Answer:
a is the _amplitude_(Length of the blades)_
The vertical shift, k, is the _Mill shaft height_
[tex]a = 15\ ft\\\\k = 40\ ft[/tex]
[tex]y = 15sin(\frac{\pi}{10}t) + 40[/tex]
Step-by-step explanation:
In this problem the amplitude of the sinusoidal function is given by the length of the blades.
[tex]a = 15\ ft[/tex]
The mill is 40 feet above the ground, therefore the function must be displaced 40 units up on the y axis. So:
[tex]k = 40\ ft[/tex]
We know that the blades have an angular velocity w = 3 rotations per minute.
One rotation = [tex]2\pi[/tex]
1 minute = 60 sec.
So:
[tex]w = \frac{3(2\pi)}{60}\ rad/s[/tex]
[tex]w = \frac{\pi}{10}\ rad/s[/tex]
Finally:
a is the _amplitude_(Length of the blades)_
The vertical shift, k, is the _Mill shaft height_
[tex]a = 15\ ft\\\\k = 40\ ft[/tex]
[tex]y = 15sin(\frac{\pi}{10}t) + 40[/tex]
