Fluxion

Time Derivative§

Newton already had the concept of the time derivative in his own formulation of calculus, which he called a fluxion. So before entering Newtonian mechanics, let us quickly review the calculus we will need.

Put simply, a time derivative tells us how quickly something changes with respect to time.

For a function $f(t)$, its time derivative at $t_0$ is defined as

$$\left.\frac{\mathrm{d}f(t)}{\mathrm{d}t}\right|_{t=t_0} := \lim_{\Delta t\to0} \frac{f(t_0+\Delta t)-f(t_0)}{\Delta t}.$$

Here, $\Delta$ indicates a finite change in a quantity. As we take the limit $\Delta t\to0$, we are asking how the quantity changes over an infinitesimally small interval of time. This limiting rate of change is what we call the time derivative.

Displacement§

For a quick review, displacement is the change in position of an object. Unlike distance, displacement is a vector, meaning that both its magnitude and direction matter.

For example, if someone moves $3\,\mathrm{m}$ forward and then $4\,\mathrm{m}$ to the right, their displacement can be represented as

$$\vec{s} = \begin{pmatrix} 3\\ 4 \end{pmatrix} \mathrm{m}.$$

The important point is that displacement describes where the object ended up relative to where it started, not the total distance travelled.

Velocity§

Velocity describes how quickly displacement changes with time.

The average velocity between $t_0$ and $t_0+\Delta t$ is

$$\vec v_{\mathrm{avg}} = \frac{\Delta\vec s}{\Delta t} = \frac{\vec s(t_0+\Delta t)-\vec s(t_0)}{\Delta t}.$$

If we shrink the time interval until it approaches zero, we obtain the instantaneous velocity:

$$\vec v(t_0) = \left. \frac{\mathrm d\vec s(t)}{\mathrm dt} \right|_{t=t_0}.$$

In other words,

$$\vec v=\frac{\mathrm d\vec s}{\mathrm dt}$$

Velocity is therefore the time derivative of displacement.

Acceleration§

We can repeat exactly the same idea one step further.

The average acceleration is the change in velocity divided by the time over which that change occurs:

$$\vec a_{\mathrm{avg}} = \frac{\Delta\vec v}{\Delta t}.$$

Taking the limit as the time interval approaches zero gives the instantaneous acceleration:

$$\vec a(t_0) = \left. \frac{\mathrm d\vec v(t)}{\mathrm dt} \right|_{t=t_0}.$$

Therefore,

$$\vec a=\frac{\mathrm d\vec v}{\mathrm dt}$$

Since velocity is itself the derivative of displacement,

$$\vec a = \frac{\mathrm d^2\vec s}{\mathrm dt^2}$$

So we now have a simple chain:

$$\vec s \;\xrightarrow{\frac{\mathrm d}{\mathrm dt}}\; \vec v \;\xrightarrow{\frac{\mathrm d}{\mathrm dt}}\; \vec a$$

Position tells us where something is.

Velocity tells us how its position changes.

Acceleration tells us how its velocity changes.

This chain will become extremely important once we begin Newtonian mechanics.

Intuition by Case Study§

Rather than leaving these definitions as abstract equations, let us see what they actually mean.

Uniform Motion§

Let's have a car moving in only one direction, so we only need one spatial coordinate.

Suppose its position is

$$s(t)=vt$$

where

$$v=3\,\mathrm{m\,s^{-1}}$$

is a constant.

This means that the car moves steadily at $3\,\mathrm{m\,s^{-1}}$.

After a time interval (\Delta t), its displacement is

$$\begin{align*} \Delta s &=s(t+\Delta t)-s(t)\\ &=v(t+\Delta t)-vt\\ &=v\Delta t. \end{align*}$$

Therefore, for every second that passes,

$$\Delta t=1\,\mathrm{s}$$

and the car travels

$$\Delta s = 3\,\mathrm{m\,s^{-1}}\times1\,\mathrm{s} = 3\,\mathrm{m}.$$

The longer the car travels, the larger its displacement becomes, but the rate at which the displacement grows never changes.

Indeed,

$$v_{\mathrm{avg}} = \frac{\Delta s}{\Delta t} = v = 3\,\mathrm{m\,s^{-1}}.$$

And because this is true for any time interval, no matter how small, the instantaneous velocity is also

$$v(t)=3\,\mathrm{m\,s^{-1}}.$$

Now look at the velocity itself.

It does not change with time:

$$\Delta v=0.$$

Therefore,

$$a_{\mathrm{avg}} = \frac{\Delta v}{\Delta t} = 0$$

And because the velocity never changes, taking the limit does not change the result:

$$a(t)=0$$

So uniform motion means exactly what the name suggests:

The object moves, but its motion does not change.

This distinction will become important later. Moving is not the same thing as accelerating.

## Uniform Acceleration§

Now let the driver press the accelerator.

Suppose the car begins from rest and its velocity increases uniformly with time. For simplicity, let

$$a=2\,\mathrm{m\,s^{-2}}.$$

This means that every second, the velocity increases by

$$2\,\mathrm{m\,s^{-1}}.$$

Therefore,

$$v(t)=at.$$

After (1,\mathrm),

$$v=2\,\mathrm{m\,s^{-1}},$$

after (2,\mathrm),

$$v=4\,\mathrm{m\,s^{-1}},$$

and after (3,\mathrm),

$$v=6\,\mathrm{m\,s^{-1}}.$$

Now we finally have a situation where average velocity and instantaneous velocity are different.

Suppose we look at the car from (t=0) to (t=3,\mathrm). Since the car starts from rest and has constant acceleration,

$$s(t)=\frac12at^2.$$

Therefore,

$$s(0)=0, \qquad s(3)=\frac12(2)(3^2)=9\,\mathrm m.$$

The average velocity over those three seconds is therefore

$$v_{\mathrm{avg}} = \frac{\Delta s}{\Delta t} = \frac{9\,\mathrm m}{3\,\mathrm s} = 3\,\mathrm{m\,s^{-1}}.$$

But what is the instantaneous velocity at (t=3,\mathrm)?

For that, we take the derivative:

$$v(3) = \left.\frac{\mathrm ds}{\mathrm dt}\right|_{t=3}.$$

Since

$$s(t)=\frac12at^2,$$

we have

$$\frac{\mathrm ds}{\mathrm dt}=at.$$

Therefore,

$$v(3)=2(3)=6\,\mathrm{m\,s^{-1}}.$$

So:

$$\frac{\Delta s}{\Delta t}=3\,\mathrm{m\,s^{-1}} \neq 6\,\mathrm{m\,s^{-1}} = \frac{\mathrm ds}{\mathrm dt}$$

The difference is important.

The average velocity asks:

Over this entire interval, how much displacement did we gain per unit time?

The instantaneous velocity asks:

At this exact moment, how quickly is the position changing?

For uniform motion, these happen to be the same because the velocity never changes. But once the velocity changes, they are generally different.

And this also gives us an intuitive picture of what the derivative is doing.

As we make the interval smaller,

$$\frac{\Delta s}{\Delta t}$$

becomes the average velocity over a shorter and shorter period. In the limit

$$\Delta t\to0,$$

that average velocity approaches the velocity at that exact instant:

$$\frac{\mathrm ds}{\mathrm dt} = \lim_{\Delta t\to0} \frac{\Delta s}{\Delta t}$$

This is why instantaneous velocity is not a completely different concept from average velocity. It is the limiting value of average velocity as the time interval shrinks to zero.

Now look at the velocity itself.

Because

$$v(t)=at,$$

the velocity changes with time:

$$\Delta v=a\Delta t.$$

Therefore,

$$\frac{\Delta v}{\Delta t}=a.$$

Taking the limit gives

$$\frac{\mathrm dv}{\mathrm dt}=a$$

So uniform acceleration means exactly this:

The velocity changes at a constant rate.

And this gives us the hierarchy we need:

$$s\xrightarrow{\frac{\mathrm d}{\mathrm dt}}v\xrightarrow{\frac{\mathrm d}{\mathrm dt}}a$$

Position tells us where the object is.

Velocity tells us how quickly its position is changing.

Acceleration tells us how quickly its velocity is changing.

And this is exactly the mathematical language we need before entering Newtonian mechanics.

Newton's question will no longer simply be "where is the object?" or "how fast is it moving?"

The question becomes:

What causes its velocity to change?