Measurement and Financial Mathematics | The Ultimate Guide to NESA’s Maths Reference Sheet

\(l=\frac{\theta}{360}\times2\pi r\)

Area

\(A= \frac{\theta}{360} \times \pi r^2\)

\(A=\frac{h}{2} (a+b) \)

\(A=2 \pi r^2 + 2 \pi r h \)

\(A = 4 \pi r^2\)

Volume

\(V= \frac{1}{3}Ah\)

\(V= \frac{4}{3} \pi r^3\)

Sequences and series

\(T_n = a+(n-1)d\)

\(S_n=\frac{n}{2} [2a+(n-1)d] = \frac{n}{2}(a+l)\)

\(T_n=ar^{n-1}\)

\(S_n=\frac{a(1-r^n)}{1-r}=\frac{a(r^n-1)}{r-1}, \ r≠1\)

\(S= \frac{a}{1-r}, \ |r|<1 \)

(using θ in degrees)

\begin{align*}
l &=\frac{115}{360}\times2\pi \times 30\\
&= \frac{115}{6} \pi \\
∴ \text{Length of the minor arc AB} &= \frac{115}{6} \pi \text{ cm}\\
\end{align*}

\begin{align*}
A &= \frac{330}{360} \times \pi \times {5}^2 \\
&= \frac{275}{12} \pi \\
∴ \text{Area of the major sector AOB} &= \frac{275}{12} \pi \text{ m}^2\\
\end{align*}

\begin{align*}
A &= \frac{12}{2} (16+20) \\
&= 216 \\
∴ \text{Area} &= 216 \text{ mm}^2\\
\end{align*}

\begin{align*}
A &=2 \pi \times 4^2 + 2 \pi \times 4 \times 9 \\
&= 32 \pi + 72 \pi \\
&= 104 \pi \\
∴ \text{Surface area of can} &= 104 \pi \text{ cm}^2\\
\end{align*}

 \(r = \text{radius}\)

Substitute \(r= \frac{0.07}{2}\) (since the radius is half of the diameter) into \(A = 4 \pi r^2\):

\begin{align*}
A &= 4 \pi \times \left( \frac{0.07}{2} \right)^2 \\
&= 4 \pi \times 0.035^2 \\
&= 0.0049 \pi \\
∴ \text{Surface area of cell} &= 0.0049 \pi \text{ mm}^2\\
\end{align*}

To find the area of the circular base, substitute \(r=5\) into \(A= \pi r^2\):

\begin{align*}
A &= \pi \times 5^2  \\
&= 25 \pi \\
∴ \text{Base area} &= 25 \pi \text{ cm}^2\\
\end{align*}

Substitute \(A= 25 \pi\) and \(h=12\) into \(V= \frac{1}{3}Ah\):

\begin{align*}
V &=\frac{1}{3} \times 25 \pi \times 12 \\
&= 100 \pi \\
∴ \text{Volume of water needed} &= 100 \pi \text{ cm}^3\\
&= 100 \pi \text{ mL}\\
\end{align*}

\begin{align*}
V &= \frac{4}{3} \pi \left( \frac{3}{2} \right) ^3 \\
&= \frac{9}{2} \pi \\
∴ \text{Amount of mince needed} &= \frac{9}{2} \pi \text{ cm}^3\\
&= \frac{9}{2} \pi \text{ g}\\
\end{align*}

\begin{align*}
T_{23} & = 3+(23-1) \times 12\\
&= 3 + 22 \times 12 \\
&= 267 \\
∴ \text{23}^{rd} \text{ term: } & 267\\
\end{align*}

\begin{align*}
632 & =\frac{16}{2} [2 \times 2 +(16-1)d]\\
632 &= 8 \times (4+ 15d) \\
632 &= 32+ 120d \\
600 &= 120d \\
∴ d &= 5\\
\end{align*}

\begin{align*}
T_1 & = a = 2 \text{ (given)}\\
\text{Since } T_n & = a+(n-1)d \text{:}\\
T_2 &= 2 + 5 \\
&= 7 \\
T_3 &= 2+ 2 \times 5 \\
&= 12 \\
T_4 &= 2 + 3 \times 5 \\
&= 17 \\
\text{∴ First 4 terms of the series: }& 2 + 7 + 12 +17\\
\end{align*}

Let’s consider what this problem would look like visually.

The height of each bounce conveniently forms a geometric sequence where the common ratio r=0.75 and the first term a=x.

Hence, \(T_n=x \times 0.75^{n-1}\)

However, the distance that the ball travels is not a simple geometric series. As you can see in the diagram above, apart from the first bounce and final n bounce (where the ball only travels downwards or upwards), the ball travels twice the distance of the bounce height as it travels up and down.

Hence, we can find the total distance d travelled by the ball before the nth bounce:

\begin{align*}
d &=  2 \times [\text{sum of the heights of the 1st, 2nd, 3rd, …, (n-2)th, (n-1)th and nth bounce}] -[\text{height of the 1st bounce}] – [\text{height of the nth bounce}]\\
&= 2 \times (T_1+T_2+T_3+…+T_{n-2}+T_{n-1}+T_n) – T_1 – T_n\\
∴d &= 2 \times S_n – T_1 – T_n\\
\end{align*}

In this question, we are given the total distance travelled by the ball before the 8th bounce is 6 metres. So, substitute d=6 and  n=8 into \( d = 2 \times S_n – T_1 – T_n \):

\begin{align*}
6 &= 2 \times S_8 – T_1 – T_8 \text{…………. (equation 1)}\\
\end{align*}

To find S₈, substitute \(a=x\), \(n=8\) and \(r=0.75\) into \(S_n=\frac{a(1-r^n)}{1-r}\):

\begin{align*}
S_8 &=\frac{x(1-0.75^8)}{1-0.75}\\
&= 4[x(1-0.75^8)] \\
\end{align*}

To find T₈, substitute \(a=x\) and \(n=8\) into \(T_n=ar^{n-1}\):

\begin{align*}
T_8 &= x \times 0.75^{8-1} \\
&= x \times 0.75^7 \\
\end{align*}

Hence, we can substitute \(S_8=4[x(1-0.75^8)] \), \(T_1= x \) and \(T_8= x \times 0.75^7 \) into equation 1: \(6 = 2 \times S_8 – T_1 – T_8\):

\begin{align*}
6 &= 2 \times 4[x(1-0.75^8)] – x – x \times 0.75^7\\
6 &= 8[x(1-0.75^8)] – x – x \times 0.75^7\\
6 &= x[8(1-0.75^8) – 1 – 0.75^7] \\
x &= \frac{6}{8(1-0.75^8) – 1 – 0.75^7} \\
x &= 0.989 \text{ (rounded to 3 decimal places)} \\
∴ \text{Height at which player began bouncing ball} &= 0.989 \text{ m}\\
\end{align*}

In part a), we were interested in the distance that the ball travels before the 8th bounce. So, we used \(n = 8\). In part b) though, we are trying to find how far the ball can travel if it was allowed to bounce forever. So here, we let \(n = ∞\).

Substitute \(n = ∞\) into \(d = 2 \times S_n – T_1 – T_n\):

\begin{align*}
d &= 2 \times S_∞ – T_1 – T_∞\\
\end{align*}

Since:

\begin{align*}
T_1 &=a \\
&=0.989 \text{ (from part a)}\\
S_∞ &= \frac{a}{1-r}, \  \text{ where a=0.989, r=0.75 and|r|<1} \\
&=  \frac{0.989}{1-0.75}\\
& = 3.956\\
T_∞ &= 0 \text{ (the ball does not lift off the ground on the ‘last’ bounce)}\\
\end{align*}

\begin{align*}
⇒ d &= 2 \times S_∞ – T_1 – T_∞\\
&= 2 \times 3.956 – 0.989 – 0\\
&= 6.92 \text{ (rounded to 2 decimal places)}\\
∴ \text{Total limiting distance} &= 6.92 \text{ m} \\
\end{align*}