# Tides are cyclical phenomena caused by the gravitational pull of the sun and the moon. On a particular retaining wall, the ocean generally reaches the 3 m mark at high tide. At low tide, the water reaches the 1 m mark. Assume that high tide occurs at 12:00 p.m. and at 12:00 a.m., and that low tide occurs at 6:00 p.m. and 6:00 a.m. What is the height of the water at 10:30 a.m.?

Since the phenomenon is cyclical, we can model with a sinusoid.

Taking 12am to be t=0, and marking the horizontal axis in 1 hour increments we have the maximum occurring at t=0, so it seems reasonable to model with the cosine.

The amplitude is the maximal distance from the midline,...

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Since the phenomenon is cyclical, we can model with a sinusoid.

Taking 12am to be t=0, and marking the horizontal axis in 1 hour increments we have the maximum occurring at t=0, so it seems reasonable to model with the cosine.

The amplitude is the maximal distance from the midline, or we can use `A=("highest"-"lowest")/2` so `A=(3-1)/2=1`

The midline is the arithmetic mean of the maximum and minimum so the midline is y=2. (`(3+1)/2=2` )

The period is given as 12 hours.

We use `y=AcosB(x-h)+k` as the model where A is the amplitude; if p is the period `B=(2pi)/p` ; h is any horizontal translation and k any vertical translation. (y=k is the midline.)

With p=12 we have `B=pi/6` , A=1,h=0, and k=2 so our model is

`y=cos((pit)/6)+1` Here is the graph:

We are asked to find the height of the water at 10:30am. This corresponds to t=10.5 (10 and 1/2 hours after 12am or t=0.)

Substituting t=10.5 into our model yields:

`y=cos((10.5pi)/6)+2=cos((7pi)/4)+2=sqrt(2)/2+2~~2.7071`

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The height of the water is `(sqrt(2)+4)/2~~2.7071` meters

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