Question

The blades of a windmill turn on an axis that is 40 feet from the ground. The blades are 15 feet long and complete 3 rotations every minute. Write a sine model, y = asin(bt) + k, for the height (in feet) of the end of one blade as a function of time t (in seconds). Assume the blade is pointing to the right when t = 0 and that the windmill turns counterclockwise at a constant rate.
a is the ___
The vertical shift, k, is the _____________
a =
k =

Answer 1

Answer:

a is the length of the blade

the vertical shift , k, is the height of the windmill

a= 15 k= 40

the period is 20 seconds

b = pi/10

y=15sin(π/10t)+40

Step-by-step explanation:

Answer 2

Answer:

a is the _amplitude_(Length of the blades)_

The vertical shift, k, is the _Mill shaft height_

$a = 15\ ft\\\\k = 40\ ft$

$y = 15sin(\frac{\pi}{10}t) + 40$

Step-by-step explanation:

In this problem the amplitude of the sinusoidal function is given by the length of the blades.

$a = 15\ ft$

The mill is 40 feet above the ground, therefore the function must be displaced 40 units up on the y axis. So:

$k = 40\ ft$

We know that the blades have an angular velocity w = 3 rotations per minute.

One rotation = $2\pi$

1 minute = 60 sec.

So:

$w = \frac{3(2\pi)}{60}\ rad/s$

$w = \frac{\pi}{10}\ rad/s$

Finally:

a is the _amplitude_(Length of the blades)_

The vertical shift, k, is the _Mill shaft height_

$a = 15\ ft\\\\k = 40\ ft$

$y = 15sin(\frac{\pi}{10}t) + 40$