Tag: calculus

WitCH 47: A Bad Inflection

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November 29, 2020

10:08 am

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The question below is from the first 2020 Specialist exam (not online). It has been discussed in the comments here, and the main issues have been noted, but we’ve decided the question is sufficiently flawed to warrant its own post.

UPDATE (10/09/21) For those who’d placed a wager, the examination report (Word-doc-VCAA-stupid) indicates that a second derivative argument was expected. Hence, thousands of VCE students no longer have any sense of what VCAA means by “hence”.

WitCH 44: Estimated Worth

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October 7, 2020

2:27 am

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This WitCH is from Cambridge’s 2020 textbook, Mathematical Methods, Unit 1 & 2. It is the closing summary of Chapter 21A, Estimating the area under a graph. (It is followed by 21B, Finding the exact area: the definite integral.)

We’re somewhat reluctant about this one, since it’s not as bad as some other WitCHes. Indeed, it is a conscious attempt to do good; it just doesn’t succeed. It came up in a tutorial, and it was sufficiently irritating there that we felt we had no choice.

WitCH 38: A Deep Hole

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June 20, 2020

3:06 pm

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This one is due to commenter P.N., who raised it on another post, and the glaring issue has been discussed there. Still, for the record it should be WitCHed, and we’ve also decided to expand the WitCHiness slightly (and could have expanded it further).

The following questions appeared on 2019 Specialist Mathematics NHT, Exam 2 (CAS). The questions are followed by sample Mathematica solutions (screenshot corrected, to include final comment) provided by VCAA (presumably in the main for VCE students doing the Mathematica version of Methods). The examination report provides answers, identical to those in the Mathematica solutions, but indicates nothing further.

UPDATE (05/07/20)

The obvious problem here, of course, is that the answer for Part (b), in both the examination report and VCAA’s Mathematica solutions, is flat out wrong: the function fk will also fail to have a stationary point if k = -2 or k = 0. Nearly as bad, and plenty bad, the method in VCAA’s Mathematica solutions to Part (c) is fundamentally incomplete: for a (twice-differentiable) function f to have an inflection point at some a, it is necessary but not sufficient to have f’’(a) = 0.

That’s all pretty awful, but we believe there is worse here. The question is, how did the VCAA get it wrong? Errors can always occur, but why specifically did the error in Part (b) occur, and why, for a year and counting, wasn’t it caught? Why was a half-method suggested for Part (c), and why was this half-method presumably considered reasonable strategy for the exam? Partly, the explanation can go down to this being a question from NHT, about which, as far as we can tell, no one really gives a stuff. This VCAA screw-up, however, points to a deeper, systemic and much more important issue.

The first thing to note is that Mathematica got it wrong: the Solve function did not return the solution to the equation fk‘ = 0. What does that imply for using Mathematica and other CAS software? It implies the user should be aware that the machine is not necessarily doing what the user might reasonably think it is doing. Which is a very, very stupid property of a black box: if Solve doesn’t mean “solve”, then what the hell does it mean? Now, as it happens, Mathematica’s/VCAA’s screw-up could have been avoided by using the function Reduce instead of Solve.* That would have saved VCAA’s solutions from being wrong, but not from being garbage.

Ask yourself, what is missing from VCAA’s solutions? Yes, yes, correct answers, but what else? This is it: there are no functions. There are no equations. There is nothing, nothing at all but an unreliable black box. Here we have a question about the derivatives of a function, but nowhere are those derivatives computed, displayed or contemplated in even the smallest sense.

For the NHT problem above, the massive elephant not in the room is an expression for the derivative function:

    \[\color{red} \boldsymbol{f'_k(x) = -\frac{x^2 + 2(k+1)x +1}{(x^2-1)^2}}\]

What do you see? Yep, if your algebraic sense hasn’t been totally destroyed by CAS, you see immediately that the values k = 0 and k = -2 are special, and that special behaviour is likely to occur. You’re aware of the function, alert to its properties, and you’re led back to the simplification of fk for these special values. Then, either way or both, you are much, much less likely to screw up in the way the VCAA did.

And that always happens. A mathematician always gets a sense of solutions not just from the solution values, but also from the structure of the equations being solved. And all of this is invisible, is impossible, all of it is obliterated by VCAA’s nuclear weapon approach.

And that is insane. To expect, to effectively demand that students “solve” equations without ever seeing those equations, without an iota of concern for what the equations look like, what the equations might tell us, is mathematical and pedagogical insanity.

 

*) Thanks to our ex-student and friend and colleague Sai for explaining some of Mathematica’s subtleties. Readers will be learning more about Sai in the very near future.

WitCH 36: Sub Standard

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April 21, 2020

8:00 pm

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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

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April 17, 2020

9:26 pm

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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

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February 24, 2020

11:02 am

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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.

Update (26/06/20)

The Examination Report is out and is basically ok; none of the nonsense and non sequiturs of the MAV solutions are included. The solution to (b)(iii) correctly focuses upon the factoring of Q”, although it needlessly worries about the sign of the denominator. There is no mention of the more natural approach to obtaining and analysing Q” but, given the question is treated by the VCAA and pretty much everyone as just another mindless exercise in pushing buttons, this is no surprise.

PoSWW 11: Pinpoint Inaccuracy

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February 7, 2020

2:19 pm

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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 31: Decomposing

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November 19, 2019

2:28 pm

8 Comments on WitCH 31: Decomposing

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We have a short Specialist post coming, and we’ll have more to write on the 2019 VCE exams once they’re online. But, for now, one more Mathematical Methods WitCH, from the 2019 (calculator-free) Exam 1:

Update (04/07/20)

The main crap here, of course, is part (f): as commenter John Friend puts it, what the hell is this question supposed to be testing? And, sure, the last part of the last question on an exam is allowed to be a little special, but one measly mark? Compared to the triviality of the rest of the question?

Of course, students bombed part (f). The examination report indicates that 19% of student correctly answered that there is one solution to the equation; as suggested by commenter Red Five, it’s also a pretty safe bet that the majority of students who got there did so with a Hail Mary guess. (It should be added, the students didn’t do swimmingly well on the rest of Question 9, the CAS-lobotomising having working its usual magic.)

OK, so what did examiners expect for that one measly mark? We’ll get to a reasonable solution below, but let’s first consider some unreasonable solutions.

Here is the examination report’s entire commentary on Part (f):

g(f(x) + f(g(x)) = 0 has exactly one solution.

This question was not well done. Few students attempted to draw a rough sketch of each equation and use addition of ordinates.

Gee, thanks. Drawing a “rough sketch” of either of these compositions is anything but trivial. For one measly mark. We’ll look at sketching aspects of these graphs below, but let’s get on with another unreasonable solution.

Given the weirdness of part (f), a student might hope that parts (a)-(e) provide some guidance. Let’s see.

Part (b) (for which the examination report contains an error), gets us to conclude that the composition

\boldsymbol{g(f(x)) = e^{\left(3+2x-x^2\right)}} has negative derivative when x > 1.

Part (c) leads us to the composition

\boldsymbol{f(g(x)) = \left(3 - e^x\right) \left(1 + e^x\right)}

having x-intercept when x = log(3).

Finally, Part (e) gives us that the composition f(g(x)) has the sole stationary point (0,4). How does this information help us with Part (f)? Bugger all.

So, what if we include the natural implications of our previous work? That gives us something like the following: Well, um, great. We’re left still hunting for that one measly mark.

OK, the other parts of the question are of little help, and the examiners are of no help, so what do else do we need? There are two further pieces of information we require (plus the Intermediate Value Theorem). First, note that

\boldsymbol{g(f(x)) = e^{\mbox{\bf THING}} > 0}.

Secondly, note that

\boldsymbol{f(g(x)) = \left(3 + e^x\right) \left(1 - e^x\right)} = -}\mbox{\bf HUGE} if x is huge.

Then, given we know the slopes of the compositions, we can finally complete our rough sketches: Now, let’s write S(x) for our sum function g(f(x)) + f(g(x)). We know S(x) > 0 unless one of our compositions is negative. So, the only place we could get S = 0 is if x > log(3). But S(log(3)) > 0, and eventually S is hugely negative. That means S must cross the x-axis (by IVT). But, since S is decreasing for x > 1, S can only cross the axis once, and S = 0 must have exactly one solution. 

We’ve finally earned our one measly mark. Yay?

WitCH 29: Bad Roots

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November 13, 2019

1:36 pm

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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 week’s 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):

Update (19/06/20)

As commenters have noted, it is very difficult to understand any purpose to these questions. They obviously suggest the inverse function theorem, testing the knowledge of and application of the formula g'(d) = 1/f'(c), where f(c) =d. The trouble is, the inverse function theorem is not part of the curriculum, appearing only implicitly as a dodgy version of the chain rule, and is typically only applied in Leibniz form.

As indicated by the solution in the first examination report, the intent seems to have been for students to have explicitly computed the inverses, although probably with their idiot machines. (The second examination report has now appeared, but is silent on the intended method.) Moreover, as JF noted below, the algebra in the first question makes the IFT approach somewhat fiddly. But, what is the point of pushing a method that is generally cumbersome, and often impossible, to apply?

To add to the nonsense, below is a sample solution for the first question, provided by VCAA to students undertaking the Mathematica version of Methods. So, the VCAA has suggested two approaches, one which is generally ridiculous and another which is outside the curriculum. That makes it all as clear as dumb mud.