April 7, 2025 · 9:09 am

Vector and Scalar Projection

The vector projection (vector resolution or vector component) of \mathbf{a} onto a non-zero vector \mathbf{b} is splitting \mathbf{a} into two vectors, one is parallel to \mathbf{b} (the vector projection) and one perpendicular to \mathbf{b}

In the above diagram \mathbf{a_1} is the vector projection of \mathbf{a} onto \mathbf{b} and \mathbf{a_2} is perpendicular to \mathbf{b}.

How do we find \mathbf{a_1} and \mathbf{a_2}?

Using right trigonometry,

cos(\theta)=\frac{|\mathbf{a_1}|}{|\mathbf{a}|}

Remember the scalar product (dot product) of vectors is

(1)   \begin{equation*}\mathbf{a}\cdot\mathbf{b}=|\mathbf{a}||\mathbf{b}|cos(\theta)\end{equation*}

Hence cos(\theta)=\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}

\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}=\frac{|\mathbf{a_1}|}{|\mathbf{a}|}

and, |\mathbf{a_1}|=\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}\times|\mathbf{a}|

|\mathbf{a_1}|=\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|}

This is the scalar projection of \mathbf{a} onto \mathbf{b}

To find the vector projection we need to multiply by \mathbf{\hat{b}}, that is find a vector with the same magnitude as \mathbf{a_1} in the direction of \mathbf{b}.

The vector projection is

\mathbf{a_1}=\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|}\times \mathbf{\hat{b}}

\mathbf{a_1}=\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|}\times \frac{\mathbf{b}}{|\mathbf{b}|}

\mathbf{a_1}=\frac{\mathbf{a}\cdot\mathbf{b}}{(|\mathbf{b}|)^2}\times \mathbf{b}

Now for \mathbf{a_2}, we know \mathbf{a}=\mathbf{a_1}+\mathbf{a_2}

Hence, \mathbf{a_2}=\mathbf{a}-\mathbf{a_1}

Example

From Cambridge Year 11 Specialist Mathematics (Chapter 3)

(a) \frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|^2}\mathbf{b}

=\frac{\begin{pmatrix}4\\1\end{pmatrix}\cdot\begin{pmatrix}1\\-1\end{pmatrix}}{2}\times(\mathbf{i}-\mathbf{j})

=\frac{3}{2}{(\mathbf{i}-\mathbf{j})

(b)4\mathbf{i}+\mathbf{j}-\frac{3}{2}{(\mathbf{i}-\mathbf{j})=\frac{5}{2}\mathbf{i}+\frac{5}{2}\mathbf{j}

(c)

The shortest distance (green vector) is the vector component of \mathbf{a} perpendicular to \mathbf{b}, i.e. \frac{5}{2}\mathbf{i}+\frac{5}{2}\mathbf{j}

|\frac{5}{2}\mathbf{i}+\frac{5}{2}\mathbf{j}|=\sqrt{\frac{25}{4}+\frac{25}{4}}=\sqrt{\frac{50}{4}}=\frac{5\sqrt{2}}{2}

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April 6, 2025 · 1:51 am

Using a Vector Method to Find an Angle Bisector

Points A and B are defined by the position vectors \mathbf{a}=3\mathbf{i}+4\mathbf{j} and \mathbf{b}=12\mathbf{i}+5\mathbf{j}.

Find a vector that bisects \angle{AOB}.

If we think about how we add vectors using the parallelogram rule

Finding the resultant vector using the parallelogram rule

we can take advantage of the geometric properties of parallelograms (or of a rhombus).

If \mathbf{a} and \mathbf{b} are unit vectors, then the parallelogram is a rhombus, and the diagonal (i.e the resultant) bisects the angle.

We need to find the sum of the unit vectors.

|\mathbf{a}|=\sqrt{3^2+4^2}=5

\therefore \hat{\mathbf{a}}=\frac{3}{5}\mathbf{i}+\frac{4}{5}\mathbf{j}

|\mathbf{b}|=\sqrt{12^2+5^2}=13

\therefore \hat{\mathbf{b}}=\frac{12}{13}\mathbf{i}+\frac{5}{13}\mathbf{j}

The vector that bisects \angle{AOB} is

\frac{3}{5}\mathbf{i}+\frac{4}{5}\mathbf{j}+\frac{12}{13}\mathbf{i}+\frac{5}{13}\mathbf{j}=\frac{99}{65}\mathbf{i}+\frac{64}{65}\mathbf{j}

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April 3, 2025 · 7:28 am

Area of Regular Polygons

Finding the area of a regular polygon when you know the side length

Find the area of an n-sided regular polygon if you know the side length, l.

An octagon for a visual reference

Find the h of the triangle in terms of l and theta.

tan(\theta)=\frac{\frac{l}{2}}{h}

h=\frac{l}{2tan(\theta)}

Remember the area of a triangle is A=\frac{1}{2}bh

Hence, A=\frac{1}{2} l \times \frac{l}{2tan(\theta)}=\frac{l^2}{4tan(\theta)}

And \theta=\frac{360}{2n}=\frac{180}{n}

Therefore A=\frac{l^2}{4tan(\frac{180}{n})}

There are n triangles in an n-sided polygon

(1)   \begin{equation*}A=\frac{nl^2}{4tan(\frac{180}{n})}\end{equation*}

Find the area of a hexagon with side length 10cm.
A=\frac{6\times10^2}{4tan(\frac{180}{6})}
A=\frac{600}{4(\frac{1}{\sqrt{3}})}
A=150\sqrt{3} cm^2

Finding the area of a polygon if you know the inradius or the apothem

The apothem and the inradius are the same. It is the radius of the incircle.

Find the area of the triangle in terms of a and theta.

tan(\theta)=\frac{\frac{l}{2}}{a}

l=2atan(\theta)

A=\frac{1}{2}bh

A=\frac{1}{2}2atan(\theta)a=a^2tan(\theta)

And \theta=\frac{180}{n}

Hence for an n-sided polygon

(2)   \begin{equation*}A=na^2tan(\frac{180}{n})\end{equation*}

Find the area of a regular pentagon with apothem 4.5cm
A=5\times 4.5^2tan(\frac{180}{5})
A=73.56cm^2

Finding the area of a regular polygon given the circumradius

The circumradius is the radius of the circumscribed circle (R in the diagram above)

Remember the area of triangle formula

A=\frac{1}{2}absin(\theta)

A=\frac{1}{2}R^2sin(\theta)

\theta=\frac{360}{n}

Hence, A=\frac{1}{2}R^2sin(\frac{360}{n})

Hence, for an n-sided polygon

(3)   \begin{equation*}A=\frac{nR^2sin(\frac{360}{n})}{2}\end{equation*}

Find the area of a regular octagon inscribed in a circle of radius 10cm.
A=\frac{8\times 10^2sin(45)}{2}
A=200\sqrt{2}cm^2

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Filed under Area, Area of Triangles (Sine), Finding an area, Non-Right Trigonometry, Regular Polygons, Right Trigonometry, Year 11 Mathematical Methods

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March 31, 2025 · 7:14 am

Deriving the Quadratic Equation formula

My year 10 students have been learning how to complete the square with the idea of then deriving the quadratic equation formula.

The general equation for a quadratic is y=ax^2+bx+c

Completing the square,

    \begin{equation*}ax^2+bx+c\end{equation}

Factorise out the leading coefficient (i.e. a)

    \begin{equation*}a(x^2+\frac{bx}{a}+\frac{c}{a})\end{equation}

Half the second term (i.e \frac{b}{a}) and subtract the square of the second term.

    \begin{equation*}a((x+\frac{b}{2a})^2-(\frac{b}{2a})^2+\frac{c}{a})\end{equation}

    \begin{equation*}a((x+\frac{b}{2a})^2-\frac{b^2}{4a^2}+\frac{c}{a})\end{equation}

Simplify

    \begin{equation*}a((x+\frac{b}{2a})^2-\frac{b^2}{4a^2}+\frac{4ac}{4a^2})\end{equation}

    \begin{equation*}a((x+\frac{b}{2a})^2+\frac{-b^2+4ac}{4a^2})\end{equation}

    \begin{equation*}a(x+\frac{b}{2a})^2+\frac{-b^2+4ac}{4a}\end{equation}

Now let’s solve

    \begin{equation*}a(x+\frac{b}{2a})^2+\frac{-b^2+4ac}{4a}=0\end{equation}

    \begin{equation*}a(x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a}\end{equation}

    \begin{equation*}(x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a^2}\end{equation}

    \begin{equation*}(x+\frac{b}{2a})=\pm \sqrt{\frac{b^2-4ac}{4a^2}}\end{equation}

    \begin{equation*}(x+\frac{b}{2a})=\frac{\pm \sqrt{b^2-4ac}}{2a}\end{equation}

    \begin{equation*}x=-\frac{b}{2a}\frac{\pm \sqrt{b^2-4ac}}{2a}\end{equation}

Which is the quadratic equation formula.

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March 26, 2025 · 12:23 pm

Logistic Growth Worked Example

A brumby is a free-roaming wild horse found in large number in parts of Australia. The culling of brumbies was banned in the year 2000. At this time the estimated population of brumbies in
Kosciuszko National Park was 1600. Scientists have modelled the population, P(t), of brumbies in
Kosciuszko National Park t years since the ban, by

    \begin{equation*}P(t)=\frac{18000}{10.25e^{0.15t}+1}\end{equation}

(a) Use the model to determine how long it will take the brumbies to increase to a number that is triple the number when the ban came into effect.

(b) From this model, determine the estimated long run number of brumbies in Kosciuszko National Park.

It can be shown that the growth rate of the population of brumbies can be expressed as

    \begin{equation*}\frac{dP}{dt}=\frac{1}{r}P(k-P)\end{equation}

(c) Determine the values of the constants r and k.

(d) Determine the greatest growth rate for the population of brumbies.

ATAR 2024 Specialist Mathematics Question 13

(a) 1600\times 3=4800
4800=\frac{18000}{10.25e^{0.15t}+1}
t=8.8 years.

(b) \lim_{\limits_{t \to \infty}\frac{18000}{10.25e^{0.15t}+1}=18000

(c)k is the carrying capacity (long run number of Brumbies), therefore k=18000.
Remember,

    \begin{equation*}\frac{dP}{dt}=rP(k-P)\Longleftrightarrow P=\frac{KP_0}{(k-P_0)e^{-rkt}+P_0}\end{equation}

We have \frac{1}{r} instead of r.

Therefore,

    \begin{equation*}0.15=\frac{1}{r}\times 18000\end{equation}

r=120000

(d) The greatest growth rate occurs when P=\frac{k}{2}=9000

    \begin{equation*}\frac{dP}{dt}=\frac{1}{r}P(k-P)\end{equation}

    \begin{equation*}\frac{dP}{dt}=\frac{1}{120000}(9000)(9000)\end{equation}

The greatest growth rate is 675 Brumbies per year.

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March 25, 2025 · 7:32 am

Deriving the Logistic Growth Equation

The logistic differential equation

    \begin{equation*}\frac{dP}{dt}=rP(k-P)\end{equation}

where r is the growth parameter and k is the carrying capacity.

And the maximum rate of increase happens when P=\frac{k}{2}

    \begin{equation*}\frac{dP}{dt}=rP(k-P)\end{equation}

    \begin{equation*}\frac{dP}{P(k-P)}=r dt{\end{equation}

    \begin{equation*}\int \frac{dP}{P(k-P)}=\int r dt{\end{equation}

I am going to separate the denominator on the left hand side

\frac{1}{P(k-P)}=\frac{A}{P}+\frac{B}{k-P}
Hence,
\frac{1}{P(k-P)}=\frac{A(k-P)+BP}{P(k-P)}
1=A(k-P)+BP
When P=0,
1=Ak\Rightarrow A=\frac{1}{k}
When P=k,
1=BK\Rightarrow B=\frac{1}{k}

So our equation is,

    \begin{equation*}\int \frac{\frac{1}{k}}{P}+\frac{\frac{1}{k}}{k-P} dP=\int r dt\end{equation}

    \begin{equation*}\frac{1}{k}\int \frac{1}{P}+\frac{1}{k-P} dP=\int r dt\end{equation}

    \begin{equation*}\int \frac{1}{P}+\frac{1}{k-P} dP=\int kr dt\end{equation}

    \begin{equation*}ln\lvert{P}\rvert-ln\lvert{k-P}\rvert=krt+c\end{equation}

    \begin{equation*}ln\lvert{\frac{P}{k-P}\rvert=krt+c\end{equation}

    \begin{equation*}\frac{P}{k-P}=e^{krt+c}\end{equation}

    \begin{equation*}\frac{P}{k-P}=e^{krt}e^{c} \end{equation}

When t=0, P=P_0,

    \begin{equation*}\frac{P_0}{k-P_0}=e^{c} \end{equation}

The equation is now

    \begin{equation*}\frac{P}{k-P}=\frac{P_0}{k-P_0}e^{krt}\end{equation}

    \begin{equation*}P=\frac{P_0}{k-P_0}e^{krt}(k-P)\end{equation}

    \begin{equation*}P=k\frac{P_0}{k-P_0}e^{krt}-P\frac{P_0}{k-P_0}e^{krt}\end{equation}

    \begin{equation*}P+P\frac{P_0}{k-P_0}e^{krt}=k\frac{P_0}{k-P_0}e^{krt}\end{equation}

    \begin{equation*}P(1+\frac{P_0}{k-P_0}e^{krt})=k\frac{P_0}{k-P_0}e^{krt}\end{equation}

    \begin{equation*}P=\frac{k\frac{P_0}{k-P_0}e^{krt}}{1+\frac{P_0}{k-P_0}e^{krt}}\end{equation}

    \begin{equation*}P=\frac{k\frac{P_0}{k-P_0}e^{krt}}{\frac{k-P_0+P_0e^{krt}}{k-P_0}}\end{equation}

    \begin{equation*}P=\frac{kP_0e^{rkt}}{k-P_0+P_0e^{rkt}}\end{equation}

Divide by e^{rkt}

    \begin{equation*}P=\frac{kP_0}{(k-P_0)e^{-rkt}+P_0}\end{equation}

    \begin{equation*}}\frac{dP}{dt}=rP(k-P)\Longleftrightarrow P=\frac{kP_0}{(k-P_0)e^{-rkt}+P_0}\end{equation}

Proving the Maximum Rate of Increase Happens When P=\frac{k}{2}

    \begin{equation*}\frac{dP}{dt}=rP(k-P)\end{equation}

    \begin{equation*}\frac{d^2P}{dt^2}=r\frac{dP}{dt}(k-P)+rP(-\frac{dP}{dt})\end{equation}

    \begin{equation*}\frac{d^2P}{dt^2}=\frac{dP}{dt}(rk-rP-rP)\end{equation}

    \begin{equation*}\frac{d^2P}{dt^2}=0\end{equation}

    \begin{equation*}\frac{dP}{dt}(rk-rP-rP)=0\end{equation}

    \begin{equation*}r\frac{dP}{dt}(k-2P)=0\end{equation}

    \begin{equation*}\frac{dP}{dt}(k-2P)=0\end{equation}

    \begin{equation*}rP(k-P)(k-2P)=0\end{equation}

Hence P=k or P=\frac{k}{2}

(1)   \begin{equation*}\frac{d^3P}{dt^3}=\frac{dP^2}{dt^2}(rk-2rP)+\frac{dP}{dt}(-2\frac{dP}{dt})\end{equation*}

Substitute P=k into equation 1

    \begin{equation*}\frac{d^3P}{dt^3}=rk(k-k)(rk-2rk)(rk-2rk)-2(rk(k-k))^2=0\end{equation}

Hence, not a maximum.

Substitute P=\frac{k}{2} into equation 1

    \begin{equation*}\frac{d^3P}{dt^3}=rk(k-\frac{k}{2})(rk-2r\frac{k}{2})(rk-2r\frac{k}{2})-2(rk(k-\frac{k}{2}))^2=0\end{equation}

    \begin{equation*}\frac{d^3P}{dt^3}=-2(rk^2-\frac{rk^2}{2})^2\end{equation}

    \begin{equation*}\frac{d^3P}{dt^3}=-2\frac{r^2k^4}{4}\end{equation}

-2\frac{r^2k^4}{4}\le 0 For all values of P, r and k.

Hence maximum when P=\frac{k}{2}

We will look at a worked example in the next post.

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Filed under Differential Equations, Differentiation, Implicit, Logistic Growth, Optimisation, Product Rule, Uncategorized, Year 12 Specialist Mathematics

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March 18, 2025 · 6:04 am

Intersecting Secant Theorem

CD is a tangent to the circle.

Prove c^2=a(a+b)

I am going to add two chords to the circle

Chord AD and BD are added

\angle{BDC}=\angle{CAD} (angles in alternate segments are congruent)

\angle{BCD}=\angle{DCA} (shared angle)

\therefore \Delta BDC\cong \Delta{DAC} (AA)

Hence

\frac{DC}{AC}=\frac{BC}{DC} (Corresponding sides in similar triangles)

\frac{c}{a+b}=\frac{a}{c}

\therefore c^2=a(a+b)

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March 17, 2025 · 12:33 pm

Circle Geometry Question 2

One of my Year 11 Specialist students had this question

Triangle ABC touches the given circle at Points P, Q, R and S only. The secant BW touches the circle at V and W.

Diagram not drawn to scale

(a) Determine the lengths of the line segments marked x, y and z, leaving your answers as exact values.

(b) If the length of the line segment QW is 4 units, determine the exact radius of the circle.

(a) We are going to use the Intersecting Secant Theorem – the tangent version

c^2=a\times(a+b)

Hence, we have

    \begin{equation*}30^2=25(25+x+6)\end{equation}

    \begin{equation*}900=25(31+x)\end{equation}

    \begin{equation*}x=5\end{equation}

Then we can use the intersecting chord theorem to find y.

    \begin{equation*}10\times y=6 \times x\end{equation}

    \begin{equation*}10y=30\end{equation}

    \begin{equation*}y=3\end{equation}

Back to the Intersecting Secant Theorem to find z

    \begin{equation*}z^2=4\times 17\end{equation}

    \begin{equation*}z=2\sqrt{17}\end{equation}

(b)


QW is part of a 3-4-5 triangle, therefore \angle{Q}=90^\circ

This is definitely the case of the diagram not being drawn to scale. If \angle{Q}=90^\circ, then the purple line must be the diameter.

We can use pythagoras to find the length of the diameter

    \begin{equation*}(2r)^2=13^2+4^2\end{equation}

    \begin{equation*}4r^2=185\end{equation}

    \begin{equation*}r=\frac{\sqrt{285}}{2}\end{equation}

The radius of the circle is \frac{\sqrt{285}}{2}

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March 9, 2025 · 11:26 am

Circle Geometry Question

In the above diagram O is the centre of the larger circle. A, B,D and E are points on the circumference of the larger circle. A, C, E and 0 are points on the circumference of the smaller circle. Show that \angle{CAB}=\angle{ABC}. AB, AC and BC are straight lines.

AO=OB (radii of the larger circle)

At a line from O to E (it is also a radius of the larger circle)

Let \angle{CAB}=\alpha.

ACEO is a cyclic quadrilateral.

Hence, \angle{CED}=180-\alpha (AECO is a cyclic quadrilateral)

As CB is a straight line \angle{OEB}=180-(180-\alpha)=\alpha.

\Delta OEB is an isosceles triangle.

Therefore, \angle{ABC}=\alpha

Therefore \angle{ABC}=\angle{CAB}

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Filed under Circle Theorems, Finding an angle, Geometry, Year 11 Specialist Mathematics

March 5, 2025 · 2:00 am

Binomial Expansion Theorem

My Year 11 Mathematics Methods students are working on the Binomial Expansion Theorem.

But before we get onto that, remember Pascal’s triangle

First 8 rows of Pascal’s triangle

Now we can use combinations to find the numbers in each row. For example, 1 4 6 4 1 is \begin{pmatrix}4\\0\end{pmatrix}=1, \begin{pmatrix}4\\1\end{pmatrix}=4, \begin{pmatrix}4\\2\end{pmatrix}=6, \begin{pmatrix}4\\3\end{pmatrix}=4, \begin{pmatrix}4\\4\end{pmatrix}=1

ExpressionExpansionCo-efficients
(x+y)^2x^2+2xy+y^21, 2, 1
(x+y)^3x^3+3x^2y+3xy^2+y^31, 3, 3, 1
(x+y)^4x^4+4x^3y+6x^2y^2+4xy^3+y^41, 4, 6, 4, 1

As you can see, the coefficients are the row of pascal’s triangle corresponding to the power. So (x+y)^6 would have co-efficients from the sixth row of the table 1, 6, 15, 20, 15, 6, 1.

To generalise

(x+y)^n=\begin{pmatrix}n\\0\end{pmatrix}x^ny^0+\begin{pmatrix}n\\1\end{pmatrix}x^{n-1}y^1+\begin{pmatrix}n\\2\end{pmatrix}x^{n-2}y^2+ ...+\begin{pmatrix}n\\n-1\end{pmatrix}x^1{y^{n-1}+\begin{pmatrix}n\\n\end{pmatrix}x^0y^n

Which we can condense to

(x+y)^n=\Sigma_{i=0}^n \begin{pmatrix}n\\i\end{pmatrix}x^{n-i}y^i

Worked Examples

(1) Expand (2x-3)^4

(2x-3)^4=\begin{pmatrix}4\\0\end{pmatrix}(2x)^4(-3)^0+\begin{pmatrix}4\\1\end{pmatrix}(2x)^3(-3)^1+\begin{pmatrix}4\\2\end{pmatrix}(2x)^2(-3)^2+\begin{pmatrix}4\\3\end{pmatrix}(2x)^1(-3)^3+\begin{pmatrix}4\\4\end{pmatrix}(2x)^0(-3)^4
(2x-3)^4=16x^4-96x^3+216x^2-216x+81

(2) Find the co-efficient of the x^3 term in the expansion of (2-5x)^5.

Remember (x+y)^n=\Sigma_{i=0}^n \begin{pmatrix}n\\i\end{pmatrix}x^{n-i}y^i, the x^3 is when i=3
\begin{pmatrix}5\\3\end{pmatrix}(2)^2(-5)^3=10\times 2\times -125=-5000

(3) Find the constant term in the expansion of (x^2+\frac{3}{x^4})^6

We need to find the term where the x‘s cancel out. Each term is \begin{pmatrix}6\\i\end{pmatrix}(x^2)^{6-i}(\frac{3}{x^4})^i.
\begin{pmatrix}6\\i\end{pmatrix}(x^{12-2i})(3^ix^{-4i}).
We need 12-2i-4i=0, hence i=2
Therefore, the co-efficient is \begin{pmatrix}6\\2\end{pmatrix}\times3^2=135

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Filed under Algebra, Binomial Expansion Theorem, Counting Techniques, Year 11 Mathematical Methods