IB Maths and GCSE Maths Resources from British International School Phuket. Theory of Knowledge (ToK). Maths explorations and investigations.  Real life maths. Maths careers. Maths videos. Maths puzzles and Maths lesson resources.

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British International School Phuket

Welcome to the British International School Phuket’s maths website. My name is Andrew Chambers and I am currently working at BISP.  I am running my site as the school’s maths resources website for both our students and students around the world.

We are a British international school located on the tropical island of Phuket in Southern Thailand. We offer a number of scholarships each year, catering for a number of national and international standard sports stars as well as for academic excellence. You can find out more about our school here.

There are a huge amount of resources to explore – especially for students doing their IAs and for students looking for revision videos.  You may also like to try our school code breaking site – where you can compete with over 10,000 students from around the world who have made it onto our school leaderboard.

 

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Spotting Asset Bubbles

Asset bubbles are formed when a service, product or company becomes massively over-valued only to crash, taking with it most of its investors’ money.  There are many examples of asset bubbles in history – the Dutch tulip bulb mania and the South Sea bubble are two of the most famous historical examples.  In the tulip mania bubble of 1636-37, the price of tulip bulbs became astronomically high – as people speculated that the rising prices would keep rising yet further.  At its peak a single tulip bulb was changing hands for around 10 times the annual wage of a skilled artisan, before crashing to become virtually worthless.

More recent bubble include the Dotcom crash of the early 2000s – where investors piled in trying to spot in what ways the internet would revolutionise businesses.  Huge numbers of internet companies tried to ride this wave by going public with share offerings.  This led to massive overvaluation and a crash when investors realised that many of these companies were worthless.  Pets.com is often given as an example of this exuberance – its stock collapsed from $11 to $0.19 in just 6 months, taking with it $300 million of venture capital.

Therefore spotting the next bubble is something which economists take very seriously.  You want to spot the next bubble, but equally not to miss out on the next big thing – a difficult balancing act!  The graph at the top of the page is given as a classic bubble.  It contains all the key phases – an initial slow take-off, a steady increase as institutional investors like banks and hedge funds get involved, an exponential growth phase as the public get involved, followed by a crash and a return to its long term mean value.

Comparing the Bitcoin graph to an asset bubble

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The above graph is charting the last year of Bitcoin growth.  We can see several similarities – so let’s try and plot this on the same axis as the model.  The orange dots represent data points for the initial model – and then I’ve fitted the Bitcoin graph over the top:

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It’s not a bad fit – if this was going to follow the asset bubble model then it would be about to crash rapidly before returning to the long term mean of around $4000.  Whether that happens or it continues to rise, you can guarantee that there will be thousands of economists and stock market analysts around the world doing this sort of analysis (albeit somewhat more sophisticated!) to decide whether Bitcoin really will become the future of money – or yet another example of an asset bubble to be studied in economics textbooks of the future.

 

Measuring the Distance to the Stars

This is a very nice example of some very simple mathematics achieving something which  for centuries appeared impossible – measuring the distance to the stars.  Before we start we need a few definitions:

  • 1  Astronomical Unit (AU) is the average distance from the Sun to the Earth.  This is around 150,000,000km.
  • 1 Light Year is the distance that light travels in one year.  This is around 9,500,000,000,000km.  We have around 63000AU = 1 Light Year.
  • 1 arc second is measurement for very small angles and is 1/3600 of one degree.
  • Parallax is the angular difference in measurement when viewing an object from different locations.  In astronomy parallax is used to mean the half the angle formed when a star is viewed from opposite sides of the Earth’s solar orbit (marked on the diagram below).Screen Shot 2017-12-09 at 8.28.33 PM

With those definitions it is easy to then find the distance to stars.  The parallax method requires that you take a measurement of the angle to a given star, and then wait until 6 months later and take the same measurement.  The two angles will be slightly different – divide this difference by 2 and you have the parallax.

Let’s take 61 Cyngi – which Friedrick Bessel first used this method on in the early 1800s.  This has a parallax of 287/1000 arc seconds.  This is equivalent to 287/1000 x 1/3600 degree or approximately 0.000080 degrees.  So now we can simply use trigonometry – we have a right angled triangle with opposite side = 1 AU and angle = 0.0000080.  Therefore the distance is given by:

tanΦ = opp/adj

tan(0.000080) = 1/d

d = 1/tan(0.000080)

d = 720000 AU

which is approximately 720000/63000 = 11 light years away.

That’s pretty incredible!  Using this method and armed with nothing more than a telescope and knowledge of the Earth’s orbital diameter,  astronomers were able to judge the distance of stars in faraway parts of the universe – indeed they used this method to prove that other galaxies apart from our own also existed.

Orion’s Belt

The constellation of Orion is one of the most striking in the Northern Hemisphere.  It contains the “belt” of 3 stars in a line, along with the brightly shining Rigel and the red super giant Betelgeuse.  The following 2 graphics are taken from the great student resource from the Royal Observatory Greenwich:

The angles marked in the picture are in arc seconds – so to convert them into degrees we need to multiply by 1/3600.  For example, Betelgeuse the red giant has a parallax of 0.0051 x 1/3600 = 0.0000014 (2sf) degrees.  Therefore the distance to Betelgeuse is:

tanΦ = opp/adj

tan(0.0000014) = 1/d

d = 1/tan(0.0000014)

d = 41,000,000 AU

which is approximately 41,000,000/63000 = 651 light years away.  If we were more accurate with our rounding we would get 643 light years.  That means that when we look into the sky we are seeing Betelgeuse as it was 643 years ago.

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The Remarkable Dirac Delta Function

This is a brief introduction to the Dirac Delta function – named after the legendary Nobel prize winning physicist Paul Dirac. Dirac was one of the founding fathers of the mathematics of quantum mechanics, and is widely regarded as one of the most influential physicists of the 20th Century.  This topic is only recommended for students confident with the idea of limits and was inspired by a Quora post by Arsh Khan.

Dirac defined the delta function as having the following 2 properties:

Screen Shot 2017-12-02 at 9.22.27 PMThe first property as defined above is that the delta function is 0 for all values of t, except for t = 0, when it is infinite.

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The second property defined above is that the integral of the delta function – and the area of the graph between 2 points (either side of 0) is 1.    We can take the bottom integral where we integrate from negative to positive infinity as this will be more useful later.

The delta function (technically not a function in a normal sense!) can be represented as the following limit:

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Whilst this looks a little intimidating, it just means that we take the limit of the function as epsilon (ε) approaches 0.  Given this definition of the delta function we can check that the 2 properties outlined above hold.

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For the first limit above we set t not equal to 0.  Then, because it is a continuous function when t is not equal to 0, we can effectively replace epsilon with 0 in the first limit above to get a limit of 0.  In the second limit when t = 0 we get a limit of infinity.  Therefore the first property holds.

To show that the second property holds, we start with the following integral identity from HL Calculus:

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Hopefully this will look similar to the function we are interested in.  Let’s play a little fast and loose with the mathematics and ignore the limit of the function and just consider the following integral:

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Therefore (using the fact that the graph of arctanx has horizontal asymptotes at positive and negative pi/2 for the final part) :

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So we have shown above that the integral of every function of this form will have an integral of 1, regardless of the value of epsilon, thus satisfying our second property.

The use of the Dirac Function

So far so good.  But what is so remarkable about the Dirac function?  Well, it allows objects to be described in terms of a single zero width (and infinitely high) spike, but despite having zero width, this spike still has an area of 1.   This then allows the representation of elementary particles which have zero size but finite mass (and other finite properties such as charge) to be represented mathematically.  With the area under the curve = 1 it can also be thought of in terms of a probability density function – i.e representing the quantum world in terms of probability wave functions.

A graphical representation:

This is easier to understand graphically.  Say for example we choose a value epsilon (ε) and gradually make it smaller (i.e we find the limit as ε approaches 0).  When ε = 5 we have the following:

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When ε = 1 we have the following:

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When ε = 0.1 we have the following:

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When ε = 0.01 we have the following:

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You can see that as  ε approaches 0 we get a function which is close to 0 everywhere except for a spike at zero.  The total area under the function remains at 1 for all ε.

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Therefore we can represent the Dirac Delta function with the above graph.  In it we have a point with zero width but with infinite height – and still with an area under the curve of 1!

 

 

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The Rise of Bitcoin

Bitcoin is in the news again as it hits $10,000 a coin – the online crypto-currency has seen huge growth over the past 1 1/2 years, and there are now reports that hedge funds are now investing part of their portfolios in the currency.   So let’s have a look at some regression techniques to predict the future price of the currency.

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Here the graph has been inserted into Desmos and the scales aligned.  1 on the y axis corresponds to $1000 and 1 on the x axis corresponds to 6 months.  2013 is aligned with  (0,0).

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Next, I plot some points to fit the curve through.

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Next, we use Desmos’ regression for y = aebx+d. This gives the line above with equation:

y = 5.10 x 10-7 e1.67x + 0.432.

I included the vertical translation (d) because without it the graph didn’t fit the early data points well.

So, If I want to predict what the price will be in December 2019, I use x = 12

y = 5.10 x 10-7 e1.67(12) + 0.432 = 258

and as my scale has 1 unit on the y axis equal to $1000, this is equal to $258,000.

So what does this show?  Well it shows that Bitcoin is currently in a very steep exponential growth curve – which if sustained even over the next 12 months would result in astronomical returns.  However we also know that exponential growth models are very poor at predicting long term trends – as they become unfeasibly large very quickly.   The two most likely scenarios are:

  1. continued growth following a polynomial rather than exponential model
  2. a price crash

Predicting which of these 2 outcomes are most likely is probably best left to the experts!  If you do choose to buy bitcoins you should be prepared for significant price fluctuations – which could be down as well as up.  I’ll revisit this post in a few months and see what has happened.

If you are interested in some more of the maths behind Bitcoin, you can read about the method that is used to encrypt these currencies (a method called elliptical curve cryptography).

 

This post is inspired by the Quora thread on interesting functions to plot.

  1. The butterfly

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This is a slightly simpler version of the butterfly curve which is plotted using polar coordinates on Desmos as:

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Polar coordinates are an alternative way of plotting functions – and are explored a little in HL Maths when looking at complex numbers. The theta value specifies an angle of rotation measured anti-clockwise from the x axis, and the r value specifies the distance from the origin. So for example the polar coordinates (90 degrees, 1) would specify a point 90 degrees ant clockwise from the x axis and a distance 1 from the origin (i.e the point (0,1) in our usual Cartesian plane).

2. Fermat’s Spiral

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This is plotted by the polar equation:

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The next 3 were all created by my students.

3.  Chaotic spiral (by Laura Y9)

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I like how this graph grows ever more tangled as it coils in on itself.  This was created by the polar equation:

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4.  The flower (by Felix Y9)

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Some nice rotational symmetries on this one.  Plotted by:

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5. The heart (by Tiffany Y9)

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Simple but effective!  This was plotted using the usual x,y coordinates:

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You can also explore how to draw the Superman and Batman logos using Wolfram Alpha here.



A geometric proof for the Arithmetic and Geometric Mean

There is more than one way to define the mean of a number.  The arithmetic mean is the mean we learn at secondary school – for 2 numbers a and b it is:

(a + b) /2.

The geometric mean on the other hand is defined as:

(x1.x2.x3…xn)1/n

So for example with the numbers 1,2,3 the geometric mean is (1 x 2 x 3)1/3.

With 2 numbers, a and b, the geometric mean is (ab)1/2.

We can then use the above diagram to prove that (a + b) /2 ≥ (ab)1/2 for all a and b. Indeed this inequality holds more generally and it can be proved that the Arithmetic mean ≥ Geometric mean.

Step (1) We draw a triangle as above, with the line MQ a diameter, and therefore angle MNQ a right angle (from the circle theorems).  Let MP be the length a, and let PQ be the length b.

Step (2) We can find the length of the green line OR, because this is the radius of the circle.  Given that the length a+b was the diameter, then (a+b) /2 is the radius.

Step (3) We then attempt to find an equation for the length of the purple line PN.

We find MN using Pythagoras:  (MN)2 = a2 +x2

We find NQ using Pythagoras:  (NQ)2 = b2 +x2

Therefore the length MQ can also be found by Pythagoras:

(MQ)2 = (MN) + (NQ)2

(MQ) = a2 +x2 + b2 +x2

But MQ = a + b.  Therefore:

(a + b) = a2 +x2 + b2 +x2

a2+ b2 + 2ab = a2 +x2 + b2 +x2

2ab = x2 +x2

ab = x2

x = (ab)1/2

Therefore our green line represents the arithmetic mean of 2 numbers (a+b) /2 and our purple line represents the geometric mean of 2 numbers (ab)1/2. The green line will always be greater than the purple line (except when a = b which gives equality) therefore we have a geometrical proof of our inequality.

There is a more rigorous proof of the general case using induction you may wish to explore as well.

Euler’s 9 Point Circle

This is a nice introduction to some of the beautiful constructions of geometry.  This branch of mathematics goes in and out of favour – back in the days of Euclid, constructions using lines and circles were a cornerstone of mathematical proof, interest was later revived in the 1800s through Poncelot’s projective geometry – later leading to the new field of non Euclidean geometry.  It’s once again somewhat out of fashion – but more accessible than ever due to programs like Geogebra (on which the below diagrams were plotted).  The 9 point circle (or at least the 6 point circle was discovered by the German Karl Wilhelm von Feuerbach in the 1820s.  Unfortunately for Feuerbach it’s often instead called the Euler Circle – after one of the greatest mathematicians of all time, Leonhard Euler.

So, how do you draw Euler’s 9 Point Circle?  It’s a bit involved, so don’t give up!

Step 1: Draw a triangle:

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Step 2: Draw the perpendicular bisectors of the 3 sides, and mark the point where they all intersect (D).

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Step 3: Draw the circle through the point D.

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Step 4: From each line of the triangle, draw the perpendicular line through its third angle.  For example, for the line AC, draw the perpendicular line that goes through both AC and angle B. (The altitudes of the triangle).  Join up the 3 altitudes which will meet at a point (E).

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Step 5:  Join up the mid points of each side of the triangle with the remaining angle.  For example, find the mid point of AC and join this point with angle B.  (The median lines of the triangle).  Label the point where the 3 lines meet as F.

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Step 6:  Remove all the construction lines.  You can now see we have 3 points in a line.  D is the centre of the circle through the points ABC, E is where the altitudes of the triangle meet (the orthoocentre of ABC) and F is where the median lines meet (the centroid of ABC).

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Step 7:  Join up the 3 points – they are collinear (on the same line).

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Step 8:  Enlarge the circle through points A B C by a scale factor of -1/2 centered on point F.

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Step 9: We now have the 9 point circle.  Look at the points where the inner circle intersects the triangle ABC.  You can see that the points M N O show the points where the feet of the altitudes (from step 4) meet the triangle.

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The points P Q R show the points where the perpendicular bisectors of the lines start (i.e the midpoints of the lines AB, AC, BC)

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We also have the points S T U on the circle which show the midpoints of the lines between E and the vertices A, B, C.

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Step 10:  We can drag the vertices of the triangle and the above relationships will still hold.

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In the second case we have both E and D outside the triangle.

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In the third case we have E and F at the same point.

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In the fourth case we have D and E on opposite sides of the triangle.

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So there we go – who says maths isn’t beautiful?

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Log Graphs to Plot Planetary Patterns

This post is inspired by the excellent Professor Stewart’s latest book, Calculating the Cosmos. In it he looks at some of the mathematics behind our astronomical knowledge.

Astronomical investigations

In the late 1760s and early 1770s, 2 astronomers Titius and Bode both noticed something quite strange – there seemed to be a relationship in the distances between the planets. There was no obvious reason as to why there would be – but nevertheless it appeared to be true. Here are the orbital distances from the Sun of the 6 planets known about in the 1760s:

Mercury: 0.39 AU
Venus: 0.72 AU
Earth: 1.00 AU
Mars: 1.52 AU
Jupiter: 5.20 AU
Saturn: 9.54 AU

In astronomy, 1 astronomical unit (AU) is defined as the mean distance from the center of the Earth to the centre of the Sun (149.6 million kilometers).

Now, at first glance there does not appear to be any obvious relationship here – it’s definitely not linear, but how about geometric? Well dividing the term above by the term below we get r values of:

1.8, 1.4, 1.5, 3.4, 1.8

4 of the numbers are broadly similar – and then we have an outlier of 3.4. So either there was no real pattern – or there was an undetected planet somewhere between Mars and Jupiter? And was there another planet beyond Saturn?

Planet X

Mercury: 0.39 AU
Venus: 0.72 AU
Earth: 1.00 AU
Mars: 1.52 AU
Planet X: x AU
Jupiter: 5.20 AU
Saturn: 9.54 AU
Planet Y: y AU

For a geometric sequence we would therefore want x/1.52 = 5.20/x. This gives x = 2.8 AU – so a missing planet should be 2.8 AU away from the Sun. This would give us r values of 1.8, 1.4, 1.5, 1.8, 1.9, 1.8. Let’s take r = 1.8, which would give Planet Y a distance of 17 AU.

So we predict a planet around 2.8 AU from the Sun, and another one around 17 AU from the Sun. In 1781, Uranus was discovered – 19.2 AU from the Sun, and in 1801 Ceres was discovered at 2.8 AU. Ceres is what is now classified as a dwarf planet – the largest object in the asteroid belt between Jupiter and Mars.

Log Plots

Using graphs is a good way to graphically see relationships. Given that we have a geometrical relationship in the form d = ab^n with a and b as constants, we can use the laws of logs to rearrange to give log d = log a + n log b.

Therefore we can plot log d on the y axis, and n on the x axis. If there is a geometrical relationship we will see us a linear relationship on the graph, with log a being the y intercept and the gradient being log b.

(n=1) Mercury: d = 0.39 AU. log d = -0.41
(n=2) Venus: d = 0.72 AU. log d = -0.14
(n=3) Earth: d = 1.00 AU. log d = 0
(n=4) Mars: d = 1.52 AU. log d = 0.18
(n=5) Ceres (dwarf): d = 2.8 AU. log d = 0.45
(n=6) Jupiter: d = 5.20 AU. log d = 0.72
(n=7) Saturn: d = 9.54 AU. log d = 0.98
(n=8) Uranus: d = 19.2 AU. log d = 1.28

 

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We can use Desmos’ regression tool to find a very strong linear correlation – with y intercept as -0.68 and gradient as 0.24.  Given that log a is the y intercept, this gives:

log a  = -0.68

a = 0.21

and given that log b is the gradient this gives:

log b = 0.24

b = 1.74

So our final formula for the relationship for the spacing of the n ordered planets is:

d = ab^n

distance = 0.21 x (1.74)^n.

Testing the formula

So, using this formula we can predict what the next planetary distance would be. When n = 9 we would expect a distance of 30.7 AU.  Indeed we find that Neptune is 30.1 AU – success! How about Pluto?  Given that Pluto has a very eccentric (elliptical) orbit we might not expect this to be as accurate.  When n = 10 we get a prediction of 53.4 AU.  The average AU for Pluto is 39.5 – so our formula does not work well for Pluto.   But looking a little more closely, we notice that Pluto’s distance from the Sun varies wildly – from 29.7 AU to 49.3 AU, so perhaps it is not surprising that this doesn’t follow our formula well.

Other log relationships

Interestingly other distances in the solar system show this same relationship.  Plotting the ordered number of the planets against the log of their orbital period produces a linear graph, as does plotting the ordered moons of Uranus against their log distance from the planet.  Why these relationships exist is still debated.  Perhaps they are a coincidence, perhaps they are a consequence of resonance in orbital periods.   Do some research and see what you find!

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This is a quick example of how using Tracker software can generate a nice physics-related exploration.  I took a spring, and attached it to a stand with a weight hanging from the end.  I then took a video of the movement of the spring, and then uploaded this to Tracker.

Height against time

The first graph I generated was for the height of the spring against time.  I started the graph when the spring was released from the low point.  To be more accurate here you can calibrate the y axis scale with the actual distance.  I left it with the default settings.

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You can see we have a very good fit for a sine/cosine curve.  This gives the approximate equation:

y = -65cos10.5(t-3.4) – 195

(remembering that the y axis scale is x 100).

This oscillating behavior is what we would expect from a spring system – in this case we have a period of around 0.6 seconds.

Momentum against velocity

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For this graph I first set the mass as 0.3kg – which was the weight used – and plotted the y direction momentum against the y direction velocity.  It then produces the above linear relationship, which has a gradient of around 0.3.  Therefore we have the equation:

p = 0.3v

If we look at the theoretical equation linking momentum:

p = mv

(Where m = mass).  We can see that we have almost perfectly replicated this theoretical equation.

Height against velocity

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I generated this graph with the mass set to the default 1kg.  It plots the y direction against the y component velocity.  You can see from the this graph that the velocity is 0 when the spring is at the top and bottom of its cycle.  We can then also see that it reaches its maximum velocity when halfway through its cycle.  If we were to model this we could use an ellipse (remembering that both scales are x100 and using x for vy):

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If we then wanted to develop this as an investigation, we could look at how changing the weight or the spring extension affected the results and look for some general conclusions for this.  So there we go – a nice example of how tracker can quickly generate some nice personalised investigations!

Predicting the UK election using linear regression

The above data is the latest opinion poll data from the Guardian.  The UK will have (another) general election on June 8th.  So can we use the current opinion poll data to predict the outcome?

Longer term data trends

Let’s start by looking at the longer term trend following the aftermath of the Brexit vote on June 23rd 2016.  I’ll plot some points for Labour and the Conservatives and see what kind of linear regression we get.  To keep things simple I’ve looked at randomly chosen poll data approximately every 2 weeks – assigning 0 to July 1st 2016, 1 to mid July, 2 to August 1st etc.  This has then been plotted using the fantastic Desmos.

Labour

You can see that this is not a very good fit – it’s a very weak correlation.  Nevertheless let’s see what we would get if we used this regression line to predict the outcome in June.  With the x axis scale I’ve chosen, mid June 2017 equates to 23 on the x axis.  Therefore we predict the percentage as

y = -0.130(23) + 30.2

y  = 27%

Clearly this would be a disaster for Labour – but our model is not especially accurate so perhaps nothing to worry about just yet.

Conservatives

As with Labour we have a weak correlation – though this time we have a positive rather than negative correlation.  If we use our regression model we get a prediction of:

y = 0.242(23) + 38.7

y = 44%

So, we are predicting a crushing victory for the Conservatives – but could we get some more accurate models to base this prediction on?

Using moving averages

The Guardian’s poll tracker at the top of the page uses moving averages to smooth out poll fluctuations between different polls and to arrive at an averaged poll figure.  Using this provides a stronger correlation:

Labour

This model doesn’t take into account a (possible) late surge in support for Labour but does fir better than our last graph.  Using the equation we get:

y = -0.0764(23) + 28.8

y = 27%

Conservatives

We can have more confidence in using this regression line to predict the election.  Putting in the numbers we get:

y = 0.411(23) + 36.48

y = 46%

Conclusion

Our more accurate models merely confirm what we found earlier – and indeed what all the pollsters are predicting – a massive win for the Conservatives.  Even allowing for a late narrowing of the polls the Conservatives could be on target for winning by over 10% points – which would result in a very large majority.  Let’s see what happens!

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