2024 Paper 1, Q19
A permanent ice glacier is in a valley in New Zealand.
Due to the temperature changes of the seasons each year, the glacier expands for six months and recedes for six months. The changing distance of a point on the front edge of the glacier to a car park can be modelled by a sine function.
During the colder months, when the glacier expands, the front edge of the glacier moves to within 270 m of the car park. However, in the warmer months, when the glacier recedes, the front edge moves to a maximum distance of 280 m away from the car park.
The erosion effects of the glacier on the ground are of most interest to geologists when the absolute value of the acceleration of the front edge is greater than $\frac{5\sqrt{3}\pi^2}{72}$ metres/month$^2$. During these times, a team of geologists sets up a camp site nearby to perform field work. Whenever the acceleration is less than this, the geologists leave camp.
The geologists will spend a total of between seven and eight months at the camp site each calendar year.
Evaluate the reasonableness of this claim.
A year of the glacier moving in and out
The front edge slides between 270 m and 280 m from the car park and back, once a year. The geologists’ camp appears only in the four short windows where the edge is changing speed fastest.
one loop = one calendar year, then a short hold · the distance strip along the top is magnified, the 268 m gap is not to scale
Camp is decided by the second derivative
The position changes smoothly and slowly, and the acceleration is a sine wave of its own. The threshold $\frac{5\sqrt3\pi^2}{72}$ is exactly $\frac{\sqrt3}{2}$ of the peak. That is why the answer comes out in whole months.
$g(t)=5\sin\frac{\pi t}{6}+275$
$g''(t)=-\frac{5\pi^2}{36}\sin\frac{\pi t}{6}$
QCAA marking guide · 6 marks
Start from a general sine function $g(t)=A\sin\!\left(B(t+C)\right)+D$ and read each parameter from the words of the question. The amplitude is $A=\frac{280-270}{2}=5$. The period is 12 months, so $B=\frac{2\pi}{12}=\frac{\pi}{6}$. There is no phase shift, so $C=0$. The midline is $D=275$, halfway between 270 and 280.
$$g(t)=5\sin\!\left(\frac{\pi t}{6}\right)+275$$
Marker: correctly determines the sine function that models the front edge of the glacier position.
$$g'(t)=\frac{5\pi}{6}\cos\!\left(\frac{\pi t}{6}\right)$$
Marker: determines the velocity of the front edge of the glacier. Follow-through marks are allowed for errors in prior working.
$$g''(t)=-\frac{5\pi^2}{36}\sin\!\left(\frac{\pi t}{6}\right)$$
Marker: determines the acceleration of the front edge of the glacier (second derivative). Equivalent forms accepted, e.g. $-\frac{5}{36}\pi^2\sin\frac{\pi t}{6}$.
Equate the acceleration to the threshold. Because it is an absolute value, both signs matter. Taking the negative branch first:
$$-\frac{5\pi^2}{36}\sin\!\left(\frac{\pi t}{6}\right)=-\frac{5\sqrt3\pi^2}{72}\ \Rightarrow\ \sin\!\left(\frac{\pi t}{6}\right)=\frac{\sqrt3}{2}$$
$\frac{\pi t}{6}=\frac{\pi}{3}$ or $\frac{2\pi}{3}$, so $t=2$ or $t=4$, a two month window.
Marker: determines a period of time in a year where the acceleration is greater than the value given. An alternative reading of the period (months 8, 9 and 10 as three months) is accepted.
Now the positive branch, which catches the other half of the year:
$$\sin\!\left(\frac{\pi t}{6}\right)=-\frac{\sqrt3}{2}\ \Rightarrow\ \frac{\pi t}{6}=\frac{4\pi}{3}\ \text{or}\ \frac{5\pi}{3}\ \Rightarrow\ t=8\ \text{or}\ 10$$
That is a second two month window, so the total time above the threshold is $2+2=4$ months a year.
Marker: determines the sum of the two possible time periods when acceleration is greater than the absolute value.
4 months, not 7 to 8, so the claim is not reasonable
The geologists are at camp for about half as long as claimed.
Marker: evaluates the reasonableness of the claim, with the decision supported by the previous working.
Why the strange-looking threshold actually helps you
Peak acceleration is $\frac{5\pi^2}{36}$. The threshold $\frac{5\sqrt3\pi^2}{72}$ is that times $\frac{\sqrt3}{2}$, so it collapses to $\left|\sin\frac{\pi t}{6}\right|>\frac{\sqrt3}{2}$. An exact-value equation, solvable without a calculator.
Solving $\sin\frac{\pi t}{6}=+\frac{\sqrt3}{2}$ and stopping gives a single two-month window. And a claim that looks merely wrong rather than half right. The absolute value is what doubles it.
Look back at the acceleration panel. The two shaded windows are at months 2 to 4 and months 8 to 10. That is four months out of twelve, or a third of the year, when the claim needed two thirds.
The question never gives you a function. You have to pull four parameters out of three sentences, and “expands for six months and recedes for six months” is the only place the period is given. After that, the hidden step is differentiating twice. If you read “acceleration” as “how fast it is moving”, you solve the wrong inequality altogether. You also have to deal with the absolute value on both sides, or the four months become two. The QCAA noted that a lot of students forgot radians here, so I would check your calculator mode before anything else.
Question wording and marking-guide steps are from the 2024 QCAA Mathematical Methods external assessment, © State of Queensland (QCAA) 2024, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.