Overview of Single-Variable Calculus

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Summary

Single-variable calculus is the study of how you go from \(f(x)\) to \(f'(x)\), and vice versa. You already know how to do this if \(f'(x)\) is a constant: just use algebra. But if \(f'(x)\) is changing, then you have to do calculation procedures which we call differential calculus and integral calculus to go from \(f(x)\) to \(f'(x)\) and from \(f'(x)\) to \(f(x)\), respectively. Single-variable calculus, which is basically a cook book or calculation procedure, is needed whenever \(f'(x)\) is changing. Professor Strang (video below) explains, through many examples, how \(x\), \(f(x)\), and \(f'(x)\) are very general and can be 'many things': they can represent time, the height of something as a function of time (i.e. height of a person when \(t\) years old), and the growth rate of that something as a function of time (i.e. how fast someone grows when they hit a growth spurt from 12-14 years old); time, and distance and speed as functions of time; time, and speed and acceleration as functions of time; and all the other problems in your calculus textbook. The main point is that going from one function to another when \(f'(x)\) is changing (which, without  differential and integral calculus, would be impossible) involves so many important problems which would be unsolvable without calculus.


Big picture: two halves of single-variable calculus

                                This video was produced by MIT OpenCourseWare.

 
                                     Figure 1

                                    Figure 1

We know from everyday experience that if I say dropped a ball from the top of the tower of Pizza, as it descended towards the Earth's surface it would keep picking up speed as time flowed forward. It would be very difficult to determine how far the ball moved away from us and how fast it is going after \(t\) seconds with no knowledge of physics or without doing a tedious experiment. For now, I'll just tell you that the function \(f(t)\)—how far the ball moved away after \(t\) seconds—was experimentally determined by Galileo to be a parabola. Why the empirically correct function is a parabola was first answered by Newton. If you are unfamiliar with physics, that is ok. For now the answer to, why a parabola?, will be a mystery which we'll address in a few separate lessons, each with a different perspective. The function \(f(t)\) is represented by the red line (a parabola, of course) in Figure 1. The tower of Pizza is 55.86 meters tall and we start our stop watch right when, say, Galileo lets go of the ball—thus, at \(t=0\), accounting for Galileo's height, the ball's initial height of \(f(0)\) will be, roughly, \(f(0)=60m\) as shown in the graph in Figure 1. 

                                      Figure 2

                                     Figure 2

Notice that from \(t_0=0\) to \(t_1\) in Figure 1, the change in the ball's height from \(f(0)\) to \(f(t_1\) isn't much—thus, the slope \(\frac{f(t_1)-f(0)}{t_1-0}\) is small. But notice that from \(t_2\) to \(t_3\), the change in height as the ball falls from \(f(t_2)\) to \(f(t_3)\) is quite big—thus, the slope \(\frac{f(t_3)-f(t_2)}{t_3-t_2}\) is pretty big in this region. The fact that the ball speeds up as it falls towards the ground is captured by the fact that the slope of \(f(t)\) at each point (which is also the speed at each point) is increasing with increasing \(t\). The speed of the ball, \(f'(t)\), is given by the straight line in Figure 2. But how do I know this? The process you have to go through to find the answer to this question is a big area of study that comprises half of the entire subject of single-variable calculus: the process one goes through to answer that question is called differential calculus. Given any function \(f(x)\), we can go through this process of differential calculus to find \(f'(x)\). For example, once I tell you that \(f(t)\) is given by the parabola in Figure 1, I can use this thing called differential calculus to determine what \(f'(t)\) is and that \(f'(t)\) is the straight line in Figure 2. I've answered that first question, but this begets another question: how does differential calculus work? How do we do differential calculus? That is something which will be answered in the next several lessons. Suppose that I ask you the opposite question: if I give you \(f'(t)\), how do you determine \(f(t)\)? What process does one have to go through to do that. This process is a series of calculations called integral calculus and will comprise the other half of our studies of single-variable calculus.

These calculation procedures can be quite complicated, but they can be summarized in a nutshell: differential calculus is entirely about doing calculations to find the slope, or steepness, of the function \(f(t)\) at each point; integral calculus is entirely about doing calculations to find the area underneath \(f(t)\)—those calculations are how you go from \(f(t)\) to \(f'(t)\), and the other way around. But notice that they are the opposites of each other: calculating the slope takes you from \(f(t)\) to \(f'(t)\) whereas calculating the area takes you from \(f'(t)\) to \(f(t)\). These two operations of going from "A to B and from B back to A" is called the fundamental theorem of integral calculus. This is all of single-variable calculus in a nutshell: but the next several lessons will be all about unraveling this nutshell.

                                              Figure 3

                                             Figure 3


Example where \(\textbf{f'(x)}\) is constant

Suppose that we're dealing with a simpler situation in which the ball is initially right next to me when I start my stopwatch at \(t=0\). In this case, the initial position is \(x_0=0\). Suppose that the ball moves away from me along the x-axis in a straight line at a constant speed \(s\). The distance that the ball has moved away is given by

$$d(t)=st.$$

                                    Figure 4

                                   Figure 4

This is easy to verify through example. If something is moving away from you at a constant speed of \(5m/s\), then after one second, it will have moved away by a distance of \(d(1s)=(5m/s)(1s)=5m\); after another second (two seconds has gone by in total), the ball will have moved a distance of \(d(2s)=(5m/s)(2s)=10m\); and so on. By plugging in many different values of \(t\) into the function \(d(t)\), we can obtain the graph of \(d(t)\) vs. \(t\) shown in Figure 4. 

                                                                                                                        Figure 5 

                                                                                                                       Figure 5 

How do we go from \(d(t)\) to \(d'(t)\)? We can do this by using algebra. The function \(d'(t)\) gives the slope of the function \(d(t)\) at each t-value and at each point along \(d(t)\). Suppose, for example, that we wanted to find the slope at \(t=1s\); the function \(d'(1s)\) is the slope of \(d(1s)\)—which is to say the slope of \(d(t)\) at the coordinate point \((1s,5m)\). We know from algebra that the slope at this point is given by \(m(1s)=f'(1s)=\frac{Δd}{Δt}=\frac{d-5m}{t-1s}\). It is easy to verify by plugging in many different values of \(d\) and \(t\) that you'll always get the same slope \(f'\) at the point \((1s,5m)\). Since the steepness of the function \(d(t)\) is the same everywhere, the slope \(f'(t)\) is the same at every value of \(t\) and every point along the curve. But I'll randomly just choose \(d=10m\) and \(t=2s\). In this case, we'll be calculating the slope \(f'(1s)=\frac{d-5m}{t-1s}=\frac{10m-5m}{2s-1s}=5m/s\). If we calculated \(f'(t)\) at every value of \(t\) (not just \(t=1s\)), we would keep getting \(5m/s\). Thus, the graph of \(f'(t)\) is given by the straight line in Figure 5. So you can see that determining \(f'(t)\) from \(f(t)\) was just a matter of calculating the slope—a concept from algebra. When \(f'(x)\) is a constant, going from \(f(x)\) to \(f'(x)\) is that easy. More generally, if the ball is moving away from you in a straight line along the \(x\)-axis at any constant speed \(s\), then from the function \(f(t)=st\), you can pull out the concept of slope from your algebra tool kit to determine that the function \(f'(t)\) is \(f'(t)=\frac{f(t)}{t}=s=m\).

How do you go from \(f'(t)\) to \(f(t)\)? To try to do this, let's multiply both sides of the equation \(f'(t)=s\) by \(t\):

$$tf'(t)=st.$$

The function \(f'(t)\) gives the height of the red line above the t-axis as illustrated in Figure 5. But notice that the time \(t\) gives the width. The product \(f'(t)t\) thus gives the area (shaded green in Figure 1) of the rectangle in Figure 5. But we also know that the product \(f'(t)t\) is equal to the distance \(f(t)\) the object traveled: \(f(t)=f'(t)t=A\). Therefore, to go from \(f'(t)\) to \(f(t)\), we just had to find the area underneath \(f'(t)\). And to recap, to go from \(f(t)\) to \(f'(t)\), we had to find the slope of \(f(t)\). Very early on, we're already starting to get the sense that slope and area are somehow opposites of each other: one can take you from A to B, whereas the other can take you from B to A, and vice verse. Going back and forth between A and B is just a matter of finding the slope and area which we did, pretty easily, just using algebra and basic math. That's all there is to it when \(f'(x)\) is a constant. But when \(f'(x)\) is changing, we'll have to use calculus to find the slope and area—only then will we be able to go back and forth between A and B. Easier said than done as we shall see in subsequent lessons.


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