WitCH 36: Sub Standard

This WitCH is a companion to our previous, MitPY post, and is a little different from most of our WitCHes. Typically in a WitCH the sin is unarguable, and it is only the egregiousness of the sin that is up for debate. In this case, however, there is room for disagreement, along with some blatant sinning. It comes, predictably, from Cambridge’s Specialist Mathematics 3 & 4 (2020).

MitPY 6: Integration by Substitution

From frequent commenter, SRK:

A question for commenters: how to explain / teach integration by substitution? To organise discussion, consider the simple case

    \[\boldsymbol{ \int \frac{2x}{1+x^2}dx \,.}\]

Here are some options.

1) Let u = 1 + x^2. This gives \frac{du}{dx} = 2x, hence dx = \frac{du}{2x}. So our integral becomes \int \frac{2x}{u}\times \frac{du}{2x} = \int \frac{1}{u}du. Benefits: the abuse of notation here helps students get their integral in the correct form. Worry: I am uncomfortable with this because students generally just look at this and think “ok, so dy/dx is a fraction cancel top and bottom hey ho away we go”. I’m also unclear on whether, or the extent to which, I should penalise students for using this method in their work.

2)  Let u = 1 + x^2. This gives \frac{du}{dx} = 2x. So our integral becomes \int \frac{1}{u}\times \frac{du}{dx}dx = \int \frac{1}{u}du. Benefit. This last equality can be justified using chain rule. Worry: students find it more difficult to get their integral in the correct form.

3) \frac{2x}{1+x^2} has the form f'(g(x))g'(x) where g(x)=1+x^2 and f'(x) = \frac{1}{x}. Hence, the antiderivative is f(g(x)) = \log (1+x^2). This is just the antidifferentiation version of chain rule.  Benefit. I find this method crystal clear, and – at least conceptually – so do the students. Worry. Students often aren’t able to recognise the correct structure of the functions to make this work.

So I’m curious how other commenters approach this, what they’ve found has been effective / successful, and what other pros / cons there are with various methods.

UPDATE (21/04)

Following on from David’s comment below, and at the risk of splitting the discussion in two, we’ve posted a companion WitCH.

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.

PoSWW 11: Pinpoint Inaccuracy

This one comes courtesy of Christian, an occasional commenter and professional nitpicker (for which we are very grateful). It is a question from a 2016 Abitur (final year) exam for the German state of Hesse. (We know little of how the Abitur system works, and how this question may fit in. In particular, it is not clear whether the question above is a statewide exam question, or whether it is more localised.)

Christian has translated the question as follows:

A specialty store conducts an ad campaign for a particular smartphone. The daily sales numbers are approximately described by the function g with \color{blue}\boldsymbol{g(t) = 30\cdot t\cdot e^{-0.1t}}, where t denotes the time in days counted from the beginning of the campaign, and g(t) is the number of sold smartphones per day. Compute the point in time when the most smartphones (per day) are sold, and determine the approximate number of sold devices on that day.

WitCH 29: Bad Roots

This one is double-barrelled. A strange multiple choice question appeared in the 2019 NHT Mathematical Methods Exam 2 (CAS). We had thought to let it pass, but a similar question appeared in last’s weeks Methods exam (no link yet, but the Study Design is here). So, here we go.

First, the NHT question:

The examination report indicates the correct answer, C, and provides a suggested solution:

\Large\color{blue} \boldsymbol{ g(x)=f^{-1}(x)=\frac{x^{\frac15}-b}{a},\ g'(x) = \frac{x^{-\frac45}}{5a},\ g'(1) = \frac1{5a}}

And, here’s last week’s question (with no examination report yet available):

WitCH 24: The Fix is In

We’ve finally found some time to take a look at VCAA’s 2019 NHT exams. They’re generally bad in the predictable ways, and they include some specific and seemingly now standard weirdness that we’ll try to address soon in a more systematic manner. WitCHwise, we were tempted by a number of questions, but we’ve decided to keep it to two or three.

Our first NHT WitCH is from the final question on Exam 2 (CAS) of Mathematical Methods:

As usual, the NHT “Report” indicates nothing of how students went, and little of what was expected. In regard to part f, the Report writes,

p(x) = q(x) = x, p'(x) = q'(x) = 1, k = 1/e

For part g, all that the Report provides is the answer, k = 1.

The VCAA also provides sample Mathematica solutions to schools trialling Methods CBE. For the questions above, these solutions are as follows:

Make of it what you will.

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.