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IB Mathematics SL 2.4 Determine key features of graphs AI HL Paper 2- Exam Style Questions- New Syllabus

IBDP Maths AI HL Paper 2 - All Topics

Question

The tidal height, \(h\), in metres, at a coastal harbour is modeled by the periodic function \(h(t) = -2.5 \cos(bt^\circ) + 4.5\), where \(t\) denotes the number of hours elapsed since midnight (\(0 \le t \le 24\)) and \(b\) is a constant. The graph of the function is provided below, showing minimum heights at midnight and midday.
Graph of the tidal height function
(a) Show that the value of the constant \(b\) is \(30\).
(b) Calculate the predicted tidal height at \(5:00\) am (\(t = 5\)).
(c) State:
  (i) the amplitude of the function \(h\).
  (ii) the equation of the principal axis.
Safety protocols indicate that vessels may only enter or exit the harbour when the water level is above \(2.65\) m. A local sailor, Finn, is aware that atmospheric pressure changes can cause the actual water level to deviate from the model by up to \(10\) cm in either direction.
Finn plans to depart for a fishing trip as soon as the conditions permit after \(12:00\) pm (noon).
(d) Determine the earliest possible time, to the nearest minute, that the water level could reach the required threshold for Finn to leave.
The transit between the harbour and the fishing grounds takes \(15\) minutes. Finn intends to return later that same day and must be certain that the tide will be high enough for safe re-entry.
(e) Calculate the maximum duration, in hours, that Finn can spend at the fishing site while ensuring a guaranteed safe return to the harbour.

Most-appropriate topic codes:

• SL 2.5: Sinusoidal models (amplitude, period, principal axis) — parts (a), (b), (c)
• SL 2.4: Finding the point of intersection and key features using technology — parts (d), (e)
▶️ Answer/Explanation
Detailed solution

(a)
From the graph/context, the tide completes a full cycle (low to low) in 12 hours (e.g., \(t=0\) to \(t=12\)).
Period \(T = 12\). The angular frequency \(b = \frac{360}{T}\) (since the angle is in degrees).
\(b = \frac{360}{12} = \mathbf{30}\) (AG).

(b)
Substitute \(t=5\) into the equation:
\(h(5) = -2.5 \cos(30 \times 5) + 4.5 = -2.5 \cos(150^\circ) + 4.5\).
Using GDC: \(h(5) \approx \mathbf{6.67 \text{ m}}\) (6.665…).

(c)
(i) Amplitude is the coefficient of the cosine term (absolute value). \(\mathbf{2.5 \text{ m}}\).
(ii) Principal axis is the vertical shift. \(\mathbf{h = 4.5}\).

(d)
Robin needs \(h > 2.65\).
To be possible to leave, he considers the best-case error (tide is 10cm higher than predicted).
So, predicted height \(h(t)\) only needs to be \(2.65 – 0.10 = 2.55\) m.
Solve \(-2.5 \cos(30t) + 4.5 = 2.55\) for \(t > 12\).
\(-2.5 \cos(30t) = -1.95 \Rightarrow \cos(30t) = 0.78\).
\(30t = \arccos(0.78)\).
Using GDC intersection for \(t > 12\):
The tide is low at 12:00 (height \(4.5 – 2.5 = 2.0\)). It rises.
\(t \approx 13.291\dots\) hours.
\(0.291 \times 60 \approx 17\) minutes.
Time \(\approx\) 13:17.

(e)
To be certain to return, he considers the worst-case error (tide is 10cm lower than predicted).
So, predicted height \(h(t)\) must be \(2.65 + 0.10 = 2.75\) m.
He leaves at 13:17 (approx 13.29h). Travel takes 15 mins (0.25h).
Arrives fishing site at \(13.29 + 0.25 = 13.54\)h.
He must return before the tide drops below 2.75 m.
Solve \(-2.5 \cos(30t) + 4.5 = 2.75\) for the falling tide (later in the day).
Using GDC: \(t \approx 22.48\) hours (approx 22:29).
Return journey takes 0.25h, so must leave fishing site at \(22.48 – 0.25 = 22.23\)h.
Time available = \(22.23 – 13.54 = \mathbf{8.69 \text{ hours}}\).

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