Question 1205924: The paddle wheel of a boat has a diameter of 6 m and makes one revolution every 16 seconds. The axle of the wheel is 1 m above the water line. Where, relative to the water line, is this point 54 seconds after being at its maximum height. How long does it take until this point first reaches the water line. How long does this point stay under water.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! at time point 0, the paddle wheel is at its maximum height.
since cos(theta) is at its highest point at time 0, we'll use that function to model this situation.
frequency = 360 / period
if the period is 16, then the frequency is 360 / 16 = 22.5
equation of y = a * cos(b * (x - c)) + d becomes y = a * cos(22.5 * (x - c)) + d.
the wheel has a radius of 3 meters.
it's height will be plus or minus 3 meters from the horizontal axis.
the horizontal axis is at y = 1.
the equation becomes y = a * cos(22.5 * (x - c)) + 1
there is no horizontal shift, so c = 0 and the equation becomes y = a * cos(22.5
* x) + 1
the amplitude of the cosine function is 3, so a = 3 and the equation becomes y = 3 * cos(22.5 * x) + 1
that should be it.
the graph of one revolution of the cosine function looks like this.
one complete revolution of the wheel takes 16 seconds.
that's equivalent to one complete cycle of the cosine function.
the value of x on the graph represents the seconds.
the cosine function starts at y = 4.
that's because the axis of rotation of the wheel is 1 meter above the water line.
since the water line is at y = 0, the axis of rotation is at y = 1.
the wheel has a radius of 3 meters, so the maximum height of the wheel is 3 meters above the axis of rotation and the minimum height of the wheel is 3 meters below the axis of rotation.
that makes the maximum height 3 + 1 = 4 meters above the water line and the minimum height = 1 - 3 = 2 meters below the water line.
the axis of rotation is the horizontal center line of the graph of the cosine function. the amplitude is 3 meters above this and 3 meters below this.
the frequency of 22.5 tells you that the cosine function goes through 22.5 complete revolutions in 360 degrees.
it starts at y = 4 and ends at y = -2 because it ends at halfway through the 23d revolution.
here's what that looks like.
you can count the high points and you'll see that there are 22 of them and 1 low point in 360 degrees.
the high points are at y = 4 and the low points are at y = -2.
the value of y at x = 4 represents where the high point of the wheel when it started to rotate at x = 0 to where is is after 54 seconds of rotation.
on the graph, that looks like this.
the coordinate point at x = 54 is (54,-1.121)
you would solve for this algebraically as follows.
the function is y = 3 * cos(22.5 * x) + 1
when x = 54, that becomes y = 3 * cos(22.5 * 54) + 1
solve for y to get y = -1.121320344.
the graph rounds that to y = -1.121.
since the water line is at y = 0, this is 1.121 meters below the water line when 54 seconds have elapsed.
based on the graph, the high point first reaches the water line 4.865 seconds after starting its rotation.
algebraically, you would solve for this as follows:
y = 3 * cos(22.5 * x) + 1.
set y = 0 and you get 3 * cos(22.5 * x) + 1 = 0
subtract 1 from both sides of the equation to get 3 * cos(22.5 * x) = -1
divide both sides of the equation by 3 to get cos(22.5 * x) = -1/3
solve for the angle of 22.5 * x to get 22.5 * x = arccos(-1/3) = 109.4712206.
solve for x to get x = 109.4712206 / 22.5 = 4.865387584.
the graph rounds that to 4.865 seconds.
the graph shows the next time the wheel is at the water line as 11.135 seconds after the wheel started its rotation.
subtract that from 4.865 and you get the paddle wheel was under water for 11.135 minus 4.865 = 6.27 seconds.
here's what that looks like on the graph.
to solve for that second point algebraically is a little tricky, but here's how i did it.
you know that the cosine function is negative in the second and third quarters.
when the period of the cosine wave is 16 seconds, then the quadrants are:
quadrant 1 is 0 to 4 seconds.
quadrant 2 is 4 to 8 seconds.
quadrant 3 is 8 to 12 seconds.
quadrant 4 is 12 to 16 seconds.
4.865 seconds is in the second quadrant because it is between 4 and 8.
the equivalent angle in the first quadrant is 8 - 4.865 = 3.135 seconds.
the equivalent in the third quadrant is 8 + 3.135 = 11.135 seconds.
that's why the graph shows the second zero point at 11.135 seconds.
that should do it.
let me know if you have any questions.
theo
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