Question

A Ferris wheel rotates such that the angle, \(\theta,\) of rotation is given by \(\theta=\frac{\pi t}{15},\) where \(t\) is the time, in seconds. A rider's height, \(h,\) in metres, above the ground can be modelled by \(h(\theta)=20 \sin \theta+22\) a) Write the equation of the rider's height in terms of time. b) Graph \(h(\theta)\) and \(h(t)\) on separate sets of axes. Compare the periods of the graphs.

Short answer

The equation in terms of time is \(h(t) = 20\sin\left(\frac{\pi t}{15}\right) + 22\). \(h(\theta)\) has a period of \(2\pi\) and \(h(t)\) has a period of 30 seconds.

Step-by-step solution

Understanding the given equations

The angle of rotation \(\theta\) is given by \(\theta=\frac{\pi t}{15}\). The height \(h\), in meters, above the ground is given by \(h(\theta)=20\sin \theta + 22\).

Substitute \(\theta=\frac{\pi t}{15}\)

We need to express \(h\) in terms of \(t\). Substitute \(\theta=\frac{\pi t}{15}\) into the height equation: \(h = 20 \sin \left( \frac{\pi t}{15} \right) + 22\).

Equation of rider's height in terms of time

After substitution, we get the equation: \(h(t) = 20 \sin \left( \frac{\pi t}{15} \right) + 22\).

Determine the period of \(h(\theta)\)

For the function \(h(\theta) = 20\sin \theta + 22\), the period of \(\sin \theta\) is \(2\pi\). Therefore, one complete cycle occurs when \(\theta\) changes from 0 to \(2\pi\).

Determine the period of \(h(t)\)

Since \(\theta = \frac{\pi t}{15}\), set \(\frac{\pi t}{15} = 2 \pi\) to find the period of \(t\). Solving this for \(t\), we get \(t = 30\) seconds. Therefore, the period of \(h(t) = 30\) seconds.

Graph \(h(\theta)\)

Graph \(h(\theta) = 20 \sin \theta + 22\). The period is \(2\pi\), the amplitude is 20, and the midline is \(y = 22\).

Graph \(h(t)\)

Graph \(h(t) = 20 \sin \left( \frac{\pi t}{15} \right) + 22\). The period is 30 seconds, the amplitude is 20, and the midline is \(y = 22\).

Compare the periods

The period of \(h(\theta)\) is \(2\pi\), while the period of \(h(t)\) is 30 seconds. They have different periods because \(\theta\) is a function of time \(t\).

Key concepts

Ferris Wheel Problem

Let's dive into the fascinating world of the Ferris Wheel problem! This type of problem is a fun and practical application of trigonometry.
Imagine a Ferris wheel rotating, and you are trying to model the height of a rider above the ground.
In our exercise, the angle of rotation \(\theta\) is given by \(\theta = \frac{\pi t}{15}\), where \(t\) is time in seconds.
The height of the rider is then modeled by the function \(h(\theta) = 20 \sin \theta + 22\).
The goal is to find how high the rider is at any given time \(t\).

Period of Sine Function

Understanding the period of the sine function is vital in this context.
The period of a basic sine function, \(\sin \theta\), is \(2\pi\). This means that the sine function completes one full cycle when \(\theta\) changes by \(2\pi\) radians.
In the context of our Ferris wheel problem, the function we're working with is \(h(\theta) = 20 \sin \theta + 22\).
This function also has a period of \(2\pi\). So, one complete cycle of the Ferris wheel's rotation corresponds to \(\theta\) changing from 0 to \(2\pi\).
This concept helps us understand how often the height function repeats itself as the wheel rotates.

Modeling Height with Trigonometry

Modeling the height of the rider involves converting the function of the angle to a function of time.
Initially, the height function is in terms of the angle: \(h(\theta) = 20 \sin \theta + 22\).
To express this in terms of time, we substitute \(\theta = \frac{\pi t}{15}\), resulting in \(h(t) = 20 \sin \left( \frac{\pi t}{15} \right) + 22\).
This equation tells us the rider's height \(h\) at any time \(t\).
This transformation helps us easily calculate the height at specific times and visualize the height changes over time.

Function Transformation

Transforming functions plays a crucial role in trigonometric modeling.
In our problem, the original function \(h(\theta) = 20 \sin \theta + 22\) is transformed into \(h(t) = 20 \sin \left( \frac{\pi t}{15} \right) + 22\).
This transformation involves replacing the variable \(\theta\) with \(\frac{\pi t}{15}\).
This changes the period of the sine function from \(2\pi\) to 30 seconds, as seen by solving \(\frac{\pi t}{15} = 2\pi\).
Understanding these transformations is key to adapting the sine function to various real-world scenarios where different inputs (like time) are involved.