Problem Statement
The temperature $T$ ($^\circ\text{C}$) in an alpine village over a $24$-hour period is modeled by $T(t) = a \cos(bt) + d$, where $t$ is the time in hours after midnight ($0 \le t \le 24$). The minimum temperature of $-2^\circ\text{C}$ occurs at $04:00$, and the maximum temperature of $16^\circ\text{C}$ occurs at $16:00$.
(a) Find the values of $a, b$, and $d$.
(b) Find the amount of time in the 24-hour period during which the temperature is above $12^\circ\text{C}$.
Verified Solution & Marking Scheme
Part (a): Find Parameters a, b, d
Amplitude $|a| = \frac{16 - (-2)}{2} = \frac{18}{2} = 9$.
Vertical shift $d = \frac{16 + (-2)}{2} = \frac{14}{2} = 7$.
Period $P = 2 \times (16 - 4) = 24\text{ hours} \implies b = \frac{2\pi}{24} = \frac{\pi}{12}$.
Since minimum occurs at $t = 4$ and maximum at $t = 16$:
$T(16) = a \cos\left(\frac{16\pi}{12}\right) + 7 = a \cos\left(\frac{4\pi}{3}\right) + 7 = a\left(-\frac{1}{2}\right) + 7 = 16 \implies -\frac{a}{2} = 9 \implies a = -9$
Thus $T(t) = -9 \cos\left(\frac{\pi t}{12}\right) + 7$ (or $9\sin$ equivalent with phase shift).
Part (b): Solve T(t) > 12
$-9 \cos\left(\frac{\pi t}{12}\right) + 7 = 12 \implies -9 \cos\left(\frac{\pi t}{12}\right) = 5 \implies \cos\left(\frac{\pi t}{12}\right) = -\frac{5}{9} \approx -0.5556$
$\frac{\pi t}{12} = \arccos(-5/9) \approx 2.1598 \implies t_1 = \frac{12 \times 2.1598}{\pi} \approx 8.25\text{ hours} \; (08:15)$
By symmetry around peak $t = 16$:
$t_2 = 24 - 8.25 = 15.75 + 8 = 23.75\text{ hours} \; (23:45)$
Duration above $12^\circ\text{C}$:
$\Delta t = t_2 - t_1 = 23.75 - 8.25 = 15.5\text{ hours}$