What Is Calculus? A Beginner's Guide to Limits and Differentiation

How to Understand Calculus

Calculus is a study of rates of change of functions and accumulation of infinitesimally small quantities. It can be broadly divided into two branches:

• Differential Calculus. This concerns rates of changes of quantities and slopes of curves or surfaces in 2D or multidimensional space.
• Integral Calculus. This involves summing infinitesimally small quantities.

 

What's Covered in This Tutorial

In this tutorial, you will learn about:

• Limits of a function
• How the derivative of a function is derived
• Rules of differentiation
• Derivatives of common functions
• What the derivative of a function means
• Working out derivatives from first principles
• 2nd and higher order derivatives
• Applications of differential calculus
• Worked examples

 

Who Invented Calculus?

Calculus was invented by the English mathematician, physicist and astronomer Isaac Newton and German mathematician Gottfried Wilhelm Leibniz independently of each other in the 17th century.

Isaac Newton (1642 - 1726) and Gottfried Wilhelm Leibniz (below) invented calculus independent of each other in the 17th century.  Image from Pixabay.

Gottfried Wilhelm von Leibniz (1646 - 1716), a German philosopher and mathematician. Public domain image via Wikimedia Commons.

What Is Calculus Used For?

Calculus is used widely in mathematics, science, in the various fields of engineering and economics.

Introduction to Limits of Functions

To understand calculus, we first need to grasp the concept of limits of a function.

Imagine we have a continuous line function with the equation f(x) = x + 1 as in the graph below.

The value of f(x) is simply the value of the x coordinate plus 1.

© Eugene Brennan.

The function is continuous which means that f(x) has a value that corresponds to all values of x, not just the integers ....-2, -1, 0, 1, 2, 3.... and so on, but all the intervening real numbers. I.e. decimal numbers like 7.23452, and irrational numbers like π, and √3.

So if x = 0, f(x) = 1

if x = 2, f(x) = 3

if x = 2.3, f(x) = 3.3

if x = 3.1, f(x) = 4.1 and so on.

Let's concentrate on the value x =3, f(x) = 4.

Scroll to Continue

As x gets closer and closer to 3, f(x) gets closer and closer to 4.

So we could make x = 2.999999 and f(x) would be 3.999999.

We can make f(x) as close to 4 as we want. In fact we can choose any arbitrarily small difference between f(x) and 4 and there will be a correspondingly small difference between x and 3. But there will always be a smaller distance between x and 3 that produces a value of f(x) closer to 4.

So What's the Limit of a Function Then?

Referring to the graph again, the limit of f(x) at x = 3 is the value f(x) approaches as x gets closer to 3. Not the value of f(x) at x=3, but the value it approaches. As we'll see later, the value of a function f(x) may not exist at a certain value of x, or it may be undefined.

This is expressed as, "The limit of f(x) as x approaches c, equals L".

Formal Definition of a Limit

The (ε, δ) Cauchy Definition of a Limit

The formal definition of a limit was specified by the mathematicians Augustin-Louis Cauchy and Karl Weierstrass

Let f(x) be a function defined on a subset D of the real numbers R.

c is a point of the set D. ( The value of f(x) at x = c may not necessarily exist)

L is a real number.

Then:

lim f(x) = L
x → c

exists if:

• Firstly for every arbritarily small distance ε > 0 there exists a value δ such that, for all x belonging to D and 0 > | x - c | < δ, then | f(x) - L | < ε
• and secondly the limit approaching from the left and right of the x coordinate of interest must be equal.

In plain English, this says that the limit of f(x) as x approaches c is L, if for every ε greater than 0, there exists a value δ, such that values of x within a range of c ± δ (excluding c itself, c + δ and c - δ) produces a value of f(x) within L± ε.

....in other words we can make f(x) as close to L as we want by making x sufficiently close to c.

This definition is known as a deleted limit because the limit omits the point x = c.

Limit of a function. 0 > |x - c| then 0 > | f(x) - L | < ϵ. © Eugene Brennan

Continuous and Discontinuous Functions

A function is continuous at a point x = c on the real line if it is defined at c and the limit equals the value of f(x) at x = c. I.e:

lim f(x) = L = f(c)
x → c

A continuous function f(x) is a function that is continuous at every point over a specified interval.

Examples of continuous functions:

• Temperature in a room versus time.
• The speed of a car as it changes over time.

A function that is not continuous, is said to be discontinuous. Examples of discontinuous functions are:

• Your bank balance. It changes instantly as you lodge or withdraw money.
• A digital signal, it's either 1 or 0 and never in between these values.

The function f(x) = sin (x) / x or sinc (x). The limit of f(x) as x approaches 0 from both sides is 1. The value of sinc (x) at x = 0 is undefined because we can't divide by zero and sinc (x) is discontinuous at this point. © Eugene Brennan

Limits of Common Functions

Common functions and their limits

Function	Limit
1/x as x tends to infinity	0
a/(a + x) as x tends to 0	a
sin x/x as x tends to 0	1

Calculating the Velocity of a Vehicle

Imagine we record the distance a car travels over a period of one hour. Next we plot all the points and join the dots, drawing a graph of the results (as shown below). On the horizontal axis, we have the time in minutes and on the vertical axis we have the distance in miles. Time is the independent variable and distance is the dependent variable. In other words, the distance travelled by the car depends on the time which has passed.

Graph of distance travelled by a vehicle at constant speed is a straight line. © Eugene Brennan

If the car travels at a constant velocity, the graph will be a line, and we can easily work out its velocity by calculating the slope or gradient of the graph. To do this in the simple case where the line passes through the origin, we divide the ordinate (vertical distance from a point on the line to the origin) by the abscissa (horizontal distance from a point on the line to the origin).

So if it travels 25 miles in 30 minutes,

Velocity = 25 miles/30 minutes = 25 miles / 0.5 hour = 50 mph

Similarly if we take the point at which it has travelled 50 miles, the time is 60 minutes, so:

Velocity is 50 miles/60 minutes = 50 miles / 1 hour = 50 mph

Note: In physics we normally speak of the "velocity" of a body. Technically, the definition of velocity is speed in a given direction, so it is a vector quantity. Speed is the magnitude of the velocity vector.

Average Velocity and Instantaneous Velocity

Ok, so this is all fine if the vehicle is travelling at a steady velocity. We just divide distance by time taken to get velocity. But this is the average velocity over the 50 mile journey. Imagine if the vehicle was speeding up and slowing down as in the graph below. Dividing distance by time still gives the average velocity over the journey, but not the instantaneous velocity which changes continuously. In the new graph below, the vehicle accelerates mid way through the journey and travels a much greater distance in a short period of time before slowing down again. Over this period, its velocity is much higher.

Graph of a vehicle travelling at a variable speed. © Eugene Brennan

In the graph below, if we denote the small distance travelled by Δs and the time taken as Δt, again we can calculate velocity over this distance by working out the slope of this section of the graph.

So average velocity over interval Δt = slope of graph = Δs/Δt

Note: The Greek letter "Δ" pronounced "delta" is often used in mathematics to represent a small quantity.

Approximate speed over a short range can be determined from slope. The average speed over the interval Δt is Δs/Δt. © Eugene Brennan

However the problem is that this still only gives us an average. It's more accurate than working out velocity over the full hour, but it's still not the instantaneous velocity. The car travels faster at the start of the interval Δt (we know this because distance changes more rapidly and the graph is steeper). Then the velocity starts to decrease midway and reduces all the way to the end of the interval Δt.

What we're aiming to do is find a way of determining the instantaneous velocity.
We can do this by making Δs and Δt smaller and smaller so we can work out the instantaneous velocity at any point on the graph.

See where this is heading? We're going to use the concept of limits we learned about before.

What Is Differential Calculus?

Differential calculus is one of the two branches of calculus which also includes integral calculus. It is a study of the rate at which quantities change.

In the example above we saw how we could attempt to determine a more accurate measurement of velocity by working out the slope of a graph over a shorter interval. We can do this using limits.

Slope of a graph

In the graph below we have a generalised function y = f (x).

x is a point on the horizontal axis and Δx is a small change in x.

The value of the function at x is f (x)

As x changes to x + Δx , f (x) changes by Δy to f (x + Δx)

You may remember from coordinate geometry if we know two points on a graph: (x₁, y₁) and (x₂, y₂), the slope of a line joining the two points is:

(y₂ - y₁) / ( x₂ - x₁)

So the slope of this line is ( f (x + Δx) - f (x) ) / (x + Δx - x) = Δy / Δx

The slope Δy / Δx is approximately the slope of a tangent to the graph for small Δx.

Over a short distance, the slope of the secant (red line) is approximately the rate of change of the function. © Eugene Brennan

Approximate slope of a function for small increments of x and f(x)

What Happens When ΔX Becomes Smaller and Smaller?

The red line that intersects the graph at two points in the diagram above is called a secant.

If we now make Δx and Δy smaller and smaller, the red line eventually becomes a tangent to the curve. The slope of the tangent is the instantaneous rate of change of f (x) at the point x.

Derivative of a Function

If we take the limit of the value of the slope as Δx tends to zero, the result is called the derivative of y = f (x).

lim (Δy / Δx) =
_{Δx → 0}

= lim ( f (x + Δx) - f (x) ) / (x + Δx - x)
_{Δx → 0}

The value of this limit is denoted as dy/dx.

Since y is a function of x, i.e. y = f(x), the derivative dy/dx can also be denoted as f '(x) or just f ' and is also a function of x. I.e. it varies as x changes.

If the independent variable is time, the derivative is sometimes denoted by the variable with a dot superimposed on top.

E.g. if a variable x represents position and x is a function of time. I.e. x(t)

Derivative of x wrt t is dx/dt or ẋ (ẋ or dx/dt is speed, the rate of change of position)

We can also denote the derivative of f (x) wrt x as d/dx(f (x))

As Δx and Δy tend to zero, the slope of the secant approaches the slope of the tangent. © Eugene Brennan, created in GeoGebra.

Slope over an interval Δx. The limit is the derivative of the function.

Differentiating Functions From First Principles

To find the derivative of a function, we differentiate it wrt to the independent variable. There are several identities and rules to make this easier, but first let's try to work out an example from first principles.

Example: Evaluate the derivative of x²

So f (x) = x²

f (x + Δx) = (x + Δx)²

d/dx( f (x)) =

lim ( f (x + Δx) - f (x) ) / ((x + Δx) - x)
_{Δx → 0}

Substitute for f (x + Δx) and f (x) giving

lim ( (x + Δx)² - x² ) / Δx)
_{Δx → 0}

Expand out (x + Δx)² giving

lim ( (x² + 2xΔx + Δx²- x² ) / Δx)
_{Δx → 0}

The two x² cancel out which gives:

lim ( (2xΔx + Δx² ) / Δx)
_{Δx → 0}

Dividing by Δx gives:

lim (2x + Δx )
_{Δx → 0}

= 2x

Using Rules to Work out Derivatives

Rather than working out the derivatives of functions from first principles, we normally use a set of rules to make things easier.

In the table below, f and g are two functions.

f ' is the derivative of f

g' is the derivative of g

Rules of Differentiation

Rules of differentiation

Rule Type	Function	Derivative
Constant factor rule	af	af '
Power rule for polynomials	xᵇ	bxᵇ ⁻ ¹
Sum rule	f + g	f ' + g '
Difference rule	f - g	f ' - g '
Product rule	fg	fg ' + gf '
Reciprocal of a function	1/f	- f ' / f ^2
Quotient rule	f / g	(f 'g - g ' f )/ g ^2
Chain rule (function of a function rule)	f(g) where g is a function of x	f ' (g) g ' (x)

Derivatives of Common Functions

Derivatives of comon functions

Function Type	Function	Derivative
Constant	c	0
Line	mx	m
Line with y axis intercept	mx + c	m
x squared	x ^ 2	2x
x cubed	x ^ 3	3x^2
Square root	x ^ (1/2)	(1/2) x ^ (-1/2)
Reciprocal	1/x	-1/x^2
Exponential function	e^x	e^x
Natural log	ln (x)	1/x
Trigonometric function	sin (x)	cos (x)
Trigonometric function	cos (x)	- sin (x)
Trigonometric function	tan (x)	1 + (tan (x))^2 = (cosec (x))^2

Examples of Working Out Derivatives

Example 1:

What is the derivative of 20?

Derivative of a constant is 0, so d/dx(20) = 0

Example 2:

What is the derivative of 3x?

Using the constant factor rule (multiplication by a constant rule)

d/dx(3x) = 3( d/dx(x) )

But using the power rule the derivative of x¹ = 1x⁰ = 1

So d/dx(3x) = 3( d/dx(x) ) = 3(1) = 3

Example 3:

What is the derivative of 6x³

Using the multiplication by a constant rule, d/dx(6x³) = 6 ( d/dx(x³) )

From the power rule, d/dx(x³) = 3x²

So d/dx(6x³) = 6 ( d/dx(x³) ) = 6 (3x²) = 18x²

Example 4:

Evaluate the derivative of 5sin (x) + 6x⁵

We use the sum rule to find the derivatives of 5sin (x) and 6x⁵ and then add the result together.

So d/dx (5sin (x)) = 5d/dx(sin (x)) = 5cos (x)

d/dx(6x⁵) = 6d/dx(x⁵) = 6(5x⁴) = 30x⁴

Adding the results together

d/dx (5sin (x) + 6x⁵) = 5cos (x) + 30x⁴

Example 5:

What is the derivative of x³sin (x) ?

Use the product rule, so:

d/dx(x³sin (x)) = x³d/dx(sin(x)) + sin(x)d/dx(x³)

d/dx (sin(x)) = cos(x)

d/dx(x³) = 3x² (from the power rule)

so d/dx(x³sin (x)) = x³d/dx(sin(x)) + sin(x)d/dx(x³) = x³cos(x) + 3x²sin(x)

Example 6:

Evaluate the derivative of tan (x)

tan (x) = sin (x) / cos (x)

We can use the quotient rule to work this out:

d/dx (f(x)/g(x)) = (f '(x)g(x) - g '(x)f (x))/ g(x)²

so d/dx(sin (x) / cos (x)) = (d/dx(sin (x))cos (x) - d/dx(cos (x))sin (x)) / cos² (x)

d/dx(sin (x)) = cos (x)

d/dx(cos (x)) = - sin x

Substituting

(d/dx(sin (x))cos (x) - d/dx(cos (x))sin (x)) / cos²(x)

= (cos (x)cos (x) - (-sin (x))sin (x)) / cos²(x)

= (cos²(x) + sin²(x)) / cos²(x)

= 1 + tan²(x) = sec²(x)

Example 7:

What is the derivative of ln(5x³) ?

We use the chain rule to work this out.

For two functions f(g) and g(x)

df/dx = (df/dg)(dg/dx)

Let g(x) = 5x³

and f(g) = ln(g)

df/dg = 1/g

dg/dx = 5(3x²) = 15x²

df/dx = (df/dg)(dg/dx)

= (1/g)(15x²)

Substituting for g:

= 1/(5x³)((15x²) = 3/x

We could also have evaluated the derivative by first using the rules of logarithms to simplify the expression.

So ln(5x³) = ln(5) + ln(x³) = ln(5) + 3ln(x) ............(product rule and power rule)

d/dx (ln(5) + 3ln(x)) = d/dx (ln(5)) + d/dx(3ln(x)) = 0 + 3d/dx(ln(x))

= 0 + 3(1/x) = 3/x

Positive and Negative Values of the Derivative

The animation below shows the function sin(Ө) and it's derivative cos(Ө). At Ө = 0, the value of the derivative is cos(Ө) = cos(0) = 1. As Ө increases, the value of cos(Ө) decreases, i.e the slope of the tangent to sin(Ө) becomes smaller. Eventually at Ө = π/2, the slope is zero. This is an important point because as we'll see later, we can use this fact to find the maxima and minima of functions.
As Ө exceeds π/2, the value of the derivative becomes negative.

How come?

Remember the definition of the derivative?

Δx is positive, but the change Δy is negative since f (x +Δx) - f (x) is negative because f (x +Δx) < f (x)......(the function is decreasing in value)

In the limit as Δx and Δy tend to zero, the derivative is also less than zero.

Sin(Ө) and Its Derivative Cos(Ө)

The derivative of sin(Ө) is cos(Ө). Notice how the value of the derivative at Ө = 0 is positive and decreases to 0 at the peak of the waveform when Ө = π/2. Then it becomes negative and reaches zero before becoming positive again. © Eugene Brennan. Created in GeoGebra

2nd and Higher Order Derivatives

What happens if we take the derivative of a derivative?

Consider the function y = f (x)

The derivative of y is f '(x) or dy/dx

The derivative of dy/dx is known as the second derivative or second order derivative and is denoted by d²y/dx² or f '' (x) ......(f with a double dash). We can have third and higher order derivatives so for instance the third order derivative of y is d³y/dx³

Examples:

(1) If y = sin (x), what is d²y/dx² ?

dy/dx = cos (x)

d²y/dx^{2 =} d/dx (cos (x)) = -sin (x)

(2) What is the second derivative of ln (x)?

y = ln (x)

So dy/dx = 1/x = x^(-1)

d²y/dx² = d/dx (x^(-1)) = -1( x^(-2) ) = -1/x²

Since dy/dx is the rate of change of a function, roughly speaking we can think of d²y/dx² as being the rate at which dy/dx itself is changing.

Returning to the car example:

s(t) is a function describing how distance travelled changes with time.
ds/dt is the rate of change of position, called speed or velocity.
d²s/dt² is the rate of change of velocity, which is called acceleration.

If v is the velocity of the vehicle and a is its acceleration:

v = ds/dt

a = dv/dt = d/dt(ds/dt) = d²s/dt²

So the second derivative of distance which is acceleration is equal to the first derivative of velocity.

We can go up to the third derivative of s, so:

d³s/dt³ = da/dt is the rate of change of acceleration, known as "jerk".

Stationary and Turning Points of a Function

A stationary point of a function is a point at which the derivative is zero. On a graph of the function, the tangent to the point is horizontal and parallel to the x-axis.

A turning point of a function is a point at which the derivative changes sign. A turning point can be either a local maxima or minima. If a function can be differentiated, a turning point is a stationary point. However the reverse is not true. Not all stationary points are turning points. For instance in the graph of f (x) = x³ below, the derivative f '(x) at x = 0 is zero and so x is a stationary point. However as x approaches 0 from the left, the derivative is positive and decreases to zero, but then increases positively as x becomes positive again. Therefore the derivative doesn't change sign and x is not a turning point.

Points A and B are stationary points and the derivative f'(x) = 0. They are also turning points because the derivative changes sign. © Eugene Brennan

Example of a function with a stationary point that is not a turning point. The derivative f'(x) at x = 0 is 0, but doesn't change sign. © Eugene Brennan

Inflection Points of a Function

An inflection point of a function is a point on a curve at which the function changes from being concave to convex. At an inflection point, the second order derivative changes sign (i.e it passes through 0. See the graph below for a visualisation).

The red circles are stationary points. The blue squares are inflection points. Self CC BY SA 3.0 via Wikimedia Commons

Explaining stationary, turning points and inflection points and how they relate to the first and second order derivatives. Cmglee, CC BY SA 3.0 unported via Wikimedia Commons

Using the Derivative to Find the Maxima, Minima and Turning Points of Functions

We can use the derivative to find the local maxima and minima of a function (the points at which the function has maximum and minimum values.) These points are called turning points because the derivative changes sign from positive to negative or vice versa. For a function f (x), we do this by:

• differentiating f (x) wrt x
• equating f ' (x) to 0
• and finding the roots of the equation, i.e. the values of x that make f '(x) = 0

Example 1:

Find the maxima or minima of the quadratic function f (x) = 3x² + 2x +7 (the graph of a quadratic function is called a parabola).

A quadratic function. © Eugene Brennan

f (x) = 3x² + 2x +7

and f '(x) = 3(2x¹) + 2(1x⁰) + 0 = 6x + 2

Set f '(x) = 0

6x + 2 = 0

Solve 6x + 2 = 0

Rearranging:

6x = -2

giving x = - ¹/₃

and f(x) = 3x² + 2x +7 = 3(-1/3)² + 2(-1/3) + 7 = 6 ²/₃

A quadratic function has a maximum when the coefficient of x² < 0 and a minimum when the coefficient > 0. In this case since the coefficient of x² was 3, the graph "opens up" and we have worked out the minimum and it occurs at the point (- ¹/₃, 6 ²/₃).

Example 2:

In the diagram below, a looped piece of string of length p is stretched into the shape of a rectangle. The sides of the rectangle are of length a and b. Depending on how the string is arranged, a and b can be varied and different areas of rectangle can be enclosed by the string. What is the maximum area that can be enclosed and what will be the relationship between a and b in this scenario?

Finding the maximum area of a rectangle that can be enclosed by a perimeter of fixed length. © Eugene Brennan

p is the length of the string

The perimeter p = 2a + 2b (the sum of the 4 side lengths)

Call the area y

and y = ab

We need to find an equation for y in terms of one of the sides a or b, so we need to eliminate either of these variables.

Let's try to find b in terms of a:

So p = 2a + 2b

Rearranging:

2b = p - 2a

and:

b = (p - 2a)/2

y = ab

Substituting for b gives:

y = ab = a(p - 2a)/2 = ap/2 - a² = (p/2)a - a²

Work out the derivative dy/da and set it to 0 (p is a constant):

dy/da = d/da((p/2)a - a²) = p/2 - 2a

Set to 0:

p/2 - 2a = 0

Rearranging:

2a = p/2

so a = p/4

We can use the perimeter equation to work out b, but it's obvious that if a = p/4 the opposite side is p/4, so the two sides together make up half the length of the string which means both of the other sides together are half the length. In other words maximum area occurs when all sides are equal. I.e when the enclosed area is a square.

So area y = (p/4)(p/4) = p²/16

Example 3 (Max Power Transfer Theorem or Jacobi's Law):

The image below shows the simplified electrical schematic of a power supply. All power supplies have an internal resistance (R_INT) which limits how much current they can supply to a load (R_L). Calculate in terms of R_INT the value of R_L at which maximum power transfer occurs.

The schematic of a power supply connected to a load, showing the supply's equivalent internal resistance Rint. © Eugene Brennan

The current I through the circuit is given by Ohm's Law:

So I = V/(R_INT+ R_L)

Power = Current squared x resistance

So power dissipated in the load R_L is given by the expression:

P = I²R_L

Substituting for I:

= (V/(R_INT+ R_L))²R_L

= V²R_L/(R_INT+ R_L)²

Expanding the denominator:

= V²R_L/(R²_INT + 2R_INTR_{L +} R²_L)

and dividing above and below by R_L gives:

P = V² / (R²_INT / R_L + 2R_INT + R_L)

Rather than finding when this is a maximum, it's easier to find when the denominator is a minimum and this gives us the point at which maximum power transfer occurs, i.e. P is a maximum.

So the denominator is R²_INT / R_L + 2R_INT + R_L

Differentiate it wrt R_L giving:

d/dR_L (R²_INT / R_L + 2R_INT + R_{L) =} -R²_INT / R²_L + 0 + 1

Set it to 0:

-R²_INT / R²_L + 0 + 1 = 0

Rearranging:

R²_INT / R²_L = 1

and solving gives R_L = R_INT.

So max power transfer occurs when R_L = R_INT.

This is called the max power transfer theorem.

A Beginner's Guide to Integration

See my other tutorial, which covers integral calculus and applications of integration:

How to Understand Calculus: A Beginner's Guide to Integration

References

Stroud, K.A., (1970) Engineering Mathematics (3rd ed., 1987) Macmillan Education Ltd., London, England.

This content is accurate and true to the best of the author’s knowledge and is not meant to substitute for formal and individualized advice from a qualified professional.

© 2019 Eugene Brennan

What Would Happen if You Fell Down a Hole That Went Through the Centre of the Earth?

Created by Grok

Well you'd be incinerated as you pass through the hot and molten core! But imagine if the core wasn't there or you had a special suit that protected you from the heat—what would happen then? Would you suddenly stop when you got to Oz?

The scenario is an example of a simple harmonic oscillator—other examples being a swinging pendulum and a weight bobbing up down, suspended from a spring. Both of the latter rely on interchange of energy and its change in form, from kinetic energy due to motion of an object, to stored potential energy. In the example of a weight suspended from a spring, the potential energy is manifested as tension in the spring. Energy in physics is defined as "the ability to do work". "Work" in physics also has a specific meaning: work is done when a force moves a body through a distance. The stored potential energy in a spring can do work, which is how a clock spring can move the hands of the clock.

Simple harmonic oscillators as the name suggests oscillate or move backwards and forwards indefinitely. For an ideal spring-mass system in a vacuum, with a perfect spring and no friction,  the mass will bounce up and down forever, with energy being interchanged as described above. Simple harmonic oscillators have a natural frequency that depends on the magnitude of the mass and the stiffness of the spring (how hard it is to stretch or compress it). In real life, systems are lossy and due to mechanical friction in a spring and drag from air, oscillations eventually die out. Technically this is known as damping. Dampers are actually used as components in your washing machine's drum suspension and also vehicle suspension systems to smooth out oscillations.

Back to the hole in the Earth that goes all the way to Australia—if you jumped in and there was no air in the hole, you would fall and accelerate, your velocity increasing as you drop downwards. In fact, you'd free fall under the influence of gravity, being continually accelerated. This differs from the situation of falling in air, because in that case, you eventually reach a constant velocity—known as terminal velocity—when the downward force of gravity is balanced by the upward drag, resulting in no net acceleration. As you continue to fall closer to the centre of the Earth, your velocity increases until you reach maximum velocity at the centre. Due to inertia however, you overshoot and pass through the centre. However, from now on, there's "less Earth" in front of you and more behind you. So now you're being decelerated and the gravitational force of the Earth is having a braking effect. Your velocity decreases after you pass through the centre, and it keeps decreasing until it falls to zero as you emerge from the hole on the other side, just like the way a ball kicked up into the air pauses momentarily when it reaches the apex of its trajectory. Once your velocity reaches zero, you start to fall back down the hole and the process is repeated as you fall back to where you started from. Without any losses due to friction, you would keep moving from one side of the planet to the other indefinitely, and it works out that the transit time would be 42 minutes. In reality, if such a hole was a civil engineering possibility, which is highly unlikely and it was filled with air, your oscillations back and forth would be damped and you'd eventually settle and float at the centre of the Earth, weightless. 

Science Friday Podcast — Jim Lovell Interview

NASA astronaut Jim Lovell, photographed in 1969. Image credit: NASA, public domain, via Wikimedia Commons

The late Jim Lovell, commander of the ill-fated Apollo 13 mission talks with Ira Flatow of SF in an interview from 1995.
Apollo 13 was the seventh crewed mission and would have been the third operation to land humans on the Moon. However an explosion in an oxygen tank caused loss of electrical power and cut short the mission. The crew had to slingshot the spacecraft around the Moon to gain enough momentum to get back to Earth. In the interview, Jim explained the details of the event, we hear recordings of radio comms between flight control and the the spacecraft and Commander Lovell answers questions from callers. 

Another Few Stats on Poulaphouca Reservoir Water Levels and Volume

An ash tree growing on a trail beside the Poulaphouca Reservoir. © Eugene Brennan

In the period from 20th to 31st January, the volume of water in the Poulaphouca Reservoir increased by 52 million m³ of water, corresponding to a 2.3 m rise in level. In the same period, discharges from Golden Falls dam totalled 16 million m³. Since the level in the Golden Falls lake didn't change appreciably over those 12 days, and its surface area is much smaller than that of the Poulaphouca lake, the amount of water that flowed into Poulaphouca must have been equal to the increase in volume plus the amount that flowed out — 68 million m³. Over the 12 days, that gives an average inflow rate of 68/12 million or approximately 5.6 million m³/day. That's about 1.6 times the peak daily flow rate through Kilcullen last week. The Poulaphouca lake is fed by two rivers: The River Liffey and Kings River and so that's understandable. Some small streams also add to the water accumulated.
The ash tree above is on a nice section of path along the lake that leads to the car park on the N81, near the turn for Ballymore Eustace .
Stats are available on the ESB hydrometric website here.
 

Making a Bit Of Space : Golden Falls Lake Level Has Risen

Water level in Golden Falls Lake, Ballymore Eustace. Image courtesy ESB hydrometrics

The Golden Falls lake rose 1.44 m since yesterday. That's a bit of a jump since the downwards trend over the last week or so, and equivalent to approximately 370,000 cubic metres (m³) of water. The Golden Falls lake doesn't really provide a large storage capacity compared to the much larger Poulaphouca Reservoir, which has a surface area of approximately 22.6 square kilometres (km²), 86 times that of the Golden Falls lake. According to Esbarchives.ie, the head of the Golden Falls dam is 17.4 m. If the walls of the 256,000 m² lake were vertical, that would give a storage capacity of about 4.4 million m³ of water. To put that into perspective, if it were empty, it could store only about a day and a quarter's worth of the water that passed through Kilcullen at peak discharge last week. So dropping the level isn't a huge advantage as regards overall storage and I guess the objective is to try and make more "headroom" available in the Poulaphouca Reservoir to cater for extended periods of heavy rain.

Image courtesy ESB hydrometrics.

Kilcullen Town Hall's Film Projector

Movie projector from Kilcullen Town Hall Cinema. © Eugene Brennan

I was examining the nameplate (rating plate) of Kilcullen Town Hall's former film projector the other day, now located in a corner of the porch of the heritage centre. The Peerless Magnarc carbon-arc lamphouse, paired-with the projector and  manufactured by the J. E. McAuley Manufacturing Company of Chicago, Illinois, ran on DC electricity. A rectifier may have been built into the projector to convert mains AC to DC, or possibly it was a separate piece of equipment.
Arc lamps were used as an intense white light source for projectors, essential for producing a bright image on the screen. They were also used of course in search lights during WWII, for spotting enemy aircraft. Arc lighting generates large amounts of harmful UV radiation and filters would presumably have been used in the optics to block this. Electric arcs also produce a lot of heat, as anyone who has ever done electric welding would know, and again, filtering would have been necessary to protect the film. Carbon rods were consumable items used in arc lamps, just like the rods used for arc welding and had to be replaced frequently. In the early days of movies, film stock was manufactured on a highly flammable nitrate base and fires were common. It's thought that a candle ignited a reel of nitrate film, resulting in the Drumcollogher tragedy of 1926, in which 46 people perished.

I'm not sure what the "motor" is, referred to in the spec. Since the lamp was an add-on unit to the projector, it was probably not the film feed motor, but rather a motor powering a cooling fan for the lamp or one that fed the electrodes as they were consumed.

The photo mounted on the projector shows actor Paul Newman standing in front of it on the occasion of his visit to Kilcullen over 20 years ago. At that stage, the projector was still located in the projection room.

In 2006, Brian Byrne published a more detailed article in the Kilcullen Diary which provided more information about the projector’s projection unit.

Arc volts 31 - 46
Arc amps 32 - 100
Motor volts 31 - 110
Motor amps 0.5 - 1.4

J.E. McAuley Mfg. Co.
Chicago, ILL 
United States of America

Rating plate attached to the projector. © Eugene Brennan

How Much Thrust?

SLS rocket. Image credit: NASA

8.8 million pounds, 4.0 million kilos or 17.3 million newtons (the SI unit of force). That's 15% more thrust than the Saturn V rocket, used for the Apollo missions to the Moon in the late '60s and early '70s.
The SLS or Space Launch System is the rocket that's being used for the Artemis missions that will once again carry astronauts to the Moon. (Artemis was Apollo's twin brother in Greek mythology).
Huge amounts of thrust are needed to accelerate spacecraft, so that they can escape Earth's gravity. The escape velocity of planet Earth is 11.2 km/s or 25,000 mph. We can't simply use rockets to travel point-to-point to the Moon as it would be inefficient and require continuous rocket firing and fuel use. By travelling fast enough initially, hence the massive engines and fuel tanks, we can "coast" there, like a car with one's foot on the clutch. The principle is based on Newton's first law of motion—so once an object starts moving, it keeps moving uniformly, unless another force acts on it, Earth's gravity in this case. (That's why the Voyager spacecraft are still moving out beyond the Solar System). Technically, a spacecraft is falling back to Earth under the influence of the latter's gravity, but in the case of lunar missions, the Moon and its gravity "take over" when the craft is close enough, allowing it to be injected into lunar orbit. 

Sewage Treatment Plant Capacities in Kildare and Brannockstown Septic Tank

Wastewater treatment plant. AI image by ChatGPT.

This is Uisce Éireann's register of available capacity at waste water treatment plants (WWTPs), published in August 2025. It lists treatment plant capacity and "does not provide an indication of network capacity", i.e. the sewerage network involved in transferring sewage to the plants. Several treatment plants are in the red, including Timolin, Nurney and Rathangan WWTPs, meaning that "no spare capacity is available" at the moment.
"Brannockstown(Grangemore) WWTP" mentioned in the Uisce Éireann register is given a "green" rating indicating that "spare capacity is available." I'm not sure whether this is the same tank mentioned in this 82-page EPA licence-application document, which calls it the "Brannockstown Septic Tank" and says that "the tank was constructed in the 1970’s to serve approximately 18 properties in an adjacent Kildare County Council housing development." The tank appears to be located behind a fence in woodland at the back of the development.
Group sewage schemes or domestic wastewater treatment systems (DWTSs) are sometimes used in rural areas for groups of houses when mains sewer connections and a full-scale treatment plant are not feasible or economical. Meanwhile, on a related sewage matter, Cllr. Tracey O'Dwyer asked a question at a September 2025 meeting of the Kildare–Newbridge MD about whether the council can "confirm the total amount paid to waste companies for the removal of sewage from Moorhill Wood Estate, Brannockstown since the service commenced to date and confirm if this expenditure can be recouped from Uisce Éireann."

OSI Benchmarks

OSI benchmarks. Image reproduced with permission from Dr. Catherine Porter, author of this RTÉ article.

There are several of these located around the town. I'm not sure which are still in existence however. The marks, resembling an arrow, were chiselled into masonry by OSI surveyors when the country was surveyed in the early 19th century to create the first detailed and accurate maps. Typically, they're located on walls and gate pillars. There should be one located on the bridge in the town and a second one on a gate pillar on New Abbey Road. (At the gate on the left hand side of the road, just before the turn for McGarry's Lane when heading towards the cemetery.) A benchmark corresponded to an elevation in feet above a sea level datum (the reference point located at Poolbeg Lighthouse). The elevation was indicated on the 25" map with a crow's foot symbol. Surveyors inserted an angle iron into a socket located near the mark, and this was used as a bench for resting a levelling rod on during the elevation measurement process.
This recent article on the RTÉ website by geographer Dr Catherine Porter and Margaret Sullivan, University of Limerick, gives a background on benchmarks and their use use in the surveying process.

Image reproduced with permission from Dr. Catherine Porter, author of the RTÉ article.

Making a New Part For My Bicycle Tail Light

© Eugene Brennan

If possible I always repair rather than replace. I've always done this before it became a trendy green thing, simply because it saves money and I've never had lots of money to spare. I remember my father fixing a hole in the silencer on the car, with a section of tin plate, cut from a Cow & Gate baby formula tin, and fixing it to the silencer with coat hanger wire. So I didn't lick it up off the ground, as the saying goes. I've got slagged off by Millennials and successive demographic cohorts, for doing such things, and asked why I don't "just but a new one". Anyway, today I was fixing the tail light for my bike.

Poulaphouca Water Level Stats Updated: Level up 30 cm

Water level in Poulaphouca reservoir

Water level has increased by 30 cm from the same time yesterday, equivalent to 6.78 million cubic metres (m³) for the 22.6 square kilometre lake. That's despite the 2.94 million m³ of water discharged in the past day from Golden Falls dam. Although discharges for Golden Falls were forecast to be 45 m³/s at four points in the day yesterday, flow stats showed that they were 34 m³/s, averaged over the whole day. Possibly this was due to lower-than-forecast discharges in the evening or at midnight today. Water level under the bridge in Kilcullen circa 8 pm last night had dropped by about 4 inches.
The ESB press office has just contacted me with details about "theoretical highest level possible" at the dams. More info later.
Screenshots courtesy ESB hydrometrics.

Higher-than-normal water levels under one of the arches of Kilcullen Bridge. © Eugene Brennan

Reduction in Poulaphouca Lake Levels

Poulaphouca Reservoir viewed from above Valleymount. © Eugene Brennan

The water level in Poulaphouca lake has dropped from yesterday's 186.64 m to 186.52 m a drop of 12 cm. It may seem small, but it corresponds to a discharge of 2.7 million cubic metres (m³) of water for the 22.6 square kilometre lake. This reduction in volume of the lake is simply due to the difference between "water in" minus "water out". The River Liffey and Kings River are feeding water into the lake, and water is discharged through the Poulaphouca dam into the Golden Falls Compensatory lake. In a previous article, we discovered the reason why there's a lake between the two dams. Some of the drop in level is accounted for by water purified into drinking water by the Ballymore Eustace Water Treatment Plant. According to the EPA, the BWTP produces 340 million litres of drinking water a day. That's equivalent to 340,000 cubic metres. So most of the reduction in water volume in the lake is accounted for by Poulaphouca discharges. Liffey releases through the Golden Falls dam today are forecast to be 45 cubic metres per second (m³/s), all day. If that rate of discharge is sustained for the day (the ESB say "flows are estimates"), it corresponds to 3.9 million m³ of water discharge versus the 2.7 million m³ reduction in volume of the Poulaphouca lake. So at the moment, the total discharge from the Golden Falls lake is greater than the rate at which the Poulaphouca lake is discharging into it, making room for high flows on rivers and lakes feeding the lakes in the coming days.

Water level stats courtesy ESB hydrometrics. 

 

Water level stats, courtesy ESB hydrometrics. 

Related Reading

• Press Release From the ESB About Potential Releases From the Poulaphouca Reservoir (this explains the setup at Golden Falls and the buffering of flow)
• What Volume Does a 1 m Change in Water Level Represent in the Golden Falls Lake? 
•  Golden Falls Headrace Level Up a Metre, Flywheels, Capacitors and Air Compressor Tanks
•  Let's Do Some Calculations: How Much Water?
•  High Volume Releases at Golden Falls Dam
•  Golden Falls Generator Assembly
•  Erosion of the River Bank in Kilcullen
•  Ballymore 5 km Loop 

 

  

What Volume Does a 1 m Change in Water Level Represent in the Golden Falls Lake?

Golden Falls lake. Image courtesy Google Maps, Airbus and Maxtar Techmologies

Distance can easily be measured on Google Maps on desktop by right clicking and selecting "Measure distance" from the context menu. Then simply click the start point and the end point, or draw a series of line segments by repeatedly left clicking and following an outline until the end point is reached. This is useful for measuring distances, but you may not have known that it's also possible to measure area by drawing a series of line segments around for instance a field, and then clicking on the start point again. You can read a complete article I wrote about how to do this here.

Types of Plastics Used in the Home: How to Identify PVC, PET, Polythene and Others

© Eugene Brennan

What Are Plastics?

"Plastic" is a term applied to any material that can be moulded or bent into shape; for instance, soft toffee, modelling clay, or red-hot iron is plastic because it can be reshaped. In general though, when we speak of plastic, we mean polymers or synthetic materials made from petrochemicals.