Sunday, 17 June 2018

ALG 19

This week we're looking at graphs that model 'real life' situations. The focus is on qualitative aspects of the graphs, though we do take a more analytic view at times. Pioneering work on school students' understanding of graphs was undertaken in the late 1970s and early 1980s by Daphne Kerslake, by Claude Janvier and by Malcolm Swan. I doubt whether one can better some of their outstanding tasks.
-
MONDAY: Mathematically, this task involves a nice context, though one that is perhaps not so familiar to today's school students - you might therefore have to tease out the crucial fact that audio tape has to travel at a constant speed (as it passes over the tape-head which picks up the sound). Most students will realise that as the left hand spool unwinds, the width (diameter) of the reel of tape decreases. This means the circumference also decreases and so each successive turn of the spool will feed out less and less tape. So if the tape passes the tape-head at a constant speed, the spool will rotate more and more quickly, and the width of the reel will decrease more and more rapidly.
This task should generate rich classroom discussion about these geometric aspects of the context and about the shape of the graph: the graph clearly slopes downwards, but is it a straight line?
As the width of the reel of tape decreases more and more rapidly (over time), the slope of the graph gets progressively steeper (the gradient is negative throughout). This raises interesting questions about the slope near points A and B. Could it be horizontal at A and vertical at B? [Well, no, not quite...] And if we continue the curve, where might it cut the x-axis, and what would this mean in terms of the context?
Note: The curve through A and B is a parabola. It turns out that it passes roughly through the point (76, 0) and that the relationship between M minutes and W mm, is given by M ≂ 76 – (W²/30) [or, more precisely, W = 78.83 – (12W²/345)*, if we assume the given measurements are accurate].
*You might like to verify this formula (please correct me if I've got it wrong).
-
Note 2: Here is an alternative wording for the task. Would this be better?
-
TUESDAY: In this task we assume that the pile of sand keeps its exact shape as it gets bigger. Students might initially think that if the time for which the sand flows is doubled (from 1 minute to 2 minutes), then the height of the pile will be doubled too. (It turns out that it takes 8 minutes rather than just 2, for the height to reach 2 metres - something we look at in Wednesday's task.)
However, it shouldn't take much discussion for students to realise that as the pile gets higher it gets wider too, so ever more sand is needed to raise the height by a given amount. This means that the graph will consist of a curve with an ever flatter positive slope. In turn this means that at the 2 minute mark, the height of the graph will be substantially less than 2 (m). However, at this juncture we wouldn't expect students to have a more precise sense of what that height will be. This is something we consider in a more analytic way in Wednesday's task.
-
WEDNESDAY: We consider what the height of the pile of sand will be after 2 minutes, and use this information to draw a more accurate graph, which we then use to interpolate the height at 2 minutes.
If we double the height of the pile of sand, from 1 m to 2 m, we also double the width (in two dimensions), so we have 2×2×2 = 8 times as much sand as before, so it will take 8 times as long to produce.
A good approximation for the value of the height at 2 minutes is 5/4. This is close to the cube-root of 2, since (5/4)-cubed is 125/64 = 1.953125.
-
THURSDAY: We step onto a travelator and think about some (nice, simple, straight line) distance-time graphs.
-
FRIDAY: We relate our distance-time graphs to speed-time graphs. This may not be easy - they look so very different!
Before thinking about the speed-time graph for Mario, it is interesting to consider how students might arrive at the red line - ie the distance-time graph for Mario while he is walking on the travelator. One way would be to find a specific point [such as (10, 14)] and join it to (2, 2) with a straight line. Another way would be to consider the distance travelled each second on the travelator - namely 0.5 m due to the travelator, plus 1 m due to the fact that Mario is walking at 1 m per second relative to the surface of the travelator - so the red line has a slope of 1.5 (m per sec).
Marco's and Mario's speeds are constant for most of the time (except for the brief moments when their speeds change). This means their speed-time graphs will consist primarily of horizontal straight lines. As such, the graphs look very different from the distance-time graphs, where the increase in distance (from Lisa) over time is shown by the upward slope of the graphs. In contrast to this, on the speed-time graphs, the increasing distance is shown by the increase in area under the graphs as the time increases. This looks far less dramatic.

Monday, 11 June 2018

ALG 18

This week we look at the rules of arithmetic, as expressed through the use of brackets. But we start informally, with a 'real life' task, specifically a sour dough problem, which may not be to everyone's liking....
-
MONDAY: A lighthearted story about pizza slices.
TUESDAY: Here we complete some numerical expressions involving brackets to describe a set of dots.
No pizzas.... No pseudo contexts...? Just dots. Simple.
Note: I have curtailed the string of expressions a bit. You might, for example, want to insert the expression 3×5×(20+?), and perhaps also 3×(5×20 + 5×?), between my second and third expressions.
-
WEDNESDAY: Here we look at expressions that give the total number of dots in a (recursive) dice pattern.
Note 1: we have used brackets more frequently than is strictly necessary here, so that the order of operations is explicit and unambiguous, ie so that the expressions can be read without knowing the conventional order of operations. Of course, it is important to learn these conventions but it means that the current expressions should be accessible to a high proportion of students. Also, as the expressions refer to familiar, concrete elements that are easy to group and count, the work should help demystify the use of expressions involving brackets - students don't need to refer to rules like 'do the brackets first' to make sense of the notation.
Note 2: you might want to ask students to annotate the diagram, or to draw a revised diagram, to show how the dots are being structured by the various expressions.
-
THURSDAY: Here we use expressions to count dots in some partitioned/compound arrays.
Again, it might be helpful to annotate the sets of dots to show how they relate to the expressions.
-
FRIDAY: We look at some neat and some not so neat ways of performing the calculation 15 × 22.

This slide is somewhat overloaded. It's really two tasks in one. And for the first task, it would be better to introduce the parts one at a time. Then, having worked through these exemplars, it should prove fruitful to devote time to students' methods of calculating 15 × 22, and to representing these as strings of expressions involving brackets.
The second task should help students take a more conscious, formal view of the distributive law and to think about when it applies.

Sunday, 3 June 2018

ALG 17

She was just seventeen, You know what I mean, And the way she looked Was way beyond compare. How could I dance with another? Oooh! When I saw her standing there.
One of the great early Beatles tracks. With ALG 17, we're standing on the number line and seeing how our position is related to multiples, especially multiples of 2 and 3.
-
MONDAY: Where do multiples of 3 occur on the number line? How can we tell whether a multiple of 3 is also a multiple of 6? And of 9??
TUESDAY: This is a classic task that I was introduced to by Lulu Healy. You might want to generate some data and look for patterns, but it is also perfectly accessible via a generic (analytic) approach by thinking about the occurance of multiples in strings of consecutive numbers.
 WEDNESDAY: more multiple fun....
This task keeps catching me out! One moment I think it's OK, the next I think I've got it wrong and it doesn't work!
Part b) can be solved by simply going through the multiples of 7 and 8 in the standard times tables, which leads us to 63, 64. However, a key feature here is the relation of these numbers to 56 (= 7×8) .... Or one can make use of 'the difference of two squares': 8×8 is a multiple of 8, 8×8 – 1  = (8–1)(8+1) is a multiple of 7. [This generalises very nicely, eg to finding consecutive numbers that are multiples of 19 and 20 respectively.]
-
THURSDAY: A classic 'think of a number ....' task. The task has links with Tuesday's version of ALG 17 and one can get quite a long way by using an empirical, data-generating approach. However, its attraction lies in the fact that one can use some fairly routine algebra to throw light on the underlying structure. [This is all the more noteworthy, given that we often use algebra to ease our path to an answer while ignoring structure!]
Note: I came across this particular task in a 1995 paper by Alan Bell (Purpose in school algebra, JMB, 14, 41-73). Sadly, Alan died this year. (9 April 1929 - 5 April 2018)
-
FRIDAY: We take an explicit look at the structure of Thursday's task.
Here the geometric and/or symbolic representation may help us see that Thursday's task results in the product of the 'outer two' of three consecutive numbers. When the middle number is odd, the outer numbers will be even so both must be a multiple of 2 and one must also be a multiple of 4 (of course, they can be multiples of other numbers too). So their product is a multiple of 8.