What is the time during which a boat can use the harbor in the following case?
A small natural harbour in Cornwall is protected from bad weather by a natural rock barrier. Unfortunately, this means that the harbour is not accessible during certain times of the day. On a particular day the depth of water, d metres, beyond the barrier is modelled by the function:
d(t) = 4sin(pi t/8) + 12
At 06:00 the depth of water beyond the barrier is 12m and the barrier is visible to a height of 0.5m. The harbour can accommodate small boats whose minimum draught is 1.5m.
What is the earliest time of day when boats of minimum draught can enter or leave the harbour?
For how long during the first part of the day is it possible for this to happen?
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The depth of water is given by the function d (t) = 4sin (pi t/8) + 12. Now, at 6:00, the depth of water beyond the barrier is 12 m and the barrier is visible to a height of 0.5m.
For a boat with a draught of 1.5 m, the height of the water level should be at least 1.5m above that of the barrier. So as the barrier is visible to a height of 0.5 m when the depth is 12m, we need a depth of 12 + 0.5 + 1.5 = 14m
According to the equation for depth of water d (t) = 4sin (pi t/8) + 12.
We need 14 = 4sin (pi t/8) + 12.
=> (14 - 12)/4 = sin (pi t/8)
=> 1/2 = sin (pi t/8)
=> sin (pi/6) = sin (pi t/8)
=> pi t/8 = pi/6
=> t = 8/6
=> t = 4/3
As t = 0 is 6:00, t = 4/3 is 7:20
Therefore boats with a draft of 1.5 m can enter after 7:20.
Boats can continue to use the harbor as long as the depth is greater than 14. We see that till t = 16, the value of 4sin (pi t/8) is increasing. Beyond t = 16, the value of 4sin (pi t/8) starts to decrease and at t = 20/3 it reaches 2.
So boats cannot use the harbor after t = 20/3. Now the time at t = 0 is 6:00. The time at t = 20/3 is 12: 40. After 12:40 boats cannot use the harbor. The harbor is accessible only between 7:20 am and 12:40 pm.
If the first half of the day is the period before 12 pm, the harbor can be used from 7: 20 to 12:00 or for 4 hours 40 minutes.
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