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August 29, 2014

August 29, 2014

Posted by **Sam** on Tuesday, April 29, 2014 at 11:25pm.

- trig -
**Steve**, Tuesday, April 29, 2014 at 11:37pmthe period is 2*6.8 = 13.6, so we have

d(t) = a*cos(2pi/13.6 t) = a*cos(.462t)

The max-min range is 18-9 = 9, so the amplitude is 4.5 and we have

d(t) = 4.5 cos(.462t)

If we call t is the hour of the day, then max is at t=6, so

d(t) = 4.5 cos(.462(t-6))

Since the average depth is (9+18)/2, we have

d(t) = 4.5 cos(.462(t-6)) + 13.5

To see the graph, visit

http://rechneronline.de/function-graphs/

enter the function as

4.5 cos(.462*(x-6))+13.5

and set the x-range at 0-20

the y-range at 0-20

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