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If you are a teacher then please also visit my new site: intermathematics.com for over 2000+ pdf pages of resources for teaching IB maths!

**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**

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/s^{2} 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:

We should note that all our units are in meters, seconds and m/s^{2} 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:

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:

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:

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:

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:

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:

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:

Next we note that this will give an answer in meters, so we can convert to light years by dividing by 9.461×10^{15}

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:

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

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?**

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.

**Essential Resources for IB Teachers**

If you are a **teacher** then please also visit my new site. This has been designed specifically for teachers of mathematics at international schools. The content now includes over **2000 pages of pdf content** for the entire SL and HL Analysis syllabus and also the SL Applications syllabus. Some of the content includes:

**Original pdf worksheets**(with full worked solutions) designed to cover all the syllabus topics. These make great homework sheets or in class worksheets – and are each designed to last between 40 minutes and 1 hour.**Original Paper 3 investigations**(with full worked solutions) to develop investigative techniques and support both the exploration and the Paper 3 examination.- Over 150 pages of
**Coursework Guides**to introduce students to the essentials behind getting an excellent mark on their exploration coursework. - A large number of
**enrichment activities**such as treasure hunts, quizzes, investigations, Desmos explorations, Python coding and more – to engage IB learners in the course.

There is also a lot more. I think this could save teachers 200+ hours of preparation time in delivering an IB maths course – so it should be well worth exploring!

**Essential Resources for both IB teachers and IB students**

1) Exploration Guides and Paper 3 Resources

I’ve put together a **168 page** Super Exploration Guide to talk students and teachers through all aspects of producing an excellent coursework submission. Students always make the same mistakes when doing their coursework – get the inside track from an IB moderator! I have also made **Paper 3 packs** for HL Analysis and also Applications students to help prepare for their Paper 3 exams. The Exploration Guides can be downloaded here and the Paper 3 Questions can be downloaded here.

One of the main benefits of flipping the classroom is allowing IB maths students to self-teach IB content. There are currently a good number of videos on youtube which allow students to self teach syllabus content, but no real opportunity to watch videos going through IB Higher Level past paper questions. So, I’ve started to put some of these together:

**Playlist, Worked Exam Solutions:**

The videos above are all around 10 minutes long and consist of talking through the solutions to 2-3 IB HL maths questions. The best way to use these videos is to pause the video at the start of the question, attempt it, then watch the video to check the answer and make notes on the method. Click on the top left hand corner to change the video being shown in the playlist.

The playlists below combine these worked solutions with the syllabus content videos, all grouped into the relevant syllabus strands:

**Playlist 1, Algebra 1: **

Sequences, Binomial, Logs, Induction, Permutations, Gaussian elimination:

**Playlist 2, Complex numbers: **

Converting from Cartesian to Polar, De Moivre’s Theorem, Roots of Unity:

**Playlist 3: Functions: **

Sketching graphs, Finding Inverses, Factor and Remainder Theorem, Sketching 1/f(x), sketching absolute f(x), translating f(x):

The Chinese Remainder Theorem is a method to solve the following puzzle, posed by Sun Zi around the 4th Century AD.

*What number has a remainder of 2 when divided by 3, a remainder of 3 when divided by 5 and a remainder of 2 when divided by 7?*

There are a couple of methods to solve this. Firstly it helps to understand the concept of modulus – for example 21 mod 6 means the remainder when 21 is divided by 6. In this case the remainder is 3, so we can write 21 ≡ 3 (mod 6). The ≡ sign means “equivalent to” and is often used in modulus questions.

Method 1:

1) We try to solve the first part of the question, *What number has a remainder of 2 when divided by 3,*

to do this we list the values of x ≡ 2 (mod 3). x = 2,5,8,11,14,17……

2) We then look at the values in this list and see which ones also satisfy the second part of the question, *What number has a remainder of 3 when divided by 5*

From the list x = 2,5,8,11,14,17…… we can see that x = 8 has a remainder of 3 when divided by 5 (i.e 8 ≡ 3 (mod 5) )

3) We now start from 8 and count up in multiples of 15 (3 x 5 because we have mod 3 and mod 5 )

8, 23, 38, 53 …..

4) We look for a number on this list which satisfies the last part of the question, *What number has a remainder of 2 when divided by 7?*

With x = 8, 23, 38, 53 ….. we can see that x = 23 has a remainder of 2 when divided by 7 (i.e 23 ≡ 2 (mod 7) )

Therefore 23 satisfies all parts of the question. When you divide it by 3 you get a remainder of 2, when you divide it by 5 you get a remainder of 3, and when you divide it by 7 you get a remainder of 2. So, 23 is our answer.

The second method is quite a bit more complicated – but is a better method when dealing with large numbers.

Method 2

1) We rewrite the problem in terms of modulus.

x ≡ 2 (mod 3)

x ≡ 3 (mod 5)

x ≡ 2 (mod 7)

We assign a = 2, b = 3, c = 2.

2) We give the values m_{1} = 3 (because the first line is mod 3), m_{2} = 5 (because the second line is mod 5), m_{3} = 2 (because the third line is mod 7).

3) We calculate M = m_{1}m_{2}m_{3} = 3x5x7 = 105

4) We calculate M_{1} = M/m_{1} = 105/3 = 35

M_{2} = M/m_{2} = 105/5 = 21

M_{3} = M/m_{3} = 105/7 = 15

4) We then note that M_{1} ≡ 2 (mod 3), M_{2} ≡ 1 (mod 5), M_{3} ≡ 1 (mod 7)

5) We then look for the multiplicative inverse of M_{1}. This is the number which when multiplied by M_{1} will give an answer of 1 (mod 3). This number is 2 because 2×2 = 4 ≡ 1 (mod 3). We assign A = 2

We then look for the multiplicative inverse of M_{2}. This is the number which when multiplied by M_{2} will give an answer of 1 (mod 5). This number is 1 because 1×1 = 1 ≡ 1 (mod 3). We assign B = 1.

We then look for the multiplicative inverse of M_{3}. This is the number which when multiplied by M_{3} will give an answer of 1 (mod 7). This number is 1 because 1×1 = 1 ≡ 1 (mod 7). We assign C = 1.

6) We now put all this together:

x = aM_{1}A + bM_{2}B + cM_{3}C (mod M)

x = 2x35x2 + 3x21x1 + 2x15x1 = 233 ≡ 23 (mod 105)

That last method may seem a lot slower – but when working with large numbers is actually a lot quicker. So there we go – that’s the method that Sun Zi noted more than 1500 years ago. This topic whilst seemingly quite abstract is a good introduction to number theory – the branch of mathematics which deals with the properties of whole numbers.