Note: GERMAN versions of the Week 19 tasks are available HERE.
In this week’s tasks we focus on a family of larger and larger squares, where the multiplicative relation between the length of side of a square and that of its immediate larger neighbour is the same for all squares. The ‘super-multiplicative’ growth in the sides of the squares as we go from one square to the next, to the next, to the next.... is called exponential growth. Some of the ideas here are quite demanding, but even if they are not thoroughly understood, students can benefit from meeting this interesting (and currently, vital) mathematical concept.
Monday: Here we meet our family of exponentially growing squares. As we go from one square to the next square on the right, it soon becomes clear that the height (for example) increases by a greater amount each time. As this amount is not constant, the relation between a square’s height (h) and its family position (p) does not have a multiplicative component (m) as is found in a linear relation (a relation of the form h = mp + c). However there is a multiplicative relation between the heights of neighbouring squares. The height of A₁ is 1.2 times the height of A. The same applies to the heights of A₁ and A₂, of A₂ and A₃, etc, and so the height of A₄ is 10cm × 1.2 × 1.2 × 1.2 × 1.2 or 10cm × 1.2⁴ = 20.736cm. This is indeed about twice that of A.
Note: Thursday’s task does involve a family of shapes where the heights increase by a constant amount, so it might help students to show it briefly here.
Some students might be happy to accept that the ×1.2 multiplicative relationship applies to all pairs of adjacent squares. However, to explain why this must be the case is not that easy. One approach is point out that the triangles above the squares are similar. An argument can be made that this means that 'sideways T-shapes', such as the red and green ones below, must be similar. Then, since the vertical limb of the red T-shape is 1.2 times the length of the horizontal limb, and equal to the horizontal limb of the next (green) T-shape, then each horizontal limb (and each vertical limb) is 1.2 times the length of its left-hand neighbour.
Tuesday: In this task we apply the idea that the ×1.2 relation holds between the sides of all pairs of adjacent squares. Using a calculator or spreadsheet, say, students can discover that 10×1.2¹² = 89.1610, so the sides of square A₁₂ are very close to being 9 times those of square A.
Wednesday: Dan would be right if the relation between square-number and side-length were linear. However, because the relation is exponential, the sides of square A₆ will √9 = 3 times those of square A rather than 9/2 = 4.5 times. Expressed formally, we can write these relations: (1.2⁶)² = 1.2¹² ≃ 9 = 3².
An empirical check of Dan and Ellie's conjectures shows that 1.2⁶ = 2.985984 ≃ 3.
Thursday: Here we provide a contrast to the week's previous tasks by having a family of shapes where the relationship between height and shape number is linear, rather than exponential: as can be seen, the height of adjacent rectangles increases steadily. Indeed, the height increases by 2 cm each time. Note that this time our shapes (rectangles) are not similar.
The relation between shape height (h cm) and shape number (p) can be written like this: h = 10 + 2p.
In the earlier tasks, we found that shape A₄ was roughly twice as tall as shape A. Here it happens slightly 'later': R₅ is twice as tall as R.
And previously we found that shape A₁₂ was roughly 9 times as tall as shape A. Here it happens much, much later: R₄₀ is twice as tall as R.
Friday: This task is challenging. It turns out that the multiplicative relation between the length of the 10mm red line and the middle red line is the same as between the middle red line and the 50mm red line.
So, if we call the multiplier m, then 10×m×m = 10m² = 50,
and so m = √5 ≃ 2.2
and so the middle red line is about 22mm long.
Students might find it hard to derive a method for finding the length of the middle red line. But nonetheless, some might hit upon the idea of multiplying 10 by the square root of 5....
We can show that 10m2 = 50, as follows.
These two triangles are similar.→
If the scale factor is m, then the longer
green line is m times the shorter one.
These two triangles are similar.→
As the longer green line is m times the
shorter green line, the scale factor is m.
So the longer red line is m times 10m.
