Answer :
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.
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.