Taking Maths Further Podcast

This week the topic was mathematics and money, and how maths is used in finance. We interviewed Sarah O’Rourke, who’s an accountant working on the problem of moving cash around to where it’s needed in cash machines. We discussed the ways she uses mathematical modelling to predict where demand for cash will be high, and also the other types of work that accountants do, and the different ways to become an accountant.

Interesting links:

• Accounting on Wikipedia
• Double entry bookkeeping at Dummies.com
• What is Financial Mathematics? at Plus Magazine
• Maths games - percentages at IXL
• Tax Matters at HMRC
• Money Talks, interactive game at the NI Curriculum website

Puzzle:

1. Using only £20 and £50 notes, what’s the largest multiple of £10 you can’t make?
2. In an imaginary scenario where the only notes are £30 and £70, again what’s the largest multiple of £10 you can’t make?
3. Why do you think we use the denominations of currency that we do use?

Solution:

1. Using only £20 and £50 notes, it’s not possible to make £10 or £30, but all other multiples of £10 are possible. This can be proven by noting that £20 x 2 = £40, and £50 x 1 = £50, and from here every other multiple of £10 can be made by adding different numbers of £20 to either of these base amounts.
2. If our notes are £30 and £70, we can’t make £50, £80 or £110, but all other multiples of £10 above £110 are possible. This can be proven by noticing that once you can make three consecutive multiples of £10, any other can be obtained by adding £30 notes - and in this case, we can make £120 = 4 x £30, £130 = £70 + 2 x £30, and £140 = 2 x £70 so we can then get £150, £160 and £170 by adding £30 to each, and so on.
3. The notes currently in use (£5, £10, £20 and (rarely) £50) have been chosen so that it’s possible to make any amount that’s a multiple of £5 using relatively few notes. We don’t need a £30, as it can be made easily using £10 + £20. The system is designed to make it as easy as possible to make any amount, while keeping the number of different types of note needed relatively small.
Show/Hide

In this episode, we talk about cellular automata - including the Game of Life - and graph theory, and interviewed Jonathan Crofts from Nottingham Trent University about his research on complex networks in neuroscience. Find out more about the Biomathematics & Bioinformatics Research Group at Nottingham Trent.

Cellular automata:

• Stephen Hawking’s introduction to the Game of Life
• Clips from lots of Game of Life configurations, cut together like an epic sci-fi movie trailer!
• Game of Life simulator on Wikipedia
• Explanation of, and uses for, cellular automata, at GCSE Bitesize Geography
• An application that uses Brian’s Brain to trigger musical notes.

Graph Theory:

• Graphs, at Wolfram Mathworld
• Seven Bridges of Konigsberg: a puzzle in graph theory

Puzzle: I have a 5 × 5 grid, in which the squares can either be empty (white) or infected (black). The four ‘neighbours’ of each square are the ones directly next to it: up, down, left and right. A square will become infected if two or more of its neighbours are infected. Can you find a set of squares to colour black (‘infect’) which will eventually spread the infection to the whole grid? What’s the smallest number of squares you need to do this?

Solution:

One way to colour the squares so they infect the whole grid is to colour the five squares on the diagonal. This will then infect the cells next to the diagonal, which will then infect the next diagonal rows, and so on until the whole grid is infected. In order to infect the whole grid, you need to colour at least 5 squares (and in general, for an n × n grid, you need to start with n squares coloured). This can be seen by looking at the perimeter of the infected area.

If a square needs two or more infected neighbours in order to become infected, then the perimeter of the infected area can never increase - if you imagine a pair of squares which are both neighbours to a third square, they will infect it. The newly infected square will have two free edges which increase the perimeter by 2, but the edges of the two squares it was neighbouring become absorbed and are no longerpart of the perimeter. The whole shape will then have the same perimeter as the two squares you started with.

If you have a square with three or four neighbours, which then becomes infected, doing so will only ever reduce the perimeter of the infected area. As time passes, and more cells become infected, the perimeter of the infected area can only decrease or stay the same.

This means that in order to infect the whole square grid, the arrangement of coloured squares you start off with must have the same perimeter as the whole square, as this will never increase - and the edge of a 5 × 5 square is 4 × 5 = 20. This means you need at least 5 squares to start with, each of which has perimeter 4.

Show/Hide

This week the topic was trigonometry. We interviewed Stephanie Yardley, who’s a solar physicist. We talked about the research Stephanie does into activity on the surface of the sun, and how she uses trigonometry to analyse data from satellites and telescopes.

Interesting links:

• Space weather information booklet
• Videos at the Space Weather Prediction Centre website
• Sine, cosine and tangent at Maths is Fun
• Shape of the sine and cosine graphs, at BBC Bitesize
• Using trigonometry to find components of vectors
• How to find vector components, at For Dummies

Puzzle: You want to calculate the height of a tall building. You set up a device for measuring angles, on a 1m high tripod, which is 200m away from the building. The angle above horizontal, when looking at the top edge of the building, is 15 degrees. What is the height of the building in metres?

Solution:

The height of a triangle with base 200m and angle 15 degrees is 53.6m. This, added to the height above the ground you are measuring from, means that the height of the building is approximately 54.6m. Show/Hide

This week the topic was calculus and differentiation. We talked to Florencia Tettamanti, who’s a mathematician working on fluid dynamics. We talked about how Flo uses calculus to study the motion of fluids like air and water, and what it’s like to be a research mathematician.

Interesting links:

• Basic differentiation, at s-cool
• Differential equations, at the University of Surrey website
• Fluid dynamics on Wikipedia
• NSF videos on Fluid Mechanics - YouTube playlist

Puzzle: If your function is given by y = x² - 6x + 13, what is the minimum value of y, and for which value of x does the function give this value?

Solution:

If you plot the points x=1, x=2, x=3 and x=4 you can clearly see the curve of this graph and that it seems to have a maximum at x=3, for which the value of y is 4. To see what the graph looks like, you can input the equation into Wolfram Alpha.

Another way to see this is to rearrange the equation: x²-6x+13 = (x-3)²+4, and by examining this equation we can see that this is just an x graph, shifted across by 3 and up by 4, so its turning point and hence the minimum will be at x=3 and y=4.

If you know how to use calculus, you can find the turning point more easily - if you differentiate x²-6x+13 you get 2x - 6, which will equal zero when x=3, and putting this value back into the original equation gives y=4.

Show/Hide

This week the topic was mechanics and friction. We interviewed Dan Hett, who works for CBBC writing computer games for their website. We talked about his work and how he uses a lot of mathematics in modelling how characters move, and making sure that’s done in a realistic way.

Interesting links:

• CBBC games website
• CBeebies story app (with pop-up book!)
• Game physics on Wikipedia
• A-level Mechanics topics at MathsRevision.net
• Friction and Coefficients of Friction at Engineering Toolbox (with some example values)
• Coefficient of friction on Wikipedia

Puzzle: Susan the Hedgehog runs at 20cm/s across the screen while the run button is held down. Once the run button is released, she slows down with constant deceleration of 8.5cm/s2. Will she stop within 32cm more of screen?

Solution:

The time taken to stop can be calculated by knowing that every second travelled, 8.5cm/s of speed is lost, so after 20/8.5=2.35 seconds, speed will be zero. We can approximate this deceleration by imagining Susan is travelling at 20cm/s for 1 second, 11.5cm/s for 1 second and 3cm/s for the remaining 0.35 seconds until she stops. This will cover more distance than the actual motion does (as your speed is lower than this for most of the time), but will cause you to travel only 31.6cm - so you will definitely stop within 32cm. (In actual fact, the distance taken to stop will be 23.53cm, because your speed continues to decrease at a constant rate for the whole time. In order to work this out, you need to use a little calculus!) Show/Hide