WitCH 79: Cos Why Not?

Yeah, we’re busy, but this one has been bugging us. It is an exercise and solution from Cambridge Specialist Mathematics 3&4. It’s basically a PoSWW, or we could simply let it pass as a textbook glitch. But we think there’s more than the obvious to say.

53 Replies to “WitCH 79: Cos Why Not?”

  1. Hmmmmmmmm….. Seems the writer got a bit too trigger happy and tried differentiating and showing at the same time, only to get confused with both. \frac{d}{dx} \left( \arccos\left(\frac{6}{x}\right)\right) = - \frac{6}{x\sqrt{x^2-36}} for x 6, it’s given by the thing above, which probably somehow functions as their show f'(x) > 0 for x>6.

  2. If a student wrote this on a VCAA exam, the examiners’ report would do their usual, “for show that questions suitable working needs to be shown”

    Beyond that though, it doesn’t help that the question fails to indicate whether a student is expected to do this by hand (which they would probably assume) or by calculator (which is what the solution writer most likely did).

    1. RF, I don’t think the writer used a CAS. This would need to be checked for other machines and software, but Mathematica gives \displaystyle \frac{6}{x^2 \sqrt{1 - \frac{36}{x^2}}} and refuses to simplify further.
      So a ‘by hand’ intervention is required to get the \displaystyle \frac{|x|}{x^2} factor and simplify this to \displaystyle \frac{|x|}{x^2} = \frac{|x|}{|x|^2} = \frac{1}{|x|}.

      Re: “If a student wrote this on a VCAA exam, the examiners’ report would do their usual, “for show that questions suitable working needs to be shown””.
      I assume you’re referring to the fact that the (wrong) derivative is found but no attempt is made (not even a “By inspection …” statement) to show that it is greater than zero for \displaystyle x > 6.

  3. This is an easy one, Marty. I’ll bet money it’s not a textbook glitch/typo etc. Cambridge simply forgot that \displaystyle \sqrt{x^2} = |x| NOT \displaystyle x. I particularly enjoy using it as an example with students.

    Cambridge also doesn’t know how to check the reasonableness of an answer. Because what’s really funny is that the correct graph in part (c) makes it clear that \dislaystyle f'(x) > 0 for \displaystyle x < -6 and yet Cambridge's derivative in part (b) gives \dislaystyle f'(x) < 0.

    I advised Cambridge of this error a long time ago – I believe that it's been fixed in the on-line copy of the textbook. It's a shame Cambridge doesn't have a link to an errata page (like many publishers do for many of their textbooks) so that the poor saps with a hard copy who can't access the on-line copy (another story told elsewhere) are not disadvantaged.

    And now for a Friendly political message:

    It's a magnificent error, one that VCAA would go into denial over. VCAA would gibber about mark attraction statistics borne out after consideration regarding the efficacy of an examination as both a metric of student competence and a method of cohort discrimination. VCAA would be content that the examination process achieved those goals in a manner that saw no student disadvantaged and achieved its intended purpose to be fit for function. In plain English (something VCAA is incapable of), errors don't matter if an out-sourced statistical analysis shows that the population as a whole wasn't disadvantaged (completely misunderstanding and misusing the statistical analysis by claiming it applies to an individual rather than a population). I've come to the conclusion that VCAA is not deceptive and does not lie – it’s simply as dumb as a rock and honestly believes its own propaganda. VCAA is a flat-Earther.

    1. Ugh. John, the VCAA bashing is gratuitous and boring. Save it for posts dedicated to VCAA bashing. It’s not like there’s a shortage of them.

      As for your non-gratuitous comment, thanks, although I think there’s quite a bit more to say. Yes, the obvious explanation is that Cambridge simply forgot the |x|, especially since they include an example that specifically deals with taking an x2 out of the square root. And, yes, the fact that their graph in (c) directly contradicts their derivative in (b) is pretty funny, and pretty sloppy. But there’s more. Also, the error may have been corrected on some online platforms, but the (possibly old) platform to which I have access still exhibits the error.

  4. By the way, an additional and petty comment is that, despite the instruction to “label endpoints and asymptotes” in part (c), the answer does not do this.

    Sorry, can’t resist: This is where a VCAA Examination Report would gleefully pontificate that “asymptotes must be labelled with an equation” and “endpoints must be labelled with coordinates”. (VCAA expects students (and teachers) to interpret the intention of its exam question and overlook its inherent errors, but VCAA refuses to give the same ‘courtesy’ to students. VCAA prefers hypocritical paternalistic pontification to common sense.)

  5. Others may disagree – but I still don’t see how the official answer satisfies the requirement to “show that…”

    1. RF, there can be no disagreement. It doesn’t. There’s not even an off-hand “By inspection it is seen that \displaystyle f'(x) > 0 when \displaystyle x > 6” comment.

      What would have been interesting is if the “show that” part of part (b) was “.. and show that \displaystyle f'(x) > 0 for \displaystyle x < -6. Or even more delicious, ".. and show that \displaystyle f'(x) < 0 for \displaystyle x < -6 …!!

      What's really disappointing to me – from an instructional point of view – is that:

      1) the answer to part (c) didn't remark on how the answer in part (b) (and part (a)) was useful in determining the shape of the graph (OK, useful only if the answer in part (b) is correct!). To me, this is an important teaching point of the example. Particularly since students need to learn how to draw the graphs of 'strange' and 'unknown' functions they've never met before … It always amazes me when students are given something like this to draw and they splutter "But we've never been taught how to draw the graph of this type of function!!"

      2) There is no attempt made to explain the existence of the – and we are left to assume this – horizontal asymptote on the graph in part (c). Now, maybe this is what Marty has in mind when he says there’s more:
      To calculate \displaystyle \lim_{x \rightarrow \infty} \cos^{-1} \left( \frac{6}{x}\right) requires using ‘the limit of a composite function theorem’:
      Consider \displaystyle \lim_{x \rightarrow a} f(g(x).

      Suppose \displaystyle \lim_{x \rightarrow a} g(x) = L and that \displaystyle f(x) is defined on an interval containing \displaystyle x = L and is continuous at \displaystyle x = L.

      Then \displaystyle \lim_{x \rightarrow a} f(g(x)) = f \left( \lim_{x \rightarrow a} g(x) \right) = f(L).

      So for the example under discussion we have

      \displaystyle \lim_{x \rightarrow \infty} \cos^{-1} \left( \frac{6}{x}\right) =  \cos^{-1} \left( \lim_{x \rightarrow \infty}\frac{6}{x} \right) = \cos^{-1}(0) = \frac{\pi}{2}

      (and the interested reader can check that all the conditions for using the theorem are met).

      But such a theorem and its use is outside of the Specialist Maths syllabus (although I suspect it’s blithely used without realising it in many classrooms). So the example either deliberately or unwittingly refuses to show ‘what’s behind the asymptote curtain’. And on the topic of shit, we could – but won’t – discuss the role of the CAS and other acroynyms in reinforcing this culture.

      1. Thanks, John. That was one of the extra things that bugged me, the semi-link between (b) and (c). Which raises the question: if your interest is really in drawing the graph of the function in (c), what is the correct question, if any, to ask as part (b)?

        I’m not so fussed about the composite limit thing, although I guess typically Specialist only has limits of algebraic functions? So, maybe I should be fussed. But, there are two other aspects that bug me and that I don’t think anybody has mentioned.

        1. I think the intent of the question was to calculate a slightly more challenging derivative and provide an interesting motivation for doing so. Part (a) was a filler and part (c) was showing off. However …

          IF the intent was the grand finale of drawing the graph, then part (a) and part (b) provide correct but not sufficient scaffolding questions. Part (b) should have asked to show that the derivative was always positive over its domain, rather than just focussing on x > 6. It should also have asked students to consider \displaystyle \lim_{x \rightarrow 6^+} f'(x) and \displaystyle \lim_{x \rightarrow -6^-} f'(x). Cambridge should have considered this too. Because then it wouldn’t have butchered the shape of the graph at the endpoints (the gradient should be approaching a vertical line to suggest a ‘vertex’ at the endpoints). Marty, I’m guessing this is something that irked you? (It irked me too).

          I would also have included a part that explicitly asked for the calculation of the horizontal asymptote, except that such a calculation ‘by hand’ is beyond the scope of the course. Such a part would need to give the ‘limit of a composite function’ theorem and then ask students to show that the necessary conditions were met for \displaystyle y = \cos^{-1} \left( \frac{6}{x}\right) before then getting them to use the theorem.

          Re: “Typically Specialist only has limits of algebraic functions?”
          Yes, at least as far as horizontal and oblique asymptotes are concerned. Which hasn’t stopped Cambridge from out-smarting itself on at least one other occasion: https://mathematicalcrap.com/2019/02/13/witch-8-oblique-reasoning/

          Re: “But, there are two other aspects that bug me and that I don’t think anybody has mentioned.”
          I’d like to leave some “aspects” for others, but I’ve already mentioned one of them (I think) above and now I’m going to mention another:
          the “… maximal domain of \displaystyle f.” when there is no definition of \displaystyle f. \displaystyle f(x) = \cos^{-1} \left( \frac{6}{x}\right) is not the same thing as the – presumably – \displaystyle function \displaystyle f (a point well made here: https://mathematicalcrap.com/2021/11/11/witch-76-cos-and-effect/ )

          1. Thanks, John. I think I’m making a slightly different point in Witch 76, but the rest I agree with, and it mostly covered my thoughts.

            As a lead-in to (c), the text’s (b) is simply the wrong (b). But also, why bother? As 6/x goes down towards zero (for x > 0) then cos-1(6/x) goes up towards π/2. So what else is there to note? The only other thing that the derivative tells you is that the gradient is going infinite at ±6. Which, as you note, the solution graph fails to indicate.

            I think there’s one final thing to note. Suppose, for whatever reason, you want to calculate f’ and then show f’ > 0. What’s the easiest way to do that?

            1. I have some thoughts but Friend and (No)Friend have unintentionally hogged this blog so I’ll leave your question as an opportunity for others.

  6. I remember this error, and others like this. The answers are unreliable and the questions in general are dodgy, so there’s not much point using the Cambridge’s exercises, especially because the exam questions are so vastly different. Really the only thing they’re useful for in my opinion is to learn how to apply basic rules and formulas, which I suppose is enough merit to use them, but I guess any other textbook would have similar exercises.

    1. Hi Anon. I disagree.

      Unreliable answers don’t invalidate the questions and don’t make the questions dodgy. I disagree that the the exam questions are vastly different (particularly the extended response questions of the sort you get on Exam 2 Section B). Out of what’s available on the market, I personally think Cambridge has the best questions in terms of coverage, richness and challenge. Many of the questions are very reminiscent of the questions from the classic Fitzpatrick and Galbraith textbooks. The other textbooks on the market don’t have a similar set of exercises (in fact, I’ve find most questions in competitor textbooks to be bland, repetitive and generally uninteresting).

      I have no vested interest in Cambridge. It has many faults. But I think it’s the best textbook currently on the market for VCE mathematics. (I think Marty once damned it with the faint praise of being a third rate textbook among fifth rate textbooks).

      1. Hello, I didn’t mean that the questions are dodgy because the answers are dodgy, I meant some questions being dodgy as a separate issue. I probably should have made that clearer.

        I don’t really think that the exam 2 section B questions are that similar to any of the questions found in the Cambridge textbook in my opinion. The exam questions are much less straight-forward, less methodical, and more reliant on CAS tricks in my opinion.

        I’ve never used another textbook so I can’t comment on them, but if I were to teach a class (and had the choice) I’d personally use exercises from worksheets and trial exams, especially if the Cambridge is the best out of the bunch.

  7. One of the issues JF and J(N)F raise which perhaps deserves more attention…

    Derivative of an inverse function: did this not come up here in a discussion about a VCAA paper?

    If not, my bad, lets (as Marty asks) leave the VCAA bashing for another day, maybe November, or won’t the NHT papers be published by then…?

    If so, I will admit not fully comprehending the awfulness of the question to a highschool student until properly thinking it through. Maybe the same curse has befallen the Cambridge author here…?

    1. Hi RF.
      I don’t think the question is awful (not withstanding the fact that getting the horizontal asymptote is outside the scope of the syllabus and so a hand-waving approach is required). Personally, I like the question. It’s the answers that are awful – incorrect on at least three counts:
      incorrect derivative,
      incorrect shape of graph at endpoints, and
      failure to show f'(x) > 0 for x > 6 (apparently it’s self-evident and requires no explanation).

      There is a solutions manual available (legitimately and also illegitimately – I have students each year who seem to get their hands on it). It would be interesting to know what it says about this question, as opposed to what the answers in the back of the book say.

      Re: “Derivative of an inverse function”.
      Discussed here: https://mathematicalcrap.com/2019/11/13/witch-29-bad-roots/
      But this was a very different type of “derivative of an inverse function” to the question under discussion here, and so not relevant (as far as I can see).

    2. Thanks, RF. I agree with John, the question is not awful, except that (b) is neither Arthur nor Martha; it’s the solution that is awful, and wrong.

      Yes, there are at least two general issues with inverse derivatives in VCE, but as John notes these are not relevant here.

  8. The graph is really bad. So (b) and (c) are wrong.

    I don’t like the “find f'(x)” part of (b). Come on, let’s get students to think that the piecewise monotonicity of \cos can be used to quickly argue that f'(x) has a sign. (If f'(x) is to be defined at all, by properly defining f, see below…)

    But I kind of think the whole question is borked. Really all of these kinds of questions suffer from this same broken-ness. Is \cos^{-1} defined canonically in the curriculum? Because, you know, there are many \cos^{-1}. For instance, \cos^{-1}(0) = \frac\pi2. Or maybe \cos^{-1}(0) = -\frac\pi2. Or \frac{5\pi}{2}. Who says I even need to make a choice of just one branch. Maybe I want to bounce around. This reminds me of the universal failure of treatments of complex arithmetic.

    In other words, I might draw the answer to (c) in a number of different ways. Maybe jumping around. Arbitrarily many discontinuities. Then it doesn’t make sense to differentiate it at those points where it jumps branches. The question doesn’t stop you from putting those points of discontinuity almost everywhere though, so you could say that actually *my* \cos^{-1} is nowhere differentiable.

    1. Thanks, Glen. I agree very much with the point that calculating f’ is pointless work for drawing the graph (except for the endpoint behaviour). I made this point above, in reply to John. You’re off base with your other criticism, at least for VCE. Indeed, cos-1 is defined “canonically” in VCE, and is not ambiguous.

      1. Oh well! They aren’t canonically defined in the greater world of math. Perhaps this is one advantage of VCE-math? *dodge*

    2. Hi Glen.

      As I commented earlier, I think the question as a whole was intended to be a vehicle for calculating a more ‘interesting’ inverse trig derivative. I think part (a), the “show” of part (b) and part (c) were intended to add ‘meat’ around the calculation.

      So I don’t think the derivative was requested simply to facilitate drawing the graph, but given that it is calculated and available, I think it should have been better exploited.

      So, although I agree with Marty that “that calculating f’ is pointless work for drawing the graph (except for the endpoint behaviour)”, I don’t think that was the intention. By the way, I like Marty’s qualifier, because if the correct endpoint behaviour is important, then calculating f’ is anything but pointless in my opinion.

      And yes, all of the inverse trig functions are indeed defined canonically in the syllabus.

      1. Thanks, John. Again, I’ll point to one (for me) final aspect, which occurred to me pretty much immediately, which I think is important, and which I don’t think anybody has noted.

        Suppose you want to calculate f’ and show from the form of f’ that f’ > 0. What is the easiest way to do that?

        1. OK, I’ll risk the sucker punch since no-one else has said anything …

          Given that “… you want to calculate f’ and show from the form of f’ that f’ > 0.”, \displaystyle I'd say the easiest way to do it is to note that:

          1) the numerator is positive, and

          2) the denominator is positive since \displaystyle \sqrt{x^2 - 36} > 0 and \displaystyle |x| > 0 (on the domain of f’)

          therefore f’ > 0 (over its entire domain, not just for x > 6).

            1. One of my better results over the journey!
              If you post \displaystyle your answer, I can give you a grade!

              By the way, every single student these days demands individual feedback on how to improve. It’s become an entitlement (despite getting assessments back with copious written feedback). So … I’m guessing my feedback is on the way …? (And be kinder than I generally am when I get some bonehead whose done no work come up and ask how they can improve).

                1. Oh ho ho! …. That’s a bold move, Marty! You’ve led a sheltered life and have a lot to learn! You’ve \displaystyle shown nothing, you’ve left it totally to the imagination (and the good will and intelligence) of the assessor.

                  F (don’t blame me, I’m just the assessor’s stooge).

                    1. Indeed. But for extra credit and my upgrade to an A:

                      The point (I think) is that it’s really dumb to ask to show something that’s so obvious.

                      (Exam questions set by an unspoken acronym have a nasty habit of doing this from time to time).

              1. JF, you have spoken the thoughts of Methods (and possibly all other Mathematics subjects) teachers everywhere…

                Teacher writes regular feedback, gives to student, assumes parent takes interest and/or reads said feedback.

                Parent teacher night: Q1: so how can he/she do better?

                Getting back to the question… I actually DO think it is a bit of an awful question because it is in a VCE Mathematics text. As a question, yeah, I can see the merits. As a VCE question… not so much.

                And Glen… the implied domain of arccos(x) is actually given these days on the VCAA formula sheet for Specialist, so we teachers assume the textbooks use the same definitions.

                1. Ha ha. Indeed. In fact, I’d broaden the net to include ALL teachers, I think.

                  Serious question, RF – What makes it a bit of an awful question for VCE Specialist Maths in your opinion (excluding my comments about finding the horizontal asymptote)?

                  1. If I was given this question by a colleague (and I do exchange the odd question with a colleague for interest/fun) then I would say it is a nice question to think about.

                    Specialist examiners (up until very recently, I will admit) seemed to be a bit more predictable than those of Methods and so questions such as this one I would not expect to see in a VCAA exam. Since the point of doing textbook exercises is to maximise one’s exam score, I think the question not great. What exactly is it testing that can’t be tested in other ways and with less of the issues raised by others about the solution?

                    That is not to say trig and calculus cannot mix well in a clever manner. I still remember in Year 12 being asked for the integral of arctan(x) (yes, my school exams used *that* notation for those playing “guess who the commenter is…”)

                    1. Aha … Back in the good old days of Pure Mathematics where integration by parts was part of its syllabus (rather than the stupid ‘integration by recognition’ that’s been forced on us for many decades now by an unnamed acronym).

                      (And \displaystyle that notation is back in favour in secondary schools – which is a good thing in my view. Currently the syllabus specifies both notations – arcwhatever(x) and whatever\displaystyle^{-1}(x) will be used interchangeably).

                      RF, I have some rebuttals:

                      Re: “questions such as this one I would not expect to see in a VCAA [Specialist Maths] exam.”

                      I would expect. I think it’s more likely than not. I’ve seen similar questions in commercial trial exams. And I know writers for the unnamed acronym look at those exams. And I’ve seen questions from those trial exams appear you-know-where, virtually word-for-word.

                      Re: “Since the point of doing textbook exercises is to maximise one’s exam score, I think the question not great.”

                      I would disagree with this. \displaystyle One of the reasons – a good and very reasonable one – for doing the textbook exercises is to “maximise one’s exam score”. But there are other – some might even say better – reasons.

                      Re: “What exactly is it testing that can’t be tested in other ways and with less of the issues raised by others about the solution?”

                      It seems like the question is being condemned for the many faults in how it’s been answered. These faults don’t make the question bad, they make the solution to the question bad. I could – and would if enticed with a suitable inducement – write solutions that would make the question look like a masterpiece. The question makes nice and natural connections to a variety of skills and knowledge, and challenges the student to identify and apply those skills and knowledge. It’s interesting and covers a lot of ground. It forces students to think beyond the routine.

                2. The domain is clear but the range is the thing that can be quite arbitrary. (Or equivalently the domain of cosine.) Anyway, my almost-complete ignorance of math in Victoria was on full display for everyone to enjoy. That’s alright, I expected it to be undefined and was wrong. (I guess it is also defined in the book?)

                  1. Trust me, Glen. Your expectation was \displaystyle very reasonable. (And no Victorian VCE maths teacher takes joy in that).

                    (And yes, it is defined in the textbook).

        2. I’ll bite at this. I answer your question by saying that the product of two negative things is positive (the product coming from the chain rule). I demonstrate it by writing it down.

          1. Pretty much. The point is the chain rule immediately gives you everything you want. There is no need to “simplify” the root, and this only makes things worse.

            1. I agree that there is no need to simplify the root in order to show f'(x) > 0.
              BUT there IS a significant simplification and the resulting answer for f'(x) is much more elegant. (The unsimplified answer is an ugly brute). Should we not encourage students to seek this answer for f'(x)? (I want my A). Simplifying makes things worse …?!?! Sacre bleu! Qu’est-ce que c’est!?

              And in encouraging students to give the simplified form, how to do so without giving away the |x| …?

              1. It’s worse than that “there’s no need to simplify the root”. Simplifying the root – if that is really what is being done here – makes the next part of the question tricky rather than trivial.

                Yes, students should be encouraged to obtain simple and elegant solutions, as a goal in itself. But this goal doesn’t trump all other goals. And including a |x| is not automatically simple or elegant.

                1. I agree. Is x^2 or |x| more simple? I’d say x^2. But, I’d expect students to “simplify” as a reflex action (unfortunately).

                  1. I think you mean is \displaystyle \sqrt{x^2} or \displaystyle |x| more simple. I say the latter. As well as clearing the ‘double-decker’ fractions.

                    1. I know the expressions are different, but the square root isn’t going anywhere, so it seems to me at least the more relevant question is x^2 vs |x|.

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