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Maths of Global Warming – Modeling Climate Change

The above graph is from NASA’s climate change site, and was compiled from analysis of ice core data. Scientists from the National Oceanic and Atmospheric Administration (NOAA) drilled into thick polar ice and then looked at the carbon content of air trapped in small bubbles in the ice. From this we can see that over large timescales we have had large oscillations in the concentration of carbon dioxide in the atmosphere. During the ice ages we have had around 200 parts per million carbon dioxide, rising to around 280 in the inter-glacial periods. However this periodic oscillation has been broken post 1950 – leading to a completely different graph behaviour, and putting us on target for 400 parts per million in the very near future.

Analysising the data


One of the fields that mathematicians are always in demand for is data analysis. Understanding data, modeling with the data collected and using that data to predict future events. Let’s have a quick look at some very simple modeling. The graph above shows a superimposed sine graph plotted using Desmos onto the NOAA data.

y = -0.8sin(3x +0.1) – 1

Whilst not a perfect fit, it does capture the general trend of the data and its oscillatory behaviour until 1950. We can see that post 1950 we would then expect to be seeing a decline in carbon dioxide rather than the reverse – which on our large timescale graph looks close to vertical.

Dampened Sine wave


This is a dampened sine wave, achieved by adding e-x to the front of the sine term.  This achieves the result of progressively reducing the amplitude of the sine function.  The above graph is:

y = e-0.06x (-0.6sin(3x+0.1) -1 )

This captures the shape in the middle of the graph better than the original sine function, but at the expense of less accuracy at the left and right.

Polynomial Regression


We can make use of Desmos’ regression tools to fit curves to points.  Here I have entered a table of values and then seen which polynomial gives the best fit:



We can see that the purple cubic fits the first 5 points quite well (with a high R² value).  So we should be able to create a piecewise function to describe this graph.

Piecewise Function


Here I have restricted the domain of the first polynomial (entered below):


Second polynomial:



Third polynomial:



Fourth polynomial:



Finished model:


Shape of model:


We would then be able to fit this to the original model scale by applying a vertical translation (i.e add 280), vertical and horizontal stretch.  It would probably have been easier to align the scales at the beginning!  Nevertheless we have the shape we wanted.

Analysing the models

Our piecewise function gives us a good data fit for the domain we were working in – so if we then wanted to use some calculus to look at non horizontal inflections (say), this would be a good model to use.  If we want to analyse what we would have expected to happen without human activity, then the sine models at the very start are more useful in capturing the trend of the oscillations.

Post 1950s


Looking on a completely different scale, we can see the general tend of carbon dioxide concentration post 1950 is pretty linear.  This time I’ll scale the axis at the start.  Here 1960 corresponds with x = 0, and 1970 corresponds with x = 5 etc.



Actually we can see that a quadratic fits the curve better than a linear graph – which is bad news, implying that the rates of change of carbon in the atmosphere will increase.  Using our model we can predict that on current trends in 2030 there will be 500 parts per million of carbon in the atmosphere.

Stern Report

According to the Stern Report, 500ppm is around the upper limit of what we need to aim to stabalise the carbon levels at (450ppm-550ppm of carbon equivalent) before the economic and social costs of climate change become economically catastrophic.  The Stern Report estimates that it will cost around 1% of global GDP to stablise in this range.  Failure to do that is predicted to lock in massive temperature rises of between 3 and 10 degrees by the end of the century.

If you are interested in doing an investigation on this topic:

  1. Plus Maths have a range of articles on the maths behind climate change
  2. The Stern report is a very detailed look at the evidence, graphical data and economic costs.

A longer look at the Si(x) function

Sinx/x can’t be integrated into an elementary function – instead we define:

Screen Shot 2016-05-21 at 9.31.12 PM

Where Si(x) is a special function.  This may sound strange – but we already come across another similar case with the integral of 1/x.  In this case we define the integral of 1/x as ln(x).  ln(x) is a function with its own graph and I can use it to work out definite integrals of 1/x.  For example the integral of 1/x from 1 to 5 will be ln(5) – ln(1) = ln(5).

The graph of Si(x) looks like this:

Screen Shot 2016-05-21 at 9.45.57 PM

Or, on a larger scale:

Screen Shot 2016-05-21 at 9.46.04 PM

You can see that it is symmetrical about the y axis, has an oscillating motion and as x gets large approaches a limit.  In fact this limit is pi/2.

Because Si(0) = 0,  you can write the following integrals as:

Screen Shot 2016-05-21 at 9.30.25 PM

How to integrate sinx/x ?

It’s all very well to define a new function – and say that this is the integral of sinx/x – but how was this function generated in the first place?

Well, one way to integrate difficult functions is to use Taylor and Maclaurin expansions.  For example the Maclaurin expansion of sinx/x for values near x=0 is:

Screen Shot 2016-05-21 at 9.31.00 PM

This means that in the domain close to x = 0, the function sinx/x behaves in a similar way to the polynomial above.  The last part of this expression O( )  just means everything else in this expansion will be x^6 or greater.

Graph of sinx/x

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Graph of 1 – x^2/6 + x^4/120

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In the region close to x=0 these functions behave in a very similar manner (this would be easier to see with similar scales so let’s look on a GDC):

So for the region above (x between 0 and 2) the 2 graphs are virtually indistinguishable.

Therefore if we want to integrate sinx/x for values close to 0 we can just integrate our new function 1 – x^2/6 + x^4/120 and get a good approximation.

Let’s try how accurate this is.  We can use Wolfram Alpha to tell us that:

Screen Shot 2016-05-22 at 6.49.08 AM

and let’s use Wolfram to work out the integral as well:

Screen Shot 2016-05-22 at 6.49.59 AM

Our approximation is accurate to 3 dp, 1.371 in both cases.  If we wanted greater accuracy we would simply use more terms in the Maclaurin expansion.

So, by using the Maclaurin expansion for terms near x = 0 and the Taylor expansion for terms near x = a we can build up information as to the values of the Si(x) function.



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This was the last question on the May 2016 Calculus option paper for IB HL.  It’s worth nearly a quarter of the entire marks – and is well off the syllabus in its difficulty.  You could make a case for this being the most difficult IB HL question ever.  As such it was a terrible exam question – but would make a very interesting exploration topic.  So let’s try and understand it!

Part (a)

First I’m going to go through a solution to the question – this was provided by another HL maths teacher, Daniel – who worked through a very nice answer.  For the first part of the question we need to try and understand what is actually happening – we have the sum of an integral – where we are summing a sequence of definite integrals.  So when n = 0 we have the single integral from 0 to pi of sint/t.  When n = 1 we have the single integral from pi to 2pi of sint/t.  The summation of the first n terms will add the answers to the first n integrals together.

Screen Shot 2016-05-21 at 9.16.26 PM

This is the plot of y = sinx/x from 0 to 6pi.  Using the GDC we can find that the roots of this function are n(pi).  This gives us the first mark in the question – as when we are integrating from 0 to pi the graph is above the x axis and so the integral is positive. When we integrate from pi to 2pi the graph is below the x axis and so the integral is negative.  Since our sum consists of alternating positive and negative terms, then we have an alternating series.

Part (b i)

This is where it starts to get difficult!  You might be tempted to try and integrate sint/t – which is what I presume a lot of students will have done.  It looks like integration by parts might work on this.  However this was  a nasty trap laid by the examiners – integrating by parts is a complete waste of time as this function is non-integrable.  This means that there is no elementary function or standard basic integration method that will integrate it.  (We will look later at how it can be integrated – it gives something called the Si(x) function).  Instead this is how Daniel’s method progresses:

Screen Shot 2016-05-21 at 9.50.36 PM

Hopefully the first 2 equalities make sense – we replace n with n+1 and then replace t with T + pi.  dt becomes dT when we differentiate t = T + pi.  In the second integral we have also replaced the limits (n+1)pi and (n+2)pi with n(pi) and (n+1)pi as we are now integrating with respect to T and so need to change the limits as follows:

t = (n+1)(pi)

T+ pi = (n+1)(pi)

T = n(pi).  This is now the lower integral value.

The third integral uses the fact that sin(T + pi) = – sin(T).

The fourth integral then uses graphical logic.  y = -sinx/x looks like this:

Screen Shot 2016-05-21 at 10.08.00 PM

This is the same as y = sinx/x but reflected in the x axis.  Therefore the absolute value of the integral of  y = -sinx/x  will be the same as the absolute integral of y = sinx/x.  The fourth integral has also noted that we can simply replace T with t to produce an equivalent integral.  The last integral then notes that the integral of sint/(t+pi) will be less than the integral of sint/t.  This then gives us the inequality we desire.

Don’t worry if that didn’t make complete sense – I doubt if more than a handful of IB students in the whole world got that in exam conditions.  Makes you wonder what the point of that question was, but let’s move on.

Part (b ii)

OK, by now most students will have probably given up in despair – and the next part doesn’t get much easier.  First we should note that we have been led to show that we have an alternating series where the absolute value of u_n+1 is less than the absolute value of u_n.  Let’s check the requirements for proving an alternating series converges:

Screen Shot 2016-05-21 at 10.22.18 PM

We already have shown it’s an absolute decreasing sequence, so we just now need to show the limit of the sequence is 0.

Screen Shot 2016-05-21 at 10.24.05 PM

OK – here we start by trying to get a lower and upper bound for u_n.  We want to show that as n gets large, the limit of u_n = 0.  In the second integral we have used the fact that the absolute value of an integral of a function is always less than or equal to the integral of an absolute value of a function.  That might not make any sense, so let’s look graphically:

Screen Shot 2016-05-21 at 9.29.06 PM

This graph above is y = sinx/x.  If we integrate this function then the parts under the x axis will contribute a negative amount.

Screen Shot 2016-05-22 at 7.36.26 AM

But this graph is y = absolute (sinx/x).  Here we have no parts under the x axis – and so the integral of absolute (sinx/x) will always be greater than or equal to the integral of y = sinx/x.

To get the third integral we note that absolute (sinx) is bounded between 0 and 1 and so the   integral of 1/x will always be greater than or equal to the integral of absolute (sinx)/x.

We next can ignore the absolute value because 1/x is always positive for positive x, and so we integrate 1/x to get ln(x). Substituting the values of the definite integral gives us a function of ln which as n approaches infinity approaches 0.  Therefore as this limit approaches 0, and this function was always greater than or equal to absolute u_n, then the limit of absolute u_n must also be 0.

Therefore we have satisfied the requirements for the Alternating Series test and so the series is convergent.

Part (c)

Part (c) is at least accessible for level 6 and 7 students as long as you are still sticking with the question.  Here we note that we have been led through steps to prove we have an alternating and convergent series.  Now we use the fact that the sum to infinity of a convergent alternating series lies between any 2 successive partial sums.  Then we can use the GDC to find the first few partial sums:

Screen Shot 2016-05-21 at 10.30.29 PMAnd there we are!  14 marks in the bag.  Makes you wonder who the IB write their exams for – this was so far beyond sixth form level as to be ridiculous.  More about the Si(x) function in the next post.


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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?



Screen Shot 2016-04-20 at 8.13.41 PM

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

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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:

Screen Shot 2016-04-20 at 7.33.54 PM


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:

Screen Shot 2016-04-20 at 7.34.15 PM

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:

Screen Shot 2016-04-20 at 7.35.29 PM

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:

Screen Shot 2016-04-20 at 7.53.58 PM

and here is the output:

Screen Shot 2016-04-20 at 7.54.12 PM

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?

Screen Shot 2015-11-27 at 3.27.40 PM

(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

We still start with:

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

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and then the motion capture software actually plots the parabola of the ball’s motion.

Screen Shot 2015-11-27 at 3.15.40 PM

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

Screen Shot 2015-11-27 at 3.19.52 PM

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.





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