Racquel Sanderson

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}

\frac{1}{P(k-P)}=\frac{A(k-P)+BP}{P(k-P)}

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

Tagged as differential equations, logistic growth, Optimisation, partial fractions

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

$\angle{BDC}=\angle{CAD}$ (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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Filed under Circle Theorems, Geometry, Year 11 Specialist Mathematics

Tagged as circle geometry, Intersecting Secants Theorem

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

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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Filed under Circle Theorems, Geometry, Pythagoras, Year 11 Specialist Mathematics

Tagged as intersecting chord theorem, intersecting secant theorem, pythagoras

March 9, 2025 · 11:26 am

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

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

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

March 3, 2025 · 2:51 am

Derangements

From Wikipedia

In combinatorial mathematics, a derangement is a permutation of the elements of a set in which no element appears in its original position.

For example, if we have the set ${A, B, C}$ , there are $6$ permutations

$ABC, ACB, BAC, BCA, CAB, CBA$

But only $2$ of them are derangements – $BCA$ and $CAB$

I did a practical question on this here. In that question I used a tree diagram, but there must be a way to determine the number of derangements given $n$ elements.

We know $3$ elements has $2$ derangements, what about $4$ elements? We know there are $4!=24$ permutations

ABCD	BACD	CABD	DABC
ACBD	BADC	CADB	DACB
ACDB	BCAD	CBAD	DBAC
ABDC	BCDA	CBDA	DBCA
ADBC	BDAC	CDAB	DCAB
ADCB	BDCA	CDBA	DCBA

I have highlighted the derangements. when $n=4$ there are $9$ derangements.

Let’s try to generalise.

In how many ways can one element be in its original position?
$\begin{pmatrix}4\\1\end{pmatrix} \times 3!=24$
Choose one element from the 4 to be in its original position, and then arrange the remaining three elements.

So far, we have $D_3=4!-\begin{pmatrix}4\\1\end{pmatrix} \times 3!=0$

Clearly we are counting some arrangements multiple times, for example ABDC has A and B in the correct position, so we need to add all of the arrangements with $2$ elements in their original position.
In how many ways can two elements be in their original position?
$\begin{pmatrix}4\\2\end{pmatrix} \times 2!=12$

So now we have, $D_3=4!-\begin{pmatrix}4\\1\end{pmatrix} \times 3!+\begin{pmatrix}4\\2\end{pmatrix} \times 2!=12$

But once again we have counted some arrangements multiple times, so we need to subtract all of the arrangements with $3$ elements in their original position.
In how many ways can three elements be in their original position?
$\begin{pmatrix}4\\3\end{pmatrix} \times 1!=4$

So now we have, $D_3=4!-\begin{pmatrix}4\\1\end{pmatrix} \times 3!+\begin{pmatrix}4\\2\end{pmatrix} \times 2!-\begin{pmatrix}4\\3\end{pmatrix} \times 1!=8$

We now need to add all of the arrangements with $4$ elements in their original position.
In how many ways can four elements be in their original position?
$\begin{pmatrix}4\\4\end{pmatrix} \times 0!=1$

So now we have, $D_3=4!-\begin{pmatrix}4\\1\end{pmatrix} \times 3!+\begin{pmatrix}4\\2\end{pmatrix} \times 2!-\begin{pmatrix}4\\3\end{pmatrix} \times 1!+\begin{pmatrix}4\\4\end{pmatrix} \times 0!=9$

Hence,

$D_n=\begin{pmatrix}n\\0\end{pmatrix}\times n!-\begin{pmatrix}n\\1\end{pmatrix}\times (n-1)!+\begin{pmatrix}n\\2\end{pmatrix}\times (n-2)!-... \mp \begin{pmatrix}n\\n\end{pmatrix}\times 0!$

Which we can simplify to

$D_n=\Sigma_{i=0}^{n}(-1)^n\begin{pmatrix}n\\i\end{pmatrix}\times(n-i)!$

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Filed under Counting Techniques, Interesting Mathematics, Year 11 Specialist Mathematics

February 25, 2025 · 9:28 am

Differentiating f(x)=e^x

We are going to differentiate $f(x)=e^x$ from first principals.

Remember the definition of a derivative is

(1) $\begin{equation*}f'(x)=\lim_{\limits{h\to 0}}\frac{f(x+h)-f(x)}{h}\end{equation*}$

If $f(x)=e^x$ , then

$\begin{equation*}f'(x)=\lim_{\limits{h \to 0}}\frac{e^{x+h}-e^x}{h}\end{equation}$

$\begin{equation*}f'(x)=\lim_{\limits{h \to 0}}\frac{e^x\times e^h-e^x}{h}\end{equation}$

$\begin{equation*}f'(x)=\lim_{\limits{h \to 0}}\frac{e^x(e^h-1)}{h}\end{equation}$

$\begin{equation*}f'(x)=e^x\lim_{\limits{h \to 0}}{\frac{e^h-1}{h}\end{equation}$

Let’s think about $\lim_{\limits{h \to 0}}\frac{e^h-1}{h}$

Remember $e$ is defined as $\lim_{\limits{n \to \infty}}(1+\frac{1}{n})^n$

(2) $\begin{equation*}\lim_{\limits{h \to 0}}{\frac{e^h-1}{h}\end{equation*}$

Let $y=e^h-1$ , as $h \to 0, e^h-1 \to 1-1=0$ hence $y \to 0$

If $y=e^h-1$ , then $h=ln(y+1)$

(3) $\begin{equation*}\lim_{\limits{y \to 0}}\frac{y}{ln(y+1)}\end{equation*}$

We are going to rewrite the equation $3$ as

(4) $\begin{equation*}\lim_{\limits{y \to 0}}\frac{\frac{1}{ln(y+1)}}{y}\end{equation*}$

And then we can write equation $4$ as

(5) $\begin{equation*}{\lim_{\limits{y \to 0}}{\frac{1}{\frac{1}{y}ln(y+1)}\end{equation*}$

Using log laws we can write equation $5$ as

(6) $\begin{equation*}\lim_{\limits{y \to 0}}\frac{1}{ln(y+1)^{\frac{1}{y}}}\end{equation*}$

Let $y=\frac{1}{n}$

As $y \to 0, n \to \infty$

equation $6$ becomes

(7) $\begin{equation*}\lim_{\limits{n \to \infty}}\frac{1}{ln(\frac{1}{n}+1)^n}\end{equation*}$

We can move the limit to inside the natural log

(8) $\begin{equation*}\frac{1}{ln(\lim_{\limits{n \to \infty}}(\frac{1}{n}+1)^n)}\end{equation*}$

And we know from the definition of $e$ that $e=\lim_{\limits{n \to \infty}}(1+\frac{1}{n})^n$

Hence, equation $8$ is

(9) $\begin{equation*}\frac{1}{ln(e)}=1\end{equation*}$

Back to our derivative

$\begin{equation*}f'(x)=e^x\lim_{\limits{h \to 0}}{\frac{e^h-1}{h}\end{equation}$

We know that $\lim_{\limits{h \to 0}}{\frac{e^h-1}{h}=1$ hence

$\begin{equation*}f'(x)=e^x\end{equation}$

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Filed under Calculus, Differentiation, Year 12 Mathematical Methods

Tagged as differentiating e

February 19, 2025 · 6:15 am

Finding a Recursive Rule for a Sequence (Year 12 Mathematics Applications)

How do we go about finding the rule for a first order linear recurrence relation?

Something like

$\begin{equation*}5, 7, 11, 19, ...\end{equation}$

There isn’t a common difference (arithmetic sequence) or a common ratio (geometric sequence). Sometimes you can just see the rule, but an algorithm will be handy.

Let’s say our relationship is

(1) $\begin{equation*}T_{n+1}=bT_n+c, T_1=a\end{equation*}$

Referring back to our sequence $5, 7, 11, 19, ...$ , we know

(2) $\begin{equation*}7=b\times5+c\end{equation*}$

and

(3) $\begin{equation*}11=b\times7+c\end{equation*}$

We can solve equation $2$ and $3$ simultaneously

equation $2-$ equation $3$

$\begin{equation*}-4=-2b\end{equation}$

Hence $b=2$

Substitute $b=2$ into equation $2$

$\begin{equation*}7=2\times 5+c\end{equation}$

$\begin{equation*}7=10+c\end{equation}$

Hence $c=-3$

$\begin{equation*}T_{n+1}=2T_n-3, T_1=5\end{equation}$

Let’s try to generalise

If $T_{n+1}=bT_n+c, T_1=a$ , then

(4) $\begin{equation*}T_2=bT_1+c\end{equation*}$

and

(5) $\begin{equation*}T_3=bT_2+c\end{equation*}$

Equation $5 -$ equation $4$

$\begin{equation*}T_3-T_2=(T_2-T_1)b\end{equation}$

$\begin{equation*} b=\frac{T_3-T_2}{T_2-T_1}\end{equation}$

Hence, $b=\frac{T_{n+2}-T_{n+1}}{T_{n+1}-T_n}$

Once you know $b$ , substitute into either equation to find $C$ .

Example

Find the recursive rule for the following

$-8, -12, -20, -36, ...$

$\begin{equation*}b=\frac{T_{n+2}-T_{n+1}}{T_{n+1}-T_n}=\frac{-20-(-12)}{-12-(-8)}=\frac{-8}{-4}=2\end{equation}$

$\begin{equation*}-12=2\times-8+c\end{equation}$

$\begin{equation*}-12=-16+c\end{equation}$

Hence $c=4$ and $T_{n+1}=2T_n+4, T_1=-8$

It is also possible to find the rule using a Classpad (if it’s in the calculator section} by using an e-activity.

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Filed under Sequences, Sequences, Year 12 Mathematics Applications

Tagged as first order linear sequences, recursive rules

February 15, 2025 · 8:35 am

Counting Techniques

Four teachers decide to swap desks at work. How many ways can this be done if no teacher sits at their previous desk?
Mathematics Specialist Units 1&2 Cambridge

I like this question as it seems easy until you start thinking about it. I think the best approach is a tree diagram.

If we think of the four teachers as A, B, C and D. Then A can no longer sit in A, so the options are B, C and D for the first desk.

For the second desk, If B is in the first desk, then A, C or D could be in the second. If C is in the first desk, then A or D could be in the second (B can’t be in the same desk). If D is in the first desk, then A or C can be in the second desk.

And so on, leaving 9 possibilities

BADC
BCDA
BDAC
CADB
CDAB
CDBA
DABC
DCAB
DCBA

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Filed under Counting Techniques, Tree Diagram, Year 11 Specialist Mathematics

Tagged as counting techniques, Year 11 Specialist Mathematics

February 4, 2025 · 9:17 am

Differentiating the Tangent Function

Remember $tan(x)=\frac{sin(x)}{cos(x)}$ .

I use the quotient rule to differentiate $f(x)=tan(x)$ .

(1) $\begin{equation*}\frac{d}{dx}(\frac{f(x)}{g(x)})=\frac{f'(x)g(x)-g'(x)f(x)}{[g(x)]^2}\end{equation*}$

If $h(x)=tan(x)=\frac{sin(x)}{cos(x)}$ then from equation $1$

(2) $\begin{equation*}h'(x)=\frac{cos(x)cos(x)-(-sin(x)sin(x))}{[cos(x)]^2}\end{equation*}$

(3) $\begin{equation*}h'(x)=\frac{cos^2(x)+sin^2(x)}{cos^2(x)}\end{equation*}$

Remember the Pythagorean identity

(4) $\begin{equation*}sin^2(x)+cos^2(x)=1\end{equation*}$

Hence

$\begin{equation*}h'(x)=\frac{1}{cos^2(x)}=sec^2(x)\end{equation}$

(5) $\begin{equation*}\frac{d}{dx}tan(x)=sec^2(x)\end{equation*}$

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Filed under Calculus, Differentiation, Differentiation, Identities, Quotient Rule, Trigonometry, Year 12 Mathematical Methods

Tagged as differentiating tan, differentiating trig

Expression	Expansion	Co-efficients
$(x+y)^2$	$x^2+2xy+y^2$	$1, 2, 1$
$(x+y)^3$	$x^3+3x^2y+3xy^2+y^3$	$1, 3, 3, 1$
$(x+y)^4$	$x^4+4x^3y+6x^2y^2+4xy^3+y^4$	$1, 4, 6, 4, 1$

Deriving the Logistic Growth Equation

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

Intersecting Secant Theorem

Circle Geometry Question 2

Circle Geometry Question

Binomial Expansion Theorem

Worked Examples

Derangements

Differentiating f(x)=e^x

Finding a Recursive Rule for a Sequence (Year 12 Mathematics Applications)

Example

Counting Techniques

Differentiating the Tangent Function

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