- #1
davedave
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At high tide, the average depth of water in a harbour is 25m and at low tide the average depth is 9cm. The tides in the harbour complete one cycle approximately every 12 hours. The first high tide occurs at 5:45am. A cosine function that relates the depth of the water in the harbour to the time in hours is given by
d(t) = 8cos(∏(t-5.75)/6) + 17
A large cruise ship needs at least 14m of water to dock safely. For how many hours per cycle, to the nearest hour. can a cruise ship dock safely.
my method
let 14 = 8cos(∏(t-5.75)/6) + 17
-3/8 = cos(∏(t-5.75)/6) equation (1)
let r = reference angle=inverse cos(3/8) = 1.186399552
Since the cosine equation (1) above is equal to a negative value, we work in quadrants 2 and 3
so, the two standard position angles are a = ∏ - r = 1.955193101
b = ∏ + r = 4.327992206
now, solving for t in equation (1) with these two angles gives the two times
t1 = 9.4841433761 and t2 = 14.01585624
hence, t2-t1 = 4.531712478 ≈ 5 hours
One of my friends says I am right. Another one of my friends says I am wrong.
Why am I right or wrong? Please help
d(t) = 8cos(∏(t-5.75)/6) + 17
A large cruise ship needs at least 14m of water to dock safely. For how many hours per cycle, to the nearest hour. can a cruise ship dock safely.
my method
let 14 = 8cos(∏(t-5.75)/6) + 17
-3/8 = cos(∏(t-5.75)/6) equation (1)
let r = reference angle=inverse cos(3/8) = 1.186399552
Since the cosine equation (1) above is equal to a negative value, we work in quadrants 2 and 3
so, the two standard position angles are a = ∏ - r = 1.955193101
b = ∏ + r = 4.327992206
now, solving for t in equation (1) with these two angles gives the two times
t1 = 9.4841433761 and t2 = 14.01585624
hence, t2-t1 = 4.531712478 ≈ 5 hours
One of my friends says I am right. Another one of my friends says I am wrong.
Why am I right or wrong? Please help