SOLUTION: Question:
A paint spot X lies on the outer rim of the wheel of a paddle-steamer. The wheel has a radius 3 m and as it rotates at a constant rate, X is seen entering the water e
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A paint spot X lies on the outer rim of the wheel of a paddle-steamer. The wheel has a radius 3 m and as it rotates at a constant rate, X is seen entering the water e
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Question 370601: Question:
A paint spot X lies on the outer rim of the wheel of a paddle-steamer. The wheel has a radius 3 m and as it rotates at a constant rate, X is seen entering the water every 4 seconds. H is the distance of X above the bottom of the boat. At time t=0, X is at its highest point.
a. Find the cosine model: H(t)=AcosB(t-C)+D
b. At what time does X first enter the water?
I have a picture of it in my book but I can't seem to paste it onto here. Do you know if there is any way I can show you the picture?
A = the amplitude
The amplitude, A, is the distance above (or below) the middle point. Since the radius of the wheel is 3m, the point on the wheel rises 3m above the water at its highest and 3m below the water. So A = 3.
B = /period
The period (the time it takes for one full cycle) is 4 seconds. So B =
C = the horizontal or "phase" shift.
The cos function has its maximum value at 0. The point on the wheel is at its highest at t=0. So there is no phase shift. C = 0.
D = the vertical shift.
The vertical shift is how much above (or below) 0 the "middle" height is. If we decide that water level is the 0 height (with below water being negative and above water being positive) then there is no vertical shift. D = 0. (Note: we could decide to make the zero height be different. FOr example we could use the bottom point of the wheel as the zero height. This would make the middle of the wheel 3m above zero and D = 3.)
We now have values fo A, B C and D. Inserting them into our model equation we get:
Simplifying this we get:
b. At what time does X first enter the water?
Since we decided that water level was the zero point for height, this question is now "b. At what time does X first have a height of zero?" We can use our equation from part a for this. We want the height to be zero so:
Now we solve for t. First we'll divide by 3:
Now we need to figure out when the cos function has zero values. Knowing the special angles tells us that cos is zero at , and at all angles coterminal with these. The way we express these infinite solutions is: (where n is any integer) (where n is any integer)
Now that we have eliminated cos, we can now solve for t. We just multiply each side by : (where n is any integer) (where n is any integer)
On the right side of each equation we need to use the Distributive Property to multiply: (where n is any integer) (where n is any integer)
A lot cancels: (where n is any integer) (where n is any integer)
leaving: (where n is any integer) (where n is any integer)
These equations tell us all the times when the point on the wheel is at water level (zero). To find the first time we try different integers for n until we figure out which resulting value for t is the smallest (without being negative since we can't have negative time). It should not take long to find that when n=0 in the first equation we end with t=1 and that 1 is the smallest non-negative value t can have when the height is zero.
So 1 second after the wheel starts will be the first time the point on the wheel enters the water.