Racquel Sanderson

November 5, 2026 · 2:23 am

3D Vector Motion Question

The particle $P$ is moving with velocity $v_{P}=(1,1,2)$ metres per second from its initial point with position vector $(-38, 31, -18)$ metres. The particle $Q$ is moving with velocity $v_{Q}=(1,\lambda,-2)$ metres per second, where $\lambda$ is a constant. At $T$ seconds, where $T>0$ , particle $Q$ passes through the point with position vector $(-35,56,\mu)$ metres, where $\mu$ is a constant.

Both particles start their motion at the same time, the paths of the particles are perpendicular, and the paths of the particles intersect at point $A$ .

(a) Determine the position vector of $A$ .

The paths of particles P and Q are perpendicular, so the dot product of their velocity vectors is zero:

$(1,1,2)\cdot(1,\lambda,-2)=0$

$1+\lambda-4=0$

$\lambda=3$

The position vectors are

$\mathbf{r}_P=\begin{pmatrix}-38+t_1\\31+t_1\\-18+2t_1\end{pmatrix}$

$\qquad\mathbf{r}_Q=\begin{pmatrix}-35+t_2\\56+3t_2\\\mu-2t_2\end{pmatrix}$

Solve simultaneously:

$-38+t_1=-35+t_2$

$31+t_1=56+3t_2$

Equation 1 subtract equation 2:

$-69=-91-2t_2$

$t_2=11$

Substitute $t_2=11$ into equation 1:

$-38+t_1=-35+11$

$t_1=14$

Hence, the position vector of A is

$\begin{pmatrix}-24\\45\\10\end{pmatrix}$

(b) Determine the value of $\mu$ and the value of $T$ so that the particles $P$ and $Q$ will collide

The $\mathbf{j}$ -component of A is 10, therefore

$\mu-2t_2=10$

$t_2=11$

$\mu=10+2(11)$

$\mu=32$

For the two particles to collide,

$t_1=t_2$

Particle Q is at the point $(-35,56,32)$ at time $T$ , and it is at point $(-24,45,10)$ 11 seconds later. Both particles start at the same time, hence

$T=3$

Both particles arrive at A at 14 seconds.

Particle $N$ is moving with constant velocity $\mathbf{v}_N)$ so that its path is perpendicular to both paths of particles $P$ and $Q$ .

(c) Determine the speed of particle $N$ , correct to 0.001 metres per second

As $\mathbf{v}_N$ is perpendicular to $P$ and $Q$ , the vector must be parallel to

$\begin{pmatrix}1\\1\\2\end{pmatrix} \times \begin{pmatrix}1\\3\\-2\end{pmatrix}=\begin{pmatrix}-8\\4\\2\end{pmatrix}$

Using the given $\mathbf{j}$ -component of $\mathbf{v}_N$ ,

$0.8=k\times2$

$k=0.4$

Therefore,

$\mathbf{v}_N=\begin{pmatrix}-3.2\\1.6\\0.8\end{pmatrix}$

and the speed of particle $N$ is

$\sqrt{(-3.2)^2+(1.6)^2+(0.8)^2}$

$\frac{2\sqrt{101}}{5}\approx 4.02\text{ metres per second}.$

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

Tagged as vector motion, vectors, year 12 mathematics specialist

September 13, 2026 · 3:15 am

Simple Harmonic Motion Question

An object in simple harmonic motion passes through the central point $x=0$ at time $t=0$ and every second thereafter.

Find an equation for the motion if $v(0)=-4.$

Because the particle passes through the central point $x=0$ when $t=0$ , it is natural to use a sine function: $x=A\sin(\omega t).$

The particle passes through the central point every second.

However, passing through the centre occurs twice during each complete cycle — once moving in one direction and once moving in the other.

Therefore, the time from one central crossing to the next is half a period: $\frac{T}{2}=1.$

Hence, $T=2\text{ seconds}.$

For simple harmonic motion, $\omega=\frac{2\pi}{T}.$

Since $T=2$ , $\omega=\frac{2\pi}{2}=\pi.$

Our equation is therefore $x=A\sin(\pi t).$

We still need to determine $A$ .

For simple harmonic motion,

$v^2=k^2(A^2-x^2)$

When $t-0, x=0$ and $v=-4$

$16=\pi^2(A^2-0)$

$A=\frac{4}{\pi}$

Hence,

$x(t)=\frac{4}{\pi}sin(\pi t)$

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Filed under Rectilinear Motion, Simple Harmonic Motion, Trigonometry, Year 12 Specialist Mathematics

Tagged as mathematics specialist, rectilinear motion, simple harmonic motion, Year 12 Specialist Mathematics

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