Answer :
To write a sine model for the height of the end of the windmill blade, we need to find the parameters of the model: [tex]\( y = a \sin(b t) + k \)[/tex]. Let's break this down step-by-step:
1. Determine the amplitude ([tex]\(a\)[/tex]):
- The amplitude of the sine wave corresponds to the length of the blade. In this case, the blade length is 15 feet. Therefore, [tex]\(a = 15\)[/tex].
2. Determine the vertical shift ([tex]\(k\)[/tex]):
- The vertical shift of the sine wave is the height of the windmill from the ground to the center point of rotation of the blades. Here, the height is 40 feet. So, [tex]\(k = 40\)[/tex].
3. Calculate the period and frequency:
- The windmill completes 3 rotations every minute, which is equivalent to 3 rotations per 60 seconds. Therefore, the frequency in rotations per second is [tex]\( \frac{3}{60} = 0.05 \)[/tex] rotations per second.
- The period of one complete rotation is the reciprocal of the frequency. So, the period [tex]\( T = \frac{1}{0.05} = 20 \)[/tex] seconds.
4. Determine the frequency coefficient ([tex]\(b\)[/tex]):
- In the sine function, [tex]\( b \)[/tex] is determined from the formula for a sine wave's period: [tex]\( T = \frac{2\pi}{b} \)[/tex].
- Given [tex]\( T = 20 \)[/tex], we can solve for [tex]\( b \)[/tex]:
[tex]\[ b = \frac{2\pi}{T} = \frac{2\pi}{20} = \frac{\pi}{10} \][/tex]
Now, substituting these values into the sine model equation, we have:
[tex]\( y = 15 \sin\left(\frac{\pi}{10} t\right) + 40 \)[/tex]
This equation models the height of the end of the windmill blade as a function of time in seconds, assuming that the blade is initially pointing to the right when [tex]\( t = 0 \)[/tex] and the windmill turns counterclockwise at a constant rate.
1. Determine the amplitude ([tex]\(a\)[/tex]):
- The amplitude of the sine wave corresponds to the length of the blade. In this case, the blade length is 15 feet. Therefore, [tex]\(a = 15\)[/tex].
2. Determine the vertical shift ([tex]\(k\)[/tex]):
- The vertical shift of the sine wave is the height of the windmill from the ground to the center point of rotation of the blades. Here, the height is 40 feet. So, [tex]\(k = 40\)[/tex].
3. Calculate the period and frequency:
- The windmill completes 3 rotations every minute, which is equivalent to 3 rotations per 60 seconds. Therefore, the frequency in rotations per second is [tex]\( \frac{3}{60} = 0.05 \)[/tex] rotations per second.
- The period of one complete rotation is the reciprocal of the frequency. So, the period [tex]\( T = \frac{1}{0.05} = 20 \)[/tex] seconds.
4. Determine the frequency coefficient ([tex]\(b\)[/tex]):
- In the sine function, [tex]\( b \)[/tex] is determined from the formula for a sine wave's period: [tex]\( T = \frac{2\pi}{b} \)[/tex].
- Given [tex]\( T = 20 \)[/tex], we can solve for [tex]\( b \)[/tex]:
[tex]\[ b = \frac{2\pi}{T} = \frac{2\pi}{20} = \frac{\pi}{10} \][/tex]
Now, substituting these values into the sine model equation, we have:
[tex]\( y = 15 \sin\left(\frac{\pi}{10} t\right) + 40 \)[/tex]
This equation models the height of the end of the windmill blade as a function of time in seconds, assuming that the blade is initially pointing to the right when [tex]\( t = 0 \)[/tex] and the windmill turns counterclockwise at a constant rate.
