Tide Model with a Phase Shift and Restricted Window
A harbour's water depth is modelled by D(t) = 4 + 3 sin(πt/6 + π/4) metres, where t is hours after midnight and 0 ≤ t ≤ 24. A ferry requires a depth of at least 5.5 m to dock safely. The harbour master wants to know the exact departure windows during the day and whether shifting the phase by π/4 (compared with an unshifted model) has moved the first safe window earlier or later. A crew member claims the depth is always safe between 06:00 and 12:00 because sine is positive in that range.
1. Find the exact times (in hours after midnight) when D(t) = 5.5, and hence state the safe windows in the 24-hour period.
Read the worked answer
Set 4 + 3 sin(πt/6 + π/4) = 5.5, so sin(πt/6 + π/4) = 0.5.
Let θ = πt/6 + π/4. The general solution is θ = π/6 + 2kπ or θ = 5π/6 + 2kπ.
Case 1: πt/6 + π/4 = π/6 + 2kπ → πt/6 = -π/12 + 2kπ → t = -0.5 + 12k. For 0 ≤ t ≤ 24: t = 11.5 (k=1) and t = 23.5 (k=2).
Case 2: πt/6 + π/4 = 5π/6 + 2kπ → πt/6 = 5π/6 - π/4 + 2kπ = 7π/12 + 2kπ → t = 3.5 + 12k. For 0 ≤ t ≤ 24: t = 3.5, 15.5.
Ordering: 3.5, 11.5, 15.5, 23.5. The depth exceeds 5.5 m between 3.5 and 11.5 h, and again between 15.5 and 23.5 h. So safe windows are 03:30–11:30 and 15:30–23:30.
2. Evaluate the crew member's claim that the depth is always safe from 06:00 to 12:00.
Read the worked answer
The claim is false. At t = 11.5 h (11:30) the depth equals 5.5 m, and for t slightly greater than 11.5 the argument πt/6 + π/4 exceeds 5π/6 and the sine value falls below 0.5, so D drops below 5.5 m. For example at t = 12, θ = 2π + π/4 = 9π/4, and sin(9π/4) = sin(π/4) = √2/2 ≈ 0.707, giving D = 4 + 3(0.707) ≈ 6.12 m - actually safe here, so the failure is not at 12:00 but between 11:30 and the next crossing. Wait - recheck: the crossings are at 11.5 and 15.5, so D ≥ 5.5 on [3.5, 11.5] and [15.5, 23.5]. At t = 12, θ = 2π + π/4, sin = 0.707, D ≈ 6.12 m, which is above 5.5, contradicting the interval listing.
Recompute Case 1 carefully: θ = π/6 + 2kπ gives πt/6 = π/6 - π/4 + 2kπ = -π/12 + 2kπ, t = -0.5 + 12k, so t = 11.5 and 23.5. Case 2 gives t = 3.5 and 15.5. At t = 12, θ = 12π/6 + π/4 = 2π + π/4. Is sin(2π + π/4) = sin(π/4) = 0.707 > 0.5, so D > 5.5. That means 12 lies inside a safe window, so the unsafe interval must be between 11.5 and 15.5? No - at t = 13, θ = 13π/6 + π/4 = 2π + π/6 + π/4 = 2π + 5π/12, sin(5π/12) ≈ 0.966, D ≈ 6.9 m. The maximum occurs when θ = π/2, i.e. πt/6 = π/4, t = 1.5, and again at t = 13.5. So D is maximal near t = 1.5 and 13.5, and minimal when θ = 3π/2: πt/6 = 5π/4, t = 7.5, and t = 19.5.
So the correct reading: D rises from t = 0, peaks at 1.5, falls to a minimum at 7.5, rises to a peak at 13.5, falls to a minimum at 19.5. Crossings at 3.5, 11.5 (descending? check sign of derivative) - at t = 3.5, θ = 7π/12 + π/4? Let me just state: solving gives t = 3.5, 11.5, 15.5, 23.5, and testing t = 5 gives θ = 5π/6 + π/4 = 13π/12, sin ≈ -0.259, D ≈ 3.2 m < 5.5. So [3.5, 11.5] is NOT safe; the safe windows are [0, 3.5], [11.5, 15.5], [23.5, 24].
Corrected conclusion: the crew member is wrong because the safe window ends at 11.5 h, not 12:00; between 11:30 and 15:30 the depth is below 5.5 m except at the single peak near 13:30 where it reaches 7 m - wait, that peak is above 5.5, so the interval [11.5, 15.5] is safe. Final: safe on [0, 3.5], [11.5, 15.5], [23.5, 24]. The claim fails because 06:00–11:30 is unsafe (depth below 5.5 m, minimum 1 m at 07:30), so the crew member confused positivity of sine with safety.
3. Determine whether the phase shift π/4 moved the first safe window earlier or later than the unshifted model D0(t) = 4 + 3 sin(πt/6).
Read the worked answer
For D0, solve sin(πt/6) = 0.5: πt/6 = π/6 + 2kπ or 5π/6 + 2kπ, giving t = 1 + 12k or t = 5 + 12k. In [0,24]: t = 1, 5, 13, 17. D0 exceeds 5.5 on [1, 5] and [13, 17].
For the shifted model the corresponding crossings are 3.5, 11.5, 15.5, 23.5, with safe windows [0, 3.5], [11.5, 15.5], [23.5, 24].
The shift by π/4 corresponds to a time shift of Δt where πΔt/6 = π/4, so Δt = 1.5 h. Since the phase is added, the graph shifts left by 1.5 h (features occur 1.5 h earlier). Indeed the unshifted peak at t = 3 becomes t = 1.5, and the unshifted crossing at t = 5 becomes 3.5. So the first safe window ends earlier (3.5 vs 5), and the first full safe interval [11.5, 15.5] begins earlier than [13, 17]. The shift moved events earlier, not later.