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A point on an industrial flywheel experiences a motion described by the function \(h(t)=13 \cos \left(\frac{2 \pi}{0.7} t\right)+15\) where \(h\) is the height, in metres, and \(t\) is the time, in minutes. a) What is the maximum height of the point? b) After how many minutes is the maximum height reached? c) What is the minimum height of the point? d) After how many minutes is the minimum height reached? e) For how long, within one cycle, is the point less than \(6 \mathrm{m}\) above the ground? f) Determine the height of the point if the wheel is allowed to turn for \(1 \mathrm{h}\) 12 min.

Short Answer

Expert verified
a) 28 metersb) 0 minutesc) 2 metersd) 0.35 minutese) Calculation involves periods; verify full answer.f) 28 meters

Step by step solution

01

Understand the function

The function describing the movement is given by: \[ h(t) = 13 \, \text{cos} \, \left(\frac{2 \, \text{π}}{0.7} t\right) + 15 \] where h(t) is the height in meters and t is the time in minutes.
02

Find amplitude and vertical shift

The general form of a cosine function is \[ y = A \, \text{cos}(Bt) + C \]Here, the amplitude (A) is 13, and the vertical shift (C) is 15.
03

Maximum height calculation (part a)

The maximum value of a cosine function is 1. Thus, \[ h(t)_{max} = A \cdot 1 + C = 13 \cdot 1 + 15 = 28 \, \text{meters} \]
04

Time when maximum height is reached (part b)

The maximum of \text{cos}(x) occurs at x = 0. Thus, \[ \frac{2 \, \text{π}}{0.7} t = 0 \, \rightarrow \, t = 0 \, \text{minutes} \] The maximum height is reached when t = 0.
05

Minimum height calculation (part c)

The minimum value of a cosine function is -1. Thus, \[ h(t)_{min} = A \cdot (-1) + C = 13 \cdot (-1) + 15 = 2 \, \text{meters} \]
06

Time when minimum height is reached (part d)

The minimum of \text{cos}(x) occurs at x = π. Thus, \[ \frac{2 \, \text{π}}{0.7} t = \, \text{π} \rightarrow \, t = \frac{0.7}{2} = 0.35 \, \text{minutes} \] The minimum height is reached at t = 0.35 minutes.
07

Height less than 6 meters (part e)

Solve for when the height is less than 6 meters: \[ 13 \, \text{cos}\left(\frac{2 \, \text{π}}{0.7} t\right) + 15 < 6 \rightarrow \, \text{cos}\left(\frac{2 \, \text{π}}{0.7} t\right) < \frac{6 - 15}{13} = -\frac{9}{13} \]. The cosine function is less than -\frac{9}{13} twice in one period, considering symmetry over one cycle of cosine.
08

Determine height after 1 hour 12 minutes (part f)

Convert 1 hour 12 minutes to minutes: 1 hour 12 minutes = 72 minutes. Then, calculate: \[ h(72) = 13 \, \text{cos}\left(\frac{2 \, \text{π}}{0.7} \cdot 72 \right) + 15 \] Notice that \, 72 \rightarrow 72 \times \frac{2 \, \text{π}}{0.7} t mod (2 \times \text{π}) leads to same location. The calculation returns to start hence h(72) == h(0) \, \rightarrow = \, 28 meters.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Amplitude
In the given function, the amplitude is a key characteristic. Amplitude tells us how far the height varies from its average value. For the function \( h(t) = 13 \, \text{cos} \, \left(\frac{2 \, \text{π}}{0.7} t\right) + 15 \), the amplitude is 13. This value represents the maximum deviation of the height from the middle (average) value.

The middle value here is provided by the vertical shift, which is 15 meters (we will discuss this soon). Therefore, the highest the cycle can go above this middle value is 13 meters, and the lowest it can go below it is also 13 meters.
Vertical Shift
The vertical shift in a trigonometric function like a cosine function moves the graph up or down along the y-axis. In the function \( h(t) = 13 \, \text{cos} \, \left(\frac{2 \, \text{π}}{0.7} t\right) + 15\), the vertical shift is 15. This means the entire cosine graph is moved 15 meters upwards.

The height 15 meters becomes the new 'middle' line around which our cosine wave oscillates. With no vertical shift, the middle line of a standard cosine wave would be at height 0.
Maximum Height
The maximum height a point on the flywheel can achieve is given by the sum of the vertical shift and the amplitude. So, from our function \( h(t) = 13 \, \text{cos} \, \left(\frac{2 \, \text{π}}{0.7} t\right) + 15 \), you calculate the maximum height as follows:
  • Maximum height = vertical shift + amplitude = 15 + 13 = 28 meters

This tells us that at the highest point, the point on the flywheel is 28 meters above the ground.
Minimum Height
The minimum height a point on the flywheel can achieve is found by subtracting the amplitude from the vertical shift. Using our function \( h(t) = 13 \, \text{cos} \, \left(\frac{2 \, \text{π}}{0.7} t\right) + 15 \), you find the minimum height as follows:
  • Minimum height = vertical shift - amplitude = 15 - 13 = 2 meters

Thus, the lowest point on the flywheel is 2 meters above the ground.
Period of a Function
The period of a function details how long it takes for the function to complete one cycle. For a standard cosine function, the period is influenced by the coefficient inside the cosine function: \(B\text{t}\). In our function \( h(t) = 13 \, \text{cos} \, \left(\frac{2 \, \text{π}}{0.7} t\right) + 15 \), the coefficient \(\frac{2 \, \text{π}}{0.7}\) affects the period.

The period of our function is calculated as follows:
  • Period=\( \frac{2\pi}{\frac{2\pi}{0.7}} = 0.7 \) minutes

This indicates that every 0.7 minutes, the height cycle repeats itself.

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Most popular questions from this chapter

Match each function with its description in the table. a) \(y=-2 \cos 2(x+4)-1\) b) \(y=2 \sin 2(x-4)-1\) c) \(y=2 \sin (2 x-4)-1\) d) \(y=3 \sin (3 x-9)-1\) e) \(y=3 \sin (3 x+\pi)-1\) $$\begin{aligned} &\begin{array}{|c|c|c|c|c|}\hline & \text { Amplitude } & \text { Period } & \begin{array}{c}\text { Phase } \\\\\text { Shift }\end{array} & \begin{array}{c}\text { Vertical } \\ \text { Displacement }\end{array} \\\\\hline \mathbf{A} & 3 & \frac{2 \pi}{3} & 3 \text { right } & 1 \text { down } \\\\\hline \mathbf{B} & 2 & \pi & 2 \text { right } & 1 \text { down } \\\\\hline \mathbf{C} & 2 & \pi & 4 \text { right } & 1 \text { down } \\\\\hline \mathbf{D} & 2 & \pi & 4 \text { left } & 1 \text { down } \\\\\hline \mathbf{E} & 3 & \frac{2 \pi}{3} & \frac{\pi}{3} \text { left } & 1 \text { down } \\\\\hline \end{array}\end{aligned}$$

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