Monthly Archives: June 2024

June 12, 2024 · 7:37 am

Year 12 Mathematics Applications Finance

Ming, a former high school student and now a successful business owner, wishes to set up a perpetuity of $6000 per year to be paid to a deserving student from her school. The perpetuity is to be paid at the start of the year in one single payment.

(a) A financial institution has agreed to maintain an account for the perpetuity paying a fixed rate of 5.9% p.a. compounded monthly. Show that an amount of $98 974, to the nearest dollar, is required to maintain this perpetuity.

(b) Ming allows herself five years to accumulate the required $98 974 by making regular quarterly payments into an account paying 5.4% p.a. compounded monthly. Determine the quarterly payment needed to reach the required amount after five years if Ming starts the account with an initial deposit of $1000.
SCSA 2017 CA 8

(a) For a perpetuity, we want the interest to equal the payment.

Remember the compound interest formula is

$A=P(1+\frac{r}{100n})^{nt}$

Where $P$ is the principal, $r$ is the interest rate (as a percent), $n$ is the number of compounding periods in a year, and $t$ is the time.

$\therefore I=A-P$

$I=P(1+\frac{r}{100n})^{nt}-P$

$6000=P(1.00492)^12-P$

$P=\frac{6000}{(1.00492^12-1)}$

$P=98 974.14$

Therefore an amount of $98 974 is required to maintain this perpetuity

For part (b) I will use the Finance Solver on a Classpad (Casio).

The quarterly payments are $4283.77.

The Logistic Equation

My year 12 Specialist students are working on logistic growth at the moment. An example might be helpful.

A new viral disease was found to spread according to the equation $\frac{dN}{dt}=kn(M-N)$ , where $M$ is the susceptible population, $N$ is the number of people infected at time $t$ months and $k=1.5\times 10^{-9}$ . In March 2010, it was thought only 100 people out of a population of 18 million were infected. Use the logistic model to find the number infected in:

(a) March 2011

(b) June 2012

(c) January 2017
Specialist 12 – Nelson Senior Maths

$\frac{dN}{dt}=kN(M-N)$

$\frac{dN}{dt}=(1.5\times10^{-9})N(18\times10^{6}-N)$

$\frac{dN}{N(18\times10^{6}-N)}=1.5\times10^{-9}dt$

Use partial fractions to separate the denominator $\frac{dN}{N(18\times10^{6}-N)}$

$\frac{1}{N(18\times10^{6}-N)}=\frac{A}{N}+\frac{B}{18\times10^{6}-N}$

$1=A(18\times10^{6}-N)+BN$

When $N=0$

$1=A(18\times10^{6})$

$A=\frac{1}{18\times10^{6}}$

When $N=18\times10^{6}$

$1=B(18\times10^{6})$

$B=\frac{1}{18\times10^{6}}$

$\frac{1}{18\times10^{6}}(\frac{1}{N}+\frac{1}{(18\times10^{6}-N)}dN)=1.5\times10^{-9}dt$

$\int\frac{1}{N}+\frac{1}{(18\times10^{6}-N)}dN=\int27\times10^{-3}dt$

$\ln|N|-\ln|18\times10^{6}-N|=(27\times10^{-3})t+c$

$\ln|\frac{N}{18\times10^{6}-N}|=(27\times10^{-3})t+c$

$\frac{N}{18\times10^{6}-N}=e^{(27\times10^{-3})t+c}$

Let $A=e^{c}$ and rearrange to make $N$ the subject.

$N=\frac{(18\times10^{6})Ae^{(27\times10^{-3})t}}{1+Ae^{(27\times10^{-3})t}}$

Divide by $Ae^{(27\times10^{-3})t}$

$N=\frac{(18\times10^{6})}{\frac{1}{A}e^{-(27\times10^{-3})t}+1}$

Initially 100 people were infected.

$100=\frac{(18\times10^{6})}{\frac{1}{A}+1}$

$A=\frac{1}{179999}$

$N=\frac{(18\times10^{6})}{179999e^{-(27\times10^{-3})t}+1}$

(a) $t=12, N=138.3$ , hence $138$

(b) $t=27, N=207.3$ , hence $207$

It is not necessary to solve the differential equation, you can use the formula

$\frac{dP}{dt}=rP(k-P)\leftrightarrowP=\frac{kP_0}{P_0+(k-P_0)e^{-rkt}}$

This formula is on the Year 12 Mathematics Specialist formula sheet for Western Australia.

For our question,

$\frac{dN}{dt}=(1.5\times10^{-9})N(18\times10^6-N)$

So, $P=\frac{18\times10^{6}\times100}{100+(18\times10^6-100)e^{-(1.5\times10^{-9})(18\times10^6)t}}$

And you can substitute values for $t$ .

Non-Right Trigonometry Problem

I worked on this question with one of my students (I don’t know where it is from).

Mike leaves the rose bush he was examining and walks 35m in the direction
S $20^{\circ}$ W towards a pond.
From there he walks 70m towards a rotunda. Mike is now 100m from the rose bush.
Find the bearing of the rotunda from the pond.

Let’s try to draw a diagram

Because we don’t the direction Mike walked from the pond, I have drawn a circle with radius 70m centred at the pond.

We know Mike is now 100m from the rose bush. As we don’t know the direction, I have drawn another circle with radius 100m centred at the rose bush. Where the two circles intersect are the possible locations of the rotunda.

First Position

Use the cosine rule to find the angle
$cos\theta=\frac{b^2+c^2-a^2}{2bc}$
$cos\theta=\frac{70^2+35^2-100^2}{2(70)(35)}$
$\theta=cos^{-1}(\frac{-3875}{5250})$
$\theta=137.6^{\circ}$

Using the fact that alternate angles in parallel lines are congruent, we can see
that the bearing from the pond to the rotunda is
$360-(137.6-20)=242.2^{\circ} T$