Summary

Code

Executable on S2

Videos

## First-Order Rate of Change

We notice that speed is the fraction of distance divided by time, or the ratio of distance to time, or the rate of change of distance per time. They all mean the same thing. In general, the concept of rate of change is fundamental to mathematics and in all higher mathematics such as Calculus. We, therefore, want to state and emphasize this again:

A rate of change is how much a quantity (A) changes per unit of another quantity (B).

$$\text{rate} = \frac{A}{B}$$

The concept ‘rate’ connects two quantities, e.g., A and B, together. It describes how one quantity A changes when the other quantity B changes. This concept is fundamental in mathematics and science in general.

In our context, speed is defined as the rate of change in distance per one unit of time. That is, the rate is speed, A is distance, and B is time.

For example, suppose a rabbit travels 10 meters in 2 seconds, then its speed is:

$$\text{speed}(v) = \frac{\text{10 meters}}{\text{2 seconds}} = 5 \text{ meter/second} = 5\text{ m/s}$$

Another example is acceleration. Acceleration is the rate of change of speed per unit time.

For example, suppose the rabbit at time 0 has a speed of $$1\text{ m/s}$$ and then in 10 seconds, it increases its speed to $$5\text{ m/s}$$. Then the acceleration during this period is:

$$\text{acceleration}(a) = \frac{5\text{ m/s} – 1\text{ m/s}}{10\text{ s}} = 0.4{\text{ m/s}}^{2}$$

There are many more rate examples. For instance, price is the the amount of money paid per item; number of people arriving at the beach each hour; number of newborn babies each year. In general, to get a rate, you will divide the total by the number of units. The total can be the lumpsum payment, the total number of people at the beach or the number of birth in the last decade.

We will see that rate (of change) is a linear function. Because of the paramount importance of rate (of change) in mathematics, physics, engineering, and pretty much all sciences, the linear function is probably the most prevalent and most important function among all functions.

## Second-Order Rate of Change

Acceleration is the rate of change of speed per unit time. Interestingly, speed itself is the rate of change of distance per unit time. We can see that acceleration must be somehow related to the change of distance and time.

Mathematically, we have,

$$\text{acceleration}(a) = \frac{\Delta v}{\Delta t} \tag{eq 1}$$,

where $$\Delta v$$ means the change or difference in speed, and $$\Delta t$$ the change or difference in time. The symbol $$\Delta$$ is a mathematical way to write change or formally difference.

Now, we know that speed

$$v = \frac{\Delta d}{\Delta t}$$,

where $$\Delta d$$ means the difference in distance and $$\Delta t$$ means the difference in time.

Plug the speed equation into the acceleration equation. That is, replace or substitute the $$v$$ in $$\text{eq 1}$$ and copy the rest of the items/symbols/terms, we have:

$$\text{acceleration}(a) = \frac{\Delta \Delta d}{\Delta t \Delta t} = \frac{\Delta^2 d}{\Delta t^2} \tag{eq 2}$$

Therefore, speed is the first-order rate of change of distance per unit time. Acceleration is the rate of change of the rate of change of distance per unit time per unit time. Alternatively, acceleration is the second-order rate of change in distance per unti time squared. Or simply, acceleration is the second-order rate of change in distance with respect to (w.r.t.) time.

We can expand $$\text{eq 1}$$ to see what this equation exactly means. To compute speed, the rate of change in distance, we need two distances to take their difference. To compute acceleration, we need three distances to compute two speeds and then the speed differences. Suppose those three distances are, in ascending order, $$d_1$$, $$d_2$$, $$d_3$$.

The first speed is:

$$v_1 = \frac{d_2 – d_1}{\Delta t}$$

The second speed is:

$$v_2 = \frac{d_3 – d_2}{\Delta t}$$

Using $$\text{eq 1}$$ to compute the acceleration,

$$\text{acceleration}(a) = \frac{\Delta v}{\Delta t} = \frac{v_2 – v_1}{\Delta t}$$$$= \frac{ \frac{d_3 – d_2}{\Delta t} – \frac{d_2 – d_1}{\Delta t}}{\Delta t}$$$$= \frac{ \frac{d_3 – 2d_2 + d_1}{\Delta t}}{\Delta t} = \frac{d_3 – 2d_2 + d_1}{\Delta t^2}$$

As we can see from the above,

$$\Delta^2 d = d_3 – 2d_2 + d_1$$

$${\Delta t^2}$$ means what it says, simply, the squared of $${\Delta t}$$.

That is, $$\text{eq 2}$$ means

$$\text{acceleration}(a) = \frac{\Delta \Delta d}{\Delta t \Delta t} = \frac{\Delta^2 d}{\Delta t^2} = \frac{d_3 – 2d_2 + d_1}{\Delta t^2}$$

Therefore, we call acceleration the second-order rate of change in distance w.r.t. time.

For example, suppose the rabbit takes 5 s to run from $$d_1 = 0\text{ m}$$ to $$d_1 = 5\text{ m}$$ and then another 5 s to from $$d_1 = 5\text{ m}$$ to $$d_1 = 30\text{ m}$$. We want to compute its speeds and acceleration.

We notice that the rabbit runs faster in the second period because during the same 5 s, it runs 20 m further than in the first period. The speeds in the two periods are different. For the first period, the speed is:

$$v_1 = \frac{5\text{ m} – 0\text{ m}}{5\text{ s}} = 1 \text{ m/s}$$

For the second period, the speed is:

$$v_2 = \frac{30\text{ m} – 5\text{ m}}{5\text{ s}} = 5 \text{ m/s}$$

The acceleration is:

$$a = \frac{5\text{ m/s} – 1\text{ m/s}}{5\text{ s}} = 0.8{\text{ m/s}}^{2}$$

We can also compute the acceleration using

$$a = \frac{30\text{ m}- 2*5\text{ m} + 0\text{ m}}{(5\text{ s})^2} = 0.8{\text{ m/s}}^{2}$$

We will see throughout this course that the concept of rate of change is fundamental to Calculus and all higher (applied) mathematics. This simple concept is one of the greatest inventions in mathematics in all time!