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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IB Revision with Revision Village

There’s a really great website been put together to help IB students with topic revision both for during the course and for the end of Year 12 school exams and Year 13 final exams.  I would strongly recommend students use this as a resource during the course (not just for final revision in Y13!) There are specific resources for HL, SL and Studies students.

You choose your subject (HL/SL/Studies if your exam is in 2020 or Applications/Analysis if your exam is in 2021) and get the following screen:

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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 like:

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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.

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The Past IB Exams section takes you to full worked solutions to each full paper.

and lastly you can also get a prediction exam for the upcoming year.

For HL students (exam in 2020) you can click here

For SL students (exam in 2020) you can click here

For Studies students (exam in 2020) you can click here

For Analysis students and Applications students (exam in 2021) you can click here

You can also download the Mathematics Studies SL Formula booklet and the Standard Level Formula booklet from here if your teachers haven’t given you a copy.

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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The Tusi couple – A circle rolling inside a circle

Numberphile have done a nice video where they discuss some beautiful examples of trigonometry and circular motion and where they present the result shown above: a circle rolling within a circle, with the individual points on the small circle showing linear motion along the diameters of the large circle.  So let’s see what maths we need to create the image above.

 Projection of points

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We can start with the equation of a unit circle centred at the origin:

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and we can then define a point on this circle parametrically by the coordinate:

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Here t is the angle measured from the horizontal.

If we then want to see the projection of this point along the y-axis we can also plot:

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and to see the projection of this point along the x-axis we can also plot:

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By then varying t from 0 to 2 pi gives the animation above – where the black dot on the circle moves around the circle and there is a projection of its x and y coordinates on the axes.

Projection along angled lines

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I can then add a line through the origin at angle a to the horizontal:

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and this time I can project so that the line joining up the black point on the edge of the large circle intersects the dotted line in a right angle.

In order to find the parametric coordinate of this point projection I can use some right angled triangles as follows:

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The angle from the horizontal to my point A is t.  The angle from the horizontal to the slanted line is a.  The length of my radius BA is 1.  This gives me the length of BC.

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But I have the identity:

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Therefore this gives:

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And using some more basic trigonometry gives the following diagram:

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Therefore the parametric form of the projection of the point can be given as:

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Adding more lines

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I can add several more slanted lines through the origin.  You can see that each dot on the line is the right angle projection between the line and the point on the circle.  As we do this we can notice that the points on the lines look as though they form a circle.  By noticing that the new smaller circle is half the size of the larger circle, and that the centre of the smaller circle is half-way between the origin and the point on the large circle, we get:

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We can the vary the position of the point on the large circle to then create our final image:

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We have a connection between both linear motion and circular motion and create the impression of a circle rolling inside another.

You can play around with this demos graph here.  All you need to do is either drag the black point around the circle, or press play for the t slider.

More ideas on projective geometry:

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Ferenc Beleznay has made this nice geogebra file here which shows a different way of drawing a connection between a moving point on a large circle and a circle half the size. Here we connect the red dot with the origin and draw the perpendicular from this line to  the other edge of the small circle.  The point of intersection of the two lines is always on the small circle.

Further exploration 

There is a lot more you can explore – start by looking into the Tusi Couple – which is what we have just drawn – and the more general case the hypocycloid.

 

IB Maths Exploration Guide (suitable for all IB students: Applications and Interpretations, Analysis and Approaches). Written by an IB teacher with an MSc. in Mathematics, 10 years experience teaching IB Standard and Higher Level and who has worked as an IB examiner (SL coursework moderation).

You can download a 70-page pdf guide to the entire IA coursework process for the new syllabus (first exam 2021) to help you get excellent marks in your maths exploration.  The guide talks you through

  1. An introduction to the essentials about the investigation,
  2. The new marking criteria,
  3. How to choose a topic,
  4. Examples of around 70 topics that could be investigated,
  5. Useful websites for use in the exploration,
  6. A student checklist for completing a good investigation,
  7. Common mistakes that students make and how to avoid them,
  8. A worked example for the maths behind correlation investigations,
  9. General stats projects advice
  10. A selection of some interesting exploration topics explored in more depth,
  11. Teacher advice for marking,
  12. Templates for draft submissions,
  13. Advice on how to use Geogebra, Desmos and Tracker in your exploration,
  14. Some examples of beautiful maths using Geogebra and Desmos.

[This guide will be sent via email the same day and is not an automatic download]:

Maths Exploration Guide

A 70 page guide to help you get top marks on your maths investigation. If you have a PayPal account click on the PayPal symbol - if not click on the relevant credit card.

$5.50

 

The Martingale system

The Martingale system was first used in France in 1700s gambling halls and remains used today in some trading strategies.  I’ll look at some of the mathematical ideas behind this and why it has remained popular over several centuries despite having a long term expected return of zero.

The scenario

You go to a fair ground and play a simple heads-or-tails game.  The probability of heads is 1/2 and tails is also 1/2.  You place a stake of counters on heads.  If you guess correctly you win that number of counters.  If you lose, you double your stake of counters and then the coin is tossed again.  Every time you lose you double up your stake of counters and stop when you finally win.

Infinitely deep pockets model:


You can see that in the example above we always have a 0.5 chance of getting heads on the first go, which gives a profit of 1 counter.  But we also have a 0.5 chance of a profit of 1 counter as long as we keep doubling up our stake, and as long as we do indeed eventually throw heads.  In the example here you can see that the string of losing throws don’t matter [when we win is arbitrary, we could win on the 2nd, 3rd, 4th etc throw].  By doubling up, when you do finally win you wipe out your cumulative losses and end up with a 1 counter profit.

This leads to something of a paradoxical situation, despite only having a 1/2 chance of guessing heads we end up with an expected value of 1 counter profit for every 1 counter that we initially stake in this system.

So what’s happening?  This will always work but it requires that you have access to infinitely deep pockets (to keep your infinite number of counters) and also the assumption that if you keep throwing long enough you will indeed finally get a head (i.e you don’t throw an infinite number of tails!)

Finite pockets model:

Real life intrudes on the infinite pockets model – because in reality there will be a limit to how many counters you have which means you will need to bail out after a given number of tosses.  Even if the probability of this string of tails is very small, the losses if it does occur will be catastrophic –  and so the expected value for this system is still 0.

Finite pockets model capped at 4 tosses:

In the example above we only have a 1/16 chance of losing – but when we do we lose 15 counters.  This gives an expected value of:

Finite pockets model capped at n tosses:

If we start with a 1 counter stake then we can represent the pattern we can see above for E(X) as follows:

Here we use the fact that the losses from n throws are the sum of the first (n-1) powers of 2. We can then notice that both of these are geometric series, and use the relevant formula to give:

Therefore the expected value for the finite pockets model is indeed always still 0.

So why does this system remain popular?

So, given that the real world version of this has an expected value of 0, why has it retained popularity over the past few centuries?  Well, the system will on average return constant linear growth – up until a catastrophic loss.  Let’s say you have 100,000 counters and stake 1 counter initially.  You can afford a total of 16 consecutive losses.  The probability of this is only:

but when you do lose, you’ll lose a total of:

So, the system creates a model that mimics linear growth, but really the small risk of catastrophic loss means that the system still has E(X) = 0.  In the short term you would expect to see the following very simple linear relationship for profit:

With 100,000 counters and a base trading stake of 1 counter, if you made 1000 initial 1 counter trades a day you would expect a return of 1000 counters a day (i.e 1% return on your total counters per day).  However the longer you continue this strategy the more likely you are to see a run of 16 tails – and see all your counters wiped out.

Computer model

I wrote a short Python code to give an idea as to what is happening. Here I started 9 people off with 1000 counters each.  They have a loss limit of 10 consecutive losses.  They made starting stakes of 1 counter each time, and then I recorded how long before they made a loss of 10 tosses in a row.

For anyone interested in the code here it is:

 

The program returned the following results.  The first number is the number of starting trades until they tossed 10 tails in a row.  The second number was their new account value (given that they had started with 1000 counters, every previous trade had increased their account by 1 counter and that they had then just lost 1023 counters).

1338, 1315
1159, 1136
243, 220
1676, 1653
432, 409
1023, 1000
976, 953
990, 967
60, 37

This was then plotted on Desmos. The red line is the trajectory their accounts were following before their loss.  The horizontal dotted line is at y = 1000 which represents the initial account value.  As you can see 6 people are now on or below their initial starting account value.  You can also see that all these new account values are themselves on a line parallel to the red line but translated vertically down.

From this very simple simulation, we can see that on average a person was left with 884 counters following hitting 10 tails.  i.e below initial starting account.  Running this again with 99 players gave an average of 869.

999 players

I ran this again with 999 players – counting what their account value would be after their first loss.  All players started with 1000 counters.  The results were:

31 players bankrupt: 3%

385 players left with less than half their account value (less than 500): 39%

600 players with less than their original account value (less than 1000): 60%

51 players at least tripled their account (more than 3000): 5%

The top player ended up with 6903 counters after their first loss.

The average account this time was above starting value (1044.68).  You can see clearly that the median is below 1000 – but that a small number of very lucky players at the top end skewed the mean above 1000.

Second iteration

I then ran the simulation again – with players continuing with their current stake.  This would have been slightly off because my model allowed players who were bankrupt from the first round to carry on [in effect being loaned 1 counter to start again].  Nevertheless it now gave:

264 players bankrupt: 26%

453 players left with less than half their account value (less than 500): 45%

573 players with less than their original account value (less than 1000): 57%

95 players at least tripled their account (more than 3000): 10%

The top player ended up with 9583 counters after their second loss.

We can see a dramatic rise in bankruptcies – now over a quarter of all players.  This would suggest the long term trend is towards a majority of players being bankrupted, though the lucky few at the top end may be able to escape this fate.

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This carries on our exploration of projectile motion – this time we will explore what happens if gravity is not fixed, but is instead a function of time.  (This idea was suggested by and worked through by fellow IB teachers Daniel Hwang and Ferenc Beleznay).   In our universe we have a gravitational constant – i.e gravity is not dependent on time.  If gravity changed with respect to time then the gravitational force exerted by the Sun on Earth would lessen (or increase) over time with all other factors remaining the same.

Interestingly time-dependent gravity was first explored by Dirac and some physicists have tried to incorporate time dependent gravity into cosmological models.  As yet we have no proof that gravity is not constant, but let’s imagine a university where it is dependent on time.

Projectile motion when gravity is time dependent

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We can start off with the standard parametric equations for projectile motion. Here v is the initial velocity, theta is the angle of launch, t can be a time parameter and g is the gravitational constant (9.81 on Earth).  We can see that the value for the vertical acceleration is the negative of the gravitational constant.  So the question to explore is, what if the gravitational constant was time dependent?  Another way to think about this is that gravity varies with respect to time.

Linear relationship

If we have the simplest time dependent relationship we can say that:

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where a is a constant.  If a is greater than 0 then gravity linearly increases as time increases, if a is less than 0 than gravity linearly decreases as time increases.  For matters of slight convenience I’ll define gravity (or the vertical acceleration) as -3at.  The following can then be arrived at by integration:

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This will produce the following graph when we fix v = 10, a = 2 and vary theta:

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Now we can use the same method as in our Projectile Motion Investigation II to explore whether these maximum points lie in a curve.  (You might wish to read that post first for a step by step approach to the method).

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therefore we can substitute back into our original parametric equations for x and y to get:

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We can plot this with theta as a parameter.  If we fix v = 4 and a =2 we get the following graph:

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Compare this to the graph from Projectile Motion Investigation II, where we did this with gravity constant (and with v fixed as 10):

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The Projectile Motion Investigation II formed a perfect ellipse, but this time it’s more of a kind of egg shaped elliptical curve – with a flat base.  But it’s interesting to see that even with time dependent gravity we still have a similar relationship to before!

Inverse relationship

Let’s also look at what would happen if gravity was inversely related to time.  (This is what has been explored by some physicists).

In this case we get the following results when we launch projectiles (Notice here we had to use the integration by parts trick to integrate ln(t)).  As the velocity function doesn’t exist when t = 0, we can define v and theta in this case as the velocity and theta value when t = 1.

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Now we use the same trick as earlier to find when the gradient is 0:

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Substituting this back into the parametric equations gives:

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The ratio v/a will therefore have the greatest effect on the maximum points.

v/a ratio negative and close to zero:

v = 40, a = -2000, v/a = -0.02

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This gives us close to a circle, radius v, centred at (0,a).

v = 1, a = -10, v/a = -0.1

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Here we can also see that the boundary condition for the maximum horizontal distance thrown is given by x = v(e).

v/a ratio negative and large:

v = 40, a = -2, v/a = -20.

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We can see that we get an egg shape back – but this time with a flatter bulge at the top and the point at the bottom.  Also notice how quickly the scale of the shape has increased.

v/a ratio n/a (i.e a = 0)

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Here there is no gravitational force, and so projectiles travel in linear motion – with no maximum.

Envelope of projectiles for the inverse relationship

This is just included for completeness, don’t worry if you don’t follow the maths behind this bit!

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Therefore when we plot the parametric equations for x and y in terms of theta we get the envelope of projectile motion when we are in a universe where gravity varies inversely to time.  The following graph is generated when we take v = 300 and a = -10.  The red line is the envelope of projectiles.

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A generalized power relationship

Lastly, let’s look at what happens when we have a general power relationship i.e gravity is related to (a)tn.  Again for matters of slight convenience I’ll look at the similar relationship -0.5(n+1)(n+2)atn.

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This gives (following the same method as above:

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As we vary n we will find the plot of the maximum points.  Let’s take the velocity as 4 and a as 2.  Then we get the following:

When n = 0:

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When n = 1:

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When n =2:

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When n = 10:

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We can see the general elliptical shape remains at the top, but we have a flattening at the bottom of the curve.

When n approaches infinity:

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We get this beautiful result when we let n tend towards infinity – now we will have all the maximum points bounded on a circle (with the radius the same as the value chosen as the initial velocity.  In the graph above we have a radius of 4 as the initial velocity is 4. Notice too we have projectiles traveling in straight lines – and then seemingly “bouncing” off the boundary!

If we want to understand this, there is only going to be a very short window (t less than 1) when the particle can upwards – when t is between 0 and 1 the effect of gravity is effectively 0 and so the particle would travel in a straight line (i.e if the initial velocity is 5 m/s it will travel 5 meters. Then as soon as t = 1, the gravity becomes crushingly heavy and the particle falls effectively vertically down.

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Projectile Motion III: Varying gravity

We can also do some interesting things with projectile motion if we vary the gravitational pull when we look at projectile motion.  The following graphs are all plotted in parametric form.

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Here t is the parameter, v is the initial velocity which we will keep constant, theta is the angle of launch which we will vary, and g is the gravitational constant which on Earth we will take as 9.81 m/s2.

Earth 

Say we take a projectile and launch it with a velocity of 10 m/s.  When we vary the angle of launch we get the folowing graphs:

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On the y axis we have the vertical height, and on the x axis the horizontal distance.  Therefore we can see that the maximum height that we achieve is around 5m and the maximum horizontal distance is around 10m.

Other planets and universal objects

We have the following values for the gravitational pull of various objects:

Enceladus (Moon of Saturn): 0.111 m/s2, The Moon: 1.62 m/s2,  Jupiter: 24.8 m/s2, The Sun: 274 m/s2, White dwarf black hole surface gravity: 7×1012m/s2.

So for each one let’s see what would happen if we launched a projectile with a velocity of 10 m/s.  Note that the mass of the projectile is not relevant (though it would require more force to achieve the required velocity).

Enceladus:

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The Moon:

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Jupiter:

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The Sun:

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Black hole surface gravity:

This causes some issues graphically!  I’ll use the equations derived in the last post to find the coordinates of the maximum point for a given launch angle theta:

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Here we have v = 10 and g = 7×1012m/s2.  For example if we take our launch angle (theta) as 45 degrees this will give the coordinates of the maximum point at:

(7.14×10-12, 3.57×10-12).

Summary:

We can see how dramatically life would be on each surface!  Whilst on Earth you may be able to throw to a height of around 5m with a launch velocity of 10 m/s., Enceladus  would see you achieve an incredible 450m.  If you were on the surface of the Sun then probably the least of your worries would be how hight to throw an object, nevertheless you’d be struggling to throw it 20cm high.  And as for the gravity at the surface of a black hole you wouldn’t get anywhere close to throwing it a nanometer high (1 billionth of a meter).

 

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Projectile Motion Investigation II

Another example for investigating projectile motion has been provided by fellow IB teacher Ferenc Beleznay.  Here we fix the velocity and then vary the angle, then to plot the maximum points of the parabolas.  He has created a Geogebra app to show this (shown above).  The locus of these maximum points then form an ellipse.

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We can see that the maximum points of the projectiles all go through the dotted elliptical line.  So let’s see if we can derive this equation.

Let’s start with the equations for projectile motion, usually given in parametric form:

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Here v is the initial velocity which we will keep constant, theta is the angle of launch which we will vary, and g is the gravitational constant which we will take as 9.81.

We can plot these curves parametrically, and for each given value of theta (the angle of launch) we will create a projectile motion graph.  If we plot lots of these graphs for different thetas together we get something like this:

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We now want to see if the maximum points are in any sort of pattern.  In order to find the maximum point we want to find when the gradient of dy/dx is 0.  It’s going to be easier to keep things in parametric form, and use partial differentiation.  We have:

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Therefore we find the partial differentiation of both x and y with respect to t.  (This simply means we can pretend theta is a constant).

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We can then say that:

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We then find when this has a gradient of 0:

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We can then substitute this value of t back into the original parametric equations for x:

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and also for y:

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We now have the parametric equations in terms of theta for the locus of points of the maximum points.  For example, with g= 9.81 and v =1 we have the following parametric equations:

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This generates an ellipse (dotted line), which shows the maximum points generated by the parametric equations below (as we vary theta):

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And here is the graph:

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We can vary the velocity to create a new ellipse.  For example the ellipse generated when v =4 creates the following graph:

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So, there we go, we have shown that different ellipses will be created by different velocities.  If you feel like a challenge, see if you can algebraically manipulate the parametric equations for the ellipse into the Cartesian form!

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Envelope of projectile motion

For any given launch angle and for a fixed initial velocity we will get projectile motion. In the graph above I have changed the launch angle to generate different quadratics.  The black dotted line is then called the envelope of all these lines, and is the boundary line formed when I plot quadratics for every possible angle between 0 and pi.

Finding the equation of an envelope for projectile motion 

Let’s start with the equations for projectile motion, usually given in parametric form:

Screen Shot 2020-04-06 at 1.17.08 PM

Here v is the initial velocity which we will keep constant, theta is the angle of launch which we will vary, and g is the gravitational constant which we will take as 9.81.

First let’s rearrange these equations to eliminate the parameter t.

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Next, we use the fact that the envelope of a curve is given by the points which satisfy the following 2 equations:

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F(x,y,theta)=0 simply means we have rearranged an equation so that we have 3 variables on one side and have made this equal to 0.  The second of these equations means the partial derivative of F with respect to theta.  This means that we differentiate as usual with regards to theta, but treat x and y like constants.

Therefore we can rearrange our equation for y to give:

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and in order to help find the partial differential of F we can write:

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We can then rearrange this to get x in terms of theta:

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We can then substitute this into the equation for F(x,y,theta)=0 to eliminate theta:

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We then have the difficulty of simplifying the second denominator, but luckily we have a trig equation to help:

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Therefore we can simplify as follows:

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and so:

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And we have our equation for the envelope of projectile motion!  As we can see it is itself a quadratic equation.  Let’s look at some of the envelopes it will create.  For example, if I launch a projectile with a velocity of 1, and taking g = 9.81, I get the following equation:

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This is the envelope of projectile motion when I take the following projectiles in parametric form and vary theta from 0 to pi:

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This gives the following graph:

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If I was to take an initial velocity of 2 then I would have the following:

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And an initial velocity of 4 would generate the following graph:

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So, there we have it, we can now create the equation of the envelope of curves created by projectile motion for any given initial velocity!

Other ideas for projectile motion

There are lots of other things we can investigate with projectile motion.  One example provided by fellow IB teacher Ferenc Beleznay is to fix the velocity and then vary the angle, then to plot the maximum points of the parabolas.  He has created a Geogebra app to show this:

Screen Shot 2020-04-06 at 3.25.43 PM

You can then find that the maximum points of the parabolas lie on an ellipse (as shown below).

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See if you can find the equation of this ellipse!

Classical Geometry Puzzle: Finding the Radius

This is another look at a puzzle from Mind Your Decisions.  The problem is to find the radius of the following circle:

We are told that line AD and BC are perpendicular and the lengths of some parts of chords, but not much more!  First I’ll look at my attempt to solve this.  It’s not quite as “nice” as the solution in the video as it requires the use of a calculator, but it still does the job.

Method 1, extra construction lines:

These are the extra construction lines required to solve this problem.  Here is the step by step thought process:

  1. Find the hypotenuse of triangle AGC.
  2. Use the circle theorem angles in the same segment are equal to show that angle CBD = angle CAG.
  3. Therefore triangle AGC and GBD are similar, so length BG = 4.  We can now use Pythagoras to find length BD.
  4. We can find length CD by Pythagoras.
  5. Now we have 3 sides of a triangle, CDB.  This allows use to find angle BDC using the cosine rule.
  6. Now we the circle theorem angles in the same segment are equal to show that angle BDC = angle BEC.
  7. Now we use the circle theorem angles in a semi circle are 90 degrees to show ECB = 90.
  8. Now we have a right angled triangle BCE where we know both an angle and a side, so can use trigonometry to find the length of BE.
  9. Therefore the radius is approximately 4.03.

Method 2, creating a coordinate system

This is a really beautiful solution – which does not require a calculator (and which is discussed in the video above).  We start by creating a coordinate system based around point G at (0,0).  Because we have perpendicular lines we can therefore create coordinates for A, B and C.  We also mark the centre of the circle as (p,q).

First we start with the equation of a circle centre (p.q):

Next we create 3 equations by substituting in our coordinates:


Next we can do equation (3) – equation (1) to give:

Next we can substitute this value for p into equations (1) and (3) and equate to get:

Lastly we can substitute both values for p and q into equation (1) to find r:

We get the same answer as before – though this definitely feels like a “cleaner” solution.  There are other ways to solve this – but some of these require the use of equations you may not already know (such as the law of sines in a circumcircle, or the equation for perpendicular chords and radius).  Perhaps explore any other methods for solving this – what are the relative merits of each approach?

Further investigation of the Mordell Equation

This post carries on from the previous post on the Mordell Equation – so make sure you read that one first – otherwise this may not make much sense.  The man pictured above (cite: Wikipedia) is Louis Mordell who studied the equations we are looking at today (and which now bear his name).

In the previous post I looked at solutions to the difference between a cube and a square giving an answer of 2.  This time I’ll try to generalise to the difference between a cube and a square giving an answer of k.  I’ll start with the same method as from the previous post:

In the last 2 lines we outline the 2 possibilities, either b = 1 or b = -1.  First let’s see what happens when b = 1:

This will only provide an integer solution for a if we have:

Which generates the following first few values for k when we run through m = 1, 2,3..:

k = 2, 11, 26, 47

We follow the same method for b = -1 and get the following:

Which generates the following first few values for k when we run through m = 1, 2,3…:

k = 4, 13, 28, 49

These are the values of k which we will be able to generate solutions to. Following the same method as in the previous post this generates the following solutions:

Let’s illustrate one of these results graphically.  If we take the solutions for k = 13, which are (17,70) and (17,-70), these points should be on the curve x cubed – y squared = 13.

This is indeed the case.  This graph also demonstrates how all solutions to these curves will have symmetrical solutions (e, f) and (e, -f).

We can run a quick computer program to show that this method does not find all the solutions for the given values of k, but it does ensure solutions will be found for the k values in these lists.

In the code solutions above, results are listed k, x, y, x cubed, y squared.  We can see for example that in the case of k = 11 our method did not find the solution x = 3 and y = 4 (though we found x = 15 and y = 58).  So, using this method we now have a way of finding some solutions for some values of k – we’ve not cracked the general case, but we have at least made a start!

 

The Mordell Equation [Fermat’s proof]

Let’s have a look at a special case of the Mordell Equation, which looks at the difference between an integer cube and an integer square.  In this case we want to find all the integers x,y such that the difference between the cube and the square gives 2.  These sorts of problems are called Diophantine problems and have been studied by mathematicians for around 2000 years.  We want to find integer solution to:

First we can rearrange and factorise, using the property of imaginary numbers.

Next we define alpha and beta such that:

For completeness we can say that alpha and beta are part of an algebraic number field:

Next we use an extension of the Coprime Power Trick, which ensures that the following 2 equations have solutions (if our original equation also has a solution). Therefore we define:

We can then substitute our definition for alpha into the first equation directly above and expand:

Next we equate real and imaginary coefficients to give:

This last equation therefore requires that either one of the following equations must be true:

If we take the case when b = 1 we get:

Therefore:

If we take the case when b = -1 we get

Therefore our solution set is (a,b): (1,1), (1,-1), (-1,1), (-1,-1.  We substitute these possible answers into our definition for y to give the following:

We can then substitute these 2 values for y into the definition for x to get:

These therefore are the only solutions to our original equation.  We can check they both work:

We can see this result illustrated graphically by plotting the graph:

and then seeing that we have our integer solutions (3,5) and (3,-5) as coordinate on this curve.

This curve also clearly illustrates why we have a symmetrical set of solutions, as our graph is symmetrical about the x axis.

This particular proof was first derived by Fermat (of Fermat’s Last Theorem fame) in the 1600s and is an elegant example of a proof in number theory.  You can read more about the Mordell Equation in this paper (the proof above is based on that given in the paper, but there is a small mistake in factorization so that y = 7 and y = -7 is erroneously obtained)

 

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