MAV’s Dangerous Inflection

This post concerns a question on the 2019 VCE Specialist Mathematics Exam 2 and, in particular, the solution and commentary for that question available through the Mathematical Association of Victoria. As we document below, a significant part of what MAV has written on this question is confused, self-contradictory and tendentious. Thus, noting the semi-official status of MAV solutions, that these solutions play a significant role in MAV’s Meet the Assessors events, and are quite possibly written by VCE assessors, there are some troubling implications.

Question 3, Section B on Exam 2 is a differential equations problem, with two independent parts. Part (a) is a routine (and pretty nice) question on exponential growth and decay.* Part (b), which is our concern, considers the differential equation

    \[\boldsymbol{\color{blue}\frac{{\rm d}Q}{{\rm d}t\ } = e^{t-Q}}\,,\]

for t ≥ 0, along with the initial condition

    \[\boldsymbol{\color{blue}Q(0) =1}\,.\]

The differential equation is separable, and parts (i) and (ii) of the question, worth a total of 3 marks, asks to set up the separation and use this to show the solution of the initial value problem is

    \[\boldsymbol{\color{blue}Q =\log_e\hspace{-1pt} \left(e^t + e -1\right)}\,.\]

Part (iii), worth 2 marks, then asks to show that “the graph of Q as a function of t” has no inflection points.**

Question 3(b) is contrived and bitsy and hand-holding, but not incoherent or wrong. So, pretty good by VCE standards. Unfortunately, the MAV solution and commentary to this problem is deeply problematic.

The first MAV misstep, in (i), is to invert the derivative, giving

    \[\boldsymbol{\color{red}\frac{{\rm d}t\ }{{\rm d}Q } = e^{Q-t}}\,,\]

prior to separating variables. This is a very weird extra step to include since, not only is the step not required here, it is never required or helpful in solving separable equations. Its appearance here suggests a weak understanding of this standard technique. Worse is to come in (iii). Before considering MAV’s solution, however, it is perhaps worth indicating an approach to (iii) that may be unfamiliar to many teachers and students and, possibly, the assessors.

If we are interested in the inflection points of Q,*** then we are interested in the second derivative of Q. The thing to note is we can naturally obtain an expression for Q” directly from the differential equation: we differentiate the equation using the chain rule, giving

    \[\boldsymbol{\color{magenta}Q'' = e^{t-Q}\left(1 - Q'\right)}\,.\]

Now, the exponential is never zero, and so if we can show Q’ < 1 then we’d have Q” > 0, ruling out inflection points. Such conclusions can sometimes be read off easily from the differential equation, but it does not seem to be the case here. However, an easy differentiation of the expression for Q derived in part (ii) gives

    \[\boldsymbol{\color{magenta}Q' =\frac{e^t}{e^t + e -1}}\,.\]

The numerator is clearly smaller than the denominator, proving that Q’ < 1, and we’re done.

For a similar but distinct proof, one can use the differential equation to replace the Q’ in the expression for Q”, giving

    \[\boldsymbol{\color{magenta}Q'' = e^{t-Q}\left(1 - e^{t-Q}\right)}\,.\]

Again we want to show the second factor is positive, which amounts to showing Q > t. But that is easy to see from the expression for Q above (because the stuff in the log is greater than \boldsymbol{e^t}), and again we can conclude that Q has no inflection points.

One might reasonably consider the details in the above proofs to be overly subtle for many or most VCE students. Nonetheless the approaches are natural, are typically more efficient (and are CAS-free), and any comprehensive solutions to the problem should at least mention the possibility.

The MAV solutions make no mention of any such approach, simply making a CAS-driven beeline for Q” as an explicit function of t. Here are the contents of the MAV solution:

Part 1: A restatement of the equation for Q from part (ii), which is then followed by 

.˙.  \boldsymbol{ \color{red}\  \frac{{\rm d}^2Q }{{\rm d}t^2\ } = \frac{e^{t+1} -e^t}{\left(e^t + e -1\right)^2} } 

Part 2: A screenshot of the CAS input-output used to obtain the conclusion of Part 1.

Part 3: The statement   

Solving  .˙.  \boldsymbol{\color{red} \  \frac{{\rm d}^2Q }{{\rm d}t^2\ } = 0} gives no solution  

Part 4: A screenshot of the CAS input-output used to obtain the conclusion of Part 3.

Part 5: The half-sentence

We can see that \boldsymbol{\color{red}\frac{{\rm d}^2Q }{{\rm d}t^2\ } > 0} for all t,

Part 6: A labelled screenshot of a CAS-produced graph of Q”.

Part 7: The second half of the sentence,

so Q(t) has no points of inflection

This is a mess. The ordering of the information is poor and unexplained, making the unpunctuated sentences and part-sentences extremely difficult to read. Part 3 is so clumsy it’s funny. Much more important, the MAV “solution” makes little or no mathematical sense and is utterly useless as a guide to what the VCE might consider acceptable on an exam. True, the MAV solution is followed by a commentary specifically on the acceptability question. As we shall see, however, this commentary makes things worse. But before considering that commentary, let’s itemise the obvious questions raised by the MAV solution:

  • Is using CAS to calculate a second derivative on a “show that” exam question acceptable for VCE purposes?
  • Can a stated use of CAS to “show” there are no solutions to Q” = 0 suffice for VCE purposes? If not, what is the purpose of Parts 3 and 4 of the MAV solutions?
  • Does copying a CAS-produced graph of Q” suffice to “show” that Q” > 0 for VCE purposes?
  • If the answers to the above three questions differ, why do they differ?

Yes, of course these questions are primarily for the VCAA, but first things first.

The MAV solution is followed by what is intended to be a clarifying comment:

Note that any reference to CAS producing ‘no solution’ to the second derivative equalling zero would NOT qualify for a mark in this ‘show that’ question. This is not sufficient. A sketch would also be required as would stating \boldsymbol{\color{red}e^t (e - 1) \neq 0} for all t.

These definitive-sounding statements are confusing and interesting, not least for their simple existence. Do these statements purport to be bankable pronouncements of VCAA assessors? If not, what is their status? In any case, given that pretty much every exam question demands that students and teachers read inscrutable VCAA tea leaves, why is it solely the solution to question 3(b) that is followed by such statements?

The MAV commentary at least makes clear their answer to our second question above: quoting CAS is not sufficient to “show” that Q” = 0 has no solutions.  Unfortunately, the commentary raises more questions than it answers:

  • Parts 3 and 4 are “not sufficient”, but are they worth anything? If so, what are they worth and, in particular, what is the import of the word “also”? If not, then why not simply declare the parts irrelevant, in which case why include those parts in the solutions at all?
  • If, as claimed, it is “required” to state \boldsymbol{e^t(e-1)\neq 0} (which is indeed the key point of this approach and should be required), then why does the MAV solution not contain any such statement, nor even the factorisation that would naturally precede this statement?
  • Why is a solution “required” to include a sketch of Q”? If, in particular, a statement such as \boldsymbol{e^t(e-1)\neq 0} is “required”, or in any case is included, why would the latter not in and of itself suffice?

We wouldn’t begin to suggest answers to these questions, or our four earlier questions, and they are also not the main point here. The main point is that under no circumstances should such shoddy material be the basis of VCAA assessor presentations. If the material was also written by VCAA assessors, all the worse.

Of course the underlying problem is not the quality or accuracy of solutions but, rather, the fundamental idiocy of incorporating CAS into proof questions. And for that the central villain is not the MAV but the VCAA, which has permitted their glorification of technology to completely destroy the appreciation of and the teaching of proof and reason.

The MAV is not primarily responsible for this nonsense. The MAV is, however, responsible for publishing it, promoting it and profiting from it, none of which should be considered acceptable. The MAV needs to put serious thought into its unhealthily close relationship with the VCAA.

 

*) We might ask, however, who refers to “The growth and decay” of an exponential function?

**) One might simply have referred to Q, but VCAA loves them their words.

***) Or, if preferred, the points of inflection of the graph of Q as a function of t.

WitCH 35: Overly Resolute

This WitCH (arguably a PoSWW) comes courtesy of Damien, an occasional commenter and an ex-student of ours from the nineteenth century. It is from the 2019 Specialist Mathematics Exam 2. We’ll confess, we completely overlooked the issue when going through the MAV solutions.

Update (16/02/20)

What a mess. Thanks to Damo for pointing out the problem, and thanks to the commenters for figuring out the nonsense.

In general form, the (intended) scenario of the exam question is

The vector resolute of \boldsymbol{\tilde{a}} in the direction of \boldsymbol{\tilde{b}} is \boldsymbol{\tilde{c}},

which can be pictured as follows: For the exam question, we have \boldsymbol{\tilde{a}} = \boldsymbol{\tilde{i}} + \boldsymbol{\tilde{j}} - \boldsymbol{\tilde{k}}, \boldsymbol{\tilde{b}} = m\boldsymbol{\tilde{i}} + n\boldsymbol{\tilde{j}} + p\boldsymbol{\tilde{k}} and \boldsymbol{\tilde{c}} = 2\boldsymbol{\tilde{i}} - 3\boldsymbol{\tilde{j}} + \boldsymbol{\tilde{k}}.

Of course, given \boldsymbol{\tilde{a}} and \boldsymbol{\tilde{b}} it is standard to find \boldsymbol{\tilde{c}}. After a bit of trig and unit vectors, we have (in must useful form)

\boldsymbol{\tilde{c} = \left(\dfrac{\tilde{a}\cdot \tilde{b}}{\tilde{b}\cdot \tilde{b}}\right)\tilde{b}}

The exam question, however, is different: the question is, given \boldsymbol{\tilde{a}} and \boldsymbol{\tilde{c}}, how to find \boldsymbol{\tilde{b}}.

The problem with that is, unless the vectors \boldsymbol{\tilde{a}} and \boldsymbol{\tilde{c}} are appropriately related, the scenario simply cannot occur, meaning \boldsymbol{\tilde{b}} cannot exist. Most obviously, the length of \boldsymbol{\tilde{c}} must be no greater than the length of \boldsymbol{\tilde{a}}. This requirement is clear from the triangle pictured, and can also be proved algebraically (with the dot product formula or the Cauchy-Schwarz inequality).

This implies, of course, that the exam question is ridiculous: for the vectors in the exam we have |\boldsymbol{\tilde{c}}| > |\boldsymbol{\tilde{a}}|, and that’s the end of that. In fact, the situation is more delicate; given the pictured vectors form a right-angled triangle, we require that \boldsymbol{\tilde{a}} - \boldsymbol{\tilde{c}} be perpendicular to \boldsymbol{\tilde{c}}. Which implies, once again, that the exam question is ridiculous.

Next, suppose we lucked out and began with \boldsymbol{\tilde{a}}- \boldsymbol{\tilde{c}} perpendicular to \boldsymbol{\tilde{c}}. (Of course it is very easy to check whether we’ve lucked out.) How, then, do we find \boldsymbol{\tilde{b}}? The answer is, as is made clear by the picture, “Well, duh”. The possible vectors \boldsymbol{\tilde{b}} are simply the (non-zero) scalar multiples of \boldsymbol{\tilde{c}}, and we’re done. Which shows that the mess in the intended solution, Answer A, is ridiculous.

There is a final question, however: the exam question is clearly ridiculous, but is the question also stuffed? The equations in answer A come from the equation for \boldsymbol{\tilde{c}} above and working backwards. And, these equations correctly return no solutions. Moreover, if the relationship between \boldsymbol{\tilde{a}} and \boldsymbol{\tilde{c}} had been such that there were solutions, then the A equations would have found them. So, completely ridiculous but still ok?

Nope.

The question is framed from start to end around definite, existing objects: we have THE vector resolute, resulting in THE values of m, n and p. If the VCAA had worded the question to find possible values, on the basis of a possible direction for the resolution, then, at least technically, the question would be consistent, with A a valid answer. Still an utterly ridiculous question, but consistent. But the VCAA didn’t do that and so the question isn’t that. The question is stuffed.

Implicit Suggestions

One of the unexpected and rewarding aspects of having started this blog is being contacted out of the blue by students. This included an extended correspondence with one particular VCE student, whom we have never met and of whom we know very little, other than that this year they undertook UMEP mathematics (Melbourne University extension). The student emailed again recently, about the final question on this year’s (calculator-free) Specialist Mathematics Exam 1 (not online). Though perhaps not (but also perhaps yes) a WitCH, the exam question (below), and the student’s comments (belower), seemed worth sharing.

Hi Marty,

Have a peek at Question 10 of Specialist 2019 Exam 1 when you get a chance. It was a 5 mark question, only roughly 2 of which actually assessed relevant Specialist knowledge – the rest was mechanical manipulation of ugly fractions and surds. Whilst I happened to get the right answer, I know of talented others who didn’t.

I saw a comment you made on the blog regarding timing sometime recently, and I couldn’t agree more. I made more stupid mistakes than I would’ve liked on the Specialist exam 2, being under pressure to race against the clock. It seems honestly pathetic to me that VCAA can only seem to differentiate students by time. (Especially when giving 2 1/2 hours for science subjects, with no reason why they can’t do the same for Maths.) It truly seems a pathetic way to assess or distinguish between proper mathematical talent and button-pushing speed writing.

I definitely appreciate the UMEP exams. We have 3 hrs and no CAS! That, coupled with the assignments that expect justification and insight, certainly makes me appreciate maths significantly more than from VCE. My only regret on that note was that I couldn’t do two UMEP subjects 🙂

 

WitCH 23: Speed Bump

Our second WitCH of the day also comes from the 2017 VCE Specialist Mathematics Exam 2. (Clearly an impressive exam, and we haven’t even gotten to the bit about using inverse trig functions to design a brooch.) It is courtesy of the mysterious SRK, who raised it in the discussion of an earlier WitCH.

Question 5 of Section B of the (CAS) exam concerns a boat and a jet ski. Though SRK was concerned with one particular aspect, the entire question is worth pondering:

The  Examiner’s Report indicates an average student score of 1.4 on part a, and comments,

Students plotted the initial positions correctly but significant numbers of students did not label the direction of motion or clearly identify the jet ski and the boat. Both requirements were explicitly stated in the question.

For part i, the Report indicates an average score of 1.3, and comments,

Most students found correct expressions for velocity vectors. The most common error was to equate these velocity vectors rather than equating speeds. 

For part ii, the Report gives the intended answer as (3,3). The Report indicates that slightly under half of students were awarded the mark, and comments,

Some answers were not given in coordinate form.

For part i, the Report suggests the answer {\sqrt{(\sin t - 2\cos t)^2 + (1 + \sin t + \cos t)^2}} (with the displayed answer adorned by a weird, extra root sign). The report indicates that a little over half of the students were awarded the mark, and comments,

A variety of correct forms was given by students; many of these were likely produced by CAS technology, including expressions involving double angles. Students should take care when transcribing expressions from technology output as errors frequently occur, particularly regarding the number and placement of brackets. Some incorrect answers retained vectors in the expression.

For Part ii, the Report indicates the intended answer of 0.33, and that 15% of students were awarded the mark for this question. The Report comments,

Many students found this question difficult. Incorrect answers involving other locally minimum values were frequent.

The Report indicates an average score of 1.3 on part d, and comments;

Most students correctly equated the vector components and solved for t . Many went on to give decimal approximations rather than supplying the exact forms. Students are reminded of the instruction saying that an exact answer is required unless otherwise specified.

Lots there. Get hunting.

WitCH 22: Inflecting the Facts

We’re back, at least sort of. Apologies for the long silence; we were off visiting The Capitalist Centre of the Universe. And yes, China was great fun, thanks. Things are still tight, but there will soon be plenty of time for writing, once we’re free of those little monsters we have to teach. (Hi, Guys!) In the meantime, we’ll try to catch up on the numerous posts and updates that are most demanding of attention.

We’ll begin with a couple new WitChes. This first one, courtesy of John the Merciless, is a multiple choice question from the 2017 VCE Specialist Mathematics Exam 2:

The Examiners’ Report indicates that 6% of students gave the intended answer of E, and a little under half the students answered C. The Report also comments that

f”(x) does not change sign at a.

Have fun.

WitCH 20: Tattletail

This one is like complaining about the deck chairs on the Titanic, but what the Hell. The WitCH is courtesy of John the Merciless. It is from the 2018 Specialist Mathematics Exam 2:

The Examiners’ Report notes the intended answer:

H0: μ = 150,   H1: μ < 150

The Report indicates that 70% of students gave the intended answer, and the Report comments on students’ answers:

The question was answered well. Common errors included: poor notation such as  H0 = 150 or similar, and not understanding the nature of a one-tailed test, evidenced by answers such as H1: μ ≠ 150.

Have fun.

WitCH 17: Compounding Our Problems

The WitCHfest is coming to an end. Our final WitCH is, once again, from Cambridge’s Specialist Mathematics 3 & 4 (2019). The section establishes the compound angle formulas, the first proof of which is our WitCH.

Update (25/08/19)

Similar to our parallel WitCH, it is difficult to know whether to focus on specific clunkiness or intrinsic absurdity, but we’ll first get the clunkiness out of the way:

  • John comments that using x and y for angles within the unit circle is irksome. It is more accurately described as idiotic.
  • The 2π*k is unnecessary and distracting, since the only possible values of k are 0 and -1. Moreover, by symmetry it is sufficient  to prove the identity for x > y, and so one can simply assume that x = y + α.
  • The spacing for the arguments of cos and sin are very strange, making the vector equations difficult to read.
  • The angle θ is confusing, and is not incorporated in the proof in any meaningful manner.
  • Having two cases is ugly and confusing and was easily avoidable by an(other) appeal to trig symmetry.

In summary, the proof could have been much more elegant and readable if the writers had bothered to make the effort, and in particular by making the initial assumption that yxy + π, relegating other cases to trig symmetry.

Now, to the general absurdity.

It is difficult for a textbook writer (or a teacher) to know what to do about mathematical proofs. Given that the VCAA doesn’t give a shit about proof, the natural temptation is to pay lip service or less to mathematical rigour. Why include a proof that almost no one will read? Commenters on this blog are better placed to answer that question, but our opinion is that there is still a place for such proofs in school texts, even if only for the very few students who will appreciate them.

The marginalisation of proof, however, means that a writer (or teacher) must have a compelling reason for including a proof, and for the manner in which that proof is presented. (This is also true in universities where, all too often, slovenly lecturers present  incomprehensible crap as if it is deep truth.) Which brings us to the above proof. Specialist 34 students should have already seen a proof of the compound angle formulas in Specialist 12, and there are much nicer proofs than that above (see below). So, what is the purpose of the above proof?

As RF notes, the writers are evidently trying to demonstrate the power of the students’ new toy, the dot product. It is a poor choice, however, and the writers in any case have made a mess of the demonstration. Whatever elegance the dot product might have offered has been obliterated by the ham-fisted approach. Cambridge’s proof can do nothing but convince students that “proof” is an incomprehensible and pointless ritual. As such, the inclusion of the proof is worse than having included no proof at all.

This is doubly shameful, since there is no shortage of very nice proofs of the compound angle formulas. Indeed, the proof in Cambridge’s Specialist 12 text, though not that pretty, is standard and is to be preferred. But the Wikipedia proof is much more elegant. And here’s a lovely proof of the formula for sin(A + B) from Roger Nelson’s Proof Without Words:

To make the proof work, just note that

x cos(A) = z = y cos(B)

Now write the area of the big triangle in two different ways, and you’re done. A truly memorable proof. That is, a proof with a purpose.

WitCH 16: The Root of Our Problem

This WitCH comes from one of our favourites, the Complex Numbers chapter from Cambridge’s Specialist Mathematics 3 & 4 (2019). It is not as deep or as beWitCHing as other aspects of the chapter. But, it’s still an impressive WitCH.

Update (11/08/19)

I guess if you’re gonna suggest a painful, ass-backwards method to solve a problem, you may as well fake the solution:

  • Checking directly that P(1 – i√2) = 0 involves expanding a cubic, and more, which the text does in one single magic line.
  • The painful multiplication of the products for part b is much more naturally and easily done as a difference of two squares: (z – 1 – i√2)(z – 1 + i√2) = (z – 1)2 + 2, etc.
  • After all that the third factor, z – 1, is determined “by inspection”? Inspection of what?

AS RF notes, it is much easier to spot that z = 1 solves the cubic. Then some easy factoring (without long division …) gives P = (z – 1)(z2 – 2z + 3). Completing the square then leads to the linear factors, answering both parts of the question in the reverse, and natural, order.

Alternatively, as John notes, the difference of two squares calculation shows that if z – 1 + i√2 is a factor of P then so is the quadratic z^2 – 2z + 3. That this is so can then be checked (without long division …), giving P = (z – 1)(z^2 – 2z + 3), and so on, as before.