speed $\left|\frac{\mathrm d\vec r}{\mathrm dt}\right|$ is always larger than $\frac{\mathrm d\,|\vec r|}{\mathrm dt}$.

Consider an object moving at constant speed $s$ and chose an origin at a distance $b$ from its trajectory. We can call $t_0$ the time at which the object is at the shortest distance from the origin. The distance between the origin and the object is $$ |\vec r(t)|=\sqrt{b^2+s^2(t-t_0)^2}$$ and its derivative is $$ \frac{\mathrm d\,|\vec r|}{\mathrm dt}=\frac{s^2t}{\sqrt{b^2+s^2(t-t_0)^2}}.$$ We observe that this depends on $b$ and $t_0$, meaning that if we choose another origin we change the value of $\frac{\mathrm d\,|\vec r|}{\mathrm dt}$ while the speed remains equal to $s$.