Pre-Calculus 12 Tide Problem: Solving Sinusoidal Functions Step by Step

The following question is a classic Pre-Calculus 12 tide problem that frequently appears on tests and exams.

On New Year’s Day in Perth, the water level (in metres) is given by:

h(t) = 5sin((π/6)(t - 5)) + 7,

where t is the number of hours after 00:00. Find every time within the first 24 hours (0 ≤ t < 24) when the water is exactly 3 metres deep.

Solution

This question requires a calculator. Make sure it's in radian mode.

Let h(t) = 3 and solve for t.

5sin((π/6)(t - 5)) + 7 = 3

sin((π/6)(t - 5)) = (-4/5)

Let θ = (π/6)(t - 5), so we have

sin(θ) = (-4/5)

Using inverse sine (also known as arcsine), we can find two solutions to this equation on [0, 2π] (radians) as follows:

Reference angle = arcsin(4/5) = 0.927295218 (enter (4/5) as a positive value).

Since sine is negative, the angles we want must be found in quadrant III and quadrant IV; remember that these are the quadrants where sine is negative.

Angle in quadrant III on [0, 2π] is π + 0.927295218 = 4.068887872

Angle in quadrant IV on [0, 2π] is 2π - 0.927295218 = 5.355890089

We then express each of these as a general solution as follows.

θ = 4.068887872 + (2π)(k), where k is an integer

θ = 5.355890089 + (2π)(k), where k is an integer

But θ = (π/6)(t - 5); we must isolate this equation for t for each of the two angles we just found.

Equation 1

(π/6)(t - 5) = 4.068887872 + (2π)(k), where k is an integer

t = 12.77100341 + 12k, where k is an integer.

Because the model has a 12-hour period, the general form t = 12.77100341 + 12k lists all matching times spaced 12 hours apart. To restrict to one day, test the nearest integers for k:

k = 0 gives t = 12.77100341 k = -1 gives t = 0.77100341

k = 1 produces t = 24.77100341, which is outside 0 ≤ t < 24, and any other k yields a value of t out of range.

So, we have two answers from this equation: t = 12.77100341 t = 0.77100341

For each of these, the integer part gives hours; the fractional part × 60 gives minutes.

Note that we will round all times to the nearest minute when we convert them to the 24-hour clock and 12-hour clock.

So, t = 12.77100341 means the 12th hour, and (0.77100341)(60) = 46 minutes approximately. Hence, t = 12.77100341 corresponds to 12:46 on the 24-hour clock, which is 12:46 pm on the 12-hour clock.

Similarly, t = 0.77100341 corresponds to 00:46 on the 24-hour clock, or 12:46 am on the 12-hour clock.

Thus, equation 1 reveals the following two times (12-hour clock; nearest minute):

t = 12:46 pm

t = 12:46 am

Equation 2

(π/6)(t - 5) = 5.355890089 + (2π)(k), where k is an integer

t = 15.22899659 + 12k, where k is an integer.

Because the model has a 12-hour period, the general form t = 15.22899659 + 12k lists all matching times spaced 12 hours apart. To restrict to one day, test the nearest integers for k:

k = 0 gives t = 15.22899659

k = -1 gives t = 3.22899659

k = 1 produces t = 27.22899659, which is outside 0 ≤ t < 24, and any other k yields a value of t out of range.

So, we have two answers from this equation:

t = 15.22899659

t = 3.22899659

On the 24-hour clock, these correspond to: t = 15:14

t = 03:14

On the 12-hour clock, these correspond to:

t = 3:14 pm

t = 3:14 am

Final Answer

All the times during a 24-hour period when the depth of the water is 3 m are 12:46 am, 3:14 am, 12:46 pm, and 3:14 pm.

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