The Nephoid – a caustic in a coffee cup

The Nephoid is the curve formed by the envelope of rays reflected on a circular surface from a set of initial parallel rays.  In other words, the curve which owns all the rays as tangents.

The geomtery of each ray path is relatively straight forward, involving properties of parallel lines and isocelels triangles.

It is interesting to note that the ray path will only form a closed path for angles of $\theta$ which divide $360^{o}$.

For a hemisphere, we would have,

For a set of parallel rays enclosed in a circle we get:

The envelope of the lines becomes clearer as the number of parallel rays are increased.

The Cardioid as the Envelope of Pencil Curves

Draw a circle and divide the circumference into and even number of equally spaced points.  In this example I have used 44:

$1,2,3, \dots , 44$

Draw following chords between numbers in such a way that when viewed as a sequemce, the second point of each pair moves twice as fast around the circle.:
$1 \rightarrow 2,\ 2 \rightarrow 4,\ 4 \rightarrow 8, \dots,\ 22 \rightarrow 44$
and
$23 \rightarrow 2,\ 24 \rightarrow 4,\ 25 \rightarrow 8, \dots,\ 44 \rightarrow 44$

and on completion, sketch the curve which owns each of the chords as a tangent:

The Parabola

Parabola definition: locus of $P$ such that $PS=PL$.

Cartesian Equations:

$(x-a)^{2}+y^{2}=(x+a)^{2}$
$x^{2}-2ax+a^{2}+y^{2} = x^{2}+2ax+a^{2}$
$y^{2} = 4ax$

$2y\dfrac{\textnormal{d}y}{\textnormal{d}x}=4a\ \implies\ \dfrac{\textnormal{d}y}{\textnormal{d}x}=\dfrac{2a}{y}$

Parametric Equation:

$x=at^{2},\ y=2at$

$\dfrac{\textnormal{d}x}{\textnormal{d}t}=2at,\ \dfrac{\textnormal{d}y}{\textnormal{d}t}=2a\ \implies\ \dfrac{\textnormal{d}y}{\textnormal{d}x}=\dfrac{1}{t}$

tangent:

$y-2at=\dfrac{1}{t}(x-at^{2})\ \implies ty=x+at^{2}$

normal:

$y-2at=-t(x-at^{2})\ \implies y+tx=2at+at^{3}$

t-Formulae and parameterisation of the circle

t-Formulae are used in integration to tackle rational expressions of tigonometric functions.  After a spell in the cold, when they were not included in some A level specifications, they are now back in sixth form lessons.

It all starts with the subsitution,

$t=\tan \frac{\theta}{2}$

from which the following functions can be derived,

$\tan \theta=\dfrac{2t}{1-t^{2}}$$\cos \theta =\dfrac{1-t^{2}}{1+t^{2}}$$\sin \theta=\dfrac{2t}{1+t^{2}}$.

These derivations can be made using compound trigonmetry fomulae.  Alternatively, there is an engaging co-ordinate geometry derivation which has the merits of doubling up as an algebraic parametrisation of the circle.

Euclid tells us that the angle subtended by the chord $PP'$ at the centre is twice the angle subtended at the circumference.  The $X$-axis providing a line of symmetry, gives the relationship between the angles $\frac{\theta}{2}$ and $\theta$ at $A$ and $O$ respectively.

Defining $t=\tan \theta$ and creating a line, $y=t(x+1)$, through $A$ with gradient $t$ gives intersections with the $Y$-axis and the circle at $R$ and $P$ respectively.

The intersection point $P$ can then be found by solving the simulataneous equations:

$y=t(x+1)$, and, $x^2+y^2=1$.

Substituting for $y$ leads to the quadratic,

$(t^2+1)x^2+2t^2x+t^2-1=0$,

which admits an easy factorisation once one acknowledges that it must have one root of $x=-1$.

$(x+1)(x-\frac{1-t^2}{1+t^2})=0$,

giving the other root, and $x$ value for $P$ as $\frac{1-t^2}{1+t^2}$.

Solving for $y$ gives the co-ordinates of $P$.  When viewed as two altenative parameterisations of the unit circle, the derivation of the $t$ formulae is complete.

$\left(\dfrac{1-t^{2}}{1+t^{2}},\dfrac{2t}{1+t^{2}}\right)=(\cos \theta,\sin \theta )$.

Sale Prices: Fifty Seven Point Five Percent Off on Streets of Canterbury!

Whilst wandering through the streets of Canterbury this sign caught my eye. With the mathematics department’s favourite coffee shop in the back-ground, the sign promises a further drop of prices for outdoor enthusiasts.

‘Half price’ and ‘15%’ are both eye catching figures which are immediately understandable to the shopper, but the sign promises something else.

Will anyone think this is a ‘65% off’ sale and that bargains can be picked up for 35% of their recommended retail price(RRP)?

In fact a 15% reduction applied to a 50% sale is compound percentage problem best dealt with by percentage multipliers:

$0.5 \times 0.85 =0.425$

that is, half price with a further 15% applied amounts to a reduction of 57.5% with shoppers paying 42.5% of the RRP rather than 35%. Are the marketeers messing with our heads?

Perhaps nicer for the maths teachers in the coffee shop to see calculate in fractioons without electronic assistance:

$\dfrac{50}{100} \times \dfrac{85}{100}= \dfrac{1}{2} \times \dfrac{17}{20}=\dfrac{17}{40}$

Much nicer perhaps, but a ‘$\frac{23}{40}$ off’ sale doesn’t have the same ring about it I suppose.

Edexcel Practice Paper G – Statistics and Mechanics

This paper felt a little rough around the edges as a finished product but good revision practice nevertheless. Only Word versions available with the parsing of symbols and format presenting small problems.

Notes:
– Statistics section seemed very chatty, had to write a lot for the marks on offer.
-Q4, not sure whether ‘p-value’ is defined anywhere for pupils who have done this course, context makes it clear what it is I think but could be confusing.
-Q3, as an ex-weather forecaster I didn’t like the premise of this question. It doesn’t seem at reasonable that same pressure ranges would relate to same weather at different locations. Use of the term hurricane could trouble the pedantic – Hurricanes in US, Typhoons in China.
-A lot of marks came very quickly in the Mechanics section, easy?
-I didn’t like the premise of Q7, two see-saws ‘joined’ seems to me to make one long rod. In any case the question assumes that they can move independently and that no moment is transferred through the ‘join’. They have to be side by side but able to move freely.

Edexcel Mathematics Practice Paper A

Practice Paper A – Questions

Practice Paper A – Mark Scheme

Practice Paper A – JPED Write Out

My write out included to show students how I would have answered the questions using the style of mathematics taught in my lessons.

This paper was one of a number of revision and preparation resources published by the board.  I could only find Word originals which did not handle the typesetting of the mathematical symbols; pdf versions of these Word documents are added here.

I wonder if this paper would have passed the quality tests of the real exams.

Notes on the paper in no particular order:

• Q6 – initially confusing.  The figure is bounded by ‘arcs’, would have preferred ‘circular arcs’, there are, after all, many types of arcs and this one looks like an ellipse.  At first sight it looks like an integration questions, but once you have figured that the arcs are circular together with the presence of $\pi$ it has to be all about radians and sector area.
• Q11 – there must be an error in the mark scheme here, I am happier with my answers than Edexcel’s.

Counting Pennies, Generating Functions and STEP

STEP I 1997, Question 1
Show that you can make up 10 pence in eleven ways using 10p, 5p, 2p and 1p coins.

In how many ways can you make up 20 pence using 20p,10p 5p, 2p and 1p coins?

There is an interesting way of approaching this question using generating functions.

A generating function is of the form:

$a_{0}+a_{1}t+a_{2}t^{2}+a_{3}t^{3}+\dots$

where the coefficients of $t$ count various problems and issues of convergence are unimportant.

Show that you can make up 10 pence in eleven ways using 10p, 5p, 2p and 1p coins.

This will be the co-efficient of $t^{10}$ in,

$(1+t+t^{2}+t^{3}+\dots+t^{10})(1+t^{2}+t^{4}+\dots+t^{10})(1+t^{5}+t^{10})(1+t^{10})$

by way of explanation:

$\overset{\textnormal{the 1p s}}{\overbrace{(1+t+t^{2}+t^{3}+\dots+t^{10})}}\overset{\textnormal{the 2p s}}{\overbrace{(1+t^{2}+t^{4}+\dots+t^{10})}}\overset{\textnormal{the 5p s}}{\overbrace{(1+t^{5}+t^{10})}}\overset{\textnormal{the 10p s}}{\overbrace{(1+t^{10})}}$

The candidate still needs to expand the function to get the answer, but having changed the problem to one of algebra, they might enjoy the benefits of practice and familiarity in the manual computation.

For those of at liberty to explore the question at leisure, a Computer Algebra System (CAS) is useful. In fact, the website Wolfram Alpha will expand the brackets for us, computing the answers to all similar counting problems in the process. The Wolfram Alpha query box understands the latex code if this is just pasted in, but only if it is presented in an explicit form, that is without the dots.  In order to simplify the question for the CAS notice that each bracket is a geometric progression, which can be summed using standard A level theory reducing the function to:

$\dfrac{(1-t^{11})}{(1-t)}\dfrac{(1-t^{12})}{(1-t^{2})}\dfrac{(1-t^{15})}{(1-t^{5})}\dfrac{(1+t^{10})}{1}$

Use of Wolfram Alpha then gives the following:

which contains our answer to the first part: 11.

Equipped with some methods and machinery we can now proceed to the second part of the question.  In fact, we can adapt the method and functions to count any similar loose change problem.

In how many ways can you make up 20 pence using 20p,10p 5p, 2p and 1p coins?

This will be the coefficient of $t^{20}$ in,

$(1+t+t^{2}+t^{3}+\dots+t^{20})(1+t^{2}+t^{4}+\dots+t^{20})(1+t^{5}+t^{10}+\dots+t^{20})(1+t^{10}+t^{20})(1+t^{20})$

$\overset{\textnormal{the 1 p s}}{\overbrace{(1+t+t^{2}+t^{3}+\dots+t^{20})}}\overset{\textnormal{the 2 p s}}{\overbrace{(1+t^{2}+t^{4}+\dots+t^{20})}}\overset{\textnormal{the 5 p s}}{\overbrace{(1+t^{5}+t^{10}+\dots+t^{20})}}\overset{\textnormal{the 10p s}}{\overbrace{(1+t^{10}+t^{20})}}\overset{\textnormal{the 20p s}}{\overbrace{(1+t^{20})}}$

which is equal to

$\dfrac{(1-t^{21})}{(1-t)}\dfrac{(1-t^{22})}{(1-t^{2})}\dfrac{(1-t^{25})}{(1-t^{5})}\dfrac{(1+t^{10}+t^{20})(1+t^{20})}{1}$

and which Wolfram Alpha gives and expansion of: