Question 10
The outside air temperature, $T$ (°C), on a particular day at a certain location is modelled by the function
$\displaystyle T=25-7\cos\left(\frac{\pi t}{12}\right),$$\displaystyle 0\le t\le24,$
where $t$ is the time (hours) since 6:00 am.
The rate of change of temperature at 11:30 am is
Not quite. 1.833 is $\frac{7\pi}{12}$, the fastest the temperature ever rises. That happens at noon, $t=6$, half an hour too late. You either used $t=6$ or dropped the sine factor. In 2025, 9.2% of Queensland students chose (A).
That’s it. Step through the hints below to see why. In 2025, 74.7% of Queensland students chose (B).
Not quite. That’s the rate at $t=11.3$. It comes from writing 11:30 as 11.3 and counting from midnight. Thirty minutes is 0.5 hours, and $t$ counts from 6:00 am. In 2025, 6.5% of Queensland students chose (C).
Close. That’s the rate at $t=11.5$, which counts from midnight. $t$ is the number of hours since 6:00 am, so 11:30 am is $t=5.5$. In 2025, 8.8% of Queensland students chose (D).
This is 11:30 am, 5.5 hours after 6:00 am. The temperature is rising at about 1.817 °C per hour.
This is noon, $t=6$. The tangent is at its steepest here, $\frac{7\pi}{12}\approx1.833$, option (A). That’s half an hour too late.
This is $t=11.3$, which is 5:18 pm. Its rate is 0.334, option (C). You get it by writing 11:30 as 11.3 and counting from midnight.
This is $t=11.5$, which is 5:30 pm. Its rate is 0.239, option (D). You get it by counting from midnight instead of 6:00 am.
The temperature rises from 6:00 am to 6:00 pm, then falls. The gradient of the orange tangent is the rate of change at that moment.
Turn the clock time into $t$. 11:30 am is 5 hours and 30 minutes after 6:00 am, and 30 minutes is half an hour.
$$t=5.5$$
Rate of change means differentiate. Use the chain rule. The derivative of $\cos$ is $-\sin$, which cancels the minus in front.
$\displaystyle \frac{dT}{dt}=-7\times\left(-\sin\left(\frac{\pi t}{12}\right)\right)$$\displaystyle \times\frac{\pi}{12}$$\displaystyle =\frac{7\pi}{12}\sin\left(\frac{\pi t}{12}\right)$
Substitute $t=5.5$ with your calculator in radian mode.
$\displaystyle \frac{dT}{dt}=\frac{7\pi}{12}\sin\left(\frac{5.5\pi}{12}\right)$$\displaystyle \approx1.817$
In degree mode you’d get about 0.046, which isn’t one of the options. That’s your sign to switch modes.
11:30 am is 5.5 hours after 6:00 am, so $t=5.5$.
$$\frac{dT}{dt}=\frac{7\pi}{12}\sin\left(\frac{\pi t}{12}\right)$$
$\displaystyle \left.\frac{dT}{dt}\right|_{t=5.5}=\frac{7\pi}{12}\sin\left(\frac{5.5\pi}{12}\right)$$\displaystyle \approx1.817\,{}^\circ\text{C per hour}$
How Queensland students answered in 2025
From the QCAA Mathematical Methods subject report for 2025. The percentages can add to less than 100 because some students didn’t answer.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording is from the 2025 QCAA Mathematical Methods external assessment, © State of Queensland (QCAA) 2025, licensed under CC BY 4.0, and has been adapted. Tangent Tuition is not affiliated with the QCAA.
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