The Madness of Crowd Models

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November 26, 2017

1:06 pm

23 Comments on The Madness of Crowd Models

VCE

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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.

Related posts:

  1. A Loss of Momentum
  2. MAV’s Dangerous Inflection
  3. Hard SEL: The Specialist Error List
  4. MELting Pot: The Methods Error List
  5. Some Special Madness
  6. Inferiority Complex

23 Replies to “The Madness of Crowd Models”

  1. You raise a good point Marty about the examiner “reports” for these NHT papers. They are not reports at all, in the sense that we have become accustomed to seeing from the November cycle exams. Despite the many failings (of which you delight in pointing out in many ways) they did report HOW some students successfully or unsuccessfully attempted questions, and what the average achievement was.

    I can only assume that by omitting the grades they are trying to hide one or more of the following horrible truths:

    1. The number of students who took the NHT exams was so pathetically small that a statistically significant result could not be reported. Unlikely.

    2. That the NHT students out-performed the November cycle students by such a large margin that if reported it would force VCAA and the AC authors to review the 7-10 Mathematics curriculum. Less unlikely, but still not believing it.

    3. That the details mentioned above were calculated, saved on a USB drive and then left on a bus or maybe in a taxi somewhere between the hotel and the airport by the poor administrator whose job it was to upload the reports… so they had to use a backup written before the actual exams were done. Makes as much sense as anything?

    Reply

    1. Thanks, Number 8. I agree, the lack of genuine reports on the NHT papers is curious. I can imagine a number of benign reasons for this. It is, however, the VCAA we’re talking about.

      Reply

  2. The concept was clearly a VCAA favourite because it appears on the 2018 November Exam 2 Question 3 (b) (iii).

    It’s worth noting that prior to the 2017 NHT exam, the concept appeared on the VCAA 1999 Exam 2 Question 4 (c) (ii) (where students are simply asked to “Explain what happens to the water level … given that initially the tank is full” without attempting to solve the DE). Everything old becomes new again …

    Speaking of which … Some of you may have heard that the both of the 2021 NHT Specialist Maths Exams contained questions involving discrete random variables. Exam 1 specifically required students to know the mean and variance of a binomial random variable. Apparently everything in Maths Methods including all of the probability and statistics (including discrete random variables) is examinable. And apparently everything in Specialist Mathematics (including sequences and series, I assume). This has been directly communicated by VCAA in response to queries about the 2021 NHT exams … Has anyone else seen those emails …? Am I the only one stunned by this ‘clarification’ from VCAA? I’ve clearly been dodging a bullet for the last 5 years by not reviewing this material with students (particularly students who did Methods the previous year and therefore will have forgotten most of their Methods discrete probability).

    Reply

    1. Funny, you must have been looking at this question recently as well, JF.

      I’m curious to see the 2021 NHT exams now.

      On the point about examinable content from other subjects, this is a major concern with 3&4 SM in the forthcoming study design. Notwithstanding other changes, I think it is an improvement that all of 1&2 SM is now a prerequisite for 3&4 SM. However, unlike MM and FM, very little of the 3&4 SM content is a rehash of 1&2 content. Now perhaps that’s fine – after all, these are meant to be the strongest students. But add to that (1) a lack of good faith from VCAA and the examiners (to what extent will circular functions, curves in the complex plane, etc. appear in 3&4 exams, since they’ve moved to 1&2?), and (2) vague descriptions in the study design about how the new proof area of study includes proofs on content from pretty much EVERY topic from Units 1&2, and this threatens to be extremely unfair on students and teachers.

      Reply

      1. Yes. And it’s a big enough problem already ( a problem I was happily ignorant of not so long ago). Unlike Methods, where all of Methods 1/2 (and Yr 10 maths) is re-hashed in Methods 3/4, there’s not a lot of re-hash of Specialist 1/2 in Specialist 3/4. OK, you have complex numbers, trigonometric identities, vectors, dynamics & statics and (Mickey Mouse) statistics but the level to which they’re covered is superficial compared to the 3/4 level (unlike Methods 1/2 leading into Methods 3/4).

        An intelligent student with only Methods 1/2 could probably get 50% on the Methods 3/4 exams (certainly for Exam 1). An intelligent student with only Specialist 1/2 would be lucky to get 20% on the Specialist 3/4 exams.

        It’s the ’boutique’ topics that get no re-hash such as Sequences and Series, Circle Theorems, Methods of Proof etc that really bothers me. They can be examined without ever being a formal part of Specialist 3/4 (so there’s the very real possibility that a complex numbers or trigonometry question involving a series might appear on Exam 1 this year …)

        And now VCAA have explicitly declared all of Maths Methods probability and statistics is also directly examinable at the 3/4 level.

        And yes, all this against the landscape of a lack of good faith from VCAA and vague descriptions in the Stupid Design. It does threaten to be extremely unfair on students and teachers and it won’t surprise me if/when enrolments in Specialist Maths drop even further. Maybe there’s a hidden agenda …?

        Then again, VCAA has defended certain trial Maths Methods exams that require an understanding of integration by substitution
        (Q3 https://mathematicalcrap.com/2020/09/25/mavs-trials-and-tribulations/)
        so it has a disgraceful history of making things up as it goes along. So the response by VCAA to Maths Methods probability appearing on the 2021 Specialist NHT exams is just the latest example of a particular person at VCAA saying whatever he can to cover his ass. Because he will NEVER admit to mistakes or incompetence.

        Re: 2021 NHT exams. I’m sure Marty knows someone that can pass on the emails that are circulating that contain screenshots of the dodgy statistics questions.

        Reply

    2. Thanks, John (and SRK), and sorry to not have provided a platform to discuss this. It is on my agenda to look at the 2021 NHT exams (and the 2020 VCAA reports …). I’m just flat out with the ACARA nonsense, and a ton of non-blog stuff. I will get to it.

      Reply

      1. Thanks, Marty. I don’t think the shit will hit the fan until those exams become publicly available, which won’t be until the start of term 4 if past experience is any guide. So take a breather on this one until at least October … Currently the emails discussing this issue probably only have a limited circulation.

        Reply

        2. Sarcasm noted JF, but I think in the past it has taken about 6 weeks from the date of the exams for them to appear on the VCAA website. (Although perhaps if there were some controversial aspects it might take longer…)

          Reply

          1. My memory is some time during the Term 3 holidays, but I might be wrong. Hopefully it \displaystyle is only about 6 weeks, which would mean middle of July. The sooner this disgrace is in the public domain the better.

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              1. Thanks, SRK. Published today. Much sooner than I expected – within 4 weeks of those exams finishing! Maybe, just maybe, a sign of things to come. I suppose it will hoping for too much that the ‘Report’ explains why probability and statistics material unique to Maths Methods is being examined in Specialist Maths Exams. Just when you think VCAA have reached peak-stupidity …

                Reply

                1. The peak stupidity of that question is that the formulas for mean and standard deviation of a linear combination of independent RVs are on the formula sheet for SM. Pity that in order to use that formula, students needed to remember a formula that is on the formula sheet for MM…

                  And as usual a bunch of other weird stuff in exam 1… I’ll work through exam 2 tomorrow, perhaps.

                  Reply

                  1. Re: Exam 1.

                    Q1 could have skipped part (a) and just been part (b).

                    Q3 is a total disgrace and VCAA should just admit the ineptitude and poor judgement of the writers and vettors. But the buck stops with the poor decision making and inept oversight at the top …

                    I hope significant pressure is applied on VCAA over this (and Exam 2 Q6 (a) and (b)). Any takers for signing an open letter to VCAA?

                    Q4: Finally. Some composite functions involving the modulus function on a VCAA exam. It’s only taken, what, 5 years …?

                    Q5: Why show? And why have part (b) – what a waste of space and a mark? Why not a stand-alone “Solve …. for y as a function of x”.

                    Q9 I hope you liked. Apart from a minor formatting change it’s identical to a question a friend of mine wrote for an updated 2015 Trial Exam. Incidentally, it also appeared on a 2020 Trial Exam (my friend had nothing to do with that exam) … exactly the same question! Clearly a very popular question!

                    Q10 I don’t know why the more natural restriction \displaystyle a \geq 0 couldn’t have been given.

                    This exam was clearly written and vetted by people lacking creativity and understanding of the Specialist course.

                    Anyway, no pressure on Marty, I’m sure we’ll be commenting at length sooner or later at a dedicated blog.

                    Reply

            2. Thanks very much, SRK, and John. I’m still up to my neck in ACARA and the draft curriculum quagmire, but I’ll try to look at the NHT exams and the issues you’ve raised very soon. (Plus ITE, plus NAPLAN, plus VCAA reports, plus whatever else I forgot is on my “short” list.)

              Reply

  3. Hi Marty.

    I have a question about your example \displaystyle {\frac{dP}{dt}= 3P - 4P^2 - \sqrt{P}.

    On your graph you show solutions with an initial condition \displaystyle P < \frac{1}{4} decaying asymptotically to P = 0. However, using \displaystyle P(0) = \frac{1}{5} as an example, it can be calculated that the solution to the DE terminates on the t-axis (at approximately t = 6.23). All solutions to the DE with an initial condition of \displaystyle P < \frac{1}{4} terminate on the t-axis rather than having the t-axis as an asymptote.

    I found this result surprising and am trying to understand why P = 0 is not a horizontal asymptote. In particular, I'm trying to understand this via a simple use of the DE itself and its equilibrium solutions. The bigger question is understanding in a simple way when a solution will decay asymptotically to an equilibrium solution and when it will decay and terminate at a point on an equilibrium solution. And will this point be a 'hole' or an included point?

    Can you or anyone shed some light on this? I'm happy to be told that I've misunderstood your graphs, but my question still stands. I think it has something to do with \displaystyle \frac{d^2P}{dt^2} being undefined at P= 0 …? (So that the graph of P versus t has a 'stationary point' at P = 0).

    Reply

    1. Hi, John. It’s three or so years ago … Possibly I made a mistake. But I think maybe I cooked up the (effective) polynomial to have a repeated root. Will check tomorrow.

      Reply

      1. Wow, that long ago. Tempus fugit.

        Anyway, I think I’ve got it figured out in a simple way (but I’m happy to be corrected):

        \displaystyle \frac{d^2P}{dt^2} \rightarrow - \infty as P \rightarrow 0 (this is due to the \displaystyle \sqrt{P} term in \displaystyle \frac{dP}{dt}).
        Therefore the gradient of the graph approaches zero at an ‘infinite rate’ as P \rightarrow 0.
        So P does not decrease asymptotically (that is, ‘slowly’) towards P = 0, it decreases ‘rapidly’ to terminate at P = 0 (and the graph has a ‘stationary point’ on the t-axis).

        Reply

        1. Thanks, John. I think you’re right, and it is related to the recent WitCH, that the √P makes things weird near P = 0, and so standard qualitative analysis needs more care.

          So, I take it the curves starting below 1/4 hit the t-axis tangentially, and can then be continued as a differentiable (but not twice-differentiable) function to be constantly zero after hitting. (And that’s the only option, although a cube root would presumably result in non-uniqueness issues.) If so, that suggests my diagram is incorrect, but the example is still basically sound.

          Reply

          1. Yes, except the VCAA restriction 0<P<1 would mean there's a 'hole' where the curve tangentially hits the t-axis and the curve has to terminate at that point.

            The example is excellent and my question is not a criticism of either it or the diagram. I would have drawn exactly the same diagram myself until 12 hours ago. I was troubled after realising that the curve actually did hit the t-axis, and it nagged at me to understand why is it so. Now I'm happy again.

            We can blame VCAA for making me write a stupid SAC and causing me to think about this stuff a bit more carefully. (It’s an ill wind that blows no good).

            Reply

            1. Well, of course it is a criticism, and of course that is fine and good. I don’t like to make mistakes, and definitely like them to be pointed out when they occur. But I’m also not kicking myself that hard, since the point was to cook up a simple example with the desired behaviour near the other equilibrium.

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        2. I’ve been cogitating on Marty’s example in light of the new Study Design. Because it appears that the explicit inclusion of the ‘Logistic equation’ implies that:

          1) Definitions of carrying capacity and growth rate and being able to identify these values from the DE will be required (if the MAV PD is credible).

          2) A qualitative understanding of the shape of the graphs of the solutions based solely on the DE will be very valuable. Which is good.

          Now, neither of these things are stated or even remotely implied or suggested in the SD. But it seems that Cambridge (as quoted in the MAV PD) devotes a whole page summarising results that are readily calculated from 2) (so it looks like the VCAA focus is on rote-learning rather than understanding the properties of the solution directly from the DE).

          Anyway, I’ve always taught 2) but the behaviour of the solutions to Marty’s example for \displaystyle 0 < P(0) < \frac{1}{4} troubled me. It was reasonable to assume asymptotic behaviour towards \displaystyle P = 0 but this is not the case.

          Anyway, at the time I tried to understand why this happened and the result was wishy-washy, hand-wavy, ‘infinite rate’ crap. (I’m happy to say something is crap, even when it’s my crap. VCAA’s crap doesn’t get special treatment). But … I think I’ve finally got it figured out (at least to my satisfaction at a Specialist Maths level). Below is the result of my cogitation. And the ‘infinite rate’ crap turns out not so much to be crap but just not the full story. Comments warmly encouraged.

          Using the chain rule it’s not hard to get

          \displaystyle \frac{d^2P}{dt^2} = \left(3 - 8P - \frac{1}{2\sqrt{P}}\right) (3P - 4P^2 - \sqrt{P}).

          It’s not hard to solve \displaystyle \frac{d^2P}{dt^2} = 0 and then confirm that there is a point of inflection at \displaystyle P = \frac{2-\sqrt{3}}{8}.

          Case 1: \displaystyle P(0) > \frac{1}{4}.
          Consider \displaystyle P(0) = 0.5, say.
          It’s not hard to show that \displaystyle \lim_{P \rightarrow \frac{1}{4}^{+}}{\frac{d^2P}{dt^2} = 0^{+} and so \displaystyle \frac{d^2P}{dt^2} > 0 for \displaystyle 0.25 < P < 0.5.
          Therefore P is always concave up but ‘flattens out’ (because the rate of change \displaystyle \frac{dP}{dt} of P approaches zero) as \displaystyle P \rightarrow \frac{1}{4}.
          Therefore P decreases asymptotically towards \displaystyle P = \frac{1}{4}.

          Case 2: \displaystyle 0< P(0) < \frac{1}{4}.
          Consider \displaystyle P(0) = 0.2, say.
          \displaystyle \lim_{P \rightarrow 0^{+}}{\frac{d^2P}{dt^2} is an indeterminate form (*) but it’s not hard to substitute \displaystyle Q = \sqrt{P}, expand, simplify (cancel out the common factor Q) and find that the limit is equal to \displaystyle \frac{1}{2}. So:

          \displaystyle \frac{d^2P}{dt^2} < 0 for \displaystyle \frac{2-\sqrt{3}}{8} < P < 0.2

          \displaystyle 0 < \frac{d^2P}{dt^2} < \frac{1}{2} for \displaystyle 0 < P < \frac{2-\sqrt{3}}{8}.

          Therefore P is concave up for \displaystyle 0 < P < \frac{2-\sqrt{3}}{8} but does NOT ‘flatten out’ (because the rate of change \displaystyle \frac{dP}{dt} of P approaches \displaystyle \frac{1}{2}) as \displaystyle P \rightarrow 0 and so does NOT decrease asymptotically towards \displaystyle P = 0. Instead, P touches the t-axis tangentially at \displaystyle P=0 and continues as \displaystyle P=0.

          * This is why the ‘infinite rate’ crap turns out not to be crap but just not the full story. The fact that \displaystyle \frac{d}{dP}\left( \frac{dP}{dt}\right) = 3 - 8P - \frac{1}{2\sqrt{P}} \rightarrow -\infty is why \displaystyle \lim_{P \rightarrow 0^{+}}{\frac{d^2P}{dt^2} = \frac{1}{2}.

          Comments warmly encouraged.

          Reply

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