2022 Paper 2, Q19
Flying foxes enter and leave a fruit-growing region every evening. The rate at which the flying foxes enter the region is modelled by the function
$$A(t)=42\sin\!\left(0.03t-\frac{\pi}{3}\right)+71,\quad 0\le t\le 240$$
The rate at which the flying foxes leave the region is modelled by the function
$$L(t)=42\sin\!\left(0.04t-\frac{\pi}{3}\right)+42,\quad 0\le t\le 240$$
Both $A(t)$ and $L(t)$ are measured in animals per minute and $t$ is measured in minutes after 7 pm. There are 100 flying foxes in the region at 7 pm.
Determine the maximum number of flying foxes in the region and the time that this occurs.
Four hours of flying foxes coming and going
Neither rate is ever zero for long, because foxes arrive and leave all evening. The colony inside peaks at the one moment when the two flows are equal.
7 pm to 11 pm in twelve seconds, then a short hold · band thickness is the rate, orange level is the population
The peak is where the two rates cross
$N(t)$ is the integral of $A-L$, so $N'=A-L$. The population turns over at the single moment $A(t)=L(t)$. Visible on the left panel as a crossing, on the right as a summit.
black: arriving · blue: leaving
one turning point in the whole evening
QCAA marking guide · 4 marks
Integrate each rate, then subtract. The number inside is what came in minus what went out:
$$\int A(t)\,dt=-1400\cos\!\left(0.03t-\frac{\pi}{3}\right)+71t+c_1$$
$$\int L(t)\,dt=-1050\cos\!\left(0.04t-\frac{\pi}{3}\right)+42t+c_2$$
With $N(0)=100$, the combined constant is 275:
$$N(t)=-1400\cos\!\left(0.03t-\frac{\pi}{3}\right)+1050\cos\!\left(0.04t-\frac{\pi}{3}\right)+29t+275$$
Marker: correctly determines the function to model the total number of flying foxes in the region.
Sketch $N(t)$ on the calculator over $0\le t\le 240$ and find the turning point. Equivalently, solve $N'(t)=A(t)-L(t)=0$. The moment the two flows balance.
Marker: uses an appropriate mathematical method to identify the maximum value. Equivalent explanations, e.g. sketching, are accepted.
$$\text{maximum at }(177.729,\ 7034.264)$$
Marker: determines the maximum number of flying foxes and time, i.e. the coordinates of the turning point.
7034 flying foxes, at 9:58 pm
177.729 minutes after 7 pm is 2 hours 57.7 minutes, so 9:58 pm to the nearest minute. Foxes are counted as whole animals.
Marker: states the number of flying foxes as a whole number and the time after 7 pm it occurs.
Two rates, and the one turning point that matters
Not where $A$ peaks (about $t=87$), and not where $L$ bottoms out (about $t=144$). It is where $A-L=0$, at $t=177.7$. The population keeps rising as long as arrivals merely exceed departures.
“There are 100 flying foxes at 7 pm” fixes the constant. Drop it and every value of $N$ is 100 too low. The peak becomes 6934 and the marks for the model go with it.
The two constants of integration also combine into one, because only their difference matters. That is why one condition is enough to find the whole function.
You are given rates, but you are asked for a number, so your work runs through a function the stem never mentions. Building $N(t)$ from $A-L$, with the starting 100 included, is the hidden step, and every mark after it depends on it. The other trap is the words “maximum number”. I see students look for the maximum of the functions they were given, when the maximum they need belongs to a function they have to build themselves. The QCAA called this the most difficult question on the 2022 exam, so do not be discouraged if it takes you a few goes.
Question wording and marking-guide steps are from the 2022 QCAA Mathematical Methods external assessment, © State of Queensland (QCAA) 2022, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.