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Articles

The impact of procedural and conceptual teaching on students' mathematical performance over time

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Pages 404-426 | Received 22 Apr 2019, Published online: 17 Nov 2019
 

ABSTRACT

The goal of the present study is to investigate the impact of procedural and conceptual teaching on students' mathematical performance over time. For this purpose, implicit differentiation and integration by parts were taught in two different ways to first-year university students of two different calculus courses: one course with Conceptual Teaching (CT) and the other course with Procedural Teaching (PT). In the CT class, the lecturer taught mathematical concepts conceptually and helped students to understand the reasons behind techniques and formulas. In the PT class, the lecturer helped students to solve problems without necessarily knowing the reasons for each step of solving problems. Both groups took a same test at the end of these lessons. The results of the test showed that there is no significant difference between the performance of the two groups. After three months, another test including the same topics had been taken by the two groups. The results of this test showed that the CT group had a better performance compared to the PT group. The findings suggest that conceptual understanding could create more sustainability in learning mathematics.

2010 MATHEMATICS SUBJECT CLASSIFICATION:

Disclosure statement

No potential conflict of interest was reported by the authors.

Notes

1 Symbolic techniques are algebraic procedures, processes, solutions, and methods that lead to solving a mathematical problem (Badillo et al., Citation2011).

2 The lecturer used Maple software to draw implicit equations, to solve y in terms of x, and to draw graphs. Students were not required to learn the Maple commands. Using Maple, the lecturer could explain in more details the reason of using formulas and procedures in implicit differentiation, and also could highlight graphical interpretations of this concept.

3 For example, to find y in the implicit equation y2+xy=2, we can replace y with f(x). So the equation will changed to f(x)2+x.f(x)=2. By differentiaing both sides of the equation with respect to x, we have 2.f(x).f(x)+f(x)+x.f(x)=0. We can easily find f(x) equals to f(x)2f(x)+x. Finally, we can substitute the symbol y for the symbol f(x), and y for f(x) to get y=y2y+x.

4 After solving an integral, the lecture differentiated the answer to check his solution as f(x)dx=F(x)F(x)=f(x). This is a good monitoring strategy to check whether the integral of a function is calculated correctly.

Additional information

Funding

Work done in the framework of the research project PGC2018-098603-B-100 (MCIU/AEI/FEDER, UE).

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