What does the derivative of velocity with respect to position mean?

According to a Physics book, for a particle undergoing motion in one dimension (like a ball in free fall) it follows that

$$\frac{dv}{ds} = \frac{dv}{dt} \frac{dt}{ds} = \frac{a}{v},$$

where $v$ is the velocity and $s$ is the position of the particle.

But I have problems understanding this, specially because of the use of Leibniz's notation.

I think of the position of a particle at time $t$ (under a frame of reference) as described by the image of a function $s$. The number $s(t)$ represents the particle's 'coordinate on the axis' at time $t$. Then the function $s$ relates instants of time to those points in space where the particle is supposed to be.

The velocity of the particle at time $t$ is then $s'(t)$ and the acceleration $s''(t)$. We denote the function $s'$ simply as $v$ and $s''$ as $a$.

It is supposed that using the chain rule yields the previous the result of the book, but formally the chain rule is stated as

$$(f \circ g)' = (f' \circ g) \cdot g'$$

for any two differentiable functions $f$ and $g$.

Then:

Why is the velocity function treated as a composition?

What function does $\frac{dv}{ds}$ represent?

• I fixed your equation, feel free to revert it if you disagree. Dec 10 '16 at 1:24
• Basically, the function $v(s)$ which maps $s$ to $v$ can be thought of as the composition of the functions $t(s)$ and $v(t)$. Dec 10 '16 at 1:26
• This is actually a MathSE question, to be formal about it......but now that you are here....
– user108787
Dec 10 '16 at 1:29
• @Gert: Actually, that's not the chain rule. Dec 10 '16 at 1:56
• $(f\times g)'=f'g+fg'$. Sorry, meant product rule, of course. My bad. OP's formula is not the chain rule.
– Gert
Dec 10 '16 at 1:58

What's confusing you is that the (perfectly standard) notation sucks, because the same letter $t$ is being used both to denote a number (that is, a particular time) and a function (that is, time as a function of position).

To reduce confusion, let $t$ be the time (a number) and let $T$ be the function that tells you, for any given position, the time at which your particle achieves that position. So at any given moment $t$, if the particle is at $s$, then $t=T(s)$.

The next thing that sucks about the notation is that $v$ is being used for two different functions: Velocity as a function of time and velocity as a function of position. To clear this up, introduce two functions: $w(t)$ tells you the velocity at time $t$ and $v(s)$ tells you the velocity when the particle is at position $s$. Then

$$v(s)=w(T(s))$$

Apply the chain rule to get $$v'(s)=w'(T(s))T'(s)$$

Because the Leibniz notation freely confuses $t$ with $T$ and $v$ with $w$, $w'(T(s))$ ends up being written as $dv/dt$ and $T'(s)$ ends up being written as $dt/ds$. So that gives you your formula.

• We know $s(t)$ exists globally (for all $t$) by standard physical axioms. However, in general it won't be the case that $T(s)$ is defined globally, and there may be multiple functions $T(s)$ for some $s$ (if $s(t)$ is non-invertible). The function $T(s)$ is only neccessarily defined on some restricted domain by the inverse function theorem. However, this doesn't obstruct the fact that you can calculate $\frac{\text{d}v}{\text{d}s}$ (this works, so long as $\frac{\text{d}s}{\text{d}t}\neq 0$, i.e. $v\neq 0$). Apr 14 '21 at 11:26

I think I may clear up the issue.

s(t) is not position it is the arc length function, it gives you the length a particle has moved along curve x(t) for a time interval t.

ds/dt is the instantaneous tangential speed of the particle also known as |v| or |dx/dt|.

So s(t) is the integral of instantaneous velocity with respect to time. And dv/ds is the rate of change of velocity with respect to arc length. The chain rule you posted checks out.

• "$s$ is the position of the particle." That's how he defines it. Changing his definition doesn't provide an answer. Authors define symbols as they see fit: answerers do not get to change that.
– Gert
Dec 10 '16 at 1:56
• You have reason but this is straight out of my calc iii book. I think OP has misunderstood the problem. Because if you'll note, arc length IS a position, it is the position along a the curve. Also note that the question absolutely does not make sense if you define it as "x(t) a vector valued function in three space", because dv/dx does not make any sense. Dec 10 '16 at 2:04
• @Gert You have reason but this is straight out of my calc iii book. I think OP has misunderstood the problem. Because if you'll note, arc length IS a position, it is the position along a the curve, bit it is a scalar valued function. Also note that the question absolutely does not make sense if you define it as "x(t) a vector valued function in three space", because dv/dx does not make any sense. Dec 10 '16 at 2:07
• $\frac{\mathbf{d}v}{\mathbf{d}x}$ is just a derivative: it doesn't 'make sense' or 'not make sense'. I don't care if it comes out of your 'calc iii' book or the Bible.
– Gert
Dec 10 '16 at 2:08
• Nor did he post the chain rule. The chain rule is $[f(g)]'=f'(g)\times g'$.
– Gert
Dec 10 '16 at 2:10