Trigonometry Word Problem

Question

The water at a local beach has an average depth of one meter at low tide and an average depth at high tide of 8m. One cycle of the tides takes approximately 12 hours. This periodic motion can be modelled by the function d(t)=3.5cos [6 t]+4.5where d(t) represents the depth of the water, in metres, at a time t, in hours. This equation assumes that the water level is at high tide at time zero (midnight). Many people dive into the water from a nearby dock. If the water must be at least 3m deep to dive safely, when during the first 24 hour period (in hours) would it be safe to dive off the dock? Please give your answer in all ranges of hours possible.

Answer 1

It appears you have an error in your function. If it were really d(t) = 3.5cos(6t) + 4.5, then the period would be 2π/6 which is a little over an hour. What it should be, for the period to be 12 hours is d(t) = 3.5cos[(π/6)t] + 4.5

That function (d(t) = 3.5cos[(π/6)t] + 4.5) looks like this:

To find the times where the depth is 3 meters (the boundary condition between good and bad depth), we can set d(t) equal to 3 and solve for t:

d(t) = 3.5cos[(π/6)t] + 4.5

3 = 3.5cos[(π/6)t] + 4.5

-1.5 = 3.5cos[(π/6)t]

-3/7 = cos[(π/6)t]

cos^-1(-3/7) = cos^-1(cos[(π/6)t])

2.0137 = (π/6)t

t = 2.0137/(π/6)

t = 3.846 hrs

We can see by the graph that that is the first place where the function crosses below the 3 meter line.

We can also see that the function is symmetrical about t = 6 so the place it crosses the second time (on it's way back up) is the same distance from t = 6. So 6 - 3.846 = 2.154 and 2.154 + 6 = 8.154. So the second place the function crosses the 3 meter line is at t = 8.154 hours.

Also, notice that after 12 hours, the function repeats, so the third place it crosses the 3 meter line will be at 12 + 3.846 hours or 15.846 hrs.

The 4th place it crosses the 3 meter line is at 12 + 8.154 hours or 20.154 hours.

So the times when it is safe to swim are: [0, 3.846] U [8.154, 15.846] U [20.154, 24] (numbers in hours) Since we are not given a "start time" these are just hours after the high tide.

Answer 2

d(t) = cos(6t) + 4.5 In this case, t = 12 and d(t) = 8

substitute these values into this equation and we get: 8 = 3.5 x cos(72) + 4.5 which becomes

3.5 x 0.28 +4.5 = 5.48

So, d(t) = 5.4 meters

Yes, this would be safe.