You are currently browsing the category archive for the ‘IB HL calculus’ category. IB HL Calculus P3 May 2016:  The Hardest IB Paper Ever?

IB HL Paper 3 Calculus May 2016 was a very poor paper.  It was unduly difficult and missed off huge chunks of the syllabus.  You can see question 5 posted above. (I work through the solution to this in the next post).  This is so far off the syllabus as to be well into undergraduate maths.  Indeed it wouldn’t look out of place in an end of first year or end of second year undergraduate calculus exam.  So what’s it doing on a sixth form paper for 17-18 year olds?   The examiners completely abandoned their remit to produce a test of the syllabus content – and instead decided that a one hour exam was the time to introduce extensions to that syllabus, whilst virtually ignoring all the core content of the course.

A breakdown of the questions

1) Maclaurin- on the syllabus.  This was reasonable.  As was using it to find the limit of a fraction.  Part (c) requires use of Lagrange error – which students find difficult and forms a very small part of the course.  If this was the upper level of the challenge in the paper then fair enough, but it was far from it.

2) Fundamental Theorem of Calculus – barely on the syllabus – and unpredictable in advance as to what is going to be asked on this.  This has never been asked before on any paper, there is no guidance in the syllabus, there was no support in the specimen paper and most textbooks do not cover this in any detail.  This seems like an all or nothing question – students will either get 7 or 0 on this question.  Part (c) for an extra 3 marks seems completely superfluous.

3) Mean Value Theorem – a small part of the syllabus given dispropotionate exam question coverage because the examiners seem to like proof questions.  This seems like an all or nothing question as well – if you get the concept then it’s 7 marks, if not it’ll likely be 0.

4) Differential equations –  This question would have been much better if they had simply been given the integrating factor /separate variables question in part (b), leaving some extra marks to test something else on part (a) – perhaps Euler’s Method?

5) An insane extension to the syllabus which took the question well into undergraduate mathematics – and hid within it a “trap” to make students try to integrate a function that can’t actually be integrated.  This really should have been nowhere near the exam.  At 14 marks this accounted for nearly a quarter of the exam.

Content unassessed

The syllabus is only 48 hours and all schools spend that time ploughing through limits and differentiability of functions, L’Hopital’s rule, Riemann sums, Rolle’s Theorem, standard differential equations, isoclines, slope fields, the squeeze theorem, absolute and conditional convergence, error bounds, indefinite integrals, the ratio test, power series, radius of convergence.  All of these went pretty much unassessed.  I would say that the exam tested around 15% of the syllabus content.  Even the assessment of alternating series convergence was buried inside question 5 – making is effectively inaccessible to all students.

The result of this is that there will be a huge squash in the grade boundaries – perhaps as low as 50-60% for a Level 6 and 25-35% for a level 4.    The last 20 marks on the paper will probably be completely useless – separating no students at all.  This then produces huge unpredictability as dropping 4-5 marks might take from from a level 5 to level 3 or level 6 to level 4.

Teachers no longer have any confidence in the IB HL examiners

One of my fellow HL teachers posted this following the Calculus exam:

At various times throughout the year I joke with my students about how the HL Mathematics examiners must be like a group of comic book villains sitting in a lair, devising new ways to form cruel questions to make students suffer and this exam leads me to believe that this is not too far fetched of a concept.

And I would tend to agree.  Who wants students to be demoralised with low scores and questions they can’t succeed on.  Surely that should not be an aim when creating an exam!

I’ve taught the HL Calculus Option for the last 4 years – I think the course is a good one.  It’s difficult but a rewarding syllabus which introduces some of the tools needed for undergraduate maths.  However I no longer have any confidence in the IB or the IB examiners to produce a fair test to examine this content.  Many other HL teachers feel the same way.  So what choice is left?  Abandon the Calculus option and start again from scratch with another option?  Or continue to put our trust in the IB, when they continue to let teachers (and more importantly the students) down? This is a nice example of using some maths to solve a puzzle from the mindyourdecisions youtube channel (screencaptures from the video).

How to Avoid The Troll: A Puzzle In these situations it’s best to look at the extreme case first so you get some idea of the problem.  If you are feeling particularly pessimistic you could assume that the troll is always going to be there.  Therefore you would head to the top of the barrier each time.  This situation is represented below:

The Pessimistic Solution: Another basic strategy would be the optimistic strategy.  Basically head in a straight line hoping that the troll is not there.  If it’s not, then the journey is only 2km.  If it is then you have to make a lengthy detour.  This situation is shown below:

The Optimistic Solution: The expected value was worked out here by doing 0.5 x (2) + 0.5 x (2 + root 2) = 2.71.

The question is now, is there a better strategy than either of these?  An obvious possibility is heading for the point halfway along where the barrier might be.  This would make a triangle of base 1 and height 1/2.  This has a hypotenuse of root (5/4).  In the best case scenario we would then have a total distance of 2 x root (5/4).  In the worst case scenario we would have a total distance of root(5/4) + 1/2 + root 2.  We find the expected value by multiply both by 0.5 and adding.  This gives 2.63 (2 dp).  But can we do any better?  Yes – by using some algebra and then optimising to find a minimum.

The Optimisation Solution: To minimise this function, we need to differentiate and find when the gradient is equal to zero, or draw a graph and look for the minimum.  Now, hopefully you can remember how to differentiate polynomials, so here I’ve used Wolfram Alpha to solve it for us.  Wolfram Alpha is incredibly powerful -and also very easy to use.  Here is what I entered: and here is the output: So, when we head for a point exactly 1/(2 root 2) up the potential barrier, we minimise the distance travelled to around 2.62 miles.

So, there we go, we have saved 0.21 miles from our most pessimistic model, and 0.01 miles from our best guess model of heading for the midpoint.  Not a huge difference – but nevertheless we’ll save ourselves a few seconds!

This is a good example of how an exploration could progress – once you get to the end you could then look at changing the question slightly, perhaps the troll is only 1/3 of the distance across?  Maybe the troll appears only 1/3 of the time?  Could you even generalise the results for when the troll is y distance away or appears z percent of the time?

Bullet Projectile Motion Experiment

This is a classic physics experiment which counter to our intuition.  We have  a situation where 1 ball is dropped from a point, and another ball is thrown horizontally from that same point.  The question is which ball will hit the ground first? (diagram from School for Champions site)

Looking at the diagram above you might argue that the ball that is dropped falls to the floor quicker as it has a shorter path.  Or, you might think that the ball thrown sideways would travel faster to the ground because of its initial horizontal velocity.  Both of these views are wrong however – as both balls will land at exactly the same time.  To understand why, let’s look at the 2 situations in turn.

The ball launched sideways

To show that both balls would hit the ground at the same time we need to split the motion into its x and y components.  We have $x = v t \cos \theta$ $y = vt \sin \theta - \frac{1}{2} g t^2$

Where the angle theta is the angle of launch, v is the initial velocity, g is the gravitational constant 9.8 m/s.  If we have a launch from the horizontal direction, then this angle is 0, which gives the simplified equations:

x = vt

y = 0.5gt2

if we relabel y as the vertical distance (d), then we have: $\ t =\ \sqrt {\frac{2d}{g}}$

which is the time taken (ignoring air resistance etc) for an object launched horizontally to fall a distance d, where g is the gravitational constant 9.8 m/s.

So if we have a ball launched at a speed of 1 m/s from a height of 1m, it would hit the ground when:

t = (2/9.8)0.5 = 0.45 seconds

So we can use this value of t to see how far in the x direction it has travelled:

x = vt

x = 1(0.45)

x = 0.45m.

The ball dropped vertically $x = v t \cos \theta$ $y = vt \sin \theta - \frac{1}{2} g t^2$

But this time we have no initial velocity as so we simply get:

x = 0

y = 0.5gt2

or as before, if we relabel y as the vertical distance (d), then we have: $\ t =\ \sqrt {\frac{2d}{g}}$

So with a ball dropped from a height of 1m, it would also hit the ground when:

t = (2/9.8)0.5 = 0.45 seconds

But this time the distance in the x direction will of course be 0.

Showing this graphically

We can also show this graphically using the tracker software.  This allows you to track the motion of objects in videos.  So using the video above we can set the axis, and the height of the table and then the motion capture software actually plots the parabola of the ball’s motion. This first graph shows the change in the y direction with respect to time for the ball launched horizontally.  We have large steps because the video was in super slow motion, so there were frames of very little movement.  Nevertheless we can clearly see the general parabola, with equation:

y = -0.43x2 -1.2x + 107 The second graph shows the change in y direction with respect to time for the ball dropped vertically down.  As before we have a clear parabola, with equation:

y = -0.43x2 -1.2x + 106

Which is a remarkably close fit.  So, there we go, we have shown that the vertical motion of our 2 objects are independent of their horizontal motion.

IB Revision If you’re already thinking about your coursework then it’s probably also time to start planning some revision, either for the end of Year 12 school exams or Year 13 final exams. There’s a really great website that I would strongly recommend students use – you choose your subject (HL/SL/Studies if your exam is in 2020 or Applications/Analysis if your exam is in 2021), and then have the following resources: The Questionbank takes you to a breakdown of each main subject area (e.g. Algebra, Calculus etc) and each area then has a number of graded questions. What I like about this is that you are given a difficulty rating, as well as a mark scheme and also a worked video tutorial.  Really useful! The Practice Exams section takes you to ready made exams on each topic – again with worked solutions.  This also has some harder exams for those students aiming for 6s and 7s and the Past IB Exams section takes you to full video worked solutions to every question on every past paper – and you can also get a prediction exam for the upcoming year.

I would really recommend everyone making use of this – there is a mixture of a lot of free content as well as premium content so have a look and see what you think.

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