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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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January 16, 2025 · 8:08 am

HSC Advanced 2024 Question 30

HSC Advanced 2024

Two circles have the same centre O. The smaller circle has a radius of 1 cm, while the larger has a radius of (x+1) cm. The circles enclose a region QRST, which is subtended by angle of {\theta} at O, as shaded.

The area of QRST is A cm2, where A is a constant and A>0

Let P cm be the perimeter of QRST

(a) By finding expressions for the area and perimeter of QRST show that P(x)=2x+\frac{2A}{x}

(b) Show that if the perimeter is minimised, then {\theta} must be less than 2.

(a) A=\frac{1}{2}\theta((x+1)^2-1^2)
A=\frac{1}{2}\theta(x^2+2x)
2A=\theta x^2 +2x \theta
\frac{2A}{x}=\theta x +2\theta

P=\theta(1)+\theta(1+x)+2x
P=2\theta +\theta x +2x
P=\frac{2A}{x}+2x

I like it when the first part requires the student to show something and the second part has them use it (that way they can still do the second part even if they couldn’t do the first part).

(b) \frac{dP}{dx}=2-\frac{2A}{x^2}
0=2-\frac{2A}{x^2}
x=\sqrt{A}

\frac{d^2P}{dx^2}=\frac{4A}{x^3}
Both x and A are greater than zero, therefore \frac{d^2P}{dx^2}>0 and x=\sqrt{A} is a minimum.

Substitute x=\sqrt{A} into the Area formula
2A=A\theta+2\sqrt{A}\theta
\theta=\frac{2A}{A+2\sqrt{A}}
\theta=\frac{2A}{\sqrt{A}(\sqrt{A}+2)}
\theta=\frac{2\sqrt{A}}{\sqrt{A}+2}

Now \frac{2\sqrt{A}}{\sqrt{A}+2}<\frac{2\sqrt{A}}{\sqrt{A}}
Hence \theta<\frac{2\sqrt{A}}{\sqrt{A}}
and \theta<2

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November 7, 2024 · 9:51 am

Mathematics Applications – Counting can be Tricky Sometimes (off by one error)

Sequences are part of the Year 12 Mathematics Applications course and sometimes it’s tricky to work out which terms the question requires.

For example, ATAR 2020 Question 11

Judith monitors the water quality in her garden pond at the same time everyday. She likes to maintain the concentration of algae between 200 and 250 unites per 100 litres (L). Her measurements show that the concentration increases daily according to the recursive rule

C_{n+1}=1.025C_n where C_1=200 units per 100 L (the minimum concentration)

When the concentration gets above the 250 units per 100 L limit, she treats the water to bring the concentration back to the minimum 200 units per 100 l.

(a) If Judith treated the water on Sunday 6 December 2020, determine

(i) the concentration on Wednesday, 9 December 2020.
(ii) the day when she next treated the water.

(b) During the first week of January 2021, Judith monitored the water and recorded the following readings

Day1234567
Concentration (C)200206212.28218.55225.10231.85238.81

(i) Determine the revised recursive rule.
(ii) If she treated the water on 10 January and went on holiday until 20 January, when she next treated the water, calculate the concentration of the water on her return. Assuming the recursive rule from (b)(i) is used.

(a)(i) If C_1 is the 6th of December, then what term is the 9th of January?

I find most students simply do 9-6=3 so C_3, but this means they are off by one.
It’s better to list them
6th C_1
7th C_2
8th C_3
9th C_4
Hence we want to find C_4



The concentration on Wednesday 9 December is 215.38 units per 100 L

a(ii) We need to find when the concentration is greater than 250


C_{11}=256.02, what day is C_{11}?
The 9th is C_4, 10th C_5, etc. 16th is C_{11}
Judith next treats the water on Wednesday 16 December

(b)
(i) r=\frac{206}{200}=1.03
C_{n+1}=1.03C_n where C_1=200
(ii) C_1 is the 10th of January, 20th of January is C_{11} (20-10+1)


The concentration of the water on Judith’s return is 268.78 units per 100 L

I get my students to count on their fingers to ensure they get the correct term or day.

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November 2, 2024 · 11:45 am

Effect of Function Transformations on Integration

My year 12 Mathematical Methods students have questions like this

Given that f(x) is continuous everywhere and that \int_{4}^{10} f(x) dx=-10, find:

(a) \int_{4}^{10}2x +f(x) dx

(b) \int_{5}^{11} f(x-1) dx

(c) \int_{1}^{3} f(3x+1) dx

(d) \int_{-10}^{-4} -f(-x) dx

(e) \int{10}^{22} f(\frac{x-2}{2}) dx

(f) \int_{-3}^{-9} f(1-x) dx

OT Lee Mathematics Methods Textbook Ex 8.3 question 6

For the most part these questions aren’t too difficult, but the horizontal dilations cause issues.

(a) \int_{4}^{10} 2x +f(x) dx
\int_{4}^{10} 2x dx +\int_{2}^{10} f(x) dx
(x^2]_4^{10} + (-10)
10^2-4^2-10
=74

(b) \int_{5}^{11} f(x-1) dx
=-10




(c) \int_{1}^{3} f(3x+1) dx
Let u=3x+1
\frac{du}{dx}=3
dx=\frac{du}{3}

When x=1, u=4 and when x=3, u=10
\int_{4}^10 f(u) \frac{du}{3}
=\frac{1}{3}\times (-10)
=\frac{-10}{3}

(d) \int_{-10}^{-4} -f(-x) dx
-\int_{-10}^{-4} f(-x) dx

Let u=-x
\frac{du}{dx}=-1
dx=-du

When x=-4, u=4 and when x=-10, u=10
-\int_{4}^{10} f(u) -dx
=-10

(e) \int_{10}^{22} f(\frac{x-2}{2} dx
Let u=f(\frac{x-2}{2})
\frac{du}{dx}=\frac{1}{2}
\du=2dx

When x=10, u=4 and when x=22, u=10
2\int_{4}^{10} f(u) du
=-20

(f) \int_{-3}^{-9} 2f(1-x) dx
Let u=1-x
\frac{du}{dx}=-1

When x=-3, u=4 and when x=-9, u=10
=-2\int_{4}^{10} f(u) du
=20


Split the integral
Integrate the first part.


This is a horizontal translation (one unit to the right) so the shape of the curve doesn’t change.
The integration bounds have also shifted one unit to the right.




This is a horizontal dilation and translation. The easiest method is to use a change of variable





































Once you get the hang of it, you can skip the change of variable and multiply the value of the definite integral by the scale factor of the horizontal dilation (only if the integration bounds are also changed).

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Filed under Definite, Integration, Uncategorized, Year 12 Mathematical Methods

October 28, 2024 · 6:51 am

Related Rates Question

A boat is moving towards the beach line at 3 metres/minute. On the boat is a rotating light, revolving at 4 revolutions/minute clockwise, as observed from the beach. There is a long straight wall on the beach line, as the boat approaches the beach, the light moves along the wall. Let x equal the displacement of the light from the point O on the wall, which faces the boat directly. See the diagram below.
Determine the velocity, in metres/minute, of the light when x=5 metres, and the distance of the boat from the beach D is 12 metres.

Mathematics Specialist Semester 2 Exam 2018


The light is rotating at 4 revolutions/minute, which means

    \begin{equation*}\frac{d\theta}{dt}=4\times\pi\end{equation}

We want to find \frac{dx}{dt} and we know \frac{d\theta}{dt} and \frac{dD}{dt}.

We need to find an equation connecting x, \theta, and D.

    \begin{equation*}tan (\theta)=\frac{x}{D}\end{equation}

Differentiate (implicitly) with respect to time.

    \begin{equation*}sec^2(\theta)\frac{d\theta}{dt}=\frac{\frac{dx}{dt}D-\frac{dD}{dt}x}{D^2}\end{equation}

Now we know x=5, and D=12, using pythagoras we can calulate the hypotenuse.


h=\sqrt{12^2+3^2}=13
sec(\theta)=\frac{13}{12}

    \begin{equation*}sec^2(\theta)\frac{d\theta}{dt}=\frac{\frac{dx}{dt}D-\frac{dD}{dt}x}{D^2}\end{equation}

    \begin{equation*}(\frac{13}{12})^2\times 4\pi=\frac{\frac{dx}{dt}12-3\times 5}{12^2}\end{equation}

    \begin{equation*}169\times4\pi=12\frac{dx}{dt}-15\end{equation}

    \begin{equation*}676\pi+15=12\frac{dx}{dt}\end{equation}

    \begin{equation*}\frac{dx}{dt}=178.2\end{equation}

The velocity of the light is 178.2 m/minute.

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October 22, 2024 · 6:55 am

Year 12 Mathematics Applications – Finance Question

What happens with an investment or a loan when the compounding periods are not the same as the payment periods?

For example,

Gerald invests $450 000 at a rate of 6.4% p.a. compounding monthly. At the end of every quarter he receives $35 000 from his investment.

(a) Write a recursive rule that will enable you to find the balance of the account, T_n after n payments.

(b) Find the balance of the account after 3 years.

(c) How long will it take for the balance in the account to reach zero?

(d) What will be the amount of Gerald’s last payment?

The interest is compounding at a different rate to the payments.
T_{n+1}=T_n(1+\frac{6.4}{100\times12})^{12\div4}-35 000, T_0=450 000

We need to raise the multiplier (1+\frac{6.4}{100\times12}) by the number of compound periods per payment, i.e. 12\div4

(a) T_{n+1}=T_n(1.0161)^3-35 000, T_0=450 000

(b) 3 years is 12 payments, \therefore n=12

T_12=245548.28

(c)

21 years

(d) The final repayment is 35 000-6535.03=28 464.97

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Filed under Classpad Skills, Finance, Finance, Uncategorized, Year 12 Mathematics Applications

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May 26, 2024 · 1:07 pm

Quadratic Rule from a Table of Values

How do you find a quadratic rule from a table of values?

For example,

x1 2 3 4
y061424

Find the first difference

First Difference6-0=614-6=824-14=10

Find the second difference (if the second difference is a constant, then it is quadratic)

Second Difference8-6=2 10-8=2

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

The second difference is 2a

Hence our equation is now y=x^2+bx+c

The c value is the vertical intercept (x=0). We can back track in the table

x01234
yc061424

As the first differences are 6, 8, 10, the one between 0 and 1 must be 4

0-c=4

c=-4

Our equation is now y=x^2+bx-4.

We can now use any other point to find the b value.

Let’s use the point (2, 6)

6=2^2+b(2)-4

6=4-2b-4

6=2b

b=3

The function is y=x^2+3x-4

Let’s try another one

x3456
y7173149

First differences

First difference101418

Second difference

Second Difference44

Hence 2a=4, therefore a=2

The equation is now y=2x^2+bx+c

Instead of back tracking, this time I am going to use two points and simultaneous equations.

Using points (3, 7) and (4, 17)

    \[7=2(3)^2+b(3)+c\]

(1)   \begin{equation*}3b+c=-11\end{equation*}

    \[17=2(4)^2+b(4)+c\]

(2)   \begin{equation*}4b+c=-15\end{equation*}

Equation 2 – Equation 1

b=-4

Substitute b=-4 into equation 1

3(-4)+c=-11

-12+c=-11

c=1

Hence the equation is y=2x^2-4x+1

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May 25, 2024 · 7:58 am

Proof of the cosine rule

    \[a^2=b^2+c^2-2bccos(A)\]

From the diagram above

    \[h^2=a^2-(b-x)^2\textnormal{ and }h^2=c^2-x^2\]

    \[\therefore a^2-(b-x)^2=c^2-x^2\]

    \[a^2-(b^2-2bx+x^2)=c^2-x^2\]

    \[a^2-b^2+2bx-x^2-c^2+x^2=0\]

(1)   \begin{equation*}a^2=b^2+c^2-2bx\end{equation*}

From the diagram above, we can see

    \[cos A=\frac{x}{c}\]

(2)   \begin{equation*}x=c cos A\end{equation*}

Substitute equation 2 into equation 1

    \[a^2=b^2+c^2-2bc cosA\]

It can also be handy to have the angle version

    \[cosA=\frac{b^2+c^2-a^2}{2bc}\]

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May 19, 2024 · 5:42 am

Right-Angled Trigonometry Question (very hard)

One of my year 10 students came with this question from his text book.

ICE_EM Mathematics Year 10 Third Edition

Here is my solution.

A pdf version of the solution

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March 31, 2024 · 8:27 am

Sketching Subsets of the Complex Plane – Problem 1

Describe the locus defined by

A pdf version.

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