The blades of a windmill turn on an axis that is 40 feet from the ground. The blades are 15 feet long and complete 3 rotations every minute.

Write a sine model, [tex]y = a \sin(b t) + k[/tex], for the height (in feet) of the end of one blade as a function of time [tex]t[/tex] (in seconds). Assume the blade is pointing to the right when [tex]t=0[/tex] and that the windmill turns counterclockwise at a constant rate.

- [tex]a[/tex] is the length of the blade.
- The vertical shift, [tex]k[/tex], is the height of the windmill.

Given:
- [tex]a = 15[/tex]
- [tex]k = 40[/tex]

Calculate the period and [tex]b[/tex]:
- The period is [tex]\frac{60}{3} = 20[/tex] seconds (since there are 3 rotations per minute).
- [tex]b = \frac{2\pi}{\text{period}} = \frac{2\pi}{20} = \frac{\pi}{10}[/tex]

Thus, the sine model is:
\[ [tex] y = 15 \sin\left(\frac{\pi}{10} t\right) + 40 \] [/tex]

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.