The Madness of Crowd Models

Our fifth and penultimate post on the 2017 VCE exam madness concerns Question 3 of Section B on the Northern Hemisphere Specialist Mathematics Exam 2. The question begins with the logistic equation for the proportion P of a petrie dish covered by bacteria:

    \[\boldsymbol{\frac{{\rm d} P}{{\rm d} t\ }= \frac{P}{2}\left(1 - P\right)\,\qquad 0 < P < 1\,.}\]

This is not a great start, since it’s a little peculiar using the logistic equation to model an area proportion, rather than a population or a population density. It’s also worth noting that the strict inequalities on P are unnecessary and rule out of consideration the equilibrium (constant) solutions P = 0 and P = 1.

Clunky framing aside, part (a) of Question 3 is pretty standard, requiring the solving of the above (separable) differential equation with initial condition P(0) = 1/2. So, a decent integration problem trivialised by the presence of the stupifying CAS machine. After which things go seriously off the rails.

The setting for part (b) of the question has a toxin added to the petri dish at time t = 1, with the bacterial growth then modelled by the equation

    \[\boldsymbol{\frac{{\rm d} P}{{\rm d} t\ }= \frac{P}{2}\left(1 - P\right) - \frac{\sqrt{P}}{20}\,.}\]

Well, probably not. The effect of toxins is most simply modelled as depending linearly on P, and there seems to be no argument for the square root. Still, this kind of fantasy modelling is par for the VCAA‘s crazy course. Then, however, comes Question 3(b):

Find the limiting value of P, which is the maximum possible proportion of the Petri dish that can now be covered by the bacteria.

The question is a mess. And it’s wrong.

The Examiners’ “Report” (which is not a report at all, but merely a list of short answers) fails to indicate what students did or how well they did on this short, 2-mark question. Presumably the intent was for students to find the limit of P by finding the maximal equilibrium solution of the differential equation. So, setting dP/dt = 0 implies that the right hand side of the differential equation is also 0. The resulting equation is not particularly nice, a quartic equation for Q = √P. Just more silly CAS stuff, then, giving the largest solution P = 0.894 to the requested three decimal places.

In principle, applying that approach here is fine. There are, however, two major problems.

The first problem is with the wording of the question: “maximum possible proportion” simply does not mean maximal equilibrium solution, nor much of anything. The maximum possible proportion covered by the bacteria is P = 1. Alternatively, if we follow the examiners and needlessly exclude = 1 from consideration, then there is no maximum possible proportion, and P can just be arbitrarily close to 1. Either way, a large initial P will decay down to the maximal equilibrium solution.

One might argue that the examiners had in mind a continuation of part (a), so that the proportion begins below the equilibrium value and then rises towards it. That wouldn’t rescue the wording, however. The equilibrium solution is still not a maximum, since the equilibrium value is never actually attained. The expression the examiners are missing, and possibly may even have heard of, is least upper bound. That expression is too sophisticated to be used on a school exam, but whose problem is that? It’s the examiners who painted themselves into a corner.

The second issue is that it is not at all obvious – indeed it can easily fail to be true – that the maximal equilibrium solution for P will also be the limiting value of P. The garbled information within question (b) is instructing students to simply assume this. Well, ok, it’s their question. But why go to such lengths to impose a dubious and impossible-to-word assumption, rather than simply asking directly for an equilibrium solution?

To clarify the issues here, and to show why the examiners were pretty much doomed to make a mess of things, consider the following differential equation:

    \[\boldsymbol{\frac{{\rm d} P}{{\rm d} t\ }= 3P - 4P^2 - \sqrt{P}\,.}\]

By setting Q = √P, for example, it is easy to show that the equilibrium solutions are P = 0 and P = 1/4. Moreover, by considering the sign of dP/dt for P above and below the equilibrium P = 1/4, it is easy to obtain a qualitative sense of the general solutions to the differential equation:

In particular, it is easy to see that the constant solution P = 1/4 is a semi-stable equilibrium: if P(0) is slightly below 1/4 then P(t) will decay to the stable equilibrium P = 0.

This type of analysis, which can readily be performed on the toxin equation above, is simple, natural and powerful. And, it seems, non-existent in Specialist Mathematics. The curriculum  contains nothing that suggests or promotes any such analysis, nor even a mention of equilibrium solutions. The same holds for the standard textbook, in which for, for example, the equation for Newton’s law of cooling is solved (clumsily), but there’s not a word of insight into the solutions.

And this explains why the examiners were doomed to fail. Yes, they almost stumbled into writing a good, mathematically rich exam question. The paper thin curriculum, however, wouldn’t permit it.

 

Some Special Madness

Our second post on the 2017 VCE exam madness concerns a question on the first Specialist Mathematics exam. Typically Specialist exams, particularly the first exams, don’t go too far off the rails; it’s usually more “meh” than madness. (Not that “meh” is an overwhelming endorsement of what is nominally a special mathematics subject.) This year, however, the Specialist exams have some notably Methodsy bits. The following nonsense was pointed out to us by John, a friend and colleague.

The final question, Question 10, on the first Specialist exam concerns the function \boldsymbol{f(x) = \sqrt{\arccos(x/2)}}, on its maximal domain [-2,2]. In part (c), students are asked to determine the volume of the solid of revolution formed when the region under the graph of f is rotated around the x-axis. This leads to the integral

    \[V \ = \ \pi \int\limits_{-2}^{2}  \arccos(x/2)\, {\rm d}x\,.\]

Students don’t have their stupifying CAS machines in this first exam, so how to do the integral? It is natural to consider integration by parts, but unfortunately this standard and powerful technique is no longer part of the VCE curriculum. (Why not? You’ll have to ask the clowns at ACARA and the VCAA.)

No matter. The VCAA examiners love to have the students to go through a faux-parts computation. So, in part (a) of the question, students are asked to check the derivative of \boldsymbol{x\arccos(x/a)}. Setting a = 2 in the resulting equation, this gives

    \[ \frac{{\rm d}\phantom{x}}{{\rm d}{x}}\left(x\arccos(x/2)\right)= \arccos(x/2) - \dfrac{x}{\sqrt{4-x^2}}\,.\]

We can now integrate and rearrange, giving

    \[ \aligned V \ &= \ \pi\left\Big[\!x\arccos(x/2)\!\!\right\Big]\limits_{-2}^{2} \quad +\quad \pi \int\limits_{-2}^{2} \dfrac{x}{\sqrt{4-x^2}}\, {\rm d}x\\[2\jot] \ &= \ 2\pi^2\quad +\quad \pi \int\limits_{-2}^{2} \dfrac{x}{\sqrt{4-x^2}}\, {\rm d}x\,.\endaligned\]

So, all that remains is to do that last integral, and … uh oh.

It is easy to integrate \boldsymbol{x/\sqrt{4-x^2}} indefinitely by substitution, but the problem is that our definite(ish) integral is improper at both endpoints. And, unfortunately, improper integrals are not part of the VCE curriculum. (Why not? You’ll have to ask the clowns at ACARA and the VCAA.) Moreover, even if improper integrals were available, the double improperness is fiddly: we are not permitted to simply integrate from some –b to b and then let b tend to 2.

So, what is a Specialist student to do? One can hope to argue that the integral is zero by odd symmetry, but the improperness is again an issue. As an example indicating the difficulty, the integral \boldsymbol{\int\limits_{-2}^2 x/(4-x^2)\,{\rm d}x} is not equal to 0. (The TI Inspire falsely computes the integral to be 0, which is less than inspiring.) Any argument which arrives at the answer 0 for integrating \boldsymbol{x/(4-x^2)} is invalid, and is thus prima facie invalid for integrating \boldsymbol{x/\sqrt{4-x^2}} as well.

Now, in fact \boldsymbol{\int\limits_{-2}^2 x/\sqrt{4-x^2}\,{\rm d}x} is equal to zero, and so \boldsymbol{V = 2\pi^2}. In particular, it is possible to argue that the fatal problem with \boldsymbol{x/(4-x^2)} does not occur for our integral, and so both the substitution and symmetry approaches can be made to work. The argument, however, is subtle, well beyond what is expected in a Specialist course.

Note also that this improperness could have been avoided, with no harm to the question, simply by taking the original domain to be, for example, [-1,1]. Which was exactly the approach taken on Question 5 of the 2017 Northern Hemisphere Specialist Exam 1. God knows why it wasn’t done here, but it wasn’t and the consequently the examiners have trouble ahead.

The blunt fact is, Specialist students cannot validly compute \boldsymbol{\int\limits_{-2}^2 x/\sqrt{4-x^2}\,{\rm d}x} with any technique they would have seen in a standard Specialist class. They must either argue incompletely by symmetry or ride roughshod over the improperness. The Examiners’ Report will be a while coming out, though presumably the examiners will accept either argument. But here is a safe prediction: the Report will either contain mealy-mouthed nonsense or blatant mathematical falsehoods. The only alternative is for the examiners to make a clear admission that they stuffed up. Which won’t happen.

Finally, the irony. Look again at the original integral for V. Though this integral arose in the calculation of a volume, it can still be interpreted as the area under the graph of the function y = arccos(x/2):

But now we can consider the corresponding area under the inverse function y = 2cos(x):

It follows that

    \[V \ = \  \pi \int\limits_{-2}^{2}  \arccos(x/2)\, {\rm d}x \ = \ \pi \int\limits_{0}^{\pi} \left[  2\cos(x) - (-2)\right]  \, {\rm d}x \ = \ 2\pi^2\,.\]

Done.

This inverse function trick is standard for Specialist (and Methods) students, and so the students can readily calculate the volume V in this manner. True, reinterpreting the integral for V as an area is a sharp conceptual shift, but with appropriate wording it could have made for a very good Specialist question.

In summary, the Specialist Examiners guided the students to calculate V with a jerry-built technique, leading to an integral that the students cannot validly compute, all the while avoiding a simpler approach well within the students’ grasp. Well played, Examiners, well played.