Modelling with Trigonometric Functions (AQA A-Level Mathematics): Revision Notes

• Amplitude $\ A$: Find the maximum and minimum values of $\ y(t)$. The amplitude is $\ A = \frac{\text{max} - \text{min}}{2} .$
• Period $\ T$: Identify how long it takes for one complete cycle. Calculate$\ B$ using $\ B = \frac{2\pi}{T} .$
• Phase Shift $\ C$: Determine where the cycle starts. If it's shifted from the standard position, find $\ C$ using the known start of the cycle.
• Vertical Shift $\ D$: Calculate the average of the maximum and minimum values to find $\ D = \frac{\text{max} + \text{min}}{2} .$

Example 1: Modelling the Height of a Ferris Wheel

• Problem: A Ferris wheel with a radius of 10 metres completes one full rotation every 30 seconds. The bottom of the wheel is 2 metres above the ground. Model the height $\ h(t)$ of a person on the wheel as a function of time $\ t$.

• Solution:
• Amplitude $\ A$: The radius of the Ferris wheel is the amplitude, so $\ A = 10$ metres.
• Period$\ T$: The wheel completes one full rotation every $30$ seconds, so $\ T = 30$ seconds. Calculate $B$: $B = \frac{2\pi}{T} = \frac{2\pi}{30} = \frac{\pi}{15}$
• Phase Shift $C$: If we assume the person starts at the bottom of the wheel, we use a cosine function with a phase shift of $\ \pi$ (since cosine starts at the maximum, and the person is at the minimum): $C = \pi$
• Vertical Shift $\ D$: The centre of the Ferris wheel is 12 metres above the ground ($10$ metres of radius plus $2$ metres from the ground). So,$\ D = 12$ metres.
• Model: $h(t) = 10 \cos\left(\frac{\pi}{15} t + \pi\right) + 12$
• Interpretation: This model describes the height of the person above the ground as they rotate on the Ferris wheel.

Example 2: Modelling Daylight Hours

• Problem: The number of daylight hours in a city varies throughout the year, with a minimum of 8 hours in winter and a maximum of 16 hours in summer. Model the daylight hours $\ H(t)$ as a function of time $\ t$ in months, where $\ t = 0$ represents January.

• Solution:
• Amplitude $\ A$: The variation in daylight hours is $\ 16 - 8 = 8$ hours, so the amplitude is $\ A = 4$ hours.
• Period $\ T$: The daylight cycle repeats annually, so $\ T = 12$ months. Calculate $\ B :$ $B = \frac{2\pi}{12} = \frac{\pi}{6}$
• Phase Shift $\ C$: If the maximum daylight occurs in June (month 6), and we assume $t = 0$ is January, the phase shift is $\ C = -\frac{\pi}{2} \ (since \ ( \cos(\frac{\pi}{6} \cdot 6) = 1 )$ occurs at $\ ( t = 6 )$.
• Vertical Shift $D$: The average daylight is $\ D = \frac{16 + 8}{2} = 12$ hours.
• Model: $H(t) = 4 \cos\left(\frac{\pi}{6} t - \frac{\pi}{2}\right) + 12$
• Interpretation: This model describes the variation in daylight hours over the course of the year.

• Modelling with trigonometric functions allows you to represent periodic phenomena in various fields.
• By identifying key parameters such as amplitude, period, phase shift, and vertical shift, you can construct a mathematical model that describes the behaviour of the system over time.
• These models are not only useful for prediction but also for understanding the underlying mechanics of periodic systems.