Solving a Trigonometric Equation on a Restricted Domain
The height of a point on a Ferris wheel above the ground is h(t) = 12 + 10 sin(πt/15), where t is in seconds and 0 ≤ t ≤ 60. A maintenance gate can only be opened when the point is at least 17 m above the ground, and the gate must stay open for a continuous interval. An operator solves sin(πt/15) = 0.5 and gets t = 2.5 only, then claims the gate can open just once. You must find all times in [0, 60] when h(t) ≥ 17 and identify the continuous interval(s).
1. Solve h(t) = 17 for all t in [0, 60].
Read the worked answer
Set 12 + 10 sin(πt/15) = 17, so sin(πt/15) = 0.5. Let θ = πt/15. The domain 0 ≤ t ≤ 60 gives 0 ≤ θ ≤ 4π. Solutions of sin θ = 0.5 in [0, 4π]: θ = π/6, 5π/6, π/6 + 2π = 13π/6, 5π/6 + 2π = 17π/6. Convert: t = 15θ/π. So t = 15/6 = 2.5; t = 75/6 = 12.5; t = 15(13)/6 = 32.5; t = 15(17)/6 = 42.5. The operator missed the second and later solutions because the sine equation has two solutions per period and the domain spans more than one period.
2. Determine the interval(s) in [0, 60] where the gate may be open (h ≥ 17).
Read the worked answer
Test a point between 2.5 and 12.5, e.g. t = 7.5: θ = π/2, sin = 1, h = 22 ≥ 17. So h ≥ 17 on [2.5, 12.5]. Between 12.5 and 32.5, e.g. t = 22.5: θ = 1.5π, sin = −1, h = 2 < 17. Not open. Between 32.5 and 42.5, e.g. t = 37.5: θ = 2.5π, sin = 1, h = 22 ≥ 17. Open on [32.5, 42.5]. After 42.5, e.g. t = 50: θ = 10π/3 ≈ 3.33π, sin negative, h < 17. So the gate can open on two intervals: [2.5, 12.5] and [32.5, 42.5]. The judgement is using periodicity and sign testing rather than trusting a single principal solution.