Developing student understanding: Contextualizing calculus concepts

School Science and Mathematics, Feb 2000 by Schwalbach, Eileen M, Dosemagen, Debra M

Results

Explanation

Wiggins & McTighe (1998) defined explanation

as the ability to "clearly, thoroughly and instructively explain how things work, what they imply, where they connect, and why they happen" (p. 46). In written responses to a series of prompts related to the concept of derivatives, students were asked to complete the statement, "You don't understand derivatives unless you understand the importance of .11 Their responses indicated a range of understanding from the more sophisticated, ("the concept of limit and the definition, algebraic and graphic, related to it") to the more simplistic, ("the different rules like power, product, chain, etc."). When asked to identify common mistakes related to derivative, student responses ranged from the more specific, ("not recognizing the product or chain rule in a problem") to the more general, ("incorrect algebra"). In response to a prompt asking them to focus on comparisons, one student identified the important distinction between continuity and differentiability, and another distinguished between the quotient rule and l'Hopital's rule. Finally, students were asked to identify connections and were able to describe the relationships between derivative and slope of a curve, limits and derivatives, graphic and algebraic forms of a function, and "derivatives and motion and change in physics." These examples illustrate that students were able to explain to a greater or lesser degree the importance of, differences between, errors in, and connections among concepts in calculus and, to some extent, physics.

One indicator of students' understanding relative to explanation is the ability to describe a problemsolving process and the rationale for the approach employed. For example, calculus students worked in groups of four to five to solve a problem with a physics context related to velocity and maximum height. Prior to a videotaped class presentation of their results, students were instructed to target their explanation at an audience who did not necessarily understand calculus. The students then collaborated on the solution and selected one or two group members to present the group's results. While all groups were able to offer feasible solutions to the problems, not all presenters were able to provide rich descriptions of the groups' process. A sample of one student's explanation follows. Throughout the explanation, the student defined terms (in italics) and explicitly connected concepts from calculus and 1)physics (in boldface).

The problem we were given is that the initial velocity is 256 feet per second, and that the equation for the position, which is s(t), where t is the time, is equal to 256t-16t2. In the problem, we have to find the maximum height by which the object is thrown. The initial velocity was 256 feet per second. That's the position graph there. [see Figure 1.1 In order to find the maximum height, we have to find when the velocity is equal to zero or where the object stopped moving, which would be at the top. So we have to find the equation for the velocity. The velocity is the rate at which the position changes. And in order to find that, we have to calculate the derivative of the position. The derivative of the position is the slope of the position equation, which is the rate at which the graph of the position changes. So the derivative of the position is equal to the velocity equation. So by finding the derivative of the position, we found the velocity equation equals 256-32t. In order to find when the velocity was zero, we set that equation equal to zero and found that at the time of eight seconds, the object was not moving. Using that time that we found, we put that into the position equation. So we found s(8), which was equal to 1,024 feet, which was the maximum height.