population models – BAD MATHEMATICS

This one’s shooting a smelly fish in a barrel, almost a POSWW. Sometimes, however, it’s easier for a tired blogger to let the readers do the shooting. (For those interested in more substantial fish, WitCH 2, WitCH 3 and Tweel’s Mathematical Puzzle still require attention.)

Our latest WitCH comes courtesy of two nameless (but maybe not unknown) Western troublemakers. Earlier this year we got stuck into Western Australia’s 2017 Mathematics Applications exam. This year, it’s the SCSA‘s Mathematical Methods exam (not online. Update: now online here and here.) that wins the idiocy prize. The whole exam is predictably awful, but Question 15 is the real winner:

The population of mosquitos, P (in thousands), in an artificial lake in a housing estate is measured at the beginning of the year. The population after t months is given by the function, $\color{blue}\boldsymbol{P(t) = t^3 + at^2 + bt + 2, 0\leqslant t \leqslant 12}$ .

The rate of growth of the population is initially increasing. It then slows to be momentarily stationary in mid-winter (at t = 6), then continues to increase again in the last half of the year.

Determine the values of a and b.

Go to it.

Update

As Number 8 and Steve R hinted at and as Damo nailed, the central idiocy concerns the expression “the rate of population growth”, which means P'(t) and which then makes the problem unsolvable as written. Specifically:

In the second paragraph, “it” has a stationary point of inflection when t = 6, which is impossible if “it” refers to the quadratic P'(t).
On the other hand, if “it” refers to P(t) then solving gives a < 0. That implies P”(0) = 2a < 0, which means “the rate of population growth” (i.e. P’) is initially decreasing, contradicting the first claim of the second paragraph.

The most generous interpretation is that the examiners intended for the population P, not the rate P’, to be initially increasing. Other interpretations are less generous.

No matter the intent, the question is inexcusable. It is also worth noting that even if corrected the question is awful, a trivial inflection problem dressed up with idiotic modelling:

Modelling population growth with a cubic is hilarious.
Months is a pretty stupid unit of time.
The ~~rate of~~ population ~~growth~~ initially increasing is irrelevant.
Why is the lake artificial? Who gives a shit?
Why is the lake in a housing estate? Who gives a shit?

Finally, it’s “latter half” or “second half”, not “last half”. Yes, with all else awful here, it hardly matters. But it’s wrong.

Further Update

The marking schemes for the exam are now up, here and here. As was predicted, “the rate of growth of the population” was intended to mean “population”. As is predictable, the grading scheme gives no indication that the question is garbled garbage.

The gutless contempt with which certain educational authorities repeatedly treat students and teachers is a wonder to behold.

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 petri 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 P = 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 P 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.

UPDATE (06/08/2021)

John Friend has pointed out an error in the graph and description of the solutions to our differential equation. If $P(0) < 1/4$ then the solution will hit the $P$ axis tangentially in finite time; that solution can then be continued differentiably to be $P\equiv 0$ from then on. The message of the example remains the same.

Tag: population models

WitCH 5: What a West

Update

Further Update

The Madness of Crowd Models