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My basic physics' knowledge is a little rusty. My apologies in advance. I know that the first derivative of position or displacement with respect to time is the instantaneous velocity. Suppose I have a distance calculated with

$$D = \sqrt{\Delta x^2 + \Delta y^2}$$

Does $\frac{dD}{dt}$ have an interpretation? For example, if it is instantaneous speed (the rate of change in distance per time), can $\frac{dD}{dt}$ be negative? I mean, can I use $\frac{dD}{dt}$ for finding critical points of $D$? And if it is 0 at $t$, does it mean an object is not moving at all at $t$?

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3 Answers 3

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As @Dvij D.C. points out, this is not the speed, because that has to be measured along the path on which the object is moving.

Instead, I'd suggest the construct to which you refer perhaps best be called a "radial speed" (or perhaps "radial velocity" as it can be negative - and the meaning of negative is that you're getting closer, not further) with regard to some reference point. Namely if your $\Delta x$ and $\Delta y$ are calculated for a moving object relative to a fixed origin, then $D$ is the distance from that origin, or "radial distance", so its derivative is how fast the radial distance is changing. That is, it is how quickly or not the object is moving farther away or closer to the origin in question, and there will be different radial speeds for different choices of origin.

For example, suppose you want to talk about a car driving on a road away from a town. The speed will be whatever is on the speedometer; that's what Dvij. D.C.'s integrals reference. But what this speed is telling you is how fast the car is getting away from the town. If the road is, say, very loopy and a lot of driving on it doesn't get you so far, the radial speed - the amount of miles you're putting between yourself and the town in any given time - will be low, because while you're moving down the road quickly, the loops are still keeping you near the town. On the other hand, if the road leads straight away, then this speed and your speedometer speed will be equal: every mile you go down the road is a mile further away from the town.

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The distance wouldn't be $\sqrt{\Delta x^2 + \Delta y^2}$ unless the particle is moving in a straight line. In general, distance would be given by $\int_{0}^{\Lambda} d\lambda \sqrt{\Big(\frac{dx}{d\lambda}\Big)^2+\Big(\frac{dy}{d\lambda}\Big)^2}$ where $\lambda$ is a parameter used to parametrize the path of the particle (it could or could not be time).

You can take this parameter $\lambda$ to be the the length of the path itself. Then the distance travelled in time $t$ is expressed by the trivial relation $D(t)=\int_0^{D(t)} ds$ ;) However, let's take this parameter to be time and see explicitly what its time derivative means. So, we write

$$D(T)=\int_{0}^{T} dt \sqrt{\Big(\frac{dx}{dt}\Big)^2+\Big(\frac{dy}{dt}\Big)^2}$$

Taking the time derivative would give us

$$\frac{D(T)}{dT}= \sqrt{\Big(\frac{dx}{dt}\Big)^2+\Big(\frac{dy}{dt}\Big)^2}\Bigg|_{t=T}$$

Thus, it's clear that the time derivative of $D$ gives us the instantaneous speed which is obvious from the physical interpretation of what the time derivative of $D$ would mean. It would mean the temporal rate of change of distance travelled by a particle and that's exactly what instantaneous speed is. As is clear from both mathematical expression and physical meaning, this derivative cannot be negative (mathematically, because square roots are not negative and physically because it's meaningless to say that you travelled a negative distance). If it's zero at a point in time, it means that the object is at instantaneous rest at that point (however, it doesn't mean that it would remain at rest afterwards).

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$D$ is just the polar coordinate $r$, so $$\frac{dD}{dt}=\frac{dr}{dt}=v_r$$ is the radial component of velocity, i.e. $v_r=\mathbf{v}\cdot\mathbf{\hat r}$.

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