Racquel Sanderson

July 30, 2026 · 11:06 am

Finding the Area formula of an Ellipse

Ellipses appear in both complex loci and vector calculus.

We are going to use calculus (integration) to derive the formula for the area of an ellipse. The equation of an ellipse is

$$ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1. $$

Our aim is to find the area under the upper half of the ellipse and then double the result to obtain the total area.

Rearrange the equation to make $y$ the subject.

$$ \frac{y^2}{b^2}=1-\frac{x^2}{a^2} $$ $$ y^2=b^2\left(1-\frac{x^2}{a^2}\right) $$ $$ y=\pm b\sqrt{1-\frac{x^2}{a^2}}. $$

We only need to consider the positive branch.

$$ y=b\sqrt{1-\frac{x^2}{a^2}}. $$

The area of the upper half of the ellipse is

$$ A=\int_0^a b\sqrt{1-\frac{x^2}{a^2}}\,dx. $$

Let

$$ x=a\sin\theta. $$

Then

$$ dx=a\cos\theta\,d\theta. $$

When $x=0$, $\theta=0$, and when $x=a$, $\theta=\frac{\pi}{2}$.

The integral becomes

$$ A =ab\int_0^{\pi/2}\cos^2\theta\,d\theta. $$

Remember the trigonometric identity

$$ \cos^2\theta=\frac{1+\cos(2\theta)}{2}. $$

Hence,

$$ A =\frac{ab}{2}\int_0^{\pi/2}\left(1+\cos(2\theta)\right)\,d\theta. $$ $$ =\frac{ab}{2}\left[\theta+\frac{1}{2}\sin(2\theta)\right]_0^{\pi/2} $$ $$ =\frac{ab}{2}\left(\frac{\pi}{2}\right) $$ $$ =\frac{\pi ab}{4}. $$

This is the area of one quarter of the ellipse. Therefore, the total area of the ellipse is

$$ 4\times\frac{\pi ab}{4}=\boxed{\pi ab}. $$

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Filed under Area, Definite, Ellipse, Identities, Integration, Trigonometric Substitution, Year 12 Specialist Mathematics

June 17, 2026 · 4:24 am

Partial Fractions

Integration Using Partial Fraction Decomposition

Partial fraction decomposition is the process of taking a rational function and decomposing it into simpler rational expressions which are easier to integrate.

We only use partial fractions if the rational function is proper. If

$f(x)=\frac{p(x)}{q(x)}$

then the degree of $p(x)$ must be less than the degree of $q(x)$ .

Types of Partial Fraction Decompositions

Factor	Term in Partial Fraction
$ax+b$	$\frac{A}{ax+b}$
$(ax+b)^n$	$\frac{A_1}{ax+b} + \frac{A_2}{(ax+b)^2} +\cdots+ \frac{A_n}{(ax+b)^n}$
Irreducible quadratic $ax^2+bx+c$	$\frac{Ax+B}{ax^2+bx+c}$

Example

$\int \frac{3x+5}{x^2-x-2}\,dx$

Find

Factorise the denominator:

$x^2-x-2=(x-2)(x+1)$

Write

$\frac{3x+5}{(x-2)(x+1)} = \frac{A}{x-2} + \frac{B}{x+1}$

Multiplying through by

$(x-2)(x+1)$ :

$3x+5=A(x+1)+B(x-2)$

Let $x=-1$ :

$2=-3B$

$B=-\frac23$

Let $x=2$ :

$11=3A$

$A=\frac{11}{3}$

Hence

$\frac{3x+5}{x^2-x-2} = \frac{11}{3(x-2)} - \frac{2}{3(x+1)}$

Integrating:

$\int\frac{3x+5}{x^2-x-2}\,dx = \int \left( \frac{11}{3(x-2)} - \frac{2}{3(x+1)} \right) dx$

$= \frac{11}{3}\ln|x-2| -\frac23\ln|x+1| +C$

Example

Find

$\int\frac{3x+2}{(x-1)^2}\,dx$

Write

$\frac{3x+2}{(x-1)^2} = \frac{A_1}{x-1} + \frac{A_2}{(x-1)^2}$

Multiplying through by $(x-1)^2$ :

$3x+2=A_1(x-1)+A_2$

Let $x=1$ :

$5=A_2$

Let $x=0$ :

$2=-A_1+5$

$A_1=3$

Therefore

$\int\frac{3x+2}{(x-1)^2}\,dx = \int \left( \frac3{x-1} + \frac5{(x-1)^2} \right) dx$

$= 3\ln|x-1| -\frac5{x-1} +C$

Example

Find

$\int \frac{2x-5} {(x-1)(x^2+1)} \,dx$

Write

$\frac{2x-5} {(x-1)(x^2+1)} = \frac{A}{x-1} + \frac{Bx+C}{x^2+1}$

Multiplying through:

$2x-5 = A(x^2+1) + (Bx+C)(x-1)$

Let $x=1$ :

$-3=2A$

$A=-\frac32$

Let $x=0$ :

$-5=-\frac32-C$

$C=\frac72$

Let $x=-1$ :

$-7=-3+\left(-B+\frac72\right)(-2)$

$-4=2B-7$

$B=\frac32$

Hence

$\int \frac{2x-5} {(x-1)(x^2+1)} \,dx = \frac12 \int \left( -\frac3{x-1} + \frac{3x}{x^2+1} + \frac7{x^2+1} \right) dx$

$= \frac12 \left( -3\ln|x-1| +\frac32\ln(x^2+1) +7\arctan(x) \right) +C$

$= -\frac32\ln|x-1| +\frac34\ln(x^2+1) +\frac72\arctan(x) +C$

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Filed under Algebra, Integration, Polynomials, Year 12 Specialist Mathematics

Tagged as Integration, integration using partial fraction decomposition, partial fractions, Year 12 Specialist Mathematics

June 4, 2026 · 11:33 am

Complex Numbers – Roots of Unity

$\begin{equation*}z^n=1, z\in\mathbb{C}\end{equation}$

In polar form $1=cis (2\pi)$

Hence,

$\begin{equation*}z^n=cis(2\pi k), k\in\mathbb{Z}\end{equation}$

$\begin{equation*}z^n=cis(\frac{2\pik}{n})\end{equation}$

For example,

$\begin{equation*}z^4=1\end{equation}$

$\begin{equation*}z^4=cis(2\pi k)\end{equation}$

$\begin{equation*}z_k=cis(\frac{2\pi k}{4})=cis(\frac{\pi k}{2})\end{equation}$

$\begin{equation*}z_0=1\end{equation}$

$\begin{equation*}z_1=cis(\frac{\pi}{2})=i\end{equation}$

$\begin{equation*}z_2=cis(\pi)=-1\end{equation}$

$\begin{equation*}z_3=cis(\frac{3\pi}{2})=-i\end{equation}$

The roots of unity are spread evenly $(\frac{2\pi}{n}$ or $\frac{360}{n}$ apart) around a circle of radius $1$ .

Properties of the roots of Unity

All the roots of unity can be generated by powers of a single root $\omega=cis(\frac{2\pi}{n})$ . The roots form the sequence $1, \omega, \omega^2, \omega^3, ..., \omega^{n-1}$ Note: $\omega \ne 1$

For example,

$\begin{equation*}z^4=1\end{equation}$

$\begin{equation*}\omega_1=cis(\frac{\pi}{2})\end{equation}$

$\begin{equation*}\omega_1^2=cis(\pi)=-1=\omega_2\end{equation}$

$\begin{equation*}\omega_1^3=cis(\frac{3\pi}{2})=-i=\omega_3\end{equation}$

$\begin{equation*}\omega_1^4=cis(2\pi)=1=\omega_0\end{equation}$

A root that can generate the remaining roots is a primitive root.

$\omega^k$ is a primitive root if $n$ and $k$ are coprime.

The sum of the $n^{th}$ roots of unity is always zero.

$\begin{equation*}\Sigma_{k=0}^{n-1}1+\omega+\omega^2+...+\omega^{n-1}=0\end{equation}$

The product of the $n^{th}$ roots of unity is

$\begin{equation*}\Pi_{k=0}^{n-1}\omega^k=(-1)^{n-1}\end{equation}$

It is $1$ when $n$ is odd and $-1$ when $n$ is even.

Example WATP 2024 Question 7a

(a) Evaluate $(4\omega^2+3)(4\omega+3)$ where $\omega$ is a complex root of unity, $\omega\ne 1$ .

$\begin{equation*}(4\omega^2+3)(4\omega+3)=16\omega^3+12\omega^2+12\omega+9\end{equation}$

$\begin{equation*}16\omega^3+12\omega^2+12\omega+9=16(1)+9(\omega^2+\omega+1)+3\omega^2+3\omega\end{equation}$

$\begin{equation*}16\omega^3+12\omega^2+12\omega+9=16+9(0)+3(\omega^2+\omega)\end{equation}$

$\begin{equation*}16\omega^3+12\omega^2+12\omega+9=16+3(-1)\end{equation}$

$\begin{equation*}=13\end{equation}$

Remember the sum of the roots is zero, $\omega^2+\omega+1=0$ hence $\omega^2+\omega=-1$

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Filed under Algebra, Complex Numbers, Roots of Unity, Year 12 Specialist Mathematics

Tagged as complex numbers, complex roots of unity, cube roots of unity

May 27, 2026 · 12:19 am

Percent

I finally put together some notes (with the help of Copilot) on percent.

Percent Notes Download

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Filed under Arithmetic, Percentages, Year 9

Tagged as amount as a percent, percent, percent decrease, percent increase, percent of an amount, repeated percent

May 18, 2026 · 7:24 am

Using Integration to find the Centroid of an Area

Finding the Centroid of a Region Download

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Filed under Algebra, Area, Calculus, Definite, Integration, Year 12 Mathematical Methods

Tagged as centroid, finding the centroid of a region, Integration

April 21, 2026 · 11:26 am

Find the Equation of a Plane Given Three Points

Find the Cartesian equation of the plane containing the points $A(1, 2, 3), B(-3, 0, 2)$ and $C(2, -4, -1)$

Find $\overrightarrow{AB}$ and $\overrightarrow{AC}$
$\overrightarrow{AB}=\begin{pmatrix}-2\\0\\2\end{pmatrix}-\begin{pmatrix}1\\2\\3\end{pmatrix}=\begin{pmatrix}-3\\-2\\-1\end{pmatrix}$
$\overrightarrow{AC}=\begin{pmatrix}2\\-4\\-1\end{pmatrix}-\begin{pmatrix}1\\2\\3\end{pmatrix}=\begin{pmatrix}1\\-6\\-4\end{pmatrix}$

Find the cross product $\overrightarrow{AB}\times \overrightarrow{AC}$
$\begin{pmatrix}-3\\-2\\-1\end{pmatrix}\times\begin{pmatrix}1\\-6\\-4\end{pmatrix}=\begin{pmatrix}2\\-13\\20\end{pmatrix}$
This is the normal, $overrightarrow{n}$ , to the plane.
We know $A$ is on the plane

$\begin{equation*}\overrightarrow{n}\cdot (\overrightarrow{r}-\overrightarrow{OA})=0\end{equation}$

Hence $\overrightarrow{n}\cdot \overrightarrow{r}=\overrightarrow{n}\cdot \overrightarrow{OA}$
$\overrightarrow{n}\cdot \overrightarrow{r}=\begin{pmatrix}2\\-13\\20\end{pmatrix} \cdot \begin{pmatrix}1\\2\\3\end{pmatrix}=38$
Therefore the Cartesian equation of the plane is

$\begin{equation*}2x-13y+20z=38\end{equation}$

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Filed under Algebra, Cross Product, Vectors, Year 12 Specialist Mathematics

April 6, 2026 · 9:23 am

The Secret Life of Numbers – Kate Kitagawa and Timothy Revell

The first place I went to buy this was sold out! Seems somewhat amazing – perhaps they only had one copy. I found it at my favourite book store – Subiaco Bookshop.

Here’s the blurb …

Mathematics shapes almost everything we do. But despite its reputation as the study of fundamental truths, the stories we have been told about it are wrong. In The Secret Lives of Numbers, historian Kate Kitagawa and journalist Timothy Revell introduce readers to the mathematical boundary-smashers who have been erased by history because of their race, gender or nationality.

From the brilliant Arabic scholars of the ninth-century House of Wisdom, and the pioneering African American mathematicians of the twentieth century, to the ”lady computers” around the world who revolutionised our knowledge of the night sky, we meet these fascinating trailblazers and see how they contributed to our global knowledge today.

This revisionist, completely accessible and radically inclusive history of mathematics is as entertaining as it is important.

This has a lovely style and is very easy to read. As part of my maths degree, I studied some history, but it was very western and I enjoyed the global approach in this book. I do think it is accessible and anyone with an interest in maths or history could read it.

A review

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Filed under Book Review

Tagged as book review, kate kitagawa, the secret life of numbers, timothy revell

March 21, 2026 · 7:47 am

Complex Loci

Sketch on an Argand diagram

$\begin{equation*}\lvert{z-6}\rvert-\lvert{z+6}\rvert=3\end{equation}$

where $z\in \mathbb{C}$

Let $z=x+yi$

$\begin{equation*}\lvert{x+yi-6}\rvert-\lvert{x+yi+6\rvert=3\end{equation}$

$\begin{equation*}\lvert{x-6+yi}\rvert-\lvert{x+6+yi\rvert=3\end{equation}$

$\begin{equation*}\sqrt{(x-6)^2+y^2}-\sqrt{(x+6)^2+y^2}=3\end{equation}$

$\begin{equation*}\sqrt{(x-6)^2+y^2}=3+\sqrt{(x+6)^2+y^2}\end{equation}$

Square both sides of the equation

$\begin{equation*}{(x-6)^2+y^2=9+6\sqrt{(x+6)^2+y^2}+(x+6)^2+y^2\end{equation}$

$\begin{equation*}x^2-12x+36+y^2-9-x^2-12x-36-y^2=6\sqrt{(x+6)^2+y^2}\end{equation}$

(1) $\begin{equation*}-24x-9=6\sqrt{(x+6)^2+y^2}\end{equation*}$

From equation $1$ we know $-24x-9\ge0$

Hence $x\le\frac{-3}{8}$ , which means we only have the left section of the hyperbola.

$\begin{equation*}-8x-3=2\sqrt{(x+6)^2+y^2}\end{equation}$

Square both sides of the equation

$\begin{equation*}(-8x-3)^2=4((x+6)^2+y^2)\end{equation}$

$\begin{equation*}64x^2+48x+9=4x^2+48x+144+y^2\end{equation}$

$\begin{equation*}60x^2-y^2=135\end{equation}$

$\begin{equation*}\frac{4x^2}{9}-\frac{y^2}{135}=1\end{equation}$

(2) $\begin{equation*}\frac{x^2}{\frac{9}{4}}-\frac{y^2}{135}=1\end{equation*}$

Remember, we have the left part of the hyperbola.

The $x-$ intercept $=-\sqrt{\frac{9}{4}}=-\frac{3}{2}$ and the asymptotes are $y=\pm \frac{\sqrt{135}}{\frac{3}{2}}x$

$y=\pm 2\sqrt{15}x$

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Filed under Algebra, Complex Numbers, Simplifying fractions, Sketching Complex Regions, Solving Equations, Year 12 Specialist Mathematics

Tagged as complex loci, hyperbola

March 11, 2026 · 5:34 am

Function Composition

If $g(f(x))=\frac{x}{x+1}$ and $f(x)=\frac{1}{1-2x}$ , find $g(x)$ .

$\begin{equation*}g(f(x))=g(\frac{1}{1-2x})=\frac{x}{x+1}\end{equation}$

Sometimes you can do this type of question by inspection, but this one is a bit harder. I am going to use a variable substitution.

Let $u=\frac{1}{1-2x}$

$\begin{equation*}u=\frac{1}{1-2x}\end{equation}$

$\begin{equation*}1-2x=\frac{1}{u}\end{equation}$

$\begin{equation*}1-\frac{1}{u}=2x\end{equation}$

$\begin{equation*}\frac{u-1}{u}=2x\end{equation}$

$\begin{equation*}x=\frac{u-1}{2u}\end{equation}$

Therefore

$\begin{equation*}g(u)=\frac{\frac{u-1}{2u}}{\frac{u-1}{2u}+1}\end{equation}$

$\begin{equation*}g(u)=\frac{\frac{u-1}{2u}}{\frac{u-1+2u}{2u}}\end{equation}$

$\begin{equation*}g(u)=\frac{u-1}{3u-1}\end{equation}$

Therefore

$\begin{equation*}g(x)=\frac{x-1}{3x-1}\end{equation}$

Let’s test it

$g(x)=\frac{x-1}{3x-1}$ and $f(x)=\frac{1}{1-2x}$

$\begin{equation*}g(f(x))=\frac{\frac{1}{1-2x}-1}{\frac{3}{1-2x}-1}\end{equation}$

$\begin{equation*}g(f(x))=\frac{\frac{1-(1-2x)}{1-2x}}{\frac{3-(1-2x)}{1-2x}}\end{equation}$

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

$\begin{equation*}g(f(x))=\frac{x}{x+1}\end{equation}$

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Filed under Algebra, Composition, Functions, Simplifying fractions, Year 12 Specialist Mathematics

Tagged as function composition

February 25, 2026 · 1:22 pm

Complex Locus Question

My Year 12 Specialist students are working on complex loci again. The following type of question always creates confusion.

$arg(z-w_1)$ is the angle the vector from $w_1$ to $z$ makes with the positive $x-$ axis, likewise for $arg(z-w_2)$ .

I am going to plot a possible $z$ and try to see the geometry that works.

We want $arg(z-w_1)-arg(z-w_2)=\frac{\pi}{6}$

I am going to take advantage of some triangle geometry

Using the External Angle Theorem, we know $\alpha=\beta+\theta$

$\begin{equation*}arg(z-w_1)-arg(z-w_2)=(\alpha+\frac{\pi}{2})-(\beta+\frac{\pi}{2}0=\alpha-\beta=\theta\end{equation}$

Therefore $\theta=\frac{\pi}{6}$

So we want all of the $z$ values that have an angle of $\frac{\pi}{6}$

Now we are going to use some circle geometry -The angle at the circumference subtended by the same arc are congruent. So we need to find a circle that has those three points ( $z, w_1$ and $w_2$ ) on the circumference.