# Tides are cyclical phenomena caused by the gravitational pull of the sun and the moon. http://postimage.org/image/k0oktyznl/Applications

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We are given a high tide of 3m at 12am,12pm; a low tide of 1m at 6am,6pm

The tides are roughly periodic, so they can be modelled by a sinusoid. In this case we can use a cos function:

`y=acos[b(x-h)]+k` a gives the amplitude, the period determines b, h is the phase shift (horizontal translation), while k is the vertical translation (it gives the midline)

(1) The amplitude is `(3-1)/2=1` (The tides vary 1m above or below the midline) so a=1

(2) The period is 12 hours. `b=(2pi)/p==>b=(2pi)/12=pi/6`

(3) Since the maximum occurs at 12am/pm, if we start the graph there it coincides with the cos which also begins at its maximum, so h=0

(4) The midline is `y=(3+1)/2=2` so k=2 (The cos function varies about 0, the model varies about y=1)

The function will be `y=cos(pi/6x)+2` where x is measured in hours

At 10:30 am x=10.5:

`y=cos(pi/6*10.5)+2~~2.7071`

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**The depth of the water at 10:30am is approximately 2.7m**

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The graph:

Since the tide is cyclical, that means that it follows a sine or cosine shape. Because the high tide is at noon and midnight, with low tide at 6am and 6pm, this means that it is following a cosine shape. Let t by the number of hours after midnight.

Also, high tide is 3m and low tide is 1m, so the amplitude is 1 and the shift up is 2, midway between high and low tide.

Finally, we know that `T={2 pi}/k` , where T is the period and k is the horizontal scaling factor. Since the period is 12 hours, we have `k={2pi}/12=pi/6` .

This means the height of the water is `h=cos({pi t}/6)+2` .

Now use substitution to find the hiehgt at t=10.5

`h(10.5)=cos({pi(10.5)}/6)+2`

`=cos({7pi}/4)+2` this is a special triangle with CAST rule

`=1/sqrt2+2`

`approx 0.707+2`

`=2.707`

**The approximate height at 10:30 is 2.707m.**