Inferiority Complex

This one is long, a real Gish gallop. Question 4, Part 2 from the 2017 VCE Specialist Mathematics Exam 2 is a mess. The Examiners’ Report is, predictably, worse.

Part (a) of Question 4 is routine, requiring students to express {-2-2\sqrt{3}i} in polar form. One wonders how a quarter of the students could muck up this easy 1-mark question, but the question is fine.

The issues begin with 4(b), for which students are required to

Show that the roots of \color{red}\boldsymbol{z^2 + 4z + 16 = 0}} are {\color{red} \boldsymbol{z=-2-2\sqrt{3}i}} and \boldsymbol{\color{red}{z=-2+2\sqrt{3}i}}.

The question can be answered with an easy application of completing the square or the quadratic formula. So, why did almost half of the students get it wrong? Were so many students really so clueless? Perhaps, but there is good reason to suspect a different source of the cluelessness.

The Examiners’ Report indicates three general issues with students’ answers. First,

students confused factors with solutions or did not proceed beyond factorising the quadratic.

Maybe the students were confused, but maybe not. Maybe some students simply thought that, once having factorised the quadratic, the microstep to then write “Therefore z = …”, to note the roots written on the exam in front of them, was too trivial in response to a 1 mark question.

Second, some students reportedly erred by

not showing key steps in their solution.

Really? The Report includes the following calculation as a sample solution:

\color{blue} \boldsymbol{z = \frac{-4\pm \sqrt{4^2 \ - \ 4 \times 1 \times 16}}{2}=\frac{-4\pm \sqrt{-48}}{2}=\frac{-4\pm 4\sqrt{3}i}{2} = -2\pm2\sqrt{3}i\, .}

Was this whole tedious, snail-paced computation required for one measly mark? It’s impossible to tell, but the Report remarks generally on ‘show that’ questions that

all steps that led to the given result needed to be clearly and logically set out.

As we have noted previously, demanding “all steps” is both meaningless and utterly mad. For a year 12 advanced mathematics student the identification of the roots is pretty much immediate and a single written step should suffice. True, in 4(b) students are instructed to “show” stuff, but it’s hardly the students’ fault that what they were instructed to show is pretty trivial.

Third, and by far the most ridiculous,

some students did not correctly follow the ‘show that’ instruction … by [instead] solely verifying the solutions given by substitution.

Bullshit.

VCAA examiners love to worry that word “show”. In true Princess Bride fashion, however, the word does not mean what they think it means.

There is nothing in standard English usage nor in standard mathematical usage, nor in at least occasional VCE usage (see Q2(a)), that would distinguish “show” from “prove” in this context. And, for 4(b) above, substitution of the given values into the quadratic is a perfectly valid method of proving that the roots are as indicated.

It appears that VCE has a special non-English code, in which “show” has a narrower meaning, akin to “derive“. This cannot alter the fact that the VCE examiners’ use of the word is linguistic and mathematical crap. It also cannot alter the fact that students being penalised for not following this linguistic and mathematical crap is pedagogical and mathematical crap.

Of course all the nonsense of 4(b) could have been avoided simply by asking the students to find the roots. The examiners declined to do so, however, probably because this would have violated VCAA’s policy of avoiding asking any mathematical question with some depth or difficulty or further consequence. The result is a question amounting to no more than an infantile and infantilising ritual, penalising any student with the mathematical common sense to answer with the appropriate “well, duh”.

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Onwards we trek to 4(c):

Express the roots of \color{red} \boldsymbol{z^2 + 4z + 16 = 0}} in terms of  \boldsymbol{{\color{red}   2 -2\sqrt{3}i}}.

Less than a third of students scored the mark for this question, and the Report notes that

Misunderstanding of the question was apparent in student responses. Many attempts at solutions were not expressed in terms of  {\color{blue} \boldsymbol{2 -2\sqrt{3}i}} as required.

Funny that. The examiners pose a question that borders on the meaningless and somehow this creates a sea of misunderstanding. Who would’ve guessed?

4(c) makes little more sense than to ask someone to write 3 in terms of 7. Given any two numbers there’s a zillion ways to “express” one number “in terms of” the other, as in 3 = 7 – 4 or whatever. Without further qualification or some accepted convention, without some agreed upon definition of “expressed in terms of”, any expression is just as valid as any other.

What was expected in 4(c)? To approach the question cleanly we can first set w = 2 - 2\sqrt{3}i, as the examiners could have and should have and did not. Then, the intended answers were -w and -\overline{w}.

These expressions for the roots are simple and natural, but even if one accepts a waffly interpretation of 4(c) that somehow requires “simple” solutions, there are plenty of other possible answers. The expressions w-4 and \overline{w-4} and w^2/4 and w^4/|w|^3 are all reasonable and natural, but nothing in the Examiners’ Report suggests that these or similar answers were accepted. If not, that is a very nasty cherry on top of an incredibly silly question.

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The pain now temporarily lessens (though the worst is yet to come). 4(d) asks for students to show that the relation |z| = |z - (2-2\sqrt{3}i)| has the cartesian form x-\sqrt{3}y - 4 = 0, and in 4(e) students are asked to draw this line on an Argand diagram, together with the roots of the above quadratic.

These questions are routine and ok, though 4(d) is weirdly aimless, the line obtained playing no role in the final parts of Q4. The Examiners’ Report also notes condescendingly that “the ‘show that’ instruction was generally followed”. Yes, people do tend to follow the intended road if there’s only one road.

The final part, 4(g), is also standard, requiring students to find the area of the major segment of the circle |z| = 4 cut off by the line through the roots of the quadratic. The question is straight-forward, the only real trick being to ignore the weird line from 4(d) and 4(e).

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Finally, the debacle of 4(f):

The equation of the line passing through the two roots of {\color{red} \boldsymbol{z^2 + 4z + 16 = 0}} can be expressed as {\color{red} \boldsymbol{|z-a| = |z-b|}}, where \color{red}\boldsymbol{a, b \in C}. Find \color{red}\boldsymbol{b} in terms of \color{red}\boldsymbol{a}.

The Report notes that

This question caused significant difficulty for students.

That’s hilarious understatement given that 99% of students scored 0/1 on the question. The further statements acknowledging and explaining and apologising for the stuff-up are unfortunately non-existent.

So, what went wrong? The answer is both obvious and depressingly familiar: the exam question is essentially meaningless. Students failed to comprehend the question because it is close to incomprehensible.

The students are asked to write b in terms of a. However, similar to 4(c) above, there are many ways to do that and how one is able to do it depends upon the initial number a chosen. The line through the two roots has equation \operatorname{Re} z = x = -2. So then, for example, with a = -4 we have b = 0 and we can write b = a + 4 or b = 0 x a or whatever. If a = -5 then b = 1 and we can write b = -a – 4, and so on.

Anything of this nature is a reasonable response to the exam question as written and none of it resembles the answer in the Report. Instead, what was expected was for students to consider all complex numbers a – except those on the line itself – and to consider all associated complex b. That is, in appropriate but non-Specialist terminology, we want to determine b as a function f(a) of a, with the domain of f being most but not all of the complex plane.

With the question suitably clarified we can get down to work (none of which is indicated in the Report). Easiest is to write a = (-2+c) + di. Since b must be symmetrically placed about the line \operatorname{Re} z = -2, it follows that b = (-2-c) + di. Then b+2 = -c + di = -\overline{(a+2)}. This gives b = -2 - \overline{(a + 2)}, and finally

\color{blue}\boldsymbol{b = -4 -\overline{a}\, ,}

which is the answer indicated in the Examiners’ Report.

In principle 4(f) is a nice question, though 1 mark is pretty chintzy for the thought required. More importantly, the exam question as written bears only the slightest resemblance to the intended question, or to anything coherent, with only the slightest, inaccurate hint of the intended generality of a and b.

99% of 2017 Specialist students have a right to be pissed off.

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That’s it, we’re done. One more ridiculous VCE exam question, and one more ridiculously arrogant Report, unsullied by an ounce of self-reflection or remorse.

The Arc Enemy

Our previous post was on good guys making a silly, funny and inconsequential mistake. This post is not.

Question B1 of Exam 2 for 2018 Northern Hemisphere Specialist Mathematics begins innocently enough. In part (a), students are required to graph the function \boldsymbol{f(x) = 10\arccos(2-2x)} over its maximal domain. Then, things begin to get stupid.

In part (b), the graph of f is rotated around the y-axis, to model a vase. Students are required to find the volume of this stupid vase, by setting up the integral and then pushing the stupid buttons on their stupid calculators. So, a reasonable integration question lost in ridiculous pseudomodelling and brainless button-pushing. Whatever. Just standard VCE crap. Then, things stay stupid.

Part (c) is a related rates question. In principle a good problem, though it’s hard to imagine anyone ever requiring dh/dt when the water depth is exactly \boldsymbol{5\pi} cm. Whatever. Standard VCE crap. Then, things get really, really stupid.

Part (d) of the problem has a bee climbing from the bottom of the vase to the top. Students are required to find the minimum distance the bee needs to travel.

Where to begin with this idiotic, 1-mark question. Let’s begin with the bee.

Why is it a bee? Why frame a shortest walk question in terms of a bug with wings? Sure, the question states that the bug is climbing, and the slight chance of confusion is overshadowed by other, much greater issues with the question. But still, why would one choose a flying bug to crawl up a vase? It’s not importantly stupid, but it is gratuitously, hilariously stupid.

Anyway, we’re stuck with our stupid bee climbing up our stupid vase. What distance does our stupid bee travel? Well, obviously our stupid, non-flying bee should climb as “up” as possible, without veering left or right, correct?

No and yes.

It is true that a bottom-to-top shortest path (geodesic) on a surface of revolution is a meridian. The proof of this, however, is very far from obvious; good luck explaining it to your students. But of course this is only Specialist Mathematics, so it’s not like we should expect the students to be inquisitive or critical or questioning assumptions or anything like that.

Anyway, our stupid non-flying bee climbs “up” our stupid vase. The distance our stupid bee travels is then the arc length of the graph of the original function f, and the required distance is given by the integral

    \[\boldsymbol{{\Huge \int\limits_{\frac12}^{\frac32}}\sqrt{1+\left[\tfrac{20}{1 - (2-2x)^2}\right]^2}}\ {\bf d}\boldsymbol{x}\]

The integral is ugly. More importantly, the integral is (doubly) improper and thus has no required meaning for Specialist students. Pretty damn stupid, and a stupidity we’ve seen not too long ago. It gets stupider.

Recall that this is a 1-mark question, and it is clearly expected to have the stupid calculator do the work. Great, sort of. The calculator computes integrals that the students are not required to understand but, apart from being utterly meaningless crap, everything is fine. Except, the calculators are really stupid.

Two brands of CAS calculators appear to be standard in VCE. Brand A will readily compute the integral above. Unfortunately, Brand A calculators will also compute improper integrals that don’t exist. Which is stupid. Brand B calculators, on the other hand, will not directly compute improper integrals such as the one above; instead, one first has to de-improper the integral by changing the limits to something like 0.50001 and 1.49999. Which is ugly and stupid. It also requires students to recognise the improperness in the integral, which they are supposedly not required to understand. Which is really stupid. (The lesser known Brand C appears to be less stupid with improper integrals.)

There is a stupid way around this stupidity. The arc length can also be calculated in terms of the inverse function of f, which avoid the improperness and then all is good. All is good, that is, except for the thousands of students who happen to have a Brand B calculator and who naively failed to consider that a crappy, 1-mark button-pushing question might require them to hunt for a Specialist-valid and B-compatible approach.

The idiocy of VCE exams is truly unlimited.

VCAA Plays Dumb and Dumber

Late last year we posted on Madness in the 2017 VCE mathematics exams, on blatant errors above and beyond the exams’ predictably general clunkiness. For one (Northern Hemisphere) exam, the subsequent VCAA Report had already appeared; this Report was pretty useless in general, and specifically it was silent on the error and the surrounding mathematical crap. None of the other reports had yet appeared.

Now, finally, all the exam reports are out. God only knows why it took half a year, but at least they’re out. We have already posted on one particularly nasty piece of nitpicking nonsense, and now we can review the VCAA‘s own assessment of their five errors:

 

So, the VCAA responds to five blatant errors with five Trumpian silences. How should one describe such conduct? Unprofessional? Arrogant? Cowardly? VCAA-ish? All of the above?

 

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.