Analysing a Sinusoidal Model for Temperature Data
The temperature (T(t)) in degrees Celsius in a town over a 24-hour period is modelled by (T(t) = 10 + 8 \sin\left(\frac{\pi}{12}(t-4)\right)), where (t) is the time in hours after midnight, (0 \le t \le 24). A student claims that the maximum temperature occurs at 4 pm ((t=16)) and the minimum at 4 am ((t=4)). Another student argues that the model is incorrect because the maximum should be at 2 pm based on typical daily patterns. The teacher asks you to evaluate the model and the students' claims.
1. Determine the maximum and minimum temperatures predicted by the model, and the times at which they occur within the 24-hour period. Evaluate the first student's claim.
Read the worked answer
The function is (T(t) = 10 + 8\sin\left(\frac{\pi}{12}(t-4)\right)). The sine function oscillates between -1 and 1. Thus, maximum (T = 10+8=18) °C when (\sin(\cdot)=1), i.e., (\frac{\pi}{12}(t-4) = \frac{\pi}{2} + 2k\pi). Solving: (t-4 = 6 + 24k), so (t = 10 + 24k). Within (0\le t\le24), (k=0) gives (t=10) (10 am). Minimum (T = 10-8=2) °C when (\sin(\cdot)=-1), i.e., (\frac{\pi}{12}(t-4) = \frac{3\pi}{2} + 2k\pi), so (t-4 = 18 + 24k), (t = 22 + 24k). Within the interval, (t=22) (10 pm). So the maximum occurs at 10 am, minimum at 10 pm. The first student's claim that max at 4 pm and min at 4 am is incorrect. (Check: at (t=16), (T=10+8\sin(\pi)=10) not max; at (t=4), (T=10+8\sin(0)=10) not min.)
2. The second student suggests the maximum should be at 2 pm (\(t=14\)). Modify the model to have a maximum at 2 pm while keeping the same amplitude and mean, and write the new equation. Then, using the new model, find the times when the temperature is 14 °C.
Read the worked answer
To have a maximum at (t=14), we need the sine argument to equal (\frac{\pi}{2}) when (t=14). The general form is (T(t)=10+8\sin\left(\frac{\pi}{12}(t-h)\right)) where (h) is the horizontal shift. Setting (\frac{\pi}{12}(14-h)=\frac{\pi}{2}) gives (14-h=6), so (h=8). Thus new model: (T(t)=10+8\sin\left(\frac{\pi}{12}(t-8)\right)). Check: at (t=14), (\sin(\pi/2)=1), (T=18) max. Now solve (10+8\sin\left(\frac{\pi}{12}(t-8)\right)=14). Then (8\sin(\cdot)=4), (\sin(\cdot)=0.5). So (\frac{\pi}{12}(t-8)=\frac{\pi}{6}+2k\pi) or (\frac{5\pi}{6}+2k\pi). Solving first: (t-8=2+24k), (t=10+24k). Within (0\le t\le24), (t=10). Second: (t-8=10+24k), (t=18+24k), gives (t=18). So temperatures of 14 °C occur at 10 am and 6 pm.