Question 12
A cockroach population is modelled by the function $P(t)=P_0\,e^{kt}$, where $P$ is the population after $t$ weeks and $k$ is a population constant. Initially, 100 cockroaches were counted. After three weeks, there were 120.
Determine the constant $k$.
[2 marks]Determine the function $P'(t)$.
[1 mark]Determine when the rate of population growth would reach 10 cockroaches per week.
[1 mark]To decrease the growth rate of the cockroach population, a pest control treatment was trialled.
The new function to model the cockroach population $t$ weeks after using the treatment is
$$I(t)=c\ln\left(t+8\right)+172,$$
where $c$ is a constant.
Three weeks after using the treatment, the rate of population growth was five cockroaches per week.
Determine the value of $c$.
[2 marks]Determine the cockroach population when the treatment began.
[1 mark]A rate of growth is a gradient
Watch two counts fix the first model, then a tangent slide along it until the growth rate is 10 per week. After the treatment, a gradient of 5 fixes the second model.
Whenever the question says “per week”, you are working with the derivative, not the population.
Is the number a population or a rate?
The top graph is the population. The bottom graph is its rate of growth, the gradient of the top one. Choose a part, then move the slider until the dashed target line in the bottom graph is met.
This is part c). The rate reaches 10 cockroaches per week at $t\approx8.19$ weeks. The population itself is about 165 then, not 10.
The population is growing, but the gradient is still under 10 cockroaches per week. Move later in time.
The gradient is already over 10 cockroaches per week here. It passed 10 earlier.
The population starts at 100 and only grows, so $P(t)$ never equals 10. “Rate” has to mean $P'(t)$.
Now $I'(3)=5$, so $c=55$. Part e) uses this same curve at $t=0$, the ringed point: about 286 cockroaches when the treatment began.
At $t=3$ the gradient is less than 5 cockroaches per week. Make $c$ bigger.
At $t=3$ the gradient is more than 5 cockroaches per week. Make $c$ smaller.
The 5 is a rate, so it goes with $I'(3)$. Setting $I(3)=5$ instead gives $c\approx-69.6$, and the population would fall.
Work through the solution one mark at a time
Try each step yourself before you reveal it. Keep the unrounded value of $k$ in your calculator until the last line.
“Initially” means $t=0$. Since $e^{0}=1$, the population at the start is just $P_0$.
$$P(0)=100\ \Rightarrow\ P_0=100$$
This mark can be implied by your next line, if you go straight to $P(t)=100e^{kt}$.
Marker: correctly identifies the value of $P_0$ in the model
“After three weeks, there were 120” gives $P(3)=120$. Substitute and solve for $k$.
$$\begin{aligned}120&=100e^{3k}\\k&=\frac{1}{3}\ln(1.2)\end{aligned}$$
On your calculator, put $120=100e^{3k}$ into the equation solver and solve for $k$. By hand, divide by 100 and take the natural log of both sides. The exact value $\frac{1}{3}\ln(1.2)$ is accepted, and so is a decimal to any number of places. Follow-through applies if $P_0$ was wrong.
Marker: determines the value of $k$
Differentiate. The derivative of $e^{kt}$ is $ke^{kt}$, so $k$ comes out the front and multiplies the 100.
$$\begin{aligned}P'(t)&=100e^{0.06077t}\\&\quad\times0.06077\end{aligned}$$
Leaving it with pronumerals, as $P'(t)=P_0ke^{kt}$ or $P'(t)=100ke^{kt}$, is also accepted, and so are decimals to any number of places.
Marker: determines the derivative function
“The rate of population growth” is $P'(t)$, not $P(t)$. Set it equal to 10 and solve for $t$ on your calculator. You can also graph $y=P'(t)$ and $y=10$ and find where they intersect.
$$10=6.077e^{0.06077t}$$
That answer uses the unrounded $k$. If you used $k=0.06077$ you get $t\approx8.1960$, which is also accepted. Any number of decimal places is accepted, rounded or truncated.
Marker: determines the time when the rate of population growth reached 10 cockroaches per week
The five cockroaches per week is a rate again, so you need $I'(t)$. The derivative of $\ln(t+8)$ is $\frac{1}{t+8}$, and the constant 172 disappears.
$$I'(t)=\frac{c}{t+8}$$
Marker: correctly differentiates the pest control function
“Three weeks after using the treatment” is $t=3$, and the rate there is 5. So $I'(3)=5$.
$$\begin{aligned}\frac{c}{3+8}&=5\\\frac{c}{11}&=5\end{aligned}$$
Follow-through applies if your derivative had a slip.
Marker: determines the value of $c$
In the new model, $t$ counts weeks after the treatment started, so “when the treatment began” is $t=0$. Substitute into $I(t)$ with $c=55$.
$$\begin{aligned}I(0)&=55\ln(8)+172\\&\approx286.3693\end{aligned}$$
The value to any number of decimal places is accepted. A whole number of cockroaches, as in the QCAA sample response, makes sense. Use $I(t)$ here, not $P(t)$. The new model has its own clock.
Marker: determines the population when the treatment began
Each phrase of the question gave you something
Every part is a routine you have practised. You find a constant from a given point, differentiate an exponential and a log, then solve or substitute. The calculator does the solving in parts a) and c).
If I had to guess where you might lose marks, it would be mixing up the population with its rate. “Cockroaches per week” is a rate, so it goes with $P'(t)$ in part c) and $I'(3)$ in part d). Solving $P(t)=10$ gives a negative time, and $I(3)=5$ gives a negative $c$. Either one is a sign you used the wrong function.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording and marking-guide steps are from the 2025 QCAA Mathematical Methods external assessment, © State of Queensland (QCAA) 2025, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.
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