The Treachery of Images

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.

4 Replies to “The Treachery of Images”

Number 8 says:

October 25, 2017 at 7:00 pm

Oh how sad and yet so true.

One need only choose a random integer in the domain [2006, 2017*] and one would find an example of such CRAP.

*2017 including the Northern Hemisphere VCE which produces its own set of problems on a variety of levels.

The 4% of success does not (one assumes) allow for the lucky guess based on the fact that the question asks for value(s) to which some students have learned to give a set rather than a single answer.

1. marty says:
  
  November 20, 2017 at 9:08 pm
  
  Thanks, Number 8. Yep, the “look at the picture” culture of VCE mathematics is phenomenal, and phenomenally stupid.
  
Anonymous says:

April 2, 2023 at 8:32 am

Is d an integer or a continuous variable (not sure what your set notation means). If it is a variable, you can just fill in t=5 and do a max/min problem differentiating d, no? Even if d is an integer, you’ll get the right insight and be able to try integers to either side of the max/min point. No? [Been a while since I actually did this stuff…like 40 years. But that’s my quick thinking.]

1. marty says:
  
  April 2, 2023 at 10:15 am
  
  d is real. And yes, the problem is not hard, but you need to be looking for the (possible) critical point for any d, and comparing the location to the endpoints. It takes time, which is not allocated for the 2-mark question. WHat was expected was quick and non-sensical.

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4 Replies to “The Treachery of Images”

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