Category Archives: Year 11 Mathematical Methods

June 4, 2025 · 5:30 am

Closest Approach (Shortest Distance) e-activity (Casio Classpad)

At 1pm, object H travelling with constant velocity \begin{pmatrix}200\\10\end{pmatrix}km/h is sighted at the point with position vector \begin{pmatrix}-90\\-100\end{pmatrix}km. At 2pm object J travelling with constant velocity \begin{pmatrix}100\\-100\end{pmatrix}km/h is sighted at the point with position vector \begin{pmatrix}20\\-120\end{pmatrix}km. Determine the minimum distance between H and J and when this occurs.

OT Lee Mathematics Specialist Year 11 Unit 1 and 2 Exercise 10.1 Question 6.

(1)   \begin{equation*}\mathbf{r_H}=\begin{pmatrix}-90\\-100\end{pmatrix}+t\begin{pmatrix}200\\10\end{pmatrix}\end{equation*}

(2)   \begin{equation*}\mathbf{r_J}=\begin{pmatrix}-80\\-20\end{pmatrix}+t\begin{pmatrix}100\\-100\end{pmatrix}\end{equation*}

\begin{pmatrix}-80\\-20\end{pmatrix} is the position vector of J at 1pm.

Find the relative displacement of H to J

    \begin{equation*}\mathbf{_H}\mathbf{r_J}=\mathbf{r_H}-\mathbf{r_J}\end{equation}

    \begin{equation*}\mathbf{_H}\mathbf{r_J}=\begin{pmatrix}-90\\-100\end{pmatrix}+t\begin{pmatrix}200\\10\end{pmatrix}-(\begin{pmatrix}-80\\-20\end{pmatrix}+t\begin{pmatrix}100\\-100\end{pmatrix})\end{equation}

    \begin{equation*}\mathbf{_H}\mathbf{r_J}=\begin{pmatrix}-10\\-80\end{pmatrix}+t\begin{pmatrix}100\\110\end{pmatrix}\end{equation}

Find the relative velocity of H to J

    \begin{equation*}\mathbf{_H}\mathbf{v_J}=\begin{pmatrix}100\\110\end{pmatrix}\end{equation}

The relative displacement is perpendicular to the relative velocity at the closest approach.

That is

(3)   \begin{equation*}\mathbf{_H}\mathbf{r_J}\cdot\mathbf{_H}\mathbf{v_J}=0\end{equation*}

    \begin{equation*}(\begin{pmatrix}-10\\-80\end{pmatrix}+t\begin{pmatrix}100\\110\end{pmatrix})\cdot(\begin{pmatrix}100\\110\end{pmatrix})=0\end{equation}

    \begin{equation*}(-10+100t)(100)+(-80+110t)(110)=0\end{equation}

    \begin{equation*}-1000+10 000t-8800+12100t=0\end{equation}

    \begin{equation*}22100t=9800\end{equation}

    \begin{equation*}t=\frac{98}{221}\end{equation}

Substitute t=\frac{98}{221} into the relative displacement and find the magnitude.

    \begin{equation*}\mathbf{_H}\mathbf{r_J}=\begin{pmatrix}-10\\-80\end{pmatrix}+\frac{98}{221}\begin{pmatrix}100\\110\end{pmatrix}\end{equation}

    \begin{equation*}\mathbf{_H}\mathbf{r_J}=\begin{pmatrix}34\frac{76}{221}\\-31\frac{49}{221}\end{pmatrix}\end{equation}

    \begin{equation*}\|\begin{pmatrix}34\frac{76}{221}\\-31\frac{49}{221}\end{pmatrix}\|=46.4\end{equation}

The closest objects H and J get to each other is 46.4km at 1:27pm.

I have made an e-activity for this.

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Filed under Classpad Skills, Closest Approach, Vectors, Year 11 Mathematical Methods

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May 6, 2025 · 7:27 am

Combinations Proof

Prove \begin{pmatrix}n+1\\r\end{pmatrix}=\begin{pmatrix}n\\r-1\end{pmatrix}+\begin{pmatrix}n\\r\end{pmatrix}

Let’s start with the right hand side.

RHS=\begin{pmatrix}n\\r-1\end{pmatrix}+\begin{pmatrix}n\\r\end{pmatrix}

RHS=\frac{n!}{(n-(r-1))!(r-1)!}+\frac{n!}{(n-r)!r!}

Simplify

RHS=\frac{n!}{(n+1-r)!}+\frac{n!}{(n-r)!r!}

There is a common denominator of (n+1-r)!r!

RHS=\frac{n!}{(n+1-r)!}\times \frac{r}{r}+\frac{n!}{(n-r)!r!}\times \frac{n+1-r}{n+1-r}

RHS=\frac{n!r+n!(n+1-r)}{(n+1-r)r!}

RHS=\frac{n!r+(n+1)n!-n!r}{(n+1-r)!r!}

RHS=\frac{(n+1)!}{(n+1-r)!r!}

RHS=\begin{pmatrix}n+1\\r\end{pmatrix}

RHS=LHS

We know this intuitively from Pascal’s triangle

Where each entry is the sum of the two entries above it – for example,

6+4=10

Remember, Pascal’s triangle can be written as combinations,

so \begin{pmatrix}5\\3\end{pmatrix}=\begin{pmatrix}4\\2\end{pmatrix}+\begin{pmatrix}4\\3\end{pmatrix}

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Filed under Algebra, Counting Techniques, Year 11 Mathematical Methods

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 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

October 7, 2024 · 7:43 am

Arithmetic Sequence

I did this question with on of my year 11 students. I think the algebra and the subscripts can be a bit tricky.

If T_m=n and T_n=m, then prove that T_{m+n}=0. Here where T_n and T_m are terms of an arithmetic sequence.
Mathematics Methods Units 1&2 – Exercise 15B Question 19

If T_m=n then,

(1)   \begin{equation*}n=a+(m-1)d\end{equation*}


And if T_n=m then,

(2)   \begin{equation*}m=a+(n-1)d\end{equation*}


Subtract equation (2) from equation (1)

    \begin{equation*}n-m=(m-1)d-((n-1)d)\end{equation*}


    \begin{equation*}n-m=md-nd\end{equation*}


(3)   \begin{equation*}n-m=d(m-n)\end{equation*}


Therefore d must equal -1
Substitute d=-1 into equation (1)

    \begin{equation*}n=a+(m-1)(-1)\end{equation*}


(4)   \begin{equation*}n=a-m+1\end{equation*}


Therefore a=n+m-1


(5)   \begin{equation*}T_{m+n}=a+(m+n-1)d\end{equation*}


Substitute a=n+m-1 and d=-1 into equation (5)

    \begin{equation*}$T_{m+n}=n+m-1+(m+n-1)(-1)$\end{equation*}


    \begin{equation*}$T_{m+n}=n+m-1-m-n+1$\end{equation*}


(6)   \begin{equation*}$T_{m+n}=0$\end{equation*}

As you can see from equation (6), T_{m+n}=0

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