WitCH 10: Malfunction

It’s a long, long time since we’ve had a WitCH. They have been not-so-slowly accumulating, however. And now, since we’re temporarily free of the Evil Mathologer, it is the WitCHing hour.

Due mostly to the hard work of Damo, all of the outstanding WitCHes have been resolved, with the exception of WitCH 8. That one will take time: it’s a jungle of half-maths. Our new WitCHes are not so tricky, although there is perhaps more to be said than indicated at first glance.

The first of our new batch of WitCHes is from the VCE 2018 Specialist Exam 1:

The Examiners’ Report gives the answer as \int_0^{\frac34}\left(2-t^2\right)dt. The Report also indicates that the average score on this question was 1.3/5, with 98% of students scoring 3 or lower, and over a third of students scoring 0.

Happy WitCHing.

48 Replies to “WitCH 10: Malfunction”

  1. I feel like this doesn’t get to the heart of the matter, but I’m not convinced that the distance travelled by the particle in “three quarters of a second” is given by an integral with lower and upper limits of, respectively, 0 and 0.75. Couldn’t it be any integral with lower and upper limits of x and x + 0.75?

    1. I felt the same thing, but was willing to throw VCAA a bone and assume that out of all the possible 3/4 second time intervals, they were specifying the one from t = 0 to t = 3/4.

    2. Thanks, SRK. Of course you are correct. Yeah, it’s not the main point and the required answer makes clear the time interval. But the wording is simply wrong, and needlessly wrong.

      And, true, one might consider, as John does, to throw the VCAA a bone. But think about it. This is lazy “Who gives a shit?” wording from the same nasty obsessives who self-righteously nitpick students to death. So screw it. Throw the bone at them.

      1. Indeed. It would have been no skin off their chin to have added two simple words and said “… in *the first* three-quarters of a second….”

        1. Exactly. I can completely understand a harried teacher writing such a sloppy expression. And at times there is a tricky balancing act to be made between accuracy and clarity. But not here. Here, it is just laziness. Or idiocy. Or both.

  2. Was thinking a bit more about this one, because this topic is coming up shortly. Again, I feel like I am not seeing the main problem Marty sees with this one, but:

    I think it is unnecessary and, for reasons given below, somewhat harsh (perhaps even malicious) to write this as as an arc-length question. Given what was required to achieve full marks for this question, I think it would have been better to ask students to find an expression for the speed of the particle. Reasons:

    (1) students weren’t expected to evaluate the definite integral. (Indeed, they were required to write their final answer as the definite integral). Maybe there’s a point here about assessing whether students know the arc-length formula, but it’s given on the formula sheet so that seems pointless.

    (2) A savvy student would realise that writing each line of working with a definite integral is asking for trouble – very easy to forget to include a terminal or a “dt”. So they would just work out the expression for the speed, and in the final line write down the answer as a definite integral in the required form. I wonder if this is what VCAA had in mind…

    Edit: actually, re-reading the question, I realised the definite integral is completely pointless, because students were only required to find the values of a, b, c.

    1. Thanks, SRK. It’s been so long since I could get back to this, I kind of forgot myself! You don’t quite raise the point I have in mind, but I agree entirely: asking for the speed would have been a much more natural question. Although, as John indicates below, the whole framing in terms of kinematics is not natural for the given function.

  3. I also don’t feel like I’ve seen the main problem that Marty has with this question. As such, I’ve gone looking and I think that I may have something, but maybe I’m overthinking it…

    The question is phrased as a Kinematics question, which means that we can find the velocity and acceleration vectors. These are:

    v(t)= and a(t)=

    Both of these are undefined when t=1, which is within the stated domain of the function. In particular, the magnitude of the acceleration goes to neg infinity as t goes to 1. Which, when you consider the force that would need to be applied for this humble particle to follow this particular path, well it’s pretty heroic.

    1. Yes, using a kinematics context for this question is pretty dumb, because then things have to make physical sense. Which they don’t.

      I think the velocity is OK: v = t^2 i + 2Sqrt[1 – t^2] j and there’s no trouble here as t –> 1.

      It’s the acceleration where things go South. It makes no physical sense (to me, anyway) that at t = 1 the object is at r = i/3 + pi/2 j, |v| = 1 m/s and yet |a| –> oo m/s^2.

      To use a kinematics context here simply to, I assume, also test whether students understood that distance travelled is length along a curve is just stupid and clumsy. The curve should just have been defined parametrically and the length between two points asked for: “A curve is defined by ….. Show that the length of the curve between …. is given by ….” The fact that the square root of a perfect square had to be simplified by noting the domain is a nice touch, but ruined by everything else.

      I find the constant desire by VCAA to contextualise their questions with irrelevant, contrived, artificial and unphysical scenarios simply to make the mathematics look ‘relevant’ (I suppose) hilarious – better to do nothing and be thought a bonehead than do something and remove all doubt.

      1. Yep, the kinematics context just makes the whole question a little absurd. The other thing that contributes to the absurdity is the manner in which they have defined the domain to be 0<=t<=1, but then asked for the length of the curve from t=0 to t=3/4. I am assuming that they were seeking to avoid an improper integral, but it adds an unnecessary and confusing element to the question.

        A quick edit of my above post – the magnitude of the acceleration goes to infinity, the j component of the acceleration goes to neg infinity. Also, I don't know what happened to my equations for velocity and acceleration, I'm mostly confident that I typed them in.

        1. The integral for length isn’t improper when calculated over the entire domain, but given past experience I don’t think checking this possibility would even occur to anyone at VCAA.

          Since the resulting integral is readily calculated, I’m surprised a value wasn’t required.

          Wouldn’t it be great if there was a ‘Meet The Exam Writing Panel’ each year ….

          1. A genuine question here: because you substitute the derivative into the arc length integral, and because the derivative is undefined when t=1, does this not make the integral improper? Even if it can be simplified, through cancelling, to a form that could be integrated?

        2. Hi Damo, quickly on your missing equations. I don’t know whether this is the issue, but I noticed that wordpress doesn’t seem to like any html code in comments: it will reject the whole comment. However, once the comment is up, I can edit the comment and insert the html code. That’s what I do to get superscripts and links etc in comments. You and everyone also now have the ability to edit your own comments (for thirty minutes after posting).

  4. Damo, I overlooked the subtlety that dy/dt is indeterminant at t = 1. So if integrating to t = 1, technically a limit *would* have to be set up first, so I guess the integral to t = 1 *is* improper. You’re right. I haven’t come across that before. Very clever. So I doubt that’s what stopped VCAA from asking.

    1. Yes, I doubt it as well, particularly given past history. But if not that, why? I second the idea of a Meet the Exam Writing Panel…

  5. Thanks to Damo and John, and to SRK to whom I replied above, for the extended discussion, and sorry to be so slow to get back to comment.

    All your criticisms are (eventually) spot on, and the question in total is half-baked weirdness in the ways you all indicate. But, interestingly, no one has picked up on the issue that first caught my eye. To me, the issue is glaring and the exam question is simply and ridiculously wrong. Perhaps teachers, however, are used to this type of thing, and would regard the issue as a nitpick, and would simply ignore it. (Versions of this error/nitpick appear in other exam questions but, to me, the issue is particularly glaring in the question above.)

    Quickly on the improperness. Yes, at t = 1 the derivatives of arcsin(t) and √(1-t2) are undefined, which implies the velocity is undefined, which implies the speed is undefined. This necessarily makes the distance integral improper at t = 1. The improperness is “removable”, in the sense that the speed function can be extended to be continuous at t = 1. (A simpler example exhibiting the same behaviour would be integrating the function x/|x| on [0,1].) But, the improperness is there, and that is undoubtedly why the examiners wisely, and with astonishing ineptness, avoided integrating over the whole interval.

      1. Hi, SRK. I haven’t had a close look at the NHT exams. I don’t see what you’re seeing on the NHT question (and am curious), but it’s not what I’m thinking of for the question above.

        1. As mentioned above, was really just fishing. I was wondering if it had something to do with either (i) saying “distances are measured in metres” (as opposed to displacements), or (ii) not explicitly mentioning that i, j are perpendicular unit vectors.

  6. I am now at the guessing stage: is it that the distance that the particle travels is actually given by the arc length integral, which simplifies to their integral?

  7. This is a nitpick , but I don’t think it’s what Marty has in mind: The use of brackets around the integrand irks me. Not so long ago VCAA said that brackets around integrands were NOT required, that the integral symbol and the ‘dx’ (or whatever) acted as defacto brackets. And yet here’s VCAA using brackets.

  8. Just a quick follow-up. Given that commenters are not seeing what I see as a central flaw in the exam question, I asked five people – four mathematicians and one (very smart) non-mathematician – to take a *one-minute* look at the exam question, and to tell me what they saw. All four mathematicians noted the flaw (plus some of the other nonsense), and the non-mathematician did not.

    This is turning out to be a very interesting WitCH.

    1. OK, so I’m going to think outside the teacher-box with all its implied assumptions and get technical:

      For the position of a particle to be specified on a curve, the particle is either point-like or the position vector specifies the location of the centre of mass, say, of the non-point-like particle. None of which is said in the question.

      I really hope this or something like this is *NOT* the perceived flaw (because there are so many ‘better’ flaws already mentioned), but I’m assuming there’s some technicality that us non-mathematicians have automatically overlooked as a matter of implied assumption. Given how much ink mathematicians have used to prove 1+1=2, I’m guessing it might be something along these lines (I just assume 1+1 = 2 and hope for the best).

      1. Hi JF, I understand the frustration non-mathematicians have with logical formalism and pedantry, but it’s definitely not that kind of flaw/nitpick.

        1. I don’y have a problem with logical formalism and pedantry – I totally understand their importance. It’s just not my cup of tea unless I’m forced to drink it (I got through 3 years of real analysis and have worked through Rudin but my teeth were gritted all the way).

  9. Is it that dy/dx = 2*sqrt(1-t^2)/t^2, which means that it is undefined (unbounded?) when t=0, which means that the integral is improper (despite what appears to be an attempt to avoid that very thing)?

    1. Nice try, Damo. But dy/dx is not needed for calculating the distance, only dx/dt and dy/dt (which are both defined over the interval of integration). dy/dx simply gives information about the gradient of the path and says that the gradient is undefined at (0, 0). This does not constitute a technical infelicity requiring methods beyond the scope of the course.

      (My money’s still on the boot studs being 1mm longer than the rules allow).

      1. Well, commenters here can eventually decide whether they think the flaw is analogous to boot stud length. But I can assure you that mathematicians think it more akin to Ted Whitten using a flick pass.

  10. Probably my final suggestion:

    Since distances are given as measured in metres, specifying that d is in metres is redundant. In fact, the introduction of d itself is redundant. Taking this and many earlier comments into account, the question could be better worded as:

    “The position vector ….. ” The distance travelled by the particle in the first three-quarters of a second is given by int….. where a, b, c e Z. Find the values of a, b and c.

    Or even better:

    A curve is defined parametrically by x = … and y = … where 0 <= t <= 3/4. Find the length of the curve.

    Mathematicians value elegance, rigour, beauty …. So I can well imagine mathematicians immediately considering this question ugly. Teachers on the other hand see this ugliness year after year after year. It's viewed with such jaded eyes that the ugliness no longer even consciously registers. Just like living next to a train line, there comes a time when you just don't hear the passing trains anymore …..

    So here's my final suggestion:

    The main flaw is the that the question is coyote ugly (and, unfortunately, you can't chew your arm off to get away from VCAA )

    1. JF, you’re right again: the “metres” is redundant. And, you’re right, that, overall, the question is appallingly ugly (and clunky). Do teachers become blind to such ugliness? I don’t know. Commenters on this blog seem to pick up the ugliness quickly.

      1. Speaking for myself, I’m not so much blind as just weary and resigned …. I might have a fleeting thought about it and then shrug my shoulders, let it pass as par for the course and forget about it. When you have muppets who refuse point-blank to acknowledge the existence of serious mathematical errors such as functions given as probability density functions that are no such thing (eg. Maths Methods 2016 Q3(h)), what hope do you have when it’s just simple ugliness because they’re a bunch of tryhards ….? It just goes without saying.

        My theory is that most commenters on your blog generally pick up ugliness quickly because they’re explicitly being invited to comment on witCH. But in this case, I think explicitly acknowledging ugliness is over-shadowed by the desire to find mathematical errors (a desire no doubt motivated by the ugliness). I know that’s the case with myself.

        1. Yeah, I guess. I was tutoring a Methods student today, going through a practice SAC. The SAC contained a question that was so poorly worded that it was effectively impossible (and the teacher’s suggested solution was wrong), and another question that was blatantly wrong. But beyond that, the whole fucking SAC was meaningless, klutzy, pointless, ugly as sin pseudo-modelling. Just complete and utter trash from beginning to end. So, which was worse: the two isolated errors, or the endemic awfulness?

          And I’d suggest the same with the exam question above. The question is clumsy, and clumsily worded, to the point of awfulness. But, as well, the question contains a (for mathematicians) blatant error. Which is worse? Which should teachers be more alert to? Tough call.

          1. That’s what happens when the instructions you’re following mandate klutzy, pointless, ugly as sin pseudo-modelling. And you have to write a different three every year under those ridiculous guidelines.

            Errors are worse than ugliness. Then again, if it’s ugly enough it will be unintelligible, in which case it doesn’t matter whether there are errors or not.

            And the problem is that SACs rarely get vetted competently, if at all. There’s simply no time, everyone assumes that the other guy(s) vetted it …. The teacher is simply the (time)poor sap who had to write it.

  11. Does the error you refer to exist in a re-worded version?:

    A curve is defined parametrically by x = … and y = … where 0 <= t <= 3/4. Find the length of the curve.

  12. OK, I haven’t responded to this post for a while because I feel that others in the discussion are far superior in their Mathematical knowledge, but I will offer just one suggestion: is there a problem with how “distance” or “length” is interpreted? A simple definition of distance would be the Euclidean distance from the start and finish points, done using a simple right-angled triangle.

    If not… any hints?

    1. Hi RF, no that’s not it. There is an interesting question there, when you generalise from straight-line distance to path-length, but that’s not an issue here.

      As for a hint, maybe later this week. I’m flat out preparing a talk. After that, if no one has gotten it by then.

  13. So, using the arc length formula from the VCAA formula sheet, the positive root is implied. But what then happens when (2-t^2)^2 and (t^2-2)^2 both have the same square? Is it possible that some students, factorised 4-4t^2+t^4 as (t^2-2)^2 and then when taking the square root wrote a=1, b=0, c=-2?

    Yes t^2-2 is negative for the domain given, but students have been caught by VCAA on this before.

    Back to the wine…

    1. Thanks, RF. There’s no choice over the square root, since we’re after a distance. But it’s natural to factorise as you indicate, forget the domain of t, and miss the answer by a minus sign. Intended or otherwise, it’s a trick in the question, and I imagine it’s why so few students scored more than 3/5.

    2. Well, Specialist Maths students should certainly know that Sqrt[a^2] = |a|, particularly since the modulus function is on the course (for more reasons I assume than just to solve integrals of the form int[1/u, u]).

      Sure, it’s a ‘trick’ in the question, but only if the falsehood Sqrt[a^2] = a has been ‘encouraged’ throughout the year.

      1. JF, by labelling it a trick I wasn’t meaning to suggest it was unfair, just that many students would have been tricked.

        1. Unfortunately, I think the reason many of them would have been tricked is that they might not have met Sqrt[a^2] = |a|. This is something that has to be explicitly taught and then reinforced (within broader examples) throughout the year and therefore requires a certain level of experience on the part of the teacher. And as we know, there is a steady exodus of experience ….

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