## 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:

## 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.

Start with the following figure:

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.

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

## Area of a circle – proof by exhuastion

Once $\pi$ has been defined as the ratio circumference to diameter, the area of a circle must be $\pi r^{2}$.

A proof relies on an infinite, limiting process which paves the way to some calculus-like ideas.

A circle is cut up into 6 sectors which are then rearranged into a near rectangle,

What doesn’t look too close to a rectangle at 6 sectors looks better at 26:

As the number of sectors becomes very big the shape becomes indistinguishable from a rectangle and the argument is complete.

I think that Archimedes used such arguments, or, proof by exhaustion, in some of his solid geometry work.