Question 1042217
For a certain day, the depth of water, h, in metres in Tofino, B.C, at time t, in hours, is given by the formula:
h(t)=7.8+3.5sin[pi/6(t-3)], tE[0,24]. Assume t=0 represents midnight. Provide an algebraic solution to determine the time(s) of day, the water reaches the depth of 10.29m. (Give the answer in hours and minutes)
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Solve::
7.8 + 3.5*sin((pi/6)(t-3) = 10.29
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3.5*sin((pi/6)(t-3)) = 2.49
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sin((pi/6)(t-3)) = 0.711 
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Take sin^-1 of both sides to get:
(pi/6)(t-3) = 0.791+-(2pi*n)  or (pi/6)(t-3) = pi-0.7911  = 2.35+-(2pi*n)
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t-3 =1.51+-(12) or (t-3) = 4.488+-(12)
Since you want t E [0,24] t = 16.488 + 3 = 19.488 = 19 hr 0.488sec
= 19hr +0.488*60min = 19hr 29min + 0.28(60)sec = 19hr 29min 17sec

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Cheers,
Stan H.
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