2021 Paper 1, Q20
The population of rabbits $(P)$ on an island, in hundreds, is given by $P(t)=t^2\ln(3t)+6,\ t>0$, where $t$ is time in years.
Determine the intervals of time when the population is increasing and the intervals when it is decreasing.
The population dips briefly, then grows
For the first ten weeks the colony is shrinking. From then on it only grows. And by year three there are nearly two thousand rabbits on the island.
three years in twelve seconds, then a short hold · the arrow is the sign of $P'(t)$
A product of two factors, and only one can change sign
$P'(t)=t\left(1+2\ln 3t\right)$. Since $t>0$ always, the whole sign question is carried by the bracket, and the bracket is negative until $t=\frac{1}{3\sqrt{e}}$.
$P=t^2\ln(3t)+6$
$P'(t)=t\left(1+2\ln 3t\right)$
QCAA marking guide · 7 marks
$$P'(t)=t^2\times\frac{3}{3t}+2t\ln(3t)=t+2t\ln(3t)=t\left(1+2\ln 3t\right)$$
Marker: correctly determines $P'(t)$. Equivalent forms accepted, e.g. $t^2\times\frac{3}{3t}+2t\ln(3t)$.
Setting $P'(t)=0$ gives $t=0$ from the first factor, but $t=0$ is outside the domain, since $\ln(3t)$ needs $t>0$. It is rejected, and saying so earns a mark.
Marker: correctly determines the rejected solution for $t$.
$$1+2\ln(3t)=0\ \Rightarrow\ \ln(3t)=-\frac12\ \Rightarrow\ 3t=e^{-1/2}\ \Rightarrow\ t=\frac{1}{3\sqrt{e}}$$
Marker: determines the $t$-ordinate of the critical point. Equivalent values accepted, e.g. $\frac{e^{-0.5}}{3}$.
$$P''(t)=3+2\ln(3t),\qquad P''\!\left(\frac{1}{3\sqrt{e}}\right)=3+2\ln\!\left(\frac{1}{\sqrt{e}}\right)=3-1=2$$
Marker: determines the value of $P''(t)$ at the critical point.
$P''>0$, so the point is a minimum.
Marker: determines the nature of the critical point. Alternative methods of identifying the nature are accepted. A sign test on $P'$ either side does the same job.
Decreasing for $0<t<\frac{1}{3\sqrt{e}}$, increasing for $t>\frac{1}{3\sqrt{e}}$
Marker: communicates when the population is increasing and when it is decreasing. Equivalent statements accepted, e.g. the population decreases until time $\frac{1}{3\sqrt e}$ and then increases.
One of the seven marks is for the writing. Inequality signs used properly, derivative notation, the domain stated, and linking sentences between steps.
Marker: shows logical organisation communicating key steps, appropriate mathematical vocabulary, symbols and conventions, e.g. use of inequality signs, connecting statements, use of derivative notation, use of everyday language.
Two routes to the same interval
Solve $P'=0$, then test with $P''$. A minimum at $\frac{1}{3\sqrt e}$ means down before, up after.
$t\left(1+2\ln 3t\right)>0$ with $t>0$ reduces to $\ln(3t)>-\frac12$, giving $t>\frac{1}{3\sqrt e}$ in one line. Both are full-mark responses.
It is worth noticing how small the dip is. At the minimum the population is about 5.98 hundred, only about 2 rabbits below where it started, which is why the turning point is so easy to miss on a graph. The question asks about the function, and the function decreases and then increases.
The word “intervals”, plural, is the only hint that anything changes direction. The turning point sits at $t=0.202$, hidden in the first ten weeks where the curve looks flat. The hidden work is the domain. $t=0$ solves $P'=0$, but $\ln(3t)$ does not allow it, and the QCAA give a whole mark for rejecting it. There is a second hidden mark for communication. This is one of the rare questions where a correct answer written carelessly does not get full marks, so I would take the time to set your working out clearly.
Question wording and marking-guide steps are from the 2021 QCAA Mathematical Methods external assessment, © State of Queensland (QCAA) 2021, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.