Illustrate the principle that surface area to volume ratios decrease with increasing size by calculating surface areas and volumes of simple 3-D shapes (as shown in the Mathematical requirements)
Illustrate the principle that surface area to volume ratios decrease with increasing size by calculating surface areas and volumes of simple 3-D shapes (as shown in the Mathematical requirements)
Answered step-by-step
To illustrate the principle that the surface area-to-volume (SA
) ratio decreases as size increases, we’ll calculate the surface area and volume for simple 3-D shapes (cubes and spheres) of varying sizes and compare the SA
ratios.
1. Cube
For a cube with side length ss:
\text{Surface Area (SA)} = 6s^2 \text{Volume (V)} = s^3 \text{SA ratio} = \frac{\text{SA}}{\text{V}} = \frac{6s^2}{s^3} = \frac{6}{s}Example Calculations
Let’s calculate the SA
ratios for cubes with side lengths of s=1s = 1 cm, s=2s = 2 cm, and s=3s = 3 cm.
- For s=1s = 1 cm: \text{SA} = 6 \times (1)^2 = 6 , \text{cm}^2 \text{V} = (1)^3 = 1 , \text{cm}^3 \text{SA ratio} = \frac{6}{1} = 6
- For s=2s = 2 cm: \text{SA} = 6 \times (2)^2 = 24 , \text{cm}^2 \text{V} = (2)^3 = 8 , \text{cm}^3 \text{SA ratio} = \frac{24}{8} = 3
- For s=3s = 3 cm: \text{SA} = 6 \times (3)^2 = 54 , \text{cm}^2 \text{V} = (3)^3 = 27 , \text{cm}^3 \text{SA ratio} = \frac{54}{27} = 2
2. Sphere
For a sphere with radius rr:
\text{Surface Area (SA)} = 4 \pi r^2 \text{Volume (V)} = \frac{4}{3} \pi r^3 \text{SA ratio} = \frac{\text{SA}}{\text{V}} = \frac{4 \pi r^2}{\frac{4}{3} \pi r^3} = \frac{3}{r}Example Calculations
Let’s calculate the SA ratios for spheres with radii of r=1r = 1 cm, r=2r = 2 cm, and r=3r = 3 cm.
- For r=1r = 1 cm: \text{SA} = 4 \pi (1)^2 = 4 \pi \approx 12.57 , \text{cm}^2 \text{V} = \frac{4}{3} \pi (1)^3 = \frac{4}{3} \pi \approx 4.19 , \text{cm}^3 \text{SA ratio} = \frac{4 \pi}{\frac{4}{3} \pi} = \frac{3}{1} = 3
- For r=2r = 2 cm: \text{SA} = 4 \pi (2)^2 = 16 \pi \approx 50.27 , \text{cm}^2 \text{V} = \frac{4}{3} \pi (2)^3 = \frac{32}{3} \pi \approx 33.51 , \text{cm}^3 \text{SA ratio} = \frac{16 \pi}{\frac{32}{3} \pi} = \frac{3}{2} = 1.5
- For r=3r = 3 cm: \text{SA} = 4 \pi (3)^2 = 36 \pi \approx 113.10 , \text{cm}^2 \text{V} = \frac{4}{3} \pi (3)^3 = 36 \pi \approx 113.10 , \text{cm}^3 \text{SA ratio} = \frac{36 \pi}{36 \pi} = 1
Analysis of Results
As the size of both cubes and spheres increases, the surface area-to-volume ratio decreases. This decrease shows that larger objects have relatively less surface area for their volume compared to smaller ones, impacting processes such as heat loss, material exchange, and efficiency in biological cells.
Sourav
BNO Team
