2021 Paper 2, Q18
The number of animals in a population (in thousands) is modelled by the function $P$ such that $P(t)=\dfrac{100}{1+4e^{-t}}$, where $t$ is in years.
Determine the number of animals in the population when the population is growing the fastest.
The population keeps growing, but the growth rate does not
The population starts at twenty thousand animals and can never quite reach a hundred thousand. Watch the bar on the right. The growth speeds up, peaks early, and then fades for the rest of the decade.
eight years in twelve seconds, then a short hold · the bar is $P'(t)$, the question is about the bar
The steepest point on the left is the peak on the right
Drag the slider. The tangent on $P$ is steepest at exactly the $t$ where $P'$ tops out. And that is where $P''$ changes sign, the point of inflection.
$P(t)=\frac{100}{1+4e^{-t}}$
$P'(t)=\frac{400e^{-t}}{\left(1+4e^{-t}\right)^2}$
QCAA marking guide · 3 marks
“Growing the fastest” is a statement about $P'$, not $P$. The population grows fastest where $P'(t)$ is at its maximum, which is the same as where $P''(t)=0$, the point of inflection of $P$.
$$P'(t)=\frac{400e^{-t}}{\left(1+4e^{-t}\right)^{2}}$$
Marker: correctly identifies the conditions for the most rapid increase. This mark may be implied by subsequent working.
QCAA’s own response graphs $P$ (dotted) and $P'$ (solid) and reads off the maximum of $P'$ at $t=1.386$. Algebraically:
$$P''(t)=\frac{400e^{-t}\left(4e^{-t}-1\right)}{\left(1+4e^{-t}\right)^{3}}=0\ \Rightarrow\ 4e^{-t}=1\ \Rightarrow\ t=\ln 4\approx 1.386\ \text{years}$$
Marker: determines when the population is growing the fastest. Accept the alternative approach of setting up and solving $P''(t)=0$; accept alternative rounding; FT marks allowed.
$$P(\ln 4)=\frac{100}{1+4e^{-\ln 4}}=\frac{100}{1+1}=50$$
Approximately 50 000 animals
Marker: determines the population at this time. $P$ is measured in thousands, so $P=49.993$ on the calculator becomes 50 000 animals. A time in years is not an answer to this question.
Three ways to get the same answer
Paper 2: plot $P'(t)$, use the maximum feature, get $t=1.386$, then evaluate $P$ there. Two calculator moves earn three marks.
Solve $P''(t)=0$. The quotient collapses to $4e^{-t}=1$, so $t=\ln 4$ exactly. And $P(\ln 4)=50$ exactly.
Every logistic curve grows fastest at half its ceiling. Here the ceiling is 100, so the answer is 50 before you differentiate anything. Use it to check. The condition still has to be shown for the mark.
At $t=\ln 4$ the population is climbing at 25 000 a year. By year 8 it has slowed to under 500 a year and there are 99 866 animals. The model approaches 100 000 but never gets there.
This is only three marks, with no diagram and one line of stem, but it hides two steps. First, growing the fastest has to become the maximum of the derivative, which means differentiating twice a function most students have only ever been asked to substitute into. Second, the question asks for a number of animals, not a time. I see a lot of students stop at $t=1.386$, and that only earns two of the three marks. The last mark is for reading your own units, because $P$ is in thousands. The QCAA love a short question like this, so do not assume three marks means easy!
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.