Harbour Tides: Calculating Safe Docking Time for Cruise Ships

[davedave]

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

 

[haruspex]

You have found two times at which the depth is 14m. Have you checked whether it's higher in (t1,t2) or in (t2,t1)? Here's a quick check: the average depth is 17m, so would you expect the ok period to be more or less than half the complete cycle?

 

[HallsofIvy]

I immediately see one thing that is either a massive error or a typo.

You say: "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"
That has a maximum of 25 and a minimum of 9 but what units? If d(t) is in meters, your maximum is correct but your minimum is 9 meters, not 9 cm. Or was that a typo and you mean 9 meters? 9 cm to 25 m would be a greater range than the Bay of Fundy!

 

[Joffan]

Think about what haruspex said, but also I'd take issue with your rounding of "≈ 5 hours" when the question clearly gives an example of time resolution down to at least quarter hours, and arguably 5-minute precision.

On the other hand, I'm, shocked - shocked, I tell you! - at the question's imprecision in the length of a tide cycle.

(In response to Halls, I reckon the formula was in the question, so I'm voting "typo").

 

[Mark44]

The 9cm low tide depth has to be a typo. The greatest tidal range in the world is at the Bay of Fundy, in Canada between New Brunswick and Nova Scotia. The maximum range is 16.3 m. If we assume that the OP meant an average depth of 9 m., the tidal range in the posted problem is about that of the Bay of Fundy.

 

[davedave]

yes 9cm is a typo. It should be 9m. I've just let my 3rd friend check my solution. He said my answer is wrong. What went wrong in my solution?

 

[haruspex]

What went wrong in my solution?

Did you understand what I asked you in post #2?