Monthly Archives: January 2025

January 16, 2025 · 8:08 am

HSC Advanced 2024 Question 30

HSC Advanced 2024

Two circles have the same centre $O$ . The smaller circle has a radius of $1$ cm, while the larger has a radius of $(x+1)$ cm. The circles enclose a region $QRST$ , which is subtended by angle of ${\theta}$ at $O$ , as shaded.

The area of $QRST$ is $A$ cm², where $A$ is a constant and $A>0$

Let $P$ cm be the perimeter of $QRST$

(a) By finding expressions for the area and perimeter of $QRST$ show that $P(x)=2x+\frac{2A}{x}$

(b) Show that if the perimeter is minimised, then ${\theta}$ must be less than $2$ .

Deriving the Quotient Rule for Differentiation

Like we did for the product rule, we are going to derive the differentiating rule for functions in the form $y=\frac{f(x)}{g(x)}$ .

Something like, $y=\frac{x^2+3x+2}{x^3-1}$

Remember the first principals limit

$\lim_{\limits h \to 0}\frac{f(x+h)-f(x)}{h}$

If $y=\frac{f(x)}{g(x)}$ , then

$y'=\lim_{\limits h \to 0}\frac{\frac{f(x+h)}{g(x+h)}-\frac{f(x)}{g(x)}}{h}$

Find a common denominator for the numerator (i.e. $g(x+h)g(x)$ )

$y'=\lim_{\limits h \to 0}\frac{\frac{f(x+h)g(x)-f(x)g(x+h)}{g(x+h)g(x)}}{h}$

To make things a bit easier I am going to multiply by $\frac{1}{h}$ rather than having $h$ as the denominator

$y'=\lim_{\limits h \to 0}\frac{f(x+h)g(x)-f(x)g(x+h)}{g(x+h)g(x)} \times \frac{1}{h}$

Now I am going to add and subtract $f(x)g(x)$

$y'=\lim_{\limits h \to 0}\frac{f(x+h)g(x)-f(x)(g(x)+f(x)g(x)-f(x)g(x+h)}{g(x+h)g(x)} \times \frac{1}{h}$

Factorise

$y'=\lim_{\limits h \to 0}\frac{g(x)(f(x+h)-f(x))+f(x)(g(x)-g(x+h))}{g(x+h)g(x)} \times \frac{1}{h}$

Change the sign in the middle

$y'=\lim_{\limits h \to 0}\frac{g(x)(f(x+h)-f(x))-f(x)(g(x+h)-g(x))}{g(x+h)g(x)} \times \frac{1}{h}$

Separate the limits

$y'=g(x)\lim_{\limits h \to 0}\frac{\frac{f(x+h)-f(x)}{h}}{g(x+h)g(x)}-f(x)\lim_{\limits h \to 0}\frac{\frac{g(x+h)-g(x)}{h}}{g(x+h)g(x)}$

which simplifies to

$y'=g(x)\frac{f'(x)}{g(x)g(x)}-f(x)\frac{g'(x)}{g(x)g(x)}$

$y'=\frac{f'(x)g(x)-g'(x)f(x)}{[g(x)]^2}$

In words

The derivative of the top times the bottom take the derivative of the bottom times the top all over the bottom squared

Example

$y=\frac{x^2+3x+2}{x^3-1}$

$y'=\frac{(2x+3)(x^3-1)-3x^2(x^2+3x+2)}{(x^3-1)^2}$

$y'\frac{2x^4-2x+3x^3-3-3x^4-9x^3-6x^2}{(x^3-1)^2}$

$y'=\frac{-x^4-6x^3-6x^2-2x-3}{(x^3-1)^2}$

Exam questions usually specify no simplifying.

1 Comment

Filed under Calculus, Differentiation, Quotient Rule, Year 12 Mathematical Methods

January 11, 2025 · 9:56 am

Deriving the Product Rule for Differentiation

In my previous post we looked at the Chain Rule for Differentiation, this post is on the Product Rule. Differentiating a function in the form $y=f(x)\times g(x)$ .

For example, $y=(3x^3+2x-1)(x^4+2x^2)$

Remember differentiating from first prinicpals:

$f'(x)=\lim_{\limits h \to 0} \frac{f(x+h)-f(x)}{h}$

$y=f(x)g(x)$

$\frac{dy}{dx}=\lim_{\limits h\to 0}\frac{f(x+h)g(x+h)-f(x)g(x)}{h}}$

$\small{ \frac{dy}{dx}=\lim_{\limits h \to 0} \frac{f(x+h)g(x+h)-g(x+h)f(x)+g(x+h)f(x)-f(x)g(x)}{h}}$

By subtracting and then adding $g(x+h)f(x)$ we haven’t changed the limit, but it means we can do some factorising.

$\frac{dy}{dx}=\lim_{\limits h \to 0}\frac{g(x+h)(f(x+h)-f(x))+f(x)(g(x+h)-g(x))}{h}$

$\small{\frac{dy}{dx}=\lim_{\limits h \to 0}g(x+h)\lim_{\limits h \to 0}\frac{f(x+h)-f(x)}{h}+\lim_{\limits h \to 0}f(x)\lim_{\limits h \to 0}\frac{g(x+h)-g(x)}{h}}$

When we evaluate the limits

$\frac{dy}{dx}=g(x)f'(x)+f(x)g'(x)$

Example

Find the derivative of $y=(3x^3+2x-1)(x^4+2x^2)$

I remember the rule in words ‘derivative of the first times the second plus the derivative of the second times the first’.

$y'=(9x^2+2)(x^4+2x^2)+(4x^3+4x)(3x^3+2x-1)$

$y'=9x^6+18x^4+2x^4+4x^2+12x^6+8x^4-4x^3+12x^4+8x^2-4x$

$y'=21x^6+40x^4-4x^312x^2-4x$

Most exam questions have ‘don’t simplify’, so the first line of working above would be enough.

Onto the Quotient Rule.

2 Comments

Filed under Calculus, Differentiation, Product Rule, Year 12 Mathematical Methods

January 9, 2025 · 9:40 am

Deriving the Chain Rule for Differentiation

How to differentiate something in the form $y=[f(x)]^n$

For example, $y=(3x^2-2x+6)^5$ , we could expand the expression, but the Chain Rule provides a quick and easy method.

Differentiate $y=[f(x)]^n$

Let $u=f(x)$ , then $y=u^n$

We want to find $\frac{dy}{dx}$ , but $\frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx}$

They’re not fractions, but limits of fractions, but they work like fractions.

$\frac{du}{dx}=f'(x)$ and $\frac{dy}{du}=nu^{n-1}$

Therefore, $\frac{dy}{dx}=f'(x)\times nu^{n-1}$

Replace $u$ with $f(x)$

(1) $\begin{equation*}\frac{dy}{dx}=n[f(x)]^{n-1}f'(x)\end{equation*}$

What about a function in the form $y=f(g(x))$ ?

We’re going to follow the same process.

Let $u=g(x)$ , then $y=f(u)$

$\frac{du}{dx}=g'(x)$ and $\frac{dy}{du}=f'(u)$

Therefore $\frac{dy}{dx}=f'(u)g'(x)$

(2) $\begin{equation*}\frac{dy}{dx}=f'(g(x))g'(x) \end{equation*}$

Equations $1$ and $2$ are versions of the Chain Rule.

Example

Find the derivative of $y=(3x^2-2x+6)^5$

$\begin{equation*}\frac{dy}{dx}=5(3x^2-2x+6)^4\times (6x-2)\end{equation}$

$\begin{equation*}\frac{dy}{dx}=5(6x-2)(3x^2-2x+6)^4\end{equation}$

$\begin{equation*}\frac{dy}{dx}=10(3x-1)(3x^2-2x+6)\end{equation}$

Next time we are going to look at the Product Rule.

1 Comment

Filed under Calculus, Chain Rule, Differentiation, Year 12 Mathematical Methods

Tagged as Calculus, chain rule, differentiation, year 12 mathematical methods

January 6, 2025 · 3:58 am

An Imaginary Tale – The story of i

An Imaginary Tale – The Story of $\sqrt{-1}$

I bought a second hand copy of this book from Abe Books (it’s possible to find cheap maths books this way).

Here’s the blurb …

Today complex numbers have such widespread practical use–from electrical engineering to aeronautics–that few people would expect the story behind their derivation to be filled with adventure and enigma. In An Imaginary Tale , Paul Nahin tells the 2000-year-old history of one of mathematics’ most elusive numbers, the square root of minus one, also known as i . He recreates the baffling mathematical problems that conjured it up, and the colorful characters who tried to solve them. In 1878, when two brothers stole a mathematical papyrus from the ancient Egyptian burial site in the Valley of Kings, they led scholars to the earliest known occurrence of the square root of a negative number. The papyrus offered a specific numerical example of how to calculate the volume of a truncated square pyramid, which implied the need for i . In the first century, the mathematician-engineer Heron of Alexandria encountered I in a separate project, but fudged the arithmetic; medieval mathematicians stumbled upon the concept while grappling with the meaning of negative numbers, but dismissed their square roots as nonsense. By the time of Descartes, a theoretical use for these elusive square roots–now called “imaginary numbers”–was suspected, but efforts to solve them led to intense, bitter debates. The notorious i finally won acceptance and was put to use in complex analysis and theoretical physics in Napoleonic times. Addressing readers with both a general and scholarly interest in mathematics, Nahin weaves into this narrative entertaining historical facts and mathematical discussions, including the application of complex numbers and functions to important problems, such as Kepler’s laws of planetary motion and ac electrical circuits. This book can be read as an engaging history, almost a biography, of one of the most evasive and pervasive “numbers” in all of mathematics.

It took me a long time to read this book – according to storygraph I started it in August (12th to be exact). I really enjoyed this book, but I do think you need to understand maths. I enjoyed working through all of the different formulae and examples. I have put some on this blog

I found the historical aspects very interesting.

Minus times minus is plus

The reason for this we need not discuss

W. H. Auden

Maybe I should use the above quote with my Year 8s who are just starting on their negative number journey.

This book covers quite complex (pun intended) ideas – particularly in Chapter 6 Wizard Mathematics and Chapter 7 The Nineteenth Century, Cauchy, and the Beginning of Complex Function Theory.

If you can do algebra and a bit of calculus and complex numbers interest you, then I think this book is for you.

A review

Volume of Rotation About a Slanting Line

Given the area in the first quadrant bounded by $x^2=12y$ , the line $y=3$ and the $y-$ axis. What is the volume generated when this area is rotated about the line $2x-y+4=0$ ?