Tag: calculus

WitCH 15: Principled Objection

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June 14, 2019

11:46 am

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OK, playtime is over. This one, like the still unresolved WitCH 8, will take some work. It comes from Cambridge’s Mathematical Methods 3 & 4 (2019). It is the introduction to “When is a function differentiable?”, the final section of the chapter “Differentiation”.

Update (12/08/19)

We wrote about this nonsense seven long years ago, and we’ll presumably be writing about it seven years from now. Nonetheless, here we go.

The first thing to say is that the text is wrong. To the extent that there is a discernible method, that method is fundamentally invalid. Indeed, this is just about the first nonsense whacked out of first year uni students.

The second thing to say is that the text is worse than wrong. The discussion is clouded in gratuitous mystery, with the long-delayed discussion of “differentiability” presented as some deep concept, rather than simply as a grammatical form. If a function has a derivative then it is differentiable. That’s it.

Now to the details.

The text’s “first principles” definition of differentiability is correct and then, immediately, things go off the rails. Why is the function f(x) = |x| (which is written in idiotic Methods style) not differentiable at 0? The wording is muddy, but example 46 makes clear the argument: f’(x) = -1 for x < 0 and f’(x) = 1 for x > 0, and these derivatives don’t match. This argument is unjustified, fundamentally distinct from first principles, and it can easily lead to error. (Amusingly, the text’s earlier, “informal” discussion of f(x) = |x| is exactly what is required.)

The limit definition of the derivative f’(a) requires looking precisely at a, at the gradient [f(a+h) – f(a)]/h as h → 0. Instead, the text, with varying degrees of explicitness and correctness, considers the limit of f’(x) near a, as x → a. This second limit is fundamentally, conceptually different and it is not guaranteed to be equal.

The standard example to illustrate the issue is the function f(x) = x2sin(1/x) (for x≠ 0 and with f(0) = 0). It is easy to to check that f’(x) oscillates wildly near 0, and thus f’(x) has no limit as x → 0. Nonetheless, a first principles argument shows that f’(0) = 0.

It is true that if a function f is continuous at a, and if f’(x) has a limit L as x → a, then also f’(a) = L. With some work, this non-obvious truth (requiring the mean value theorem) can be used to clarify and to repair the text’s argument. But this does not negate the conceptual distinction between the required first principles limit and the text’s invalid replacement.

Now, to the examples.

Example 45 is just wrong, even on the text’s own ridiculous terms. If a function has a nice polynomial definition for x ≥ 0, it does not follow that one gets f’(0) for free. One cannot possibly know whether f’(x) exists without considering x on both sides of 0. As such, the “In particular” of example 46 is complete nonsense. Further, there is the sotto voce claim but no argument that (and no illustrative graph indicating) the function f is continuous; this is required for any argument along the text’s lines.

Example 46 is wrong in the fundamental wrong-limit manner described above. it is also unexplained why the magical method to obtain f’(0) in example 45 does not also work for example 46.

Example 47 has a “solution” that is wrong, once again for the wrong-limit reason, but an “explanation” that is correct. As discussed with Damo in the comments, this “vertical tangent” example would probably be better placed in a later section, but it is the best of a very bad lot.

And that’s it. We’ll be back in another seven years or so.

WitCH 10: Malfunction

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June 13, 2019

2:13 am

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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.

Signs of the Times

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May 5, 2019

4:02 pm

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Our second sabbatical post concerns, well, the reader can decide what it concerns.

Last year, diagnostic quizzes were given to a large class of first year mathematics students at a Victorian tertiary institution. The majority of these students had completed Specialist Mathematics or an equivalent. On average, these would not have been the top Specialist students, nor would they have been the weakest. The results of these quizzes were, let’s say, interesting.

It was notable, for example, that around 2/5 of these students failed to simplify the likes of 81-3/4. And, around 2/3 of the students failed to solve an inequality such as 2 + 4x ≥ x2 + 5. And, around 3/5 of the students failed to correctly evaluate \boldsymbol {\int_0^{\pi} \sin 5x \,{\rm d}x}\, or similar. There were many such notable outcomes.

Most striking for us, however, were questions concerning lists of numbers, such as those displayed above. Students were asked to write the listed numbers in ascending order. And, though a majority of the students answered correctly, about 1/4 of the students did not.

What, then, does it tell us if a quarter of post-Specialist students cannot order a list of common numbers? Is this acceptable? If not, what or whom are we to blame? Will the outcome of the current VCAA review improve things, or will it make matters worse?

Tricky, tricky questions.

WitCH 5: What a West

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November 29, 2018

10:08 am

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This one’s shooting a smelly fish in a barrel, almost a POSWW. Sometimes, however, it’s easier for a tired blogger to let the readers do the shooting. (For those interested in more substantial fish, WitCH 2, WitCH 3 and Tweel’s Mathematical Puzzle still require attention.)

Our latest WitCH comes courtesy of two nameless (but maybe not unknown) Western troublemakers. Earlier this year we got stuck into Western Australia’s 2017 Mathematics Applications exam. This year, it’s the SCSA‘s Mathematical Methods exam (not online. Update: now online here and here.) that wins the idiocy prize. The whole exam is predictably awful, but Question 15 is the real winner:

The population of mosquitos, P (in thousands), in an artificial lake in a housing estate is measured at the beginning of the year. The population after t months is given by the function, \color{blue}\boldsymbol{P(t) = t^3 + at^2 + bt + 2, 0\leqslant t \leqslant 12}.

The rate of growth of the population is initially increasing. It then slows to be momentarily stationary in mid-winter (at t = 6), then continues to increase again in the last half of the year. 

Determine the values of a and b.

Go to it.

Update

As Number 8 and Steve R hinted at and as Damo nailed, the central idiocy concerns the expression “the rate of population growth”, which means P'(t) and which then makes the problem unsolvable as written. Specifically:

  • In the second paragraph, “it” has a stationary point of inflection when t = 6, which is impossible if “it” refers to the quadratic P'(t).
  • On the other hand, if “it” refers to P(t) then solving gives a < 0. That implies P”(0) = 2a < 0, which means “the rate of population growth” (i.e. P’) is initially decreasing, contradicting the first claim of the second paragraph.

The most generous interpretation is that the examiners intended for the population P, not the rate P’, to be initially increasing. Other interpretations are less generous.

No matter the intent, the question is inexcusable. It is also worth noting that even if corrected the question is awful, a trivial inflection problem dressed up with idiotic modelling:

  • Modelling population growth with a cubic is hilarious.
  • Months is a pretty stupid unit of time.
  • The rate of population growth initially increasing is irrelevant.
  • Why is the lake artificial? Who gives a shit?
  • Why is the lake in a housing estate? Who gives a shit?

Finally, it’s “latter half” or “second half”, not “last half”. Yes, with all else awful here, it hardly matters. But it’s wrong.

Further Update

The marking schemes for the exam are now up, here and here.  As was predicted, “the rate of growth of the population” was intended to mean “population”. As is predictable, the grading scheme gives no indication that the question is garbled garbage.

The gutless contempt with which certain educational authorities repeatedly treat students and teachers is a wonder to behold.

Little Steps for Little Minds

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May 27, 2018

12:09 pm

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Here’s a quick but telling nugget of awfulness from Victoria’s 2017 VCE maths exams. Q9 of the first (non-calculator) Methods Exam is concerned with the function

    \[\boldsymbol {f(x) = \sqrt{x}(1-x)\,.}\]

In Part (b) of the question students are asked to show that “the gradient of the tangent to the graph of f” equals \boldsymbol{ \frac{1-3x}{2\sqrt{x}} } .

A normal human being would simply have asked for the derivative of f, but not much can go wrong, right? Expanding and differentiating, we have

    \[\boldsymbol {f'(x) = \frac{1}{2\sqrt{x}} - \frac32\sqrt{x}=\frac{1-3x}{2\sqrt{x}}\,.}\]

Easy, and done.

So, how is it that 65% of Methods students scored 0 on this contrived but routine 1-point question? Did they choke on “the gradient of the tangent to the graph of f” and go on to hunt for a question written in English?

The Examiners’ Report pinpoints the issue, noting that the exam question “required a step-by-step demonstration …“. And, “[w]hen answering ‘show that’ questions, students should include all steps to demonstrate exactly what was done“ (emphasis added). So the Report implies, for example, that our calculation above would have scored 0 because we didn’t explicitly include the step of obtaining a common denominator.

Jesus H. Christ.

Any suggestion that our calculation is an insufficient answer for a student in a senior maths class is pedagogical and mathematical lunacy. This is obvious, even ignoring the fact that Methods questions way too often are flawed and/or require the most fantastic of logical leaps. And, of course, the instruction that “all steps” be included is both meaningless and utterly mad, and the solution in the Examiners’ Report does nothing of the sort. (Exercise: Try to include all steps in the computation and simplification of f’.)

This is just one 1-point question, but such infantilising nonsense is endemic in Methods. The subject is saturated with pointlessly prissy language and infuriating, nano-step nitpicking, none of which bears the remotest resemblance to real mathematical thought or expression.

What is the message of such garbage? For the vast majority of students, who naively presume that an educational authority would have some expertise in education, the message is that mathematics is nothing but soulless bookkeeping, which should be avoided at all costs. For anyone who knows mathematics, however, the message is that Victorian maths education is in the clutches of a heartless and entirely clueless antimathematical institution.

The Treachery of Images

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October 25, 2017

12:36 pm

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Harry scowled at a picture of a French girl in a bikini. Fred nudged Harry, man-to-man. “Like that, Harry?” he asked.

“Like what?”

“The girl there.”

“That’s not a girl. That’s a piece of paper.”

“Looks like a girl to me.” Fred Rosewater leered.

“Then you’re easily fooled,” said Harry. It’s done with ink on a piece of paper. That girl isn’t lying there on the counter. She’s thousands of miles away, doesn’t even know we’re alive. If this was a real girl, all I’d have to do for a living would be to stay at home and cut out pictures of big fish.”

                       Kurt Vonnegut, God Bless you, Mr. Rosewater

 

It is fundamental to be able to distinguish appearance from reality. That it is very easy to confuse the two is famously illustrated by Magritte’s The Treachery of Images (La Trahison des Images):

The danger of such confusion is all the greater in mathematics. Mathematical images, graphs and the like, have intuitive appeal, but these images are mere illustrations of deep and easily muddied ideas. The danger of focussing upon the image, with the ideas relegated to the shadows, is a fundamental reason why the current emphasis on calculators and graphical software is so misguided and so insidious.

Which brings us, once again, to Mathematical Methods. Question 5 on Section Two of the second 2015 Methods exam is concerned with the function V:[0,5]\rightarrow\Bbb R, where

\phantom{\quad} V(t) = de^{\frac{t}3} + (10-d)e^{\frac{-2t}3}\,.

Here, d \in (0,10) is a constant, with d=2 initially; students are asked to find the minimum (which occurs at t = \log_e8), and to graph V. All this is par for the course: a reasonable calculus problem thoroughly trivialised by CAS calculators. Predictably, things get worse.

In part (c)(i) of the problem students are asked to find “the set of possible values of d” for which the minimum of V occurs at t=0. (Part (c)(ii) similarly, and thus boringly and pointlessly, asks for which d the minimum occurs at t=5). Arguably, the set of possible values of d is (0,10), which of course is not what was intended; the qualification “possible” is just annoying verbiage, in which the examiners excel.

So, on to considering what the students were expected to have done for (c)(ii), a 2-mark question, equating to three minutes. The Examiners’ Report pointedly remarks that “[a]dequate working must be shown for questions worth more than one mark.” What, then, constituted “adequate working” for 5(c)(i)? The Examiners’ solution consists of first setting V'(0)=0 and solving to give d=20/3, and then … well, nothing. Without further comment, the examiners magically conclude that the answer to (c)(i) is 20/3 \leqslant d< 10.

Only in the Carrollian world of Methods could the examiners’ doodles be regarded as a summary of or a signpost to any adequate solution. In truth, the examiners have offered no more than a mathematical invocation, barely relevant to the question at hand: why should V having a stationary point at t=0 for d=20/3 have any any bearing on V for other values of d? The reader is invited to attempt a proper and substantially complete solution, and to measure how long it takes. Best of luck completing it within three minutes, and feel free to indicate how you went in the comments.

It is evident that the vast majority of students couldn’t make heads or tails of the question, which says more for them than the examiners. Apparently about half the students solved V'(0)=0 and included d = 20/3 in some form in their answer, earning them one mark. Very few students got further; 4% of students received full marks on the question (and similarly on (c)(ii)).

What did the examiners actually hope for? It is pretty clear that what students were expected to do, and the most that students could conceivably do in the allotted time, was: solve V'(0)=0 (i.e. press SOLVE on the machine); then, look at the graphs (on the machine) for two or three values of d; then, simply presume that the graphs of V for all d are sufficiently predictable to “conclude” that 20/3 is the largest value of d for which the (unique) turning point of V lies in [0,5]. If it is not immediately obvious that any such approach is mathematical nonsense, the reader is invited to answer (c)(i) for the function W:[0,5]\rightarrow\Bbb R where W(t) = (6-d)t^2 + (d-2)t.

Once upon a time, Victorian Year 12 students were taught mathematics, were taught to prove things. Now, they’re taught to push buttons and to gaze admiringly at pictures of big fish.