To derive a model for the height of the end of one blade of a windmill as a function of time [tex]\( t \)[/tex] in seconds, let’s go through this step-by-step:
1. Amplitude ( [tex]\(a\)[/tex] ):
The amplitude, [tex]\(a\)[/tex], represents the maximum deviation of the blade’s tip from the axis. Since the blades are 15 feet long, the amplitude of the sine wave, which represents this maximum deviation, is 15 feet. Hence,
[tex]\[
a = 15
\][/tex]
2. Vertical Shift ( [tex]\(k\)[/tex] ):
The vertical shift [tex]\(k\)[/tex] represents the height of the axis from the ground. According to the problem, this height is 40 feet. Hence,
[tex]\[
k = 40
\][/tex]
3. Angular Frequency ( [tex]\(b\)[/tex] ):
The blades complete 3 rotations every minute. Since there are 60 seconds in a minute, each rotation is completed in [tex]\( \frac{60}{3} = 20 \)[/tex] seconds.
The angular frequency [tex]\( b \)[/tex] of the sine function is determined by the formula:
[tex]\[
b = \frac{2\pi}{\text{period}}
\][/tex]
The period of our blade’s rotation is 20 seconds, so:
[tex]\[
b = \frac{2\pi}{20} = 0.3141592653589793
\][/tex]
Combining all these results, we get:
- Amplitude [tex]\(a\)[/tex] = 15
- Vertical Shift [tex]\(k\)[/tex] = 40
- Angular Frequency [tex]\(b\)[/tex] = 0.3141592653589793
The sine model for the height [tex]\(y\)[/tex] of the end of one blade as a function of time [tex]\(t\)[/tex] in seconds is thus:
[tex]\[
y = 15 \sin(0.3141592653589793 \cdot t) + 40
\][/tex]
Summarizing:
- [tex]\(a\)[/tex] is 15.
- The vertical shift, [tex]\(k\)[/tex], is 40.
- Hence, [tex]\(a = 15\)[/tex], and [tex]\(k = 40\)[/tex].
