Calculus is the mathematics of change and motion. Calculus is a branch of mathematics which provides methods for solving two large classes of problems. The first of these is differential calculus. It involves finding the rate at which a variable quantity is changing. In our case, we will start with a person on a bicycle at a starting point. We will have timers set up at specified distances. By measuring the change in distance with respect to the change in time, we can determine how fast the bicycle is traveling at any instant.
The second branch of calculus is integral calculus. This branch deals with finding a function when its rate of change is known. In our case, we will start with the graphs for velocity versus time. We will take the area under the curve at specified time intervals. This area should be the distance the rider has traveled.
In this section, I hope to give a person a feel for what the derivative and integral are. I will do this by first plotting points of Kasey pedaling. I will then connect the points. This graph should be a curve. However, as noted in the previous section, the submitted graphs will be done with straight lines. This is not all that bad because in Calculus, you assume that as the interval becomes smaller and smaller that the graph approximates the straight line segment in each interval. I will then draw tangents to the distance versus time graph. I will show how the slope of the tangents is the velocity at the respective points. I will then take the velocity versus time graph and find the area under the curve. I will show how this area is the distance that the rider has traveled. I will not present algorithms to find the derivative or integral.
The section starts with coordinates. The basic idea of analytic geometry is the establishment of a one-to-one correspondence between a pair of numbers and the points on a plane. Some students have difficulty with which coordinate is presented first. I tell them that it is alphabetical order; the x-coordinate is presented first.
Next, we will discuss the slope of a straight line, where slope is given by the change in y-direction divided by the change in the x-direction. In finding the slope of graphs in a mathematics book, I tell my students to start at a point which is on the grid of x and y values. Let’s stay away from fractions. I then tell them to walk forward. It is important to move in the positive direction since this is then one less negative number to worry about. When you hit the next intersection, look up or down the y-axis and see if the graph goes through an intersection. If it does, count how many blocks away the graph is. This is why it is important to move positively along the x-axis, if you move up to meet the graph, you have a positive slope; if you move down, you have a negative slope.
Unfortunately, my graph of the tangents to the distance versus velocity didn’t work out as neatly as a textbook example. Then I noticed something. If I continued the tangent so that it crossed the x-axis, I only had one unknown to estimate. I drew the tangent at a known point; that’s one coordinate pair known. When I cross the x-axis, the y-value is known; it’s zero. That only leaves one x-value to find. This technique worked fine. We will then proceed to the equation for a straight line. We will start with the slope intercept method: y = mx + b, where m is the slope and b is the y-intercept. From there we will go to a more general equation Ax + By + C = O.
In an advanced class, I would then move to functions. A function is a set of ordered pairs of numbers (x,y) such that to each value of the first variable (x) there corresponds a unique value of the second variable (y). We will then graph the functions.
We now go from the slope of a straight line to the slope of a curve. We can show how if we make the intervals small enough, a curve can be approximated by a set of straight lines. This leads to finding the derivative of a function. By the derivative we mean the slope of the tangent to the curve at a particular point.
I drew five tangent lines to the distance versus time graph. Looking at the graph, the slopes from left to right are 10.7, 17.1, 21.0, 30.5 and 32.1. This matches up to computed velocities of 8.5, 14.0, 22.7, 29.4 and 30.5. So, we’re in the ball park. From these results I was able to show how the slope of the line increases as the velocity increases. Also, how when a state of constant velocity is reached, the lines are parallel.
Although it is not in the scope of this paper, I did a regression analysis on my observed points. I used the Data Regression Menu found in Lotus 1-2-3. This command is straight forward and simple. You do not have to have any knowledge of degrees of freedom or standard error of coefficients. You just plug in the numbers. Using a cubic regression, this method gave me velocities of 9.2, 16.0, 20.2, 28.4, 32.5 and 30.7. I liked this curve fitting method better, but I didn’t want to confuse my students. After all, I had been breaking it on them to make sure that their numbers were exact.
To show integration, I took the graph of instantaneous velocity versus time and drew lines parallel to the y-axis through the given points. I then used the trapezoid rule for approximation of the area under a curve. The formula for area of a trapezoid is:
Area = 1/2 x (basel + base2) x height
Our results were extremely accurate once the constant velocity state was reached. For our other values, we should have had smaller intervals. However, they were close enough for a Friday in May demonstration.
We could have gone on with the derivative of the velocity to find acceleration. We didn’t because, as noted, I wanted to have more measurements. I still view this section as a success because I was able to graphically show the derivative and the integral.