Mathematics: Applications and Interpretation
Higher Level
Investigating shapes that light illuminates at the bottom of a cup
Candidate code: jrk720
1
1.0 Introduction
It is all too common to see intriguing shapes in the world around us. The symmetric shape of a
starfish, the pristine structure of a snowflake or the rings of a planet like Saturn, all have the
power to catch the eye. There happens to be one set of kind of shapes that have always
fascinated me. As a person who has coffee every other day, I have always noticed a sparkle of
light taking a heart shape at the bottom of my cup. If we look inside a coffee cup with sufficient
light, we can see a heart form. But what exactly is this shape? And how does it form? After
doing some research, I learnt that one such shape that forms at the base of the cup is called a
“cardioid”, which means heart in Greek. I learnt that a cardioid can be traced by a point on the
circumference of a circle rolling completely around another circle of the same radius. However,
this was clearly not the case with my coffee cup. I wanted to know how the light reflected in
the cup. So, I played around and tilted the cup at different angles to understand it. Alas, it didn’t
help. In fact, it illumined other arcane shapes. Hence, I thought it would be appropriate to take
this chance and investigate the math behind this enigma.
2.0 Aim of the investigation
This study aims to investigate the mathematical model behind the shapes formed by light
reflecting off the bottom of a coffee cup. To understand this, the scheme of work shall be:
First, to understand the geometric method of modelling cardioids through polar equations. Then
to look at the principle of light behaviour to understand how light behaves when reflected.
Next, to investigate the concept of mathematical envelopes which helps explain the shapes we
see in our cup, particularly: nephroid and cardioid.
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Shapes that light illuminate at the bottom of a coffee cup when kept at different angles
Fig 1.0 Cardioid (left image) and nephroid (right image)
3.0 Understanding the representation of a cardioid through polar equations
Before we investigate how cardioids form in mugs, I thought it was important to understand
the geometric perspective first to build a foundation of cardioids. To look at the geometric
perspective of cardioids, we must use polar equations. Equations involving polar coordinates
show a relationship between r and 𝜃, where r denotes a point's distance from the origin and 𝜃
represents the angle formed between a curve's point, the origin, and the positive x-axis in
anticlockwise direction (Hayes, Pao, & Ellinor, n.d.).
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To derive the polar equation of the cardioid, let us first understand how it is drawn.
A cardioid is drawn by setting a point on a circle's circumference and “rolling” it externally
without slipping on the boundary of another circle with the same radius (Hayes, Pao, & Ellinor,
n.d.). This can be shown in the diagram below:
Fig 2.0 Cardioid drawn by rolling two circles
Source: (Proofwiki, 2020)
“Rolling” essentially means that:
1)
The circle (centred A and B) must be touching: 𝐴𝑄 = 𝑄𝐵
2)
The circle should roll an equal distance, so, 𝑎𝑟𝑐𝑂𝑄 = 𝑎𝑟𝑐𝑄𝑃
The cardioid derived from a polar equation is:
𝑟 = 2𝑎 (1 + 𝑐𝑜𝑠𝜃)

Polar coordinates are expressed (r, θ) rather than (x, y).

r is the distance of a line segment with one endpoint on the cardioid and the other at the
point of origin, O.

The angle, θ, is measured anticlockwise. It is the measure of the central angle created
between the polar axis, x, and the line segment.

𝑎 is the radius of the circles.
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To arrive at this equation:
Let P= (r,θ) be any point along the cardioid, C.
Let points A and B be the centres of the two circles, stator and rotor, separately.
Let Q be the point where the rotor and stator contact each other.
Note that the arc OQ of the stator equals the arc P of the rotor.
Thus:
∠OAQ = ∠PBQ
and it follows that AB is parallel to OP.
With reference to the diagram above, we have:
𝑟 = 𝑂𝑅 + 𝑅𝑆 + 𝑆𝑃
= 𝑎𝑐𝑜𝑠𝜃 + 2𝑎 + 𝑎𝑐𝑜𝑠𝜃
= 2𝑎 (1 + 𝑐𝑜𝑠𝜃)
This shows that coordinate geometry and trigonometry will be extensively used to understand
epicycloids. But, to understand how they form in our mugs, first, light behaviour must be
understood.
4.0 Understanding how light behaves in the bottom of the mugs to form cardioids
Background
Observing the glowing light in the cup, I realised the shapes are caused by how the light
reflects. These shapes are called “caustics” and are caused by reflecting light from a surface.
Caustics comes from the Greek word for 'burned,' because they could potentially start a fire
since light was highly focused at these points to form the outline of the shapes. When light
rays reflect from a curved surface (a two-dimensional curve), there is a point where light is
the brightest. The curve that has the highest intensity of the light is called the caustic of the
curve. It defines a boundary of an envelope of rays as a curve of concentrated light.
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4.1 Reflection of light
To truly understand caustics, we must first understand the fundamentals of how light reflects.
This will help explain envelopes. Reflection occurs when a light ray is reflected by a surface
and bounces off at the same angle as it hits it.
This leads to another important idea: the law of reflection.
Which states, angle of incidence (i)= angle of reflection (r)
Fig 3.0 The law of reflection
To depict this, the following diagram shows what happens when a light ray hits a smooth
surface. The incoming ray which is above the normal is the incident ray and has an angle of
incidence ‘i’. The normal perpendicular line from the surface at the point where the incident
ray strikes. Lastly, reflected ray is the one below the normal which travel with an angle of
reflection ‘r’ after striking the surface.
Another interesting insight that I could generate was the fact the height of the cup (cylinder)
seemed to be irrelevant while to trying to explain the observed caustics. This is because even
when I changed the height of the reflecting cylinder, the observed shapes did not change. An
empty plastic bucket or a thin bangle yielded the same results as a coffee cup.
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Now, to understand how the light reflects in the cup, a closed and curved surface, we can
summarize our assumptions as:

Firstly, the cup will be a perfect circle with a smooth surface that reflects the incoming
light completely. The incoming light will be taken on the same plane as the mug’s base,
hence, a 2D plane.

Secondly, the base of the cup will be modelled as a unit circle, hence, with a radius of
one unit.

Thirdly, to obtain a plane wavefront we will assume the point source to be at infinity.
This will mean all light rays will be assumed to be straight lines and are parallel to each
other.
With these assumptions in mind, let us investigate the cup and understand how the light
behaves in it. The light rays dispersing parallel to the x-axis will have an angle of incidence =
to the angle of reflection after hitting the inside edge of the cup. Since the light reflects within
a circular shape, the angle at which each incoming (incident) light ray reflects in the cup is
slightly different compared to another ray. This causes the incoming light rays to intersect with
the reflected light rays. The point at which they intersect is where the light is the brightest and
is called the focal point. Connecting all the focal points, bright light emerges and forms a curve.
But how the cardioid-shaped light forms would be explained by something called envelopes.
Fig 4.0 Paths of two parallel light rays and their respective reflection angles, where x and y axis
number the point where the light ray hits the rim of the cup
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4.2 Exploring envelopes
Mathematically, an envelope is itself a curve created by a family of curves that are related by
a specific parameter, say, 𝑎. It is also important to note that it is tangent to all members of the
family of curve at some point. A parameter is a quantity that influences the output or
behaviour of a mathematical object but is viewed as being held constant (Q.). Just for an
example, it is like the gradient function we get when differentiating a function of a curve.
Whereas, a family of lines is a set of curves, each of which is given by a function in which one
or more of the parameters is variable (Weisstein, Family of Curves). The function for the set
of lines for the cardioid in our cup would be a set of straight lines with a parameter this will
represent the incoming light rays that enter the cup. For each value of 𝑎 plugged in, we get a
line corresponding to the value of 𝑎.
So how do we find the equation of the envelope?
To determine the family of curves' equation, we must first develop a standard equation for the
curves by formulating them in relation to a parameter. By differentiating with respect to the
parameter, we can derive the equation of all their tangents. Because we need both equations to
match at every point on the envelope, we solve them both simultaneously to arrive at the
envelope's equation.
We can summarize the process mathematically as:
Step – 1: Given 𝑓(𝑥, 𝑦, 𝑎) = 0… (1), when plotting the caustic, this would be the function of
the reflected light rays.
Step – 2: Differentiating equation one with respect to the parameter 𝑎, taking x and y constants
𝑑𝑓
We get 𝑑𝑎 = 0 … (2)
Step – 3: Eliminating 𝜃 from equation (1) and (2) we get the required envelope.
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4.3 Example 1
Let us find the envelope of the family of straight lines which make a triangle of unit area with
the positive X and Y axes.
The standard straight-line equation that represents both axis intercepts is, where 𝑎 is the xintercept and 𝑏 is y-intercept of the straight line:
𝑥 𝑦
+ =1
𝑎 𝑏
Here, 𝑎 and 𝑏 are both parameters that describe the family of curve. Next, we use the fact that
the area, A, under the line is A =
𝑎𝑏
2
and is constant. This is better depicted in the diagram
below:
1
Area of triangle = × 𝑏𝑎𝑠𝑒 × ℎ𝑒𝑖𝑔ℎ𝑡
2
∴ 𝑏𝑎𝑠𝑒 = 𝑎 𝑎𝑛𝑑 ℎ𝑒𝑖𝑔ℎ𝑡 = 𝑏
∴ 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒 = 𝐴 =
Fig 5.0 Diagram of straight line and shaded area under the line
To simplify, we shall consider only that family of straight lines which form a triangle of a
unit area with the positive X and Y axes.
𝑎𝑏
2
2
And 𝑏 = 𝑎
𝐴=1=
∴
𝑥 𝑎𝑦
+
=1
𝑎
2
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(1)
𝑎𝑏
2
Differentiating equation (1) with respect to 𝑎 results in:
−
𝑥 𝑦
+ =0
𝑎2 2
⇒
𝑥
𝑦
=
2
𝑎
2
⇒
2𝑥
= 𝑎2
𝑦
(2)
Equation of the envelope is then found after solving equation (1) and (2) simultaneously and
2𝑥
eliminating 𝑎 between the two equations in the process. ∴ putting a = √
𝑦
𝑖𝑛 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 (1)
2𝑥
𝑦 𝑦
+
=1
2
2𝑥
√
𝑦
√
𝑥
⇒
√𝑥𝑦
√2
+
√𝑥𝑦
√2
=1
1
⇒ √𝑥𝑦 = √2
Which gives us an equation for the first quadrant as:
𝑥𝑦 =
.
1
2
Fig 6.0 The envelope of the family of straight lines, which make a triangle of unit area with the
positive axes.
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4.4 Example 2
To understand this better, I also tried to find the envelope of the family of all circles of unit
radius whose centre lie on the line y=x.
The standard equation that represents circles of unit radius (where 𝑎 and 𝑏 are the 𝑋 and
𝑌 coordinates of the centre of the circle respectively) can be written as:
(𝑥 − 𝑎)2 + (𝑦 − 𝑏)2 = 1
Now, 𝑎 and 𝑏 are both parameters that define the family of curves and keep the centre of the
circle on the line y=x. Hence, we shall consider only that family of circles which have the
centre of the form (𝑎, 𝑎).
(𝑥 − 𝑎)2 + (𝑦 − 𝑎)2 = 1
(3)
Differentiating equation (3) with respect to 𝑎 results in,
−2(𝑥 − 𝑎) − 2(𝑦 − 𝑎) = 0
⇒ −(𝑥 − 𝑎) = (𝑦 − 𝑎)
⇒ 2𝑎 = 𝑥 + 𝑦
⇒𝑎=
𝑥+𝑦
2
(4)
Equation of the envelope is then found after solving equation (3) and (4) simultaneously and
eliminating 𝑎 between the two equations in the process. This gives:
⇒ (𝑥 −
𝑥+𝑦 2
𝑥+𝑦 2
) + (𝑦 −
) =1
2
2
⇒ (𝑥 −
𝑥 𝑦 2
𝑥 𝑦 2
− ) + (𝑦 − − ) = 1
2 2
2 2
⇒
2(𝑦 − 𝑥)2
=1
4
⇒ (𝑦 − 𝑥)2 = 2, ⇒ 𝑦 − 𝑥 = ±√2
⇒ 𝑦 = 𝑥 ± √2.
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The resulting solution is the required equation of the envelope. It consists of two straight lines:
𝑦 = 𝑥 − √2 and 𝑦 = 𝑥 + √2
Fig 7.0 The envelope of all circles of unit radius whose centre lies on the line y=x,
Now, that we are well familiar with the steps to find the equation of an envelope, we can use
this to model nephroids and cardioids.
5.0 Investigating the shapes illuminated by light at the bottom of a cup
5.1 Nephroid
Using the equations found in the previous section, we can model the incoming light rays
coming from a point source at infinity. But we need to find a way to model the reflected light
beams as well to understand how the envelope forms to make a nephroid.
Firstly, we will model the incoming light ray as a straight line coming from the right and
striking the cup at a point A. The beam is at an angle θ to the normal of the surface, so using
angle of incidence = angle of reflection we can deduce that it will reflect at the same angle
shown in Fig 8.0.
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Fig 8.0 Angle of reflection of the light rays.
̂ the
Furthermore, we can reason out that ∆𝐴𝐵𝑂 should be an isosceles triangle. at angle 𝐸𝐴𝐵
angle can be bisected at the normal and are therefore split into equal parts. And as alternating
interior angles are equal, angle theta at O is equal to angle theta at A inside the triangle.
Therefore, the two base angles of ∆𝐴𝐵𝑂 are equal. As shown in Fig. 8.0, ∆𝐴𝐵𝑂 is isosceles so
that point A is at an angle 𝜃 from the x-axis. Now, we can label the point as (−𝑐𝑜𝑠𝜃, 𝑠𝑖𝑛𝜃)
instead of (𝑥, 𝑦) since the radius of the cup is 1. Hence, we are parametrizing the equations to
introduce the additional independent variable of 𝜃 in our equations.
We can use right angled trigonometry to show this further:
𝑐𝑜𝑠𝜃 =
𝑎𝑑𝑗
𝑥
= =𝑥
ℎ𝑦𝑝
1
∴ −𝑐𝑜𝑠𝜃 is labelled as the displacement in the x-axis on the left quadrant. And,
𝑠𝑖𝑛𝜃 =
𝑜𝑝𝑝 𝑦
= =𝑦
ℎ𝑦𝑝 1
∴ 𝑠𝑖𝑛𝜃 is labelled as the displacement in the y-axis.
̂ = 180° − 2𝜃, slope of the reflected line shall be given by:
Also, since 𝐴𝐵𝑂
m= −𝑡𝑎𝑛2𝜃
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Step-1: The equation for the ray reflected at the point (−𝑐𝑜𝑠𝜃, sin 𝜃) is,
𝑦 − 𝑠𝑖𝑛𝜃
= −𝑡𝑎𝑛2𝜃
𝑥 + 𝑐𝑜𝑠𝜃
⇒ 𝑦 − sin 𝜃 = −𝑡𝑎𝑛2𝜃 (𝑥 + 𝑐𝑜𝑠𝜃)
⇒ 𝑦 − 𝑠𝑖𝑛𝜃 =
−𝑠𝑖𝑛2𝜃
(𝑥 + 𝑐𝑜𝑠𝜃)
𝑐𝑜𝑠2𝜃
⇒ 𝑦𝑐𝑜𝑠2𝜃 − 𝑠𝑖𝑛𝜃𝑐𝑜𝑠2𝜃 = −𝑥𝑠𝑖𝑛2𝜃 − 𝑠𝑖𝑛2𝜃𝑐𝑜𝑠𝜃
⇒ 𝑥𝑠𝑖𝑛2𝜃 + 𝑦𝑐𝑜𝑠2𝜃 = 𝑠𝑖𝑛𝜃𝑐𝑜𝑠2𝜃 − 𝑠𝑖𝑛2𝜃𝑐𝑜𝑠𝜃
⇒ 𝑥𝑠𝑖𝑛2𝜃 + 𝑦𝑐𝑜𝑠2𝜃 = 𝑠𝑖𝑛𝜃(2𝑐𝑜𝑠 2 𝜃 − 1) − 2𝑠𝑖𝑛𝜃𝑐𝑜𝑠 2 𝜃
(∵ 𝑠𝑖𝑛2𝜃 = 2𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃 𝑎𝑛𝑑 𝑐𝑜𝑠2𝜃 = 2𝑐𝑜𝑠 2 𝜃 − 1)
⇒ 𝑥𝑠𝑖𝑛2𝜃 + 𝑦𝑐𝑜𝑠2𝜃 = 2𝑠𝑖𝑛𝜃𝑐𝑜𝑠 2 𝜃 − 2𝑠𝑖𝑛𝜃𝑐𝑜𝑠 2 𝜃 − 𝑠𝑖𝑛𝜃
⇒ 𝑥𝑠𝑖𝑛2𝜃 + 𝑦𝑐𝑜𝑠2𝜃 = −𝑠𝑖𝑛𝜃 (5)
Step 2: Differentiating equation (5) with respect to θ gives: 2𝑥𝑐𝑜𝑠2𝜃 − 2𝑦𝑠𝑖𝑛2𝜃 = −𝑐𝑜𝑠𝜃
Using chain rule,
𝑑
(𝑠𝑖𝑛𝑥(2𝜃)) =
𝑑𝜃
𝑑𝑠𝑖𝑛𝑢 𝑑𝑢
𝑑𝑢 𝑑𝜃
𝑐𝑜𝑠(𝑢)(𝑤ℎ𝑒𝑟𝑒 𝑢 = 2𝜃)
𝑑
Similarly, 2𝑥𝑐𝑜𝑠2𝜃 𝑑𝜃 𝑦𝑐𝑜𝑠2𝜃 = −2𝑦𝑠𝑖𝑛2𝜃
𝑑
And 𝑑𝜃 − 𝑠𝑖𝑛𝜃 = 𝑐𝑜𝑠𝜃
= 2𝑥𝑐𝑜𝑠2𝜃 − 2𝑦𝑠𝑖𝑛2𝜃 = −𝑐𝑜𝑠𝜃 (6)
If we plot these curves by putting in different values of 𝜃, the reflected rays of light will appear
where the points of intersection look almost like the envelope of the nephroid.
Fig. 9.0 The envelope calculated from equation (4) is shown in a plot. The x and y axis represent the
point where the light rays strike the cup.
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Step 3: Now we would need to eliminate 𝜃 to get an equation in terms of x and y. But it would
be easier to express equation (4) and (5) in matrix form to solve for x and y. To convert into
matrix form, first we make two matrices A and B, where A includes all variables in the LHS
of equations 4 and 5, and B includes all variables in the RHS of equations 4 and 5. Therefore:
𝐴= [
𝑥𝑠𝑖𝑛2𝜃
2𝑥𝑐𝑜𝑠2𝜃
𝐵= [
𝑦𝑐𝑜𝑠2𝜃
]
−2𝑦𝑠𝑖𝑛 2𝜃
− sin 𝜃
]
−𝑐𝑜𝑠𝜃
𝑥
𝑋=[ ]
𝑦
𝑥
𝑐𝑜𝑠2𝜃
−𝑠𝑖𝑛𝜃
] [𝑦] = [
]
−2𝑠𝑖𝑛2𝜃
−𝑐𝑜𝑠𝜃
𝑠𝑖𝑛2𝜃
∴[
2𝑐𝑜𝑠2𝜃
∴ 𝐴×𝑋 = 𝐵
∴ 𝑋 = 𝐴−1 × 𝐵
𝐴−1 =
1
𝑑𝑒𝑡𝐴
(
𝑑
−𝑐
−𝑏
) 𝑤ℎ𝑒𝑟𝑒 𝐴 = (𝑎 𝑏 )
𝑐 𝑑
𝑎
𝑑𝑒𝑡𝐴 = |𝐴| = 𝑎𝑑 − 𝑏𝑐
= (𝑠𝑖𝑛2𝜃 × −2𝑠𝑖𝑛2𝜃) − (𝑐𝑜𝑠2𝜃 × 2𝑐𝑜𝑠2𝜃) = 2𝑠𝑖𝑛2 2𝜃 − 2𝑐𝑜𝑠2 2𝜃 = −2(𝑠𝑖𝑛2 2𝜃 + 𝑐𝑜𝑠2 2𝜃) = −2
1 −2𝑦𝑠𝑖𝑛2𝜃
∴ 𝐴−1 = − [
2 −2𝑥𝑐𝑜𝑠2𝜃
1 −2𝑠𝑖𝑛2𝜃
∴ 𝑋 = − [
2 −2𝑐𝑜𝑠2𝜃
1 2𝑠𝑖𝑛2𝜃
−𝑐𝑜𝑠2𝜃 −𝑠𝑖𝑛𝜃
][
]= − [
𝑠𝑖𝑛2𝜃 −𝑐𝑜𝑠𝜃
2 2𝑐𝑜𝑠2𝜃
1 −4𝑠𝑖𝑛2𝜃𝑠𝑖𝑛𝜃
= ( )[
4 −4𝑐𝑜𝑠2𝜃𝑠𝑖𝑛𝜃
1 −4𝑠𝑖𝑛2𝜃𝑠𝑖𝑛𝜃
= (− ) [
4 −4𝑐𝑜𝑠2𝜃𝑠𝑖𝑛𝜃
1
−𝑦𝑐𝑜𝑠2𝜃
]
𝑥𝑠𝑖𝑛2𝜃
= (− 4) [
−2𝑠𝑖𝑛2𝜃𝑠𝑖𝑛𝜃
−2𝑐𝑜𝑠2𝜃𝑠𝑖𝑛𝜃
1 −2𝑐𝑜𝑠𝜃
= ( )[
2𝑠𝑖𝑛𝜃
4
−
+
2𝑐𝑜𝑠2𝜃𝑐𝑜𝑠𝜃
]
2𝑠𝑖𝑛2𝜃𝑐𝑜𝑠𝜃
− 2𝑐𝑜𝑠2𝜃𝑐𝑜𝑠𝜃
]
+ 2𝑠𝑖𝑛2𝜃𝑐𝑜𝑠𝜃
−
+
2(𝑐𝑜𝑠(2𝜃 − 𝜃))
]
2(𝑠𝑖𝑛(2𝜃 − 𝜃))
− (𝑐𝑜𝑠𝜃 − 𝑐𝑜𝑠3𝜃)
]
− (𝑠𝑖𝑛3𝜃 − 𝑠𝑖𝑛𝜃)
𝑥
1 𝑐𝑜𝑠3𝜃 − 3𝑐𝑜𝑠𝜃
∴[ ]= [
]
𝑦
4 3𝑠𝑖𝑛𝜃 − 𝑠𝑖𝑛3𝜃
15
+
−
𝑐𝑜𝑠2𝜃𝑐𝑜𝑠𝜃
]
𝑠𝑖𝑛2𝜃𝑐𝑜𝑠𝜃
If we graph the points, we get the envelope of the nephroid. A nephroid is like a cardioid but
with an extra cusp. A cusp is a point where two branches of a curve, which are in opposite
direction, meet (Glen). Since all values of theta are considered, an extra cusp is present in the
nephroid. A cardioid is almost like a nephroid but with only 1 cusp.
5.2 Cardioid
Cardioid is another caustic we can observe in a cup if we change the setup slightly. The method
of modelling the equation is the same as we did for modelling the nephroid, only the location
of the point source is different. Since the light reflects off the rim of the cup to form the cardioid,
we can assume the rim to be the point source of the light instead of having a point source at
infinity as we did for the nephroid. As the edges of the cup will stop nearly half of the light, it
makes sense to see only half of the nephroid in our cup to form a cardioid.
Below is a diagram that shows the angle of the incoming and outgoing rays.
Fig. 10.0 Setting up a point on a rim. The angles of incidence and reflection have been defined
differently to maintain the parameterisation from the last case.
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Step-1: Writing in parametric form, we get the family of curves of light rays reflecting from
the rim of the cup as,
𝑦(1 + 𝑐𝑜𝑠3𝜃) + 𝑥𝑠𝑖𝑛3𝜃 = 𝑠𝑖𝑛𝜃 − 𝑠𝑖𝑛2𝜃
Step-2: Differentiating the previous equation gives us,
−3𝑦𝑠𝑖𝑛3𝜃 + 3𝑥𝑐𝑜𝑠3𝜃 = 𝑐𝑜𝑠𝜃 − 2𝑐𝑜𝑠2𝜃
Step-3: Making 𝑥𝑦 the subject gives us the equation of the envelope as,
𝑥
1 𝑐𝑜𝑠2𝜃 − 2𝑐𝑜𝑠𝜃
[ ]= [
]
𝑦
3 2𝑠𝑖𝑛𝜃 − 𝑠𝑖𝑛2𝜃
And finally, we get a cardioid!
Fig.11 The final cardioid. The x and y axis represent the point where x and y point of the cardioid lies.
6.0 Conclusion
This math project was a modest step toward understanding caustics. The aim was to model the
nephroid, and cardioid shapes formed by light reflection in the mug. The final finding a was a
method to find the envelope of a nephroid and a cardioid, which finally resulted in the equation:
𝑥
1 𝑐𝑜𝑠2𝜃 − 2𝑐𝑜𝑠𝜃
[ ]= [
]
𝑦
3 2𝑠𝑖𝑛𝜃 − 𝑠𝑖𝑛2𝜃
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Though, a few assumptions were made to construct the model of these caustics. To model a
nephroid, the light source was assumed to be infinite, whereas to model a cardioid, the light
source was assumed to be at the cup's rim.
The assumptions employed to construct the caustics equations, however, do not hold true in
reality. The nephroid in a cup is unlikely to be an exact replica of the nephroid's modelled
equation. The point source for nephroids was assumed to be at infinity, but the caustics show
that the light source is not too far away. Furthermore, when modelling a cardioid, the point
source was assumed to be at the cup's rim, when in fact it isn't. So, we can safely assume the
light source is nearby, like our light bulb. Additionally, the cup is assumed to be a twodimensional circle, though the cup is 3D. However, the same principles apply to 3D since we
assumed the edges of the circle to be reflecting off the light it just reflects on the bottom of
the cup as a 2D image.
Overall, this study only looks at the mathematical model of a nephroid and a cardioid, but
caustic engineering is a huge field of study. Using new milling techniques on glass and
plexiglass, this study shows how to create windows that project holographic images from
natural light. Controlled caustics, an optical phenomenon that appears chaotic at first, allows
this (Manaugh).
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References
Glen, Stephanie. Cusps in Graphs & Corners in Graphs. n.d.
<https://www.calculushowto.com/cusps-in-graphs-corners/>.
Hayes, Andy, Hobart Pao and Andrew Ellinor. Polar Curve. n.d. Website. 22 March 2022.
<https://brilliant.org/wiki/polar-curves/#polar-equations>.
Manaugh, Geoff. Caustic Engineering. 6 February 2013.
<https://bldgblog.com/2013/02/caustic-engineering/ >.
ProofWiki. Equation of Cardioid/ Polar. 29 September 2020.
<https://proofwiki.org/wiki/Equation_of_Cardioid/Polar>.
Vasilkova, Dominika. Cardioids in Coffee Cups. 18 October 2017. 19 November 2021.
<https://chalkdustmagazine.com/features/cardioids-coffee-cups/>.
Weisstein, Eric W. Cirlce Catacaustic. n.d.
<https://mathworld.wolfram.com/CircleCatacaustic.html>.
—. Family of Curves. n.d. <https://mathworld.wolfram.com/FamilyofCurves.html>.
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