Question

The Canadian National Historic Windpower Centre, at Etzikom, Alberta, has various styles of windmills on display. The tip of the blade of one windmill reaches its minimum height of \(8 \mathrm{m}\) above the ground at a time of 2 s. Its maximum height is \(22 \mathrm{m}\) above the ground. The tip of the blade rotates 12 times per minute. a) Write a sine or a cosine function to model the rotation of the tip of the blade. b) What is the height of the tip of the blade after \(4 \mathrm{s} ?\) c) For how long is the tip of the blade above a height of \(17 \mathrm{m}\) in the first \(10 \mathrm{s} ?\)

Short answer

a) \( y(t) = 15 - 7 \cos\left( \frac{2\pi}{5} (t - 2) \right) \) b) 17.16 meters. c) Approximately 2.86 seconds.

Step-by-step solution

Determine Amplitude

The amplitude is half the difference between the maximum and minimum heights. \[ \text{Amplitude} = \frac{22 - 8}{2} = 7 \text{ meters} \]

Determine Midline

The midline is the average of the maximum and minimum heights.\[ \text{Midline} = \frac{22 + 8}{2} = 15 \text{ meters} \]

Determine Period

The period is the time it takes for the windmill blade to complete one rotation. Since the blade rotates 12 times per minute, the period in seconds is:\[ \text{Period} = \frac{60 \text{ seconds}}{12} = 5 \text{ seconds} \]

Determine Frequency

Frequency is the reciprocal of the period. \[ \text{Frequency} = \frac{1}{\text{Period}} = \frac{1}{5} \text{ cycles per second} \]

Determine Horizontal Shift

The minimum height is at 2 seconds. Since cosine starts at the maximum, a horizontal shift of 2 seconds is needed to match the given data. For a cosine function: \[ y = 15 + 7 \cos\left( \frac{2\pi}{5} (t - 2) \right) \]

Part (a) - Write the Function

The function to model the tip of the blade's height is: \[ y(t) = 15 - 7 \cos\left( \frac{2\pi}{5} (t - 2) \right) \]

Part (b) - Model at 4 seconds

To find the height at 4 seconds, substitute 4 for t in the function: \[ y(4) = 15 - 7 \cos\left( \frac{2\pi}{5} (4 - 2) \right) = 15 - 7 \cos\left( \frac{4\pi}{5} \right) \approx 15 - 7(-0.309) = 17.16 \text{ meters} \]

Part (c) - Solve Inequality for 10 second Interval

Find the time intervals when the height is above 17 meters. Solve \[ y > 17 \Rightarrow 15 - 7 \cos\left( \frac{2\pi}{5} (t - 2) \right) > 17 \Rightarrow -7 \cos\left( \frac{2\pi}{5} (t - 2) \right) > 2 \Rightarrow \cos\left( \frac{2\pi}{5} (t - 2) \right) < -\frac{2}{7} \]

Solve Cosine Inequality

\( \cos\left( \frac{2\pi}{5} (t - 2) \right) = -0.286 \text{ at } \frac{2\pi}{5} (t - 2) = \arccos(-0.286) \) and \(2\pi - \arccos(-0.286)\). This gives solution intervals in the first period from \(2s\) to \(7s\).

Determine Total Time Above 17m in First 10 seconds

Identify the total time span within the first 10 seconds where the blade's tip is above 17 meters. The interval within the first 10 seconds remains the same as one period matches exactly 10 seconds. Thus, calculate the total time spent in the intervals > 17 meters.

Key concepts

Sine and Cosine Functions

Sine and cosine functions are essential in trigonometry. They describe repetitive oscillations, like a windmill's spinning blades or waves in water. These functions are very useful in modeling real-world cyclical phenomena.

Both sine \( \text{sin}(x) \) and cosine \( \text{cos}(x) \) functions vary between -1 and 1. They repeat their values every \(2\text{π} \). This means after \(2\text{π} \) units along the x-axis, the function will start its cycle again. This is called the period of the function.

For a windmill's blade, we can use a cosine function to model its rotation because cosine starts at its maximum value. If the minimum height of the windmill occurs 2 seconds after the start, our cosine function will help accurately depict this motion with an appropriate horizontal shift.

Amplitude

Amplitude represents the height from the midpoint (or midline) of the sine or cosine wave to its peak. It's essentially the maximum deviation from the average value of the wave.

In our windmill example, the amplitude is calculated as half the difference between the maximum and minimum heights. This is determined as follows:

\( \text{Amplitude} = \frac{22 - 8}{2} = 7 \text{ meters} \)

This means the blade's tip oscillates 7 meters above and below the midline of the function.

Period

The period of a function is the time it takes to complete one full cycle. For our windmill, the blade completes 12 rotations every minute.

To find the period in seconds, use the formula:

\( \text{Period} = \frac{60 \text{ seconds}}{12} = 5 \text{ seconds} \)

This tells us every 5 seconds, the blade starts a new rotation.

Frequency

Frequency signifies how many cycles occur in a unit time interval, often seconds. It's the inverse of the period, represented as \( \text{frequency} = \frac{1}{\text{period}} \).

For our windmill blade, the period is 5 seconds, so the frequency is:

\( \text{Frequency} = \frac{1}{5} = 0.2 \text{ cycles per second} \)

This means the blade completes 0.2 rotations every second.

Horizontal Shift

A horizontal shift translates the function along the x-axis. This is useful when the starting point of the function is not at the origin. For a cosine function starting at a minimum, we adjust accordingly.

In our example, the minimum height of the blade occurs at 2 seconds. To shift the entire function by 2 seconds, we use a horizontal shift. The equation for our windmill blade height then becomes:

\( y(t) = 15 - 7 \text{cos}\bigg(\frac{2\text{π}}{5}(t - 2)\bigg) \)

This ensures our model accurately depicts the windmill blade's height changes over time.