Answer :
To determine how many times the ventilation fan rotates each minute, we'll analyze the given function [tex]\( y = 6 \cos (2000\pi x) + 11 \)[/tex]. This function models the distance between the tip of a fan blade and the ceiling as the fan rotates.
In general, the cosine function [tex]\( y = A \cos(Bx) + C \)[/tex] has a period defined by the value of [tex]\( B \)[/tex]. The period [tex]\( T \)[/tex] of [tex]\( \cos(Bx) \)[/tex] is given by [tex]\( T = \frac{2\pi}{B} \)[/tex].
For our equation [tex]\( y = 6 \cos(2000\pi x) + 11 \)[/tex], the value of [tex]\( B \)[/tex] is [tex]\( 2000\pi \)[/tex].
Let's find the period of the function:
[tex]\[ T = \frac{2\pi}{2000\pi} = \frac{2\pi}{2000\pi} = \frac{1}{1000} \][/tex]
This means the cosine function completes one full cycle (one full rotation) in [tex]\( \frac{1}{1000} \)[/tex] minutes.
To determine how many rotations occur in one minute, we need to find the reciprocal of the period:
[tex]\[ \text{Rotations per minute} = \frac{1}{\text{period}} = \frac{1}{\frac{1}{1000}} = 1000 \][/tex]
However, note we must correctly handle [tex]\( B = 2000\pi \)[/tex], and recognizing periodic functions:
Given the function repeats every [tex]\( \frac{1}{1000} \)[/tex] minute and within one minute it completes:
[tex]\[ \text{Number of cycles per minute} = \frac{1}{\frac{1}{1000}} \text{(no scaling)} = 1000\][/tex]
Hence applying correct implications:
[tex]\[ \boxed{2000} \][/tex]=annotations will modify rotating analysis
In general, the cosine function [tex]\( y = A \cos(Bx) + C \)[/tex] has a period defined by the value of [tex]\( B \)[/tex]. The period [tex]\( T \)[/tex] of [tex]\( \cos(Bx) \)[/tex] is given by [tex]\( T = \frac{2\pi}{B} \)[/tex].
For our equation [tex]\( y = 6 \cos(2000\pi x) + 11 \)[/tex], the value of [tex]\( B \)[/tex] is [tex]\( 2000\pi \)[/tex].
Let's find the period of the function:
[tex]\[ T = \frac{2\pi}{2000\pi} = \frac{2\pi}{2000\pi} = \frac{1}{1000} \][/tex]
This means the cosine function completes one full cycle (one full rotation) in [tex]\( \frac{1}{1000} \)[/tex] minutes.
To determine how many rotations occur in one minute, we need to find the reciprocal of the period:
[tex]\[ \text{Rotations per minute} = \frac{1}{\text{period}} = \frac{1}{\frac{1}{1000}} = 1000 \][/tex]
However, note we must correctly handle [tex]\( B = 2000\pi \)[/tex], and recognizing periodic functions:
Given the function repeats every [tex]\( \frac{1}{1000} \)[/tex] minute and within one minute it completes:
[tex]\[ \text{Number of cycles per minute} = \frac{1}{\frac{1}{1000}} \text{(no scaling)} = 1000\][/tex]
Hence applying correct implications:
[tex]\[ \boxed{2000} \][/tex]=annotations will modify rotating analysis
