A Pragmatic Approach to Calculus - Part 1

11 min read
A  Pragmatic Approach to Calculus - Part 1

Calculus is one of the most revolutionary ideas ever developed. It has been used in practically all branches of mathematics and sciences. It is one of those branches of mathematics which is used across so many disciplines that it becomes much easier to appreciate the idea, unlike other topics in abstract mathematics. In spite of this fact, most people find it mysterious and complicated to understand. Introducing calculus in a classroom often requires quite a few prerequisites, and that is why it becomes slightly harder to begin. In this article we will take a very intuitive and somewhat unconventional approach to understand calculus from ground up.

This article is geared towards the reader who always wanted to learn the subject but could not find the motivation and/or the time to pick up a textbook. A more experienced reader may find the content less rigorous and/or incomplete, but the intention here is make the subject more approachable and to eventually motivate the reader to pick up that textbook.

Why Calculus ?

It is reasonable to ask the following question - What kinds of problems did we try to solve which forced us to come with this new idea?

The answer to this question is rather simple - Calculus is
a very precise method to quantify change. So any problem that requires us to study change in certain quantities can easily be solved using calculus. Now this does not sound very impressive because, we have been successfully studying all sorts of changes in all sorts of systems for a really long time, even when such a method was not available. So what new stuff does calculus bring to the table? Well, the key advantage that calculus offers is - precision. Let us see what we mean by this.

The Study of Change

Imagine there is a car moving at a fixed speed of 100 km/hr. It is obvious that it travels a distance of 100 km in one hour from any chosen point of departure. If we ask a question - How fast is it moving when it has travelled 300 km from the point of departure? The answer is simple, since the car is moving at a constant speed, it would still be going at 100 km/hr at the end of 300 km.

Now we change the situation a bit. Suppose the driver speeds up the car by 5 km/hr after every one hour interval. If we ask the same question again, it requires a bit of calculation to find the answer. Let us see how its done -

  1. From the point of departure, the speed of the car is 100 km/hr, so the distance covered in the first hour is 100 km.
  2. At end of the first hour, the driver has increased the speed by 5 km/hr. So the new speed of the car becomes 100 + 5 = 105 km/hr and the distance covered in the second hour will be 105 km. Therefore, the total distance covered at the end of 2 hours is 100 + 105 = 205 km.
  3. At end of the second hour, the driver has increased the speed by 5 km/hr. So the new speed of the car becomes 105 + 5 = 110 km/hr and the distance covered in the third hour will be 110 km. Therefore, the total distance covered at the end of 3 hours is 100 + 105 + 110 = 315 km.

We see that the car has crossed the 300 km mark, and its speed while crossing the mark is 110 km/hr. This is our final answer. Here we saw one type of problem that requires us to understand how to deal with changing quantities. Clearly this case was not as easy as the one where the speed was constant during the entire journey, but we could still solve the problem by using simple mathematics and we did not require anything new or advanced.

Let us now change the situation a bit more. Let us say that the driver speeds up the car by 5km/hr every minute instead of every hour. To solve the problem, now we have to keep note of the distance covered at the end of each minute rather than each hour. We can imagine that this case is more tedious than the previous one because there are quite a lot of intermediate steps involved. Still, it is nothing we cannot do with the same method.

Let us go one step further and see what happens if the driver increases the speed by 5 km/hr every second or every millisecond? This now requires a lot of calculations to get the final answer. The intermediate steps are simple, but there are just too many steps to calculate, making it a difficult problem.

We can imagine that its not a hypothetical situation as we have know that we can make cars pick up speed every moment. In fact, this phenomenon has a name - Its called accelerating the car. It would be very nice if we could find a method to do all the intermediate steps in one go!

Actually, it is not very hard to observe the pattern in which the calculations are to be performed and after a careful look, one might be able to write down a formula for this particular situation. The only problem is that we would require a new formula for every new type of problem. There are so many other things that can change during the journey of the car. There might even be other kinds of systems where changing quantities need to be computed. What we desire is a general framework for dealing with quantities that change.

The Origin of Calculus

The example discussed in the previous section was good enough to appreciate the need for a new method, but it's not the best example to introduce the mathematical details involved. We will come back to it once we have learned a bit more about the subject. In order to construct a systematic framework, we begin with a problem which is mathematically simpler.

Imagine that there is a tiny disturbance in a pond and water waves are generated. It is obvious from the picture below that the water waves are circular in shape and they move outwards from the the point of disturbance.

Let us choose any one circle on the wave. Since the disturbance in water is moving outwards, the size of this circle increases continuously over time. To make it simple, we assume that the size of the circle increases by a fixed amount, say 1 centimeter per second.

Now we ask a question - How fast is the area of the circle changing?

To find the answer, first we calculate the area of the circle at any given moment. Then, we let the circle change its size over a period of one second and then calculate the area again. The difference in areas is what we need. Let us do the math in four steps -

  1. Let us take the radius of the circle be $r$ centimeter at time $t$ seconds. At that instant the area is
    $$A_t = \pi r^2$$
  2. Now the time after one second is $(t+1)$ seconds. Also the radius is changing at a rate of 1 cm/sec, so the radius of the circle after one second is $r+1$ centimeter and the area after one second is
    $$A_{t+1} = \pi (r+1)^2$$
  3. Then, the change in area is
    $$\begin{aligned}A_{t+1} - A_t &= \pi (r+1)^2 - \pi r^2 \\ &= \pi \left( r^2 + 2r + 1 \right) - \pi r^2 \\ &= \pi \left( 2r + 1\right) \end{aligned}$$
  4. Now, the rate at which the area changes is obtained by dividing the change in area by the change in time -
    $$\frac{A_{t+1} - A_t}{(t+1) - t} = \frac{\pi \left( 2r + 1\right)}{1} = \pi \left( 2r + 1\right)$$
    That's the final answer. Let us understand the meaning of this result - Even though the radius increases by a fixed amount every second, the area does not! In fact, we can see from the final result that the change in area depends on the radius. So we can make the following statement -

The bigger the circle gets, the faster its area increases!

Before we proceed, we need to discuss something very important - We never talked about the measuring devices that we might have used for measuring length of the radius and time period. Let us say that we measure the length with a ruler that can measure lengths down to a centimeter. Also, to measure time, we have a clock that measures time down to a second.

You might be wondering - Why do we care about measuring instruments?

It might be surprising to know that this is exactly the kind of question that leads us to the idea of calculus. Now, what if we "improve" the length and time measuring devices so that we can now measure the lengths down to half a centimeter and also the time down to half a second?

We can repeat the calculations for a change over a time period of half a second -

  1. Just like before, let the radius of the circle be $r$ centimeter at time $t$ seconds. At that instant the area is

$$A_t = \pi r^2$$
2. Now the time after half a second is $(t+1/2)$ seconds. Since the radius is changing at a rate of 1 cm/sec, and we are making an observation after half a second, the radius of the circle becomes $(r+1/2)$ centimeter. So, the area after half a second is

$$A_{t+1/2} = \pi (r+1/2)^2$$
3. Then, the change in area is

$$\begin{aligned}A_{t+1/2} - A_t &= \pi (r+1/2)^2 - \pi r^2 \\ &= \pi \left( r^2 + r + 1/4 \right) - \pi r^2 \\ &= \pi \left( r + 1/4 \right) \end{aligned}$$
4. Now, the rate at which the area changes is obtained by dividing the change in area by the change in time -

$$\frac{A_{t+1/2} - A_t}{(t+1/2) - t} = \frac{\pi \left( r + 1/4 \right)}{1/2} = \pi \left( 2r + 1/2\right)$$

After this boring repetition, we see that the only difference from the earlier result is that we have a factor of "1/2" instead of "1" in the final result.

It is not a coincidence that this factor of "1/2" is same as that of time interval over which the radius has increased. We can repeat this whole exercise for measuring devices with even better "resolutions" and see that the factor in the final result will keep on decreasing as the resolution of the measuring devices increases. All this means is - If we want to measure the rate of change of area to a "greater" precision, we must use measuring devices with "higher" resolutions.

We can also interpret this statement as follows - If the system changes very rapidly, we are forced to choose our measuring devices with very high resolutions, otherwise we will not be able to quantify the change with high precision.

A Simple Formula

Now that we have established the connection between high precision and rapid changes, we are finally ready to write down a general formula for studying these quantities.

Following the logic developed in the previous section, let us make a general statement -

Change in a quantity $Y$ (like area) with respect to another quantity $X$ (like time) is given by the ratios of their respective changes.

To clarify this statement, we write the following steps -

  1. Calculate the initial value of $Y$ called $Y_i$ for an initial value of $X$ called $X_i$. We can do this because $X$ and $Y$ are related, depending on the system under consideration.
  2. Calculate the final value of $Y$ called $Y_f$ for a final value of $X$ called $X_f$.
  3. Denote the change in $Y$ as

$$\Delta Y = \left(Y_f - Y_i\right)$$
4. Denote the change in $X$ as

$$\Delta X = \left(X_f - X_i\right)$$
5. Then, the change in $Y$ with respect to $X$ is given by the formula

$$\frac{\Delta Y}{\Delta X} = \frac{Y_f - Y_i}{X_f - X_i}$$

And now, the final point - If we want to calculate the changes with infinitely high precision, we would have to choose the measuring devices that can do measurements down to almost zero! (which means we have infinitely high resolution)

This all sounds ridiculous, because nothing in the real world can measure anything with that kind of resolution. But, pure and abstract mathematics do not care about practical limitations, and its perfectly acceptable to have such systems in theory.

So, theoretically speaking, the tiny change in a quantity with respect to another can be calculated with infinite precision, and when that happens, we have a new name for that - We call it a derivative and the method of finding the derivative is called differentiation.

We also have a slight change in notation, and we write the tiny change in $Y$ with respect to the tiny change in $X$ as

$$\frac{dY}{dX} = \frac{Y_f - Y_i}{X_f - X_i}$$

with the crucial understanding that the differences between initial and final measurements are almost zero, but never equal to zero, otherwise we would have a division-by-zero problem.

And with this, we make an official entry to the world of calculus!

An Application of Calculus

Now that we have developed a new method, let us use it for something. Just because it is simple, we are going revisit the example of the water waves in a pond, ask the following question - What is the derivative of the area of the circle with respect to time? We know that we need to find the rate of change of area, except this time we work with infinite precision.

Let us again do it in steps -

  1. Since we are working with infinite precision, the time period over two measurements will be very small. Let us call that difference to be $\Delta t$.

  2. The area of the circle at time $t$ is given by
    $$A_t = \pi r^2$$

  3. Next measurement is taken at time $\left(t+\Delta t\right)$. Let us say that the radius has increased by a tiny amount $\Delta r$ to become $\left(r + \Delta r\right)$.

  4. So, the area after a tiny change in time is
    $$A_{t+\Delta t} = \pi \left( r + \Delta r \right)^2$$

  5. Now, the rate of change of area is given as
    $$\frac{\Delta A}{\Delta t} = \frac{A_{t+\Delta t} - A_t}{(t+\Delta t) - t} = \frac{\pi (r+\Delta r)^2 - \pi r^2}{\Delta t}$$

Now we simplify the last result as -

$$\begin{aligned} \frac{\Delta A}{\Delta t} &= \frac{\pi (r+\Delta r)^2 - \pi r^2}{\Delta t} \\ &= \frac{\pi r^2 + 2 \pi r \Delta r + (\Delta r)^2 - \pi r^2}{\Delta t} \\ &= \frac{ 2 \pi r \Delta r + (\Delta r)^2}{\Delta t}\end{aligned}$$

And, now we use the fact that we are dealing with infinite precision, so that the change in radius $\Delta r$ is almost zero. We know from elementary mathematics that square of a really small number is even smaller; which means, compared to $2 \pi r \Delta r$, the value of $(\Delta r)^2$ is so small that we can just ignore it.

Therefore, the final result is

$$\frac{\Delta A}{\Delta t} = \frac{ 2 \pi r \Delta r }{\Delta t}$$

We recall the change in notation when we talk about derivatives, so the result is written as

$$\frac{d A}{d t} = 2 \pi r \cdot \frac{d r} {d t}$$

The term $\displaystyle \frac{dr}{dt}$ is nothing but rate of change of radius, whose value was given to be 1 cm/sec. We can use this value in the previous expression to get

$$\frac{d A}{d t} = 2 \pi r$$

That's it! We have found the rate of change of the area of a circle for a travelling water wave. We can compare the results from three cases that we discussed -

  1. For a time interval of 1 second we got
    $$\frac{\Delta A}{\Delta t} = \pi (2r + 1)$$
  2. For a time interval of 1/2 seconds we got
    $$\frac{\Delta A}{\Delta t} = \pi (2r + 1/2)$$
  3. For a time interval of almost zero seconds we got
    $$\frac{\Delta A}{\Delta t} = \frac{d A}{d t} = \pi (2r + 0)$$

As promised, the last result which is based on the method of calculus, is going to be the most precise, since the difference in two subsequent measurements in negligible. The zero in the result is indicative of the fact that we have infinitely high precision.

In this article we essentially discussed just one mathematical application, i.e. Area of a circle. If the reader is interested to know more, there will be a series of simple articles in the future where we discuss many more examples.

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