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Answer :
We are given that the windmill blades are 15 feet in length and that the axis is located 40 feet above the ground. This tells us that the sinusoidal motion has:
- An amplitude [tex]$a = 15$[/tex] (the length of the blade).
- A vertical shift (midline) [tex]$k = 40$[/tex] (the height of the windmill’s axis).
The windmill makes 3 complete rotations every minute. Since 1 minute is 60 seconds, one complete rotation (the period) is
[tex]$$
\text{period} = \frac{60 \text{ seconds}}{3} = 20 \text{ seconds}.
$$[/tex]
The general sine function is given by
[tex]$$
y = a \sin(b t) + k.
$$[/tex]
For a sine function, the period [tex]$T$[/tex] is related to [tex]$b$[/tex] by the formula
[tex]$$
T = \frac{2\pi}{b}.
$$[/tex]
Setting [tex]$T = 20$[/tex] seconds, we can solve for [tex]$b$[/tex]:
[tex]$$
b = \frac{2\pi}{T} = \frac{2\pi}{20} = \frac{\pi}{10}.
$$[/tex]
So, the sine model for the height of the end of one blade as a function of time [tex]$t$[/tex] (in seconds) is:
[tex]$$
y = 15 \sin\left(\frac{\pi}{10} t\right) + 40.
$$[/tex]
This completes the step-by-step solution.
- An amplitude [tex]$a = 15$[/tex] (the length of the blade).
- A vertical shift (midline) [tex]$k = 40$[/tex] (the height of the windmill’s axis).
The windmill makes 3 complete rotations every minute. Since 1 minute is 60 seconds, one complete rotation (the period) is
[tex]$$
\text{period} = \frac{60 \text{ seconds}}{3} = 20 \text{ seconds}.
$$[/tex]
The general sine function is given by
[tex]$$
y = a \sin(b t) + k.
$$[/tex]
For a sine function, the period [tex]$T$[/tex] is related to [tex]$b$[/tex] by the formula
[tex]$$
T = \frac{2\pi}{b}.
$$[/tex]
Setting [tex]$T = 20$[/tex] seconds, we can solve for [tex]$b$[/tex]:
[tex]$$
b = \frac{2\pi}{T} = \frac{2\pi}{20} = \frac{\pi}{10}.
$$[/tex]
So, the sine model for the height of the end of one blade as a function of time [tex]$t$[/tex] (in seconds) is:
[tex]$$
y = 15 \sin\left(\frac{\pi}{10} t\right) + 40.
$$[/tex]
This completes the step-by-step solution.
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