Questions on trigonometric modelling

Question 1

Write down the amplitude and period of y = 3 sin 2t, with t in seconds.

Accepted Answer

The amplitude is 3 and the period is 2π/2 = π seconds. B1 for the amplitude, B1 for the period. Quoting the period as 2 is the reflex to resist; the coefficient divides into 2π rather than being the answer.

Question 2

A model gives h = 5 + 2 cos t. Write down the maximum and minimum values of h.

Accepted Answer

The cosine swings between ±1, so h runs from 5 − 2 = 3 up to 5 + 2 = 7. B1 for the maximum, B1 for the minimum. The 5 sets the centre line and the 2 the swing about it.

Question 3

The depth of water at a jetty is modelled by d = 5 + 2.4 sin (πt/6), with d in metres and t in hours after midnight. Write down the greatest depth, and find the number of hours between one high water and the next.

Accepted Answer

The sine swings between ±1, so the depth runs from 5 − 2.4 = 2.6 m up to 5 + 2.4 = 7.4 m. B1 for 7.4 m. High water comes once per period, and the period is 2π ÷ (π/6) = 12 hours. B1 for 12 hours.

Question 4

An alternating current is modelled by I = 15 sin (100πt), with I in amperes and t in seconds. Write down the amplitude, and find the period and the number of complete cycles per second.

Accepted Answer

The amplitude is 15 A, the greatest size the current reaches in either direction. B1 for 15 A. The period is 2π ÷ 100π = 0.02 s. B1 for 0.02 s. One cycle every 0.02 s means 1 ÷ 0.02 = 50 cycles per second. B1 for 50.

Question 5

A wheel's seat height is modelled by h = 10 − 8 cos (πt/20), in metres and seconds. Write down the height of the axle and the radius of the wheel, and find the maximum seat height and the period of the ride.

Accepted Answer

The axle sits on the centre line, 10 m up, and the radius is the amplitude, 8 m. The seat tops out at 10 + 8 = 18 m and bottoms at 2 m. The period is 2π ÷ (π/20) = 40 s. B1 for the axle height, B1 for the radius, B1 for the maximum height, B1 for the period. The minus sign starts the seat at the bottom of the wheel at t = 0, where a ride starts.

Question 6

A wheel's seat height is modelled by h = 10 − 8 cos (πt/20), in metres and seconds. Find the first time the seat is 14 m high.

Accepted Answer

Setting 10 − 8 cos (πt/20) = 14 gives cos (πt/20) = −½. The first positive angle with that cosine is 2π/3, so πt/20 = 2π/3 and t = 40/3 = 13.3 s to 3 significant figures. M1 for cos (πt/20) = −½, M1 for πt/20 = 2π/3, A1 for 13.3 s, B1 for taking the first crossing. Later crossings exist at 4π/3 and beyond, but the question asks for the first. Keep the calculator in radians throughout.

Question 7

A buoy's height above mean sea level is y = 0.9 sin 0.6t, in metres and seconds. Find the amplitude, the period, and the first positive time at which the buoy is at mean sea level.

Accepted Answer

The amplitude is 0.9 m and the period is 2π/0.6 = 10.5 s. Mean sea level means y = 0. The buoy starts there at t = 0, and the sine next vanishes when 0.6t = π, so t = 5.24 s, half a period later. B1 for the amplitude, B1 for the period, M1 for 0.6t = π, A1 for 5.24 s.

Question 8

The temperature in a greenhouse is modelled by T = 20 − 8 cos (πt/12), with T in °C and t hours after 6 a.m., for 0 ≤ t ≤ 18. Find the time of day at which the model gives the greatest temperature, and find how long the model takes to rise from its lowest temperature to 26 °C.

Accepted Answer

T is greatest when cos (πt/12) is at its least, −1, so πt/12 = π. M1 for setting the cosine to −1. That gives t = 12, twelve hours after 6 a.m., so the warmest hour is 6 p.m., and it lies inside 0 ≤ t ≤ 18. A1 for 6 p.m.
The lowest temperature is at t = 0, where cos 0 = 1 and T = 12 °C, so the rise starts at 6 a.m. Setting 20 − 8 cos (πt/12) = 26 gives cos (πt/12) = −6/8 = −3/4, and the first angle with that cosine is 2.4189 radians. M1 for cos (πt/12) = −3/4 and the angle in radians. Then t = (12/π)(2.4189) = 9.2394, so the rise takes 9.24 hours to 3 significant figures, about 9 hours 14 minutes. A1 for 9.24 hours. The cosine here is negative, so the angle is obtuse and the calculator gives it directly; the model comes back down through 26 °C at t = 24 − 9.24 = 14.76, which is the symmetry of the curve about its warmest hour rather than a second calculation.

Question 9

A piston moves so that its displacement x cm from the centre of its travel is modelled by x = a cos (bt), with t in seconds. The displacement varies between −6 cm and 6 cm, the piston completes 20 cycles every second, and x = −6 when t = 0. Find a and b, and find the displacement when t = 0.05.

Accepted Answer

The displacement swings 6 cm either side of the centre, so the amplitude is 6 and a is 6 or −6. At t = 0 the cosine is 1, and the displacement there is −6, so a = −6. B1 for a = −6. Twenty cycles a second means a period of 1/20 = 0.05 s, and the period is 2π/b, so b = 2π ÷ 0.05 = 40π. M1 for 2π divided by the period, A1 for 40π. At t = 0.05 the angle is 40π × 0.05 = 2π, so x = −6 cos 2π = −6 cm. A1 for −6 cm. That is where the piston started, as it has to be: 0.05 s is exactly one complete cycle. Reading 20 cycles per second as the value of b, rather than as the reciprocal of the period, would put the model out by a factor of 2π.

Question 10

A child on a swing has horizontal displacement x metres from the lowest point of the swing modelled by x = 0.6 cos (πt/2), with t in seconds. Find the period, the displacement when t = 1, and the total distance the swing travels in one complete oscillation.

Accepted Answer

The period is 2π ÷ (π/2) = 4 s. B1 for 4 s. At t = 1 the angle is π/2 and cos (π/2) = 0, so x = 0 m: the swing is passing through its lowest point. B1 for 0 m. In one complete oscillation the swing goes from x = 0.6 across to x = −0.6 and back again, covering the 1.2 m span twice, so the distance travelled is 4 × 0.6 = 2.4 m. M1 for four amplitudes, A1 for 2.4 m. Distance travelled and displacement are different quantities: after a full cycle the displacement has changed by nothing at all, while the distance travelled is 2.4 m.

Question 11

The number of adult birds on an island is modelled by P = 1200 + 300 sin (πt/6), with t in months after 1 January. Write down the greatest and least numbers the model gives, find the two values of t in 0 ≤ t < 12 at which P = 1350, and give one reason the model could not describe a population that is growing overall.

Accepted Answer

The sine swings between ±1, so P runs from 1200 − 300 = 900 up to 1200 + 300 = 1500. B1 for both. Setting P = 1350 gives sin (πt/6) = 150/300 = ½. M1 for sin (πt/6) = ½. As t runs over 0 ≤ t < 12 the angle πt/6 runs over 0 ≤ πt/6 < 2π, and the sine is ½ at π/6 and 5π/6, so t = 1 and t = 5. A1 for t = 1, A1 for t = 5. The model repeats exactly every 12 months and returns to 1200 at the start of every year, so it can carry a seasonal rise and fall but no trend at all. A population growing overall would need a term that increases with t, a straight line or an exponential added to the sine. B1 for the reason.

Question 12

A tide gauge fits the data y = 2 sin 0.5t + 1.5 cos 0.5t, in metres and hours. Express y as a single sine wave, and find the height of high tide and the first time it occurs.

Accepted Answer

R = √(4 + 2.25) = 2.5 and tan α = 1.5/2 gives α = 0.6435, so y = 2.5 sin (0.5t + 0.6435). High tide is 2.5 m, where the sine reaches 1: 0.5t + 0.6435 = π/2, so t = 1.85 hours. M1 for the harmonic form, A1 for R = 2.5, M1 for tan α, A1 for α = 0.6435, B1 for the high tide, A1 for t = 1.85 hours.

Question 13

A student claims that doubling the amplitude of a periodic model makes it oscillate twice as fast. Untangle the claim.

Accepted Answer

Amplitude and period are separate dials. Doubling the amplitude doubles how far the oscillation swings and leaves its timing alone; oscillating faster needs a bigger coefficient inside the bracket, which shortens the period. B1 for amplitude and period being independent, B1 for what does change the speed.

Question 14

The number of customers in a cafe during the hour beginning at time t is modelled by N = 40 + 25 sin (π(t − 8)/6), where t is the number of hours after midnight and 6 ≤ t ≤ 20. Find the time at which the model gives the greatest number of customers, and that number. Find also the values of t between which N exceeds 55, and give one reason the model should not be used outside 6 ≤ t ≤ 20.

Accepted Answer

N is greatest when the sine equals 1, that is when π(t − 8)/6 = π/2, so t − 8 = 3. M1 for setting the angle to π/2. That gives t = 11, the hour beginning at 11 a.m., which lies inside the stated range. A1 for t = 11. The number there is 40 + 25 = 65 customers. B1 for 65. For N to exceed 55 the sine must exceed 15/25 = 0.6. The sine first reaches 0.6 at 0.6435 radians and last holds it at π − 0.6435 = 2.4981 radians, so π(t − 8)/6 must lie strictly between those two. M1 for both angles. Then t − 8 = (6/π)(0.6435) = 1.229 and t − 8 = (6/π)(2.4981) = 4.771, so N exceeds 55 for 9.23 < t < 12.77, a stretch of about three and a half hours. A1 for the interval. Outside the stated range the model keeps rising and falling with a period of 12 hours, so it would put 65 customers in the cafe at 11 p.m. as well; a cafe that closes overnight is not described by a function that never stops oscillating. B1 for the reason. The inequality is strict at both ends because N equals 55 exactly at 9.23 and again at 12.77 rather than exceeding it there.

Question 15

Two models are proposed for the depth of water d metres at a harbour, with t hours after midnight: model P gives d = 5.5 + 2.9 cos (π(t − 3)/6), and model Q gives d = 5.5 + 2.9 cos (πt/6). Find the first time of high water under each model. High water is observed at 03:00. State with a reason which model is consistent with that observation, and use it to find the depth at 06:00.

Accepted Answer

Either model is greatest when its cosine equals 1, and the first time that happens is when the angle inside is zero. M1 for setting each angle to zero. For model P that needs π(t − 3)/6 = 0, so t = 3, high water at 03:00; for model Q it needs πt/6 = 0, so t = 0, high water at midnight. A1 for both times. The observation is high water at 03:00, so model P is the consistent one; model Q has the same mean depth, the same amplitude and the same period, and is wrong only in its phase, by three hours. B1 for choosing P with the reason. At 06:00, t = 6, and model P gives d = 5.5 + 2.9 cos (3π/6) = 5.5 + 2.9 cos (π/2) = 5.5 m. M1 for substituting t = 6 into model P, A1 for 5.5 m.