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, [tex]a = 15[/tex].
- The vertical shift, [tex]k[/tex], is the height of the windmill axis, [tex]k = 40[/tex].

Determine the period and the value of [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}[/tex].

Complete the model using these parameters.

We start by identifying the key pieces of information:

1. The length of the windmill blade gives the amplitude, so
[tex]$$a = 15.$$[/tex]

2. The windmill’s central axis is 40 feet above the ground, which is the vertical shift:
[tex]$$k = 40.$$[/tex]

3. The windmill completes 3 rotations every minute. Since there are 60 seconds in a minute, the period (the time for one complete rotation) is:
[tex]$$T = \frac{60 \text{ seconds}}{3} = 20 \text{ seconds}.$$[/tex]

4. The parameter [tex]$b$[/tex] in the sine function is related to the period by the formula:
[tex]$$T = \frac{2\pi}{b}.$$[/tex]
Solving for [tex]$b$[/tex], we have:
[tex]$$b = \frac{2\pi}{T} = \frac{2\pi}{20} \approx 0.31416.$$[/tex]

Thus, the sine model that gives the height [tex]$y$[/tex] (in feet) of the end of one blade, as a function of time [tex]$t$[/tex] (in seconds), is:
[tex]$$
y = 15 \sin(0.31416\, t) + 40.
$$[/tex]

This model meets the conditions: when [tex]$t = 0$[/tex], the blade is pointing to the right, and since the sine function passes through zero with a positive slope at [tex]$t = 0$[/tex], it accurately represents the initial position.