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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:

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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.

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:

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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!

Rational Approximations to Irrational Numbers

This year two mathematicians (James Maynard and Dimitris Koukoulopoulos) managed to prove a long-standing Number Theory problem called the Duffin Schaeffer Conjecture.  The problem is concerned with the ability to obtain rational approximations to irrational numbers.  For example, a rational approximation to pi is 22/7.  This gives 3.142857 and therefore approximates pi to 2 decimal places.  You can find ever more accurate rational approximations and the conjecture looks at how efficiently we can form these approximation, and to within what error bound.

Finding Rational Approximations for  pi

The general form of the inequality I want to solve is as follows:

Here alpha is an irrational number, p/q is the rational approximation, and f(q)/q can be thought of as the error bound that I need to keep my approximation within.

If I take f(q) = 1/q then I will get the following error bound:

So, the question is, can I find some values of q (where p and q are integers) such that the error bound is less than 1/(q squared)?

Let’s see if we can solve this for when our irrational number is pi, and when we choose q = 6.

We can see that this returns a rational approximation, 19/6 which only 0.02507… away from  pi. This is indeed a smaller error than 1/36.  We won’t be able to find such solutions to our inequality for every value of q that we choose, but we will be able to find an infinite number of solutions, each getting progressively better at approximating pi.

The General Case (Duffin Schaeffer Conjecture)

The general case of this problem states that there will be infinite solutions to the inequality for any given irrational number alpha if and only if the following condition holds:

For:

We will have infinitely many solutions (with p and q as integers in their lowest terms) if and only if:

Here the new symbol represents the Euler totient.  You can read about this at the link if you’re interested, but for the purposes of the post we can transform into something else shortly!

Does f(q) = 1/q provide infinite solutions?

When f(q) = 1/q we have:

Therefore we need to investigate the following sum to infinity:

Now we can make use of an equivalence, which shows that:

 

Where the new symbol on the right is the Zeta function.  The Zeta function is defined as:

So, in our case we have s = 2.  This gives:

But we know the limit of both the top and the bottom sum to infinity.  The top limit is called the Harmonic series, and diverges to infinity. Therefore:

Whereas the bottom limit is a p-series with p=2, this is known to converge. In fact we have:

Therefore because we have a divergent series divided by a convergent one, we will have the following result:

This shows that our error bound 1/(q squared) will be satisfied by infinitely many values of q for any given irrational number.

Does f(q) = 1/(q squared) provide infinite solutions?

With f(q) = 1/(q squared) we follow the same method to get:

But this time we have:

Therefore we have a convergent series divided by a convergent series which means:

So we can conclude that f(q) = 1/(q squared) which generated an error bound of 1/(q cubed) was too ambitious as an error bound – i.e there will not be infinite solutions in p and q for a given irrational number.  There may be solutions out there but they will be rare.

Understanding mathematicians 

You can watch the Numberphile video where James Maynard talks through the background of his investigation and also get an idea what a mathematician feels like when they solve a problem like this!

Soap Bubbles and Catenoids

Soap bubbles form such that they create a shape with the minimum surface area for the given constraints.  For a fixed volume the minimum surface area is a sphere, which is why soap bubbles will form spheres where possible.  We can also investigate what happens when a soap film is formed between 2 parallel circular lines like in the picture below: [Credit Wikimedia Commons, Blinking spirit]


In this case the shape formed is a catenoid – which provides the minimum surface area (for a fixed volume) for a 3D shape connecting the two circles.  The catenoid can be defined in terms of parametric equations:

Where cosh() is the hyperbolic cosine function which can be defined as:

For our parametric equation, t and u are parameters which we vary, and c is a constant that we can change to create different catenoids.  We can use Geogebra to plot different catenoids.  Below is the code which will plot parametric curves when c =2 and t varies between -20pi and 20 pi.

 

We then need to create a slider for u, and turn on the trace button – and for every given value of u (between 0 and 2 pi) it will plot a curve.  When we trace through all the values of u it will create a 3D shape – our catenoid.

Individual curve (catenary)


Catenoid when c = 0.1

Catenoid when c = 0.5

Catenoid when c = 1

Catenoid when c = 2

Wormholes

For those of you who know your science fiction, the catenoids above may look similar to a wormhole.  That’s because the catenoid is a solution to the hypothesized mathematics of wormholes.  These can be thought of as a “bridge” either through curved space-time to another part of the universe (potentially therefore allowing for faster than light travel) or a bridge connecting 2 distinct universes.

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Above is the Morris-Thorne bridge wormhole [Credit The Image of a Wormhole].

Further exploration:

This is a topic with lots of interesting areas to explore – the individual curves (catenary) look similar to, but are distinct from parabola.  These curves appear in bridge building and in many other objects with free hanging cables.  Proving that catenoids form shapes with minimum surface areas requires some quite complicated undergraduate maths (variational calculus), but it would be interesting to explore some other features of catenoids or indeed to explore why the sphere is a minimum surface area for a given volume.

If you want to explore further you can generate your own Catenoids with the Geogebra animation I’ve made here.

Normal Numbers – and random number generators

Numberphile have a nice new video where Matt Parker discusses all different types of numbers – including “normal numbers”.  Normal numbers are defined as irrational numbers for which the probability of choosing any given 1 digit number is the same, the probability of choosing any given 2 digit number is the same etc.  For example in the normal number 0.12345678910111213141516… , if I choose any digit in the entire number at random P(1) = P(2) = P(3) = … P(9) = 1/10.  Equally if I choose any 2 digit number at random I have P(10) = P(11) = P(12) = P(99) = 1/100.

It is incredibly hard to find normal numbers, but there is a formula to find some of them.

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In base 10, we are restricted to choosing a value of c such that 10 and c are relatively prime (i.e share no common factors apart from 1).  So if we choose c = 3 this gives:

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We can now put this into Wolfram Alpha and see what number this gives us:

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So we can put the first few digits into an online calculator to find the distributions

0.000333333444444444444448148148148148148148148148148148148148148149382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049382716049827160493827160493827160479423863312 7572016460905349794238683127572016460905349794238683127572016460 9053497942386831275720164609053497942386831275720164609053497942

4: 61
1: 41
8: 40
3: 38
0: 36
2: 33
7: 33
9: 33
6: 32
5: 10

We can see that we are already seeing a reasonably similar distribution of single digits, though with 4 and 5 outliers.  As the number progressed we would expect these distributions to even up (otherwise it would not be a normal number).

One of the potential uses of normal numbers is in random number generators – if you can use a normal number and specify a digit (or number of digits) at random then this should give an equal chance of returning each number.

To finish off this,  let’s prove that the infinite series:

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does indeed converge to a number (if it diverged then it could not be used to represent a real number).  To do that we can use the ratio test (only worry about this bit if you have already studied the Calculus Option for HL!):

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We can see that in the last limit 3 to the power n+1 will grow faster than 3 to the power n, therefore as n increases the limit will approach 0.  Therefore by the ratio test the series converges to a real number.

Is pi normal?

Interestingly we don’t know if numbers like e, pi and ln(2) are normal or not.  We can analyse large numbers of digits of pi – and it looks like it will be normal, but as yet there is no proof.  Here are the distribution of the first 100,000 digits of pi:

1: 10137
6: 10028
3: 10026
5: 10026
7: 10025
0: 9999
8: 9978
4: 9971
2: 9908
9: 9902

Which we can see are all very close to the expected value of 10,000 (+/- around 1%).

So, next I copied the first 1 million digits of pi into a character frequency counter which gives the following:

5: 100359
3: 100230
4: 100230
9: 100106
2: 100026
8: 99985
0: 99959
7: 99800
1: 99758
6: 99548

This is even closer to the expected values of 100,000 with most with +/- 0.25 %.

Proving that pi is normal would be an important result in number theory – perhaps you could be the one to do it!

 IB Revision

If you’re already thinking about your coursework then it’s probably also time to start planning some revision, either for the end of Year 12 school exams or Year 13 final exams. There’s a really great website that I would strongly recommend students use – you choose your subject (HL/SL/Studies if your exam is in 2020 or Applications/Analysis if your exam is in 2021), and then have the following resources:

Screen Shot 2018-03-19 at 4.42.05 PM.pngThe Questionbank takes you to a breakdown of each main subject area (e.g. Algebra, Calculus etc) and each area then has a number of graded questions. What I like about this is that you are given a difficulty rating, as well as a mark scheme and also a worked video tutorial.  Really useful!

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The Practice Exams section takes you to ready made exams on each topic – again with worked solutions.  This also has some harder exams for those students aiming for 6s and 7s and the Past IB Exams section takes you to full video worked solutions to every question on every past paper – and you can also get a prediction exam for the upcoming year.

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

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Cartoon from here

The Gini Coefficient – Measuring Inequality 

The Gini coefficient is a value ranging from 0 to 1 which measures inequality. 0 represents perfect equality – i.e everyone in a population has exactly the same wealth.  1 represents complete inequality – i.e 1 person has all the wealth and everyone else has nothing.  As you would expect, countries will always have a value somewhere between these 2 extremes.  The way its calculated is best seen through the following graph (from here):

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The Gini coefficient is calculated as the area of A divided by the area of A+B.  As the area of A decreases then the curve which plots the distribution of wealth (we can call this the Lorenz curve) approaches the line y = x.  This is the line which represents perfect equality.

Inequality in Thailand

The following graph will illustrate how we can plot a curve and calculate the Gini coefficient.  First we need some data.  I have taken the following information on income distribution from the 2002 World Bank data on Thailand where I am currently teaching:

Thailand:

The bottom 20% of the population have 6.3% of the wealth
The next 20% of the population have 9.9% of the wealth
The next 20%  have 14% of the wealth
The next 20% have 20.8% of the wealth
The top 20% have 49% of the wealth

I can then write this in a cumulative frequency table (converting % to decimals):

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Here the x axis represents the cumulative percentage of the population (measured from lowest to highest), and the y axis represents the cumulative wealth.  This shows, for example that the the bottom 80% of the population own 51% of the wealth.  This can then be plotted as a graph below (using Desmos):

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From the graph we can see that Thailand has quite a lot of inequality – after all the top 20% have just under 50% of the wealth.  The blue line represents how a perfectly equal society would look.

To find the Gini Coefficient we first need to find the area between the 2 curves.  The area underneath the blue line represents the area A +B.  This is just the area of a triangle with length and perpendicular height 1, therefore this area is 0.5.

The area under the green curve can be found using the trapezium rule, 0.5(a+b)h.  Doing this for the first trapezium we get 0.5(0+0.063)(0.2) = 0.0063.  The second trapezium is 0.5(0.063+0.162)(0.2) and so on.  Adding these areas all together we get a total trapezium area of 0.3074.  Therefore we get the area between the two curves as 0.5 – 0.3074  ≈ 0.1926

The Gini coefficient is then given by 0.1926/0.5  = 0.3852.

The actual World Bank calculation for Thailand’s Gini coefficient in 2002 was 0.42 – so we have slightly underestimated the inequality in Thailand.  We would get a more accurate estimate by taking more data points, or by fitting a curve through our plotted points and then integrating.  Nevertheless this is a good demonstration of how the method works.

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In this graph (from here) we can see a similar plot of wealth distribution – here we have quintiles on the x axis (1st quintile is the bottom 20% etc).  This time we can compare Hungary – which shows a high level of equality (the bottom 80% of the population own 62.5% of the wealth) and Namibia – which shows a high level of inequality (the bottom 80% of the population own just 21.3% of the wealth).

How unequal is the world?

Screen Shot 2016-02-09 at 10.43.23 PM

We can apply the same method to measure world inequality.  One way to do this is to calculate the per capita income of all the countries in the world and then to work out the share of the total global per capita income the (say) bottom 20% of the countries have.  This information is represented in the graph above (from here).  It shows that there was rising inequality (i.e the richer countries were outperforming the poorer countries) in the 2 decades prior to the end of the century, but that there has been a small decline in inequality since then.

If you want to do some more research on the Gini coefficient you can use the following resources:

The intmaths site article on this topic – which goes into more detail and examples of how to calculate the Gini coefficient

The ConferenceBoard site which contains a detailed look at world inequality

The World Bank data on the Gini coefficients of different countries.

IB Revision

Screen Shot 2018-03-19 at 4.35.19 PM

If you’re already thinking about your coursework then it’s probably also time to start planning some revision, either for the end of Year 12 school exams or Year 13 final exams. There’s a really great website that I would strongly recommend students use – you choose your subject (HL/SL/Studies if your exam is in 2020 or Applications/Analysis if your exam is in 2021), and then have the following resources:

Screen Shot 2018-03-19 at 4.42.05 PM.pngThe Questionbank takes you to a breakdown of each main subject area (e.g. Algebra, Calculus etc) and each area then has a number of graded questions. What I like about this is that you are given a difficulty rating, as well as a mark scheme and also a worked video tutorial.  Really useful!

Screen Shot 2019-07-27 at 10.02.40 AM

The Practice Exams section takes you to ready made exams on each topic – again with worked solutions.  This also has some harder exams for those students aiming for 6s and 7s and the Past IB Exams section takes you to full video worked solutions to every question on every past paper – and you can also get a prediction exam for the upcoming year.

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

Screen Shot 2019-02-09 at 9.30.23 PM.png

Is Intergalactic space travel possible?

The Andromeda Galaxy is around 2.5 million light years away – a distance so large that even with the speed of light at traveling as 300,000,000m/s it has taken 2.5 million years for that light to arrive.  The question is, would it ever be possible for a journey to the Andromeda Galaxy to be completed in a human lifetime?  With the speed of light a universal speed limit, it would be reasonable to argue that no journey greater than around 100 light years would be possible in the lifespan of a human – but this remarkably is not the case.  We’re going to show that a journey to Andromeda would indeed be possible within a human lifespan.   All that’s needed (!) is a rocket which is able to achieve constant acceleration and we can arrive with plenty of time to spare.

Time dilation

Screen Shot 2019-02-09 at 9.40.06 PM

To understand how this is possible, we need to understand that as the speed of the journey increases, then time dilation becomes an important factor.  The faster the rocket is traveling the greater the discrepancy between the internal body clock of the astronaut on the rocket and an observer on Earth.  Let’s see how that works in practice by using the above equation.

Here we have

t(T): The time elapsed from the perspective of an observer on Earth

T: The time elapsed from the perspective of an astronaut on the rocket

c: The speed of light approx 300,000,000 m/s

a: The constant acceleration we assume for our rocket.  For this example we will take a = 9.81 m/s2 which is the same as the gravity experienced on Earth. This would be the most natural for a human environment.  The acceleration is measured relative to an inert observer.

Sinh(x): This is the hyperbolic sine function which can be defined as:

Screen Shot 2019-02-09 at 9.40.13 PM

We should note that all our units are in meters, seconds and m/s2 therefore when the astronaut experiences 1 year passing on this rocket, we first need to convert this to seconds:  1 year = 60 x 60 x 24 x 365 = 31,536,000 seconds.  Therefore T = 31,536,000 and:

Screen Shot 2019-02-09 at 9.40.06 PM

Screen Shot 2019-02-09 at 10.32.27 PM

which would give us the time experienced on Earth in seconds, therefore by dividing by (60 x 60 x 24 x 365) we can arrive at the time experienced on Earth in years:

Screen Shot 2019-02-09 at 10.31.33 PM

Using either Desmos or Wolfram Alpha this gives an answer  of 1.187.  This means that 1 year experienced on the rocket is experienced as 1.19 years on Earth.  Now we have our formula we can easily calculate other values.  Two years is:

Screen Shot 2019-02-09 at 10.32.08 PM

which gives an answer of 3.755 years.  So 2 years on the rocket is experienced as 3.76 years on Earth.  As we carry on with the calculations, and as we see the full effects of time dilation we get some startling results:

Screen Shot 2019-02-09 at 10.11.18 PM

After 10 years on the space craft, civilization on Earth has advanced (or not) 15,000 years.  After 20 years on the rocket, 445,000,000 years have passed on Earth and after 30 years 13,500,000,000,000 years, which around 1000 times greater than the age of the Universe post Big Bang.  So, as we can see, time is no longer a great concern.

Distance travelled

Next let’s look at how far we can reach from Earth.  This is given by the following equation:

Screen Shot 2019-02-09 at 9.40.03 PM

Here we have

x(T): The distance travelled from Earth

T, c and a as before.

Cosh(x): This is the hyperbolic cosine function which can be defined as:

Screen Shot 2019-02-09 at 9.40.17 PM

Again we note that we are measuring in meters and seconds.  Therefore to find the distance travelled in one year we convert 1 year to seconds as before:

Screen Shot 2019-02-09 at 10.37.38 PM

Next we note that this will give an answer in meters, so we can convert to light years by dividing by 9.461×1015

Screen Shot 2019-02-09 at 10.37.02 PM

Again using Wolfram Alpha or Desmos we find that after one year the spacecraft will be 0.563 light years from Earth.  After two years we have:

Screen Shot 2019-02-09 at 10.36.27 PM

which gives us 2.91 light years from Earth.  Calculating the next values gives us the following table:

Screen Shot 2019-02-09 at 10.45.34 PM

We can see that as our spacecraft approaches the speed of light, we will travel the expected number of light years as measured by an observer on Earth.

So we can see that we would easily reach the Andromeda Galaxy within 20 years on a spacecraft and could have spanned the size of the observable universe within 30 years.   Now, all we need is to build a spaceship capable of constant acceleration, and resolve how a human body could cope with such forces and we’re there!

How likely is this?

Screen Shot 2019-02-10 at 7.31.07 AM.png

Well, the technology needed to build a spacecraft capable of constant acceleration to get close to light speed is not yet available – but there are lots of interesting ideas about how these could be designed in theory.  One of the most popular ideas is to make a “solar sail” – which would collect light from the Sun (or any future nearby stars) to propel it along on its journey.  Another alternative would be a laser sail – which rather than relying on the Sun, would receive pin-point laser beams from the Earth.

Equally we are a long way from being able to send humans – much more likely is that the future of spaceflight will be carried out by machines.  Scientists have suggested that if the spacecraft payload was around 1 gram (say either a miniaturized robot or digital data depending on the mission’s aim), a solar sail or laser sail could be feasibly built which would be sufficient to achieve 25% the speed of light.

NASA have begun launching continuous acceleration spacecraft powered by the Sun.  In 2018 they launched the  Near-Earth Asteroid Scout.  This will unfurl a solar sail and be propelled to a speed of 28,600 m/s.  Whilst this is a long way from near-light speeds, it is a proof of concept and does show one potential way that interstellar travel could be achieved.

You can read more about the current scientific advances on solar sails here, and some more on the mathematics of space travel here.

IB Revision

Screen Shot 2018-03-19 at 4.35.19 PM

If you’re already thinking about your coursework then it’s probably also time to start planning some revision, either for the end of Year 12 school exams or Year 13 final exams. There’s a really great website that I would strongly recommend students use – you choose your subject (HL/SL/Studies if your exam is in 2020 or Applications/Analysis if your exam is in 2021), and then have the following resources:

Screen Shot 2018-03-19 at 4.42.05 PM.pngThe Questionbank takes you to a breakdown of each main subject area (e.g. Algebra, Calculus etc) and each area then has a number of graded questions. What I like about this is that you are given a difficulty rating, as well as a mark scheme and also a worked video tutorial.  Really useful!

Screen Shot 2019-07-27 at 10.02.40 AM

The Practice Exams section takes you to ready made exams on each topic – again with worked solutions.  This also has some harder exams for those students aiming for 6s and 7s and the Past IB Exams section takes you to full video worked solutions to every question on every past paper – and you can also get a prediction exam for the upcoming year.

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

The Folium of Descartes

The folium of Descartes is a famous curve named after the French philosopher and mathematician Rene Descartes (pictured top right).  As well as significant contributions to philosophy (“I think therefore I am”) he was also the father of modern geometry through the development of the x,y coordinate system of plotting algebraic curves.  As such the Cartesian plane (as we call the x,y coordinate system) is named after him.

Screen Shot 2018-02-25 at 6.59.40 PM

Pascal and Descartes

Descartes was studying what is now known as the folium of Descartes (folium coming from the Latin for leaf) in the first half of the 1600s.  Prior to the invention of calculus, the ability to calculate the gradient at a given point was a real challenge.  He placed a wager with Pierre de Fermat, a contemporary French mathematician (of Fermat’s Last Theorem fame) that Fermat would be unable to find the gradient of the curve – a challenge that Fermat took up and succeeded with.

Calculus – implicit differentiation:

Today, armed with calculus and the method of implicit differentiation, finding the gradient at a point for the folium of Descartes is more straightforward.  The original Cartesian equation is:

Screen Shot 2018-02-25 at 6.59.46 PM

which can be differentiated implicitly to give:

Screen Shot 2018-02-25 at 7.25.54 PM

Therefore if we take (say) a =1 and the coordinate (1.5, 1.5) then we will have a gradient of -1.

Parametric equations

It’s sometimes easier to express a curve in a different way to the usual Cartesian equation.  Two alternatives are polar coordinates and parametric coordinates.  The parametric equations for the folium are given by:

Screen Shot 2018-02-25 at 6.59.50 PM

In order to use parametric equations we simply choose a value of t (say t =1) and put this into both equations in order to arrive at a coordinate pair in the x,y plane.  If we choose t = 1 and have set a = 1 as well then this gives:

x(1) = 3/2

y(1) = 3/2

therefore the point (1.5, 1.5) is on the curve.

You can read a lot more about famous curves and explore the maths behind them with the excellent “50 famous curves” from Bloomsburg University.

IB Revision

Screen Shot 2018-03-19 at 4.35.19 PM

If you’re already thinking about your coursework then it’s probably also time to start planning some revision, either for the end of Year 12 school exams or Year 13 final exams. There’s a really great website that I would strongly recommend students use – you choose your subject (HL/SL/Studies if your exam is in 2020 or Applications/Analysis if your exam is in 2021), and then have the following resources:

Screen Shot 2018-03-19 at 4.42.05 PM.pngThe Questionbank takes you to a breakdown of each main subject area (e.g. Algebra, Calculus etc) and each area then has a number of graded questions. What I like about this is that you are given a difficulty rating, as well as a mark scheme and also a worked video tutorial.  Really useful!

Screen Shot 2019-07-27 at 10.02.40 AM

The Practice Exams section takes you to ready made exams on each topic – again with worked solutions.  This also has some harder exams for those students aiming for 6s and 7s and the Past IB Exams section takes you to full video worked solutions to every question on every past paper – and you can also get a prediction exam for the upcoming year.

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

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