Calculus is the mathematical study of change. In precise terms, it attempts to model systems of instantaneous change. Geometry, on the other hand, is concerned with the properties of structures such as points, lines, solids, etc. The former and latter branches of mathematics thus appear to be seemingly unconnected. But is this impression correct?
The answer is a keen no and this is solidly supported by examples of two calculus problems I had solved for a student; I have shared their images from my virtual teaching board.
In both cases, the objective was to find the volume of two three-dimensional (3D) structures – a cylinder and a pyramid. The attempt to find such volume through calculus requires integration – or essentially, the summing of “tiny” volumes of infinitesimal areas “A(x)” multiplied by the corresponding infinitesimal thickness of dx. This process, however, is preceded by a geometrical assessment of the two-dimensional (2D) “base” shape, which is the foundation for the constructed, 3D cylindrical structure. This assessment of whether the base is a square, circle, triangle, etc, determines what the appropriate, formulated area or A(x) would be for the problem at hand.
In the case of a volume of the pyramid (in part A), the base shape was a square, which would have base area A equal to the product of its side, “s”, with itself. However, since we needed this area, A(y) to be simplified in terms of only one variable – the independent variable y, we had to resort to a geometry principle of similar triangles to link the variable of side “s” with the variable of height “y” so as to have the area be completely in terms of “y” instead of “s”. Therefore, two similar triangles were accordingly identified by setting up the figure (as shown in the image) for this problem. The “big” triangle had constant height h, and base l, whereas the “small” triangles had variable sides and heights of s and y accordingly. The ratio of these proportional “big” and “small” corresponding sides was set up as s/l = y/h. Subsequently, the variable of side “s” was found in terms of the variable of height y (multiplied by the constant l/h ), as we preferred. This expression was then incorporated in the base area formula and subsequent integral to find volume. As such, this problem showed how calculus and geometry had a critical interplay. This connection needed to be identified and applied in order to attempt the problem correctly at several steps of its solving.
In the case of a volume of a cylinder (in part B), the base shape was a circle. Accordingly, one had to assign variables of radius “r” and diameter “d” and assess the area of the circle to be pi* r^2 which was incorporated as the base area formula and subsequent integral to find volume. This problem again demonstrated the need to recall the geometrical concept of the area of the particular mathematical structure of a square.
In conclusion, I seek to reiterate the idea that, to solve problems in higher-level mathematics such as calculus, a fundamental knowledge of preceding math areas such as geometry and algebra is often quite fundamental and needs to be recalled. There is often an interplay between different areas of mathematics, and, as such, this awareness is key for the correct application of the problem technique required by the student. I emphasized this understanding to my student; additionally, I have hereby discussed its importance by sharing two case demonstrations of this nature of problem-solving.