*TUTOR POST* Trig Water Wheel Problem

PROBLEM: The top of a bucket 0.5 m high is attached to a waterwheel of diameter 2m, from which the bucket always hangs downward. The wheel sits above the river so that half of the bucket dips below the surface of the water at its lowest position. Write a function for the height of the center of the bucket (in meters) above the river as a function, f, of the angle t as measured counterclockwise from the 3 o'clock position.

The first thing I like to do is draw a picture. We have a water wheel with a diameter of 2, which means the radius is 1. The bucket is .5 tall, but only .25 is between the water wheel and where the bucket meets the water.

As our problem stands, if the water is our x-axis, then the center of our water wheel circle is 1.25 meters above the water. If the radii are all 1, then our water wheel circle coordinates look like this:

The problem is asking about where the center of the bucket is at any given point in its rotation. Well, the bucket is in an entirely different circle than the water wheel. Since the bucket always dangles downward, the positions for the bucket center look like the red marks:

 

Since the center of the bucket is always .25m lower than the water wheel, we can set up the new bucket center circle with the following coordinates:

We have our circle set up, but the instructions say that we need to start at the 3 o' clock position, and calculate our theta angles in a counter-clockwise direction from 3 o' clock. This means that we have to shift our circle so that the origin is at (0, 0), and the x-axis cuts the circle horizontally so that we can start in the 3 o' clock position. Notice that our radius is still 1.

 Well, hmm. Now we have a familiar looking image:

 

At this point we need to reflect on what we know about our sine function. The sine of an angle is the ratio of the opposite side over the hypotenuse (SOH-CAH-TOA). Since our hypotenuse (radius) is 1, then the sine of our angle is actually the measurement of our opposite side. In our model this will give us the distance above the x-axis, but we will need to add 1 to find the actual height from the bottom of the circle, or where the bucket middle meets the water.

To set up an equation, we say "y equals what?" In this case, y = sin(Θ) + 1. We now substitute f(x) for y to create our function f(x) = sin(Θ) + 1. Now we can test the function.
We know that when our bucket is at the bottom of the circle, it is 0 meters above the water. The angle for this point on the circle from the 3 o' clock position, going counter-clockwise, is 270 degrees.When we plug 270 in for theta, we have f(270) = sin(270) + 1. The answer is 0, which is what we expected.
We know that when the bucket is at the top of the circle, it is 2 meters above the water. This puts our theta measurement at 90 degrees. Does sin(90) + 1 equal 2? Yes.
We know that when our radius is even with our x-axis, our bucket should be one meter above the water. There are two places where this occurs - at sin(0) and sin(180). When we plug this into our function, it gives us the expected answer of 1.

Good luck! Find me at WyzAnt if you need tutoring.


Stephen L. Wilson
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