Course Notes for Calculus I

So far, in Section 3.1 and elsewhere in the course, I’ve relied heavily on biological examples. In this section, I’m going to instead rely on physics. Coming from physics, the velocity problem is one of the basic motivating problems of calculus. Assume there is an object moving in one dimension. I can describe its position as a function \(p(t)\) where \(p\) is position in terms of time \(t\text{.}\) I want to know its velocity.

If \(p(t) = at + b\) is a linear function, then algebra can answer this question. With this linear function, for each unit of time the object travels \(a\) units of distance. The value \(a\) is the velocity. Geometrically, the velocity is measured by the slope of the straight line graph of \(p(t)\text{.}\) The slope is calculated by the ratio of the change in \(p\text{,}\) \(\Delta p\text{,}\) to the change in \(t\text{,}\) \(\Delta t\text{.}\)

If slope is the way to measure velocity, then I need a notion of slope for non-linear functions as well. The notion comes from the idea of tangent lines. A tangent line to a graph is a line which touches the graph at one point without crossing it (as opposed to a secant line, which crosses the graph twice).

The slope of a graph is defined to be the slope of its tangent line, should such a line exist. The velocity problem is reduced to the problem of finding the tangent line (or, more particularly, its slope). How do I calculate a tangent line? Algebra has trouble with this. The best that algebra can do is find secant lines which approximate a tangent line. I can adjust the approximation by letting the two points of the secant line come closer and closer together, as in the Figure 6.1.4. In this way, I can build an approximation process which improves as I move the points. However, algebra an never finish the process — it can just supply improved approximations.

Now I will consider the opposite problem. Consider a function \(v(t)\) which tells me the velocity of an object at any point of time. How can I determine the distance the object has covered over a period of time?

Again, algebra can only answer this question for very simple situations. Assume that the velocity \(v(t) = c\) is constant. In this case, if the constant velocity is \(c\) units of distance for each unit of time and the object has travelled for \(t_0\) units of time, the distance is just the product \(ct_0\text{.}\) Graphically, the situation is summarized in Figure 6.1.5. The distance travelled under constant velocity is the area under the velocity graph, as in Figure 6.1.6. I can extend this idea to non-constant velocity: the distance should be the area under the velocity graph. Therefore, to solve the distance travelled problem, I need to find areas under curves. Like tangent lines, this is not a problem that algebra can easily tackle. However, it can approximate areas. Algebra is good at areas of rectangles, so I can use rectanges to approximate areas.

The sum of the areas of all the rectangles is a reasonable approximation to the area under the curve. If I want a better approximation, I can divide the area into smaller rectangles. In this way, I set up an approximation process to understand areas under curves. Algebra can never completely answer the question, but it can get better and better approximations by using more and more rectangles in its approximation.