Solving Venn diagrams where the intersection is unknown

n=40

Today in 2C MAT we came across that old chestnut, the Venn diagram with the missing value in the intersection with a number in A, B and the outside region.

In many cases the easiest way is to use a guess and check approach and a lot of the time the answer will fall out by substituting into the intersection and revising your result based on the values
A union B + the outside region = n.

n=40











Another approach is to name the segments and solve a series of equations:

a = 20-b
c = 30-b
a + b + c + 5 = 40

By substitution (20-b) + b + (30 - b) + 5 = 40
Therefore b=15

Once the intersection(b) is known, finding "A only"(a) and "B only"(b) is trivial.

I was asked the question "why teach this technique?" and my response was that it was not formally taught, it was a logical answer for a question given. We have some unknowns, we have some equations, why not solve for them? This sort of problem solving "setting up of equations" technique is common in optimisation and linear programming - why not use it in a probability setting?

I remember a particular student that was renowned for having solutions of this nature where his answers always deviated from the answer key and he had the right answer (or was on the right track) more often than not. We still call intuitive answers like this after "that" student as they forced the marker to find the underlying logic rather than application of a given method (if that student is reading this - get offline and study for your uni courses, scallywag!)


Anyhow, a third and more common approach is to rearrange the property:
A U B = A + B - A intersection B

By rearranging the equation
A intersection B = A + B - AUB

Since we know that:
AUB = U - (the outside region)

to find AUB is fairly simple:
AUB = 40-5
= 35

Therefore:
A intersection B = 20 + 30 - 35
= 15 (as before)

This approach does have the advantage that you can talk about the intersection being counted twice when the union is calculated by adding A + B where A and B aren't mutually exclusive.

I can't really see how this problem could be classed complex given the second method exists. Perhaps, if combined with a wordy explanation, a question of this sort could be made complex but to my mind that would defeat the purpose of the syllabus points in defining complexity. After all, why should something be classed a "complex question" if the only reason was that the question was worded to be understood by students with strong English comprehension?

Further exploring the properties of one

To find an equivalent fraction of a decimals, one way to explain it is to take the decimal part of the original number and place it over the lowest place value. Leave any whole numbers in front. (This only works for non-recurring decimals)

eg 0.123

The lowest place value is thousandths, the decimal part is 123.

therefore:

0.123 = 123/1000


An alternative way to explain it is using properties of one. The idea is that
a) numerators of fractions should be whole numbers and;
b) the fraction should be equivalent to the decimal.

We can ensure the fraction is equivalent if we only multiply or divide by 1 or more importantly a fraction that is equivalent to 1.

To satisfy part a)
To make 0.123 a whole number we have to multiply it by a power of 10 - 1000 (10^3). This was a concept we had investigated earlier.

..but if we multiply by 1000 we will change the original number from 0.123 to 123 - it will no longer be equivalent.

So to satisfy part b)
We multiply by 1000/1000 (or 1!)

Thus:
.0123 = .123/1 x 1000/1000
= 123 / 1000

I like this because it continues to explore how fractions are constructed, the connection between decimals and fractions and why decimal conversion works. I wouldn't try it in classes with low ability due to the possibility for high levels of confusion if understandings of multiplication and commutative properties are not properly understood.

An earlier article exploring one and fractions can be found here.

Viola.

PD Days & Collegiality

One of the bugbears of PD days is the difficulty of engaging 60-70 university trained professionals of widely diverse interests, usually during times of high stress with timelines bearing down on you.

One idea is to use this time for learning area planning. This is usually unsuccessful and the planning time instead used for a wide variety of other tasks (general discussion, marking, personal planning). Why?

Some suggested reasons:
a) No deliverables are defined
b) Time frame for deliverables are unrealistic, ill defined or aspirational
c) Require sharing of resources that are thought of as proprietary (such as programmes developed in own time)
d) Require interaction between staff members that are oppositional
e) Processes are poorly lead and easily high jacked
f) Deliverables are not measured
g) No consequences for not meeting deliverables

Most of these are just indicators of poor school based management but many are problems that have arisen due to systemic ineptness. The lack of collegiality is a growing phenomenon that is occurring as competitiveness between teachers for promotional positions is rising and teaching moves from a vocational profession to an occupation. If schools do not actually manage the transfer of information and the information loss as teachers move between positions and schools, the school loses knowledge and effectiveness (especially cohort or area knowledge) with each transfer. Teachers tend to gain knowledge working in schools such as ours (on their path to effective teaching in low SES schools) rather than the other way around. Those entering these schools can encounter strong resistance to new ideas (especially if it is thought the ideas have been tried before), underestimate implementation issues or be unwilling to share until quid-pro-quo is found.

It should also be recognised that with the rapid changes in syllabus, the ability for a school to develop a working curriculum (that can be further developed over a number of years) has been made significantly harder. The weight of curriculum development has been placed on many occasions in the hands of the incompetent through no fault of their own (teaching out of area, beginning teachers, sole practitioners rather than team members, those lacking analytical skills but are fantastic teachers, administration staff that cannot measure effectiveness of a programme etc)

PD days are one opportunity to stop this information loss but it needs people that can define clearly a task to be done that would serve a real long term purpose and then measure the effectiveness of it. It is just another aspect of change management.

Drawing the first derivative

Teaching students how to visualise the first derivative in 3B MAT has been problematic over the last two years. This morning I had a bit of a breakthrough in that students weren't looking at me as if I was speaking Alien.

The major difference was that I didn't use the arrow approach. Here's what I did.

I drew a positive cubic on the board and identified the turning points. I identified clearly the x axis and the y axis and identified the coordinates for each TP. I drew their attention to (x,y)

Then I drew a second pair coordinate plane directly underneath and identified/labelled the x axis. I then deliberately (as in made a big song and dance) labelled the other axis y' asking students to think what this might mean.

I then went to the first turning point on the x,y plane and asked students what the gradient was at this point. They said zero straight away.

I then went to the second axis and said coordinates on this plane were (x,y'). Given that the TP we were examining was at (0.25) and y'(0.25) = 0, the coordinate(x,y') that we needed was at (0.25,0). We repeated this for the other turning point.

I then drew vertical dotted lines through both coordinate planes. We then looked at the slope to the left of the TP. Being a cubic (with a positive coefficient of x cubed) the slope was +ve. On the second plane I wrote +ve above the x axis to the left of the TP above the x axis. We then examined the second area and noted the slope was negative (making special note of where the point of inflection was - it wasn't mandated by the course but made sense in the context). I labelled the graph -ve underneath the x axis to the right of the TP. I then wrote +ve in the third area above the x axis.

<- It looked like this.

















Once the areas were labelled it was trivial to join the dots starting where y' was positive (y' at +ve infinity), leading to where y' was negative and then changing direction midway between the x intercepts on y', back towards to the x axis until y' was +ve again (again until y' at +ve infinity). It was also a good time to discuss the type of function produced (eg a concave up quadratic) if you differentiate a cubic with a +ve coefficient of the cubed term and how that related to our y' graph.



















We then repeated the process for a quartic.

yay!