Essence of Calculus

A Little Calculus To Get Started

     Image Credit: 3Blue1Brown

An Overview 

Science and Technology have taken the world by storm and it’s a matter of surprise to none about how much human civilization has developed over centuries. Although there is no denying that the first traces of science considered to have originated in the Greek of today, Science has vastly spread all over the world with many contributing a serious part about how much of the world we know today. One man who has to be accredited for non-arguably one of the greatest pieces of work ‘Philosophiæ Naturalis Principia Mathematica’ is Sir Isaac Newton. Newton is one of the greatest thinkers mankind has ever seen and the book he wrote would be so ahead of his time that the foundation of Classical Mechanics he laid is acknowledged and widely studied even to date. There’s no doubt that the approaches and methods mentioned in the very book are of hundred percent accuracy but as readers of now claim to realize that his Laws of Physics were over the base of Calculus, it is very difficult to really find traces of Calculus and its notation in the book. Rightfully, the man who published the first paper on Calculus and its notation has to be Gottfried Wilhelm Leibniz who independently discovered Calculus and labeled it to be “A new method for maxima and minima, and for tangents, that is not hindered by fractional or irrational quantities, and a singular kind of calculus for the above mentioned”.

Leibniz used the first recorded notation for calculus using a dy and dx to represent an infinitesimal change in y and x respectively. As the supporters of Newton claim Calculus to be an unpublished work of Newton and Leibniz plagiarising it, the first notation he suggested for Calculus came nearly two decades after what the former did in 1675. Both of them did it independently, so there’s no need for another ‘Newton vs Leibniz notation war’ but instead praise for both of the maestros for their work which would turn out to be the base for the engineering of today.

On the common part of things, both of the mathematicians saw three major things in calculus: derivatives, integrals, and an inverse relationship between those. 

Differentiation is simply a study of how an infinitesimal changes. The study of changes in one variable with respect to another shows up a lot in Physics as well. For example, to determine the velocity of an object, we look at how its displacement changes with time. Now, if the object displaces by equal amounts each second, its velocity remains constant throughout its motion - we could work out this without calculus. But when the rate of change of displacement is not constant, it becomes challenging to find its velocity at a particular instant without the use of calculus. Below, I have graphed three of the many possible ways the displacement of a body may vary with time. In all three of these graphs, the y-axis is displacement and the x-axis is time.

In this graph, if you divide the time axis into four different intervals, the corresponding change in displacement for each interval is constant. Although the displacement increases with time, the rate of its increase in constant. This means the velocity is constant because we define velocity as the change in displacement divided by the change in time, or simply, the rate of change of displacement. We need no knowledge of derivatives to compute the velocity in this case.

Here too, we have divided the time axis into four equal intervals. But contrary to the first case, the change in displacement with respect to the change in time is not equal at each interval. Instead, the change in displacement increases with time at every instant. This implies that the velocity is increasing. But to compute the instantaneous change in velocity accurately is difficult without the knowledge of calculus. We can never compute the velocity at an instant because to measure change, we need the information regarding at least two instances. However, if we measure the change in displacement at an infinitesimally small interval of time, which we can achieve using limits, we can accurately determine the velocity of the object at each instant. This has been made possible due to calculus, more specifically derivatives.

In this case too, the displacement does not vary uniformly with time. But unlike in the previous case where the displacement was increasing at an increasing rate, here the displacement also increases but at a decreasing rate.    

Now with what we have discussed above, let’s try formulating an expression to find the rate of change of a function at each instant. Keep in mind, what we did for the displacement-time graph also applies for any function. So let’s assume a function f(x) and try finding its derivative. We first sample the f(x) value at two instances say x and x+h; h is the difference between the two x values. The function then takes the values f(x) and f(x+h) at these two instances. Now what we do is find the change in these two function values, i.e. f(x+h) - f(x) and divide it with the change in the x-values, i.e. h. But what we also want to achieve is an infinitesimally small value of h. In mathematical terms, we express it using limits as:

 Now we can ready to write down our expression for the derivative of a function at a point. We denote that by using the notation dy/dx. 

                                 

Mathematicians love to call this ‘The First Principle’. All the formula we use to compute the derivative of a function, e.g. the power rule that says (dxn /dx)= nxn-1, are derived through this very principle.  

Similarly, in the simplest terms, integrals deal with finding exact areas under a curve. So let’s start with a curve y = 3sinx, and try finding its area from 0 to π.   

This is the graph of y=3sinx for x ranging from 0 to . Had the graph been a straight line or a combination of straight lines, we could have easily computed the area below it using our knowledge of areas of triangles and quadrilaterals. However, how can we go about finding the area under this sin graph and doing so at a high degree of accuracy?

 

                                                                      Why not start by trying to represent the area under it using shapes whose area we can easily compute. Here I have drawn 10 rectangles that, to a certain degree, approximate the area under the sin curve. However, we can clearly see this is nowhere near exact. What can we do to fit in the rectangles in an even better way? As you might have thought, we can increase the number of rectangles. Let’s increase it to 50, and then to a whooping 1000.

You can easily guess which one has 50 rectangles and which one has 1000. The 1000 one already seems near perfect. But the challenge lies in finding the area of each of those 1000 rectangles.

Thus, we need some more efficient method to not just compute the area of 1000 rectangles but an infinite number of those. As you might have well noticed, increasing the number of rectangles in a fixed bound means decreasing the width of each rectangle. And when we talk about drawing an infinite number of rectangles, it means that we want the width to be approaching 0. When we start talking about approaching infinity and zeroes, limits come into play. 

Let us call the width of each of the infinite rectangles dx. The length of the rectangle is the function value it takes at a certain x value, i.e. f(x) = y. Now what’s the area of each rectangle then? Simple, it’s f(x)dx. Now what we are left to do is sum up all the individual areas. One option would be to use the sigma notation ∑ for summation. But the downside is that it is usually indexed by natural numbers and hence a discrete sum. So to deal with continuous sums, the fathers of calculus came up with a new way of summation, using a symbol that looks like a long S, ∫.  

Now, similar to how we use the sigma notation, we write the integral symbol first and then the area of each rectangle so that the end result is the sum of infinitely many rectangles. We should also let know of the bounds we are interested in. So the way we would represent the required area is 

Although I have explained this through areas, do not think as if this is all that integrals mean. There’s much more to this, volumes of revolution, arc lengths, line integrals and all sorts of things. Now onto a simple real-life problem that can be solved using basic knowledge of differentiation.

Suppose you are on your farm during the lockdown and you need to help your father in building separate shelters for your cattle. Your father brings 480 metres of fencing material to build 8 separate shelters of equal size for the cattle. Suggest how the length and breadth should be determined so that the whole farm covers the maximum area?

The total fencing material available is 480m which when written in terms of x and y is 12x + 10y. 

               480 = 12x + 10y y = 48 - 1.2x

The total area of the shelter, denoted by A, is 8xy. 

                               A = 8xy 

Substituting y in terms of x from the first equation, the Area function can be written only in terms of x as

                    A = 8x(48-1.2x) = 384x - 9.6x2

Differentiating A with respect to x gives, 

dA/dx  = 384 - 19.2x

The maximum or minimum value of the area occurs at the stationary points of function A. Stationary point is the point on the curve where the slope of the tangent is 0.

So equating dA/dxwith 0, we get

384 - 19.2x = 0 x = 20 m

However, we aren’t done yet. We now need to see if this combination of x and y values gives the maximum or the minimum area. You may try with other x and y combinations to see if the area using that combination is greater and smaller than 3840. But let us prove that it’s the maximum area by differentiating the first derivative once again. That’s called the second derivative of the function A and is denoted by d2A/dx2, which in our case is -19.2. Whenever the second derivative at a point is negative, it means the curve is concave downwards, meaning that the function at that point takes a (local) maximum value. Thus, the maximum area we can have with the given constraints is 3840 m2.

-By Deepak Woli