arc length – BAD MATHEMATICS

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.

Update (16/02/20)

This problem is ridiculous and, more importantly, it is wrong. First, the wrongness.

As indicated by the examination report, the examiners imagined that they were, in essence, asking for students to determine the speed function $\boldsymbol{v(t)}$ of the particle. The distance is given by $\boldsymbol{d=\int\limits_0^{3/4} v(t)\,{\rm d}t}$ , and a non-trivial calculation gives $\boldsymbol{v(t) = 2 +0t - t^2}$ . Then, the coefficients $\boldsymbol{a = -1, b = 0, c = 2}$ can be read off.

That is not, however, the question the examiners asked. What did the examiners really ask? They asked for integers $\boldsymbol{a,b,c}$ for which $\boldsymbol{d=\int\limits_0^{3/4} at^2+bt+c\,{\rm d}t}$ . But

$\boldsymbol{d=\int\limits_0^{3/4} 2 - t^2\,{\rm d}t = \dfrac{87}{64}\,.}$

and

$\boldsymbol{\int\limits_0^{3/4} at^2+bt+c\,{\rm d}t = \frac{9}{64}a+\frac{9}{32}b+\frac{3}{4}c \,.}$

So, multiplying out the fractions and cancelling out a 3, what the examiners really asked for were integer solutions to the equation

$\boldsymbol{3a+ 6b+ 16c = 29\,.}$

This equation has infinitely many integer solutions, meaning the examination report is missing infinity minus one valid solutions.

This is a flat out, undeniable error (which the Trumpian VCAA will never concede), but is it a problem? As commenters here have noted, there is little chance of a VCE student being actively misled to chase the infinitely many solutions. In, particular, the method to find all solutions requires first finding the particular solution the examiners had in mind. We are not convinced such direct concerns should be so quickly dismissed, and we discuss this further below. Still, the extra solutions require thought to even contemplate, and significant work to compute, which is an important point.

Whatever the immediate practical concerns, however, mathematicians are aghast at this error. They are aghast because the exam question is simply not testing mathematics. Yes, the students went through the ritual and attempted to compute what was intended and were graded accordingly. And, yes, teachers can now coach current and future students on the required ritual. But none of that is mathematics and, indeed, it is worse: it is antimathematics. It is teaching students to ignore mathematical meaning, to see no value in mathematical precision, to respect only ritual.

OK, that is the awful wrongness of the exam question. Now, the sundry ridiculousnesses:

The question is badly and needlessly opaque. There is no a priori reason to imagine the distance as being given by the integral of a quadratic. Asking for (more accurately, attempting to ask for) the speed function in this overly cute manner adds no value, only confusion. The confusion is enhanced by the arbitrariness of the 3/4 limit and, especially, by the pointless specification that the coefficients of the quadratic be integers.
Independent of the opacity, the wording of the question is lazy and clumsy. The distanced travelled “in three-quarters of a second” is not the same as the distance travelled in the first three-quarters of a second and, indeed, is not anything. The phrases “moving along a curve” and “travels along a curve” are just verbiage. The units are pointless.
The question would be much more natural as an arc-length question, rather than a distance question.
The answer in the examination report is incorrect, even in the intended terms. The question asked for the values of the coefficients, not the integral. Yes, this is a nitpick, but it is exactly the kind of nitpick that the examiners routinely employ in their sanctimonious whacking of VCE students. So screw ’em. Sauce for the gander.
Last, and far from least, there is something very strange about the score distribution for the question. The average score was 1.3/5, which is depressing, although not surprising: computing the speed (without CAS) requires a level of care and facility beyond most CAS-drunk students, and the question contains a hidden absolute value to negotiate. What is strange is that, whereas 2% of students received the full 5/5 for the question, apparently 0% of students received 4/5. It is difficult to see how that could occur with any sensible grading scheme.

Tag: arc length

WitCH 10: Malfunction

Update (16/02/20)