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Volume optimization of a cuboid

This is an extension of the Nrich task which is currently live – where students have to find the maximum volume of a cuboid formed by cutting squares of size x from each corner of a 20 x 20 piece of paper.  I’m going to use an n x 10 rectangle and see what the optimum x value is when n tends to infinity.

First we can find the volume of the cuboid:

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Next we want to find when the volume is a maximum, so differentiate and set this equal to 0.

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Next we use the quadratic formula to find the roots of the quadratic, and then see what happens as n tends to infinity (i.e we want to see what the optimum x values are for our cuboid when n approaches infinity).  We only take the negative solution of the + – quadratic solutions because this will be the only one that fits the initial problem.

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Next we try and simplify the square root by taking out a factor of 16, and then we complete the square for the term inside the square root (this will be useful next!)

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Next we make a u substitution.  Note that this means that as n approaches infinity, u approaches 0.

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Substituting this into the expression gives us:

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We then manipulate the surd further to get it in the following form:

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Now, the reason for all that manipulation becomes apparent – we can use the binomial expansion for the square root of 1 + u2 to get the following:

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Therefore we have shown that as the value of n approaches infinity, the value of x that gives the optimum volume approaches 2.5cm.

So, even though we start with a pretty simple optimization task, it quickly develops into some quite complicated mathematics.  We could obviously have plotted the term in n to see what its behavior was as n approaches infinity, but it’s nicer to prove it.  So, let’s check our result graphically.

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As we can see from the graph, with n plotted on the x axis and x plotted on the y axis we approach x = 2.5 as n approaches infinity – as required.

An m by n rectangle.

So, we can then extend this by considering an n by m rectangle, where m is fixed and then n tends to infinity.  As before the question is what is the value of x which gives the maximum volume as n tends to infinity?

We do the same method.  First we write the equation for the volume and put it into the quadratic formula.

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Next we complete the square, and make the u substitution:

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Next we simplify the surd, and then use the expansion for the square root of 1 + u2

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This then gives the following answer:

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So, we can see that for an n by m rectangle, as m is fixed and n tends to infinity, the value of x which gives the optimum volume tends to m/4.  For example when we had a 10 by n rectangle (i.e m = 10) we had x = 2.5.  When we have a 20 by n rectangle we would have x = 5 etc.

And we’ve finished!  See what other things you can explore with this problem.

 

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Projective Geometry

Geometry is a discipline which has long been subject to mathematical fashions of the ages. In classical Greece, Euclid’s elements (Euclid pictured above) with their logical axiomatic base established the subject as the pinnacle on the “great mountain of Truth” that all other disciplines could but hope to scale. However the status of the subject fell greatly from such heights and by the late 18th century it was no longer a fashionable branch to study. The revival of interest in geometry was led by a group of French mathematicians at the start of the 1800s with their work on projective geometry. This then paved the way for the later development of non-Euclidean geometry and led to deep philosophical questions as to geometry’s links with reality and indeed just what exactly geometry was.

projective 1Projective geometry is the study of geometrical properties unchanged by projection. It strips away distinctions between conics, angles, distance and parallelism to create a geometry more fundamental than Euclidean geometry. For example the diagram below shows how an ellipse has been projected onto a circle. The ellipse and the circle are therefore projectively equivalent which means that projective results in the circle are also true in ellipses (and other conics).

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Projective geometry can be understood in terms of rays of light emanating from a point. In the diagram above, the triangle IJK drawn on the glass screen would be projected to triangle LNO on the ground. This projection does not preserve either angles or side lengths – so the triangle on the ground will have different sized angles and sides to that on the screen. This may seem a little strange – after all we tend to think in terms of angles and sides in geometry, however in projective geometry distinctions about angles and lengths are stripped away (however something called the cross-ratio is still preserved).

projective3We can see in the image above that a projection from the point E creates similar shapes when the 2 planes containing IJKL and ABCD are parallel. Therefore the Euclidean geometrical study of similar shapes can be thought of as a subset of plane positions in projective geometry.

projective4Taking this idea further we can see that congruent shapes can be achieved if we have the centre of projection, E, “sent to infinity:” In projective geometry, parallel lines do indeed meet – at this point at infinity. Therefore with the point E sent to infinity we have a projection above yielding congruent shapes.

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Projective geometry can be used with conics to associate every point (pole) with a line (polar), and vice versa. For example the point A had the associated red line, d. To find this we draw the 2 tangents from A to the conic. We then join the 2 points of intersection between B and C. This principle of duality allowed new theorems to be discovered simply by interchanging points and lines.

An example of both the symmetrical attractiveness and the mathematical potential for duality was first provided by Brianchon. In 1806 he used duality to discover the dual theorem of Pascal’s Theorem – simply by interchanging points and lines. Rarely can a mathematical discovery have been both so (mechanically) easy and yet so profoundly
beautiful.

Brianchon’s Theorem

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Pascal’s Theorem

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Poncelet

Poncelet was another French pioneer of projective geometry who used the idea of points and lines being “sent to infinity” to yield some remarkable results when used as a tool for mathematical proof.

Another version of Pascal’s Theorem:

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Poncelet claimed he could prove Pascal’s theorem (shown above) where 6 points on a conic section joined to make a hexagon have a common line. He did this by sending the line GH to infinity. To understand this we can note that the previous point of intersection G of lines AB’ and A’B is now at infinity, which means that AB’ and A’B will now be parallel. This means that H being at infinity also creates the 2 parallel lines AC’. Poncelet now argued that because we could prove through geometrical means that B’C and BC’ were also parallel, that this was consistent with the line HI also being at infinity. Therefore by proving the specific case in a circle where line GHI has been sent to infinity he argued that we could prove using projective geometry the general case of Pascal’s theorem in any conic .

Pascal’s Theorem with intersections at infinity:

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This branch of mathematics developed quickly in the early 1800s, sparking new interest in geometry and leading to a heated debate about whether geometry should retain its “pure” Euclidean roots of diagrammatic proof, or if it was best understood through algebra. The use of points and lines at infinity marked a shift away from geometry representing “reality” as understood from a Euclidean perspective, and by the late 1800s Beltrami, Poincare and others were able to incorporate the ideas of projective geometry and lines at infinity to provide their Euclidean models of non-Euclidean space. The development of projective geometry demonstrated how a small change of perspective could have profound consequences.

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Modeling hours of daylight

Desmos has a nice student activity (on teacher.desmos.com) modeling the number of hours of daylight in Florida versus Alaska – which both produce a nice sine curve when plotted on a graph.  So let’s see if this relationship also holds between Phuket and Manchester.

First we can find the daylight hours from this site, making sure to convert the times given to decimals of hours.

Phuket

Phuket has the following distribution of hours of daylight (taking the reading from the first of each month and setting 1 as January)

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Manchester 

Manchester has much greater variation and is as follows:

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Therefore when we plot them together (Phuket in green and Manchester in blue) we get the following 2 curves:

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We can see that these very closely fit sine curves, indeed we can see that the following regression lines fit the curves very closely:

Manchester:

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Phuket:

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For Manchester I needed to set the value of b (see what happens if you don’t do this!) Because we are working with Sine graphs, the value of d will give the equation of the axis of symmetry of the graph, which will also be the average hours of daylight over the year.  We can see therefore that even though there is a huge variation between the hours of daylight in the 2 places, they both get on average the same amount of daylight across the year (12.3 hours versus 12.1 hours).

Further investigation:

Does the relationship still hold when looking at hours of sunshine rather than daylight?  How many years would we expect our model be accurate for?  It’s possible to investigate the use of sine waves to model a large amount of natural phenomena such as tide heights and musical notes – so it’s also possible to investigate in this direction as well.

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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?

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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.

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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

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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:

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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:

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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:

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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:

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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:

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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:

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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:

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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:

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

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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:

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

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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?

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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.

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This is a nice example of using some maths to solve a puzzle from the mindyourdecisions youtube channel (screencaptures from the video).

How to Avoid The Troll: A Puzzle

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In these situations it’s best to look at the extreme case first so you get some idea of the problem.  If you are feeling particularly pessimistic you could assume that the troll is always going to be there.  Therefore you would head to the top of the barrier each time.  This situation is represented below:

The Pessimistic Solution:

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Another basic strategy would be the optimistic strategy.  Basically head in a straight line hoping that the troll is not there.  If it’s not, then the journey is only 2km.  If it is then you have to make a lengthy detour.  This situation is shown below:

The Optimistic Solution:

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The expected value was worked out here by doing 0.5 x (2) + 0.5 x (2 + root 2) = 2.71.

The question is now, is there a better strategy than either of these?  An obvious possibility is heading for the point halfway along where the barrier might be.  This would make a triangle of base 1 and height 1/2.  This has a hypotenuse of root (5/4).  In the best case scenario we would then have a total distance of 2 x root (5/4).  In the worst case scenario we would have a total distance of root(5/4) + 1/2 + root 2.  We find the expected value by multiply both by 0.5 and adding.  This gives 2.63 (2 dp).  But can we do any better?  Yes – by using some algebra and then optimising to find a minimum.

The Optimisation Solution:

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To minimise this function, we need to differentiate and find when the gradient is equal to zero, or draw a graph and look for the minimum.  Now, hopefully you can remember how to differentiate polynomials, so here I’ve used Wolfram Alpha to solve it for us.  Wolfram Alpha is incredibly powerful -and also very easy to use.  Here is what I entered:

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and here is the output:

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So, when we head for a point exactly 1/(2 root 2) up the potential barrier, we minimise the distance travelled to around 2.62 miles.

So, there we go, we have saved 0.21 miles from our most pessimistic model, and 0.01 miles from our best guess model of heading for the midpoint.  Not a huge difference – but nevertheless we’ll save ourselves a few seconds!

This is a good example of how an exploration could progress – once you get to the end you could then look at changing the question slightly, perhaps the troll is only 1/3 of the distance across?  Maybe the troll appears only 1/3 of the time?  Could you even generalise the results for when the troll is y distance away or appears z percent of the time?

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Give your university applications a headstart on other students with Coursera.  

Applying for university as an international student is incredibly competitive – for the top universities you’ll be competing with the best students from around the world, and so giving yourself a competitive advantage to make your university application stand out is really important.   One way to do this is by completing a course run by some of the world’s top Universities.

Universities offering courses:

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Examples of the top universities offering courses include: The University of Tokyo, Caltec, University of Manchester, Imperial College London, Duke, Stanford, Yale, University of Sydney, National University of Singapore, amongst many others, alongside major companies such as IBM, Google, Intel and Goldman Sachs.

Courses on offer

You can sign up for free and access modules run by these universities and companies, with the possibility of obtaining a certificate at the end of the course which can then go towards your university application.

Some of the courses on offer include:

Biological science courses such as Genetics and Evolution from Duke University, Understaning the brain from the University of Chicago, Astrobiology and the search for Extraterrestrial life from Edinburgh University and Medical Neuroscience from Duke University,

Business courses such as Business foundations from University of Pennsylvania, Digital Marketing from the University of Illinois, Viral Marketing and How to Create Contagious Content with the University of Pennsylvania, the Math Behind Moneyball with the University of Houston and studying the Global Financial Crisis with Yale University.

Physical science courses such as Welcome to Game Theory with Tokyo University, Science Literacy with Erasmus University Rotterdam and The Journey of the Universe with Yale University.

Arts and humanities courses such as Creative writing from Wesleyan University, Graphics design from Californian Institute of Arts, Music Production from Berklee College of Music, Introduction to Philosophy from the University of Edinburgh.

Overall there are over 3000 courses from 170 universities and partners – so almost certainly there’ll be something worth investigating.  Have a look and give your University application a boost over everyone else!

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

  1. The butterfly

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

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

2. Fermat’s Spiral

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This is plotted by the polar equation:

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The next 3 were all created by my students.

3.  Chaotic spiral (by Laura Y9)

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

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4.  The flower (by Felix Y9)

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Some nice rotational symmetries on this one.  Plotted by:

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5. The heart (by Tiffany Y9)

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

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

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Log Graphs to Plot Planetary Patterns

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

Astronomical investigations

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

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

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

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

1.8, 1.4, 1.5, 3.4, 1.8

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

Planet X

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

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

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

Log Plots

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

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

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

 

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

log a  = -0.68

a = 0.21

and given that log b is the gradient this gives:

log b = 0.24

b = 1.74

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

d = ab^n

distance = 0.21 x (1.74)^n.

Testing the formula

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

Other log relationships

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

Screen Shot 2017-06-15 at 10.54.40 AM

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

Height against time

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

Screen Shot 2017-06-15 at 9.06.25 AM

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

y = -65cos10.5(t-3.4) – 195

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

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

Momentum against velocity

Screen Shot 2017-06-15 at 10.31.20 AM

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

p = 0.3v

If we look at the theoretical equation linking momentum:

p = mv

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

Height against velocity

Screen Shot 2017-06-15 at 10.35.43 AM

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

Screen Shot 2017-06-15 at 11.45.41 AM

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

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