# A group of students decided to study the sinusoidal nature of tides. Values for the depth of the water level were recorded at various times. At t=3 hours low tide was recorded at a depth of 1.1 m. At t = 9 hours, high tide was recorded at a depth of 3.7m. Write the equation expressing depth in terms of time.

## mean = ( 1.1+ 3.7 ) / 2 = 2.4

amplitude = (3.7 -1.1)/2 = 1.3

time of mean level tide = (9-3)/2 = 3

tide period T = 2 (9-3) = 12 hours

so

h = 2.4 + 1.3 sin (2 pi (t-3) /12)

## To write the equation expressing depth in terms of time, we can use the sinusoidal function, specifically the sine function. The general form of a sinusoidal function is:

y = A * sin(B * (x - C)) + D

where:

A is the amplitude

B is the period

C is the horizontal shift

D is the vertical shift

First, we need to find the amplitude, which is half the difference between the high tide and low tide depths:

Amplitude (A) = (3.7 m - 1.1 m) / 2 = 1.3 m

Next, we need to find the period (B), which is the time it takes for one complete cycle from low tide to low tide or high tide to high tide. In this case, the period will be 12 hours:

Period (B) = 9 hours - 3 hours = 6 hours

The horizontal shift (C) is the time at which the function starts oscillating. Since low tide was recorded at t = 3 hours, the horizontal shift will be -3:

Horizontal shift (C) = -3

Finally, the vertical shift (D) is the average of the high and low tide depths:

Vertical shift (D) = (3.7 m + 1.1 m) / 2 = 2.4 m

Putting it all together, the equation expressing depth in terms of time is:

depth = 1.3 * sin((2π/6)(time - 3)) + 2.4

where time is measured in hours.

## To write the equation expressing depth in terms of time, we can assume a sinusoidal function, as tides follow a periodic pattern. A standard equation for a sinusoidal function is:

y = A * sin(B(x - C)) + D

Where:

A represents the amplitude (half of the difference between the maximum and minimum values).

B represents the frequency (how many complete cycles occur in a given period).

C represents the horizontal shift (phase shift).

D represents the vertical shift.

In this case, we have the following information:

At t = 3 hours, low tide was recorded at a depth of 1.1m (minimum value).

At t = 9 hours, high tide was recorded at a depth of 3.7m (maximum value).

From this, we can deduce the following information:

Amplitude (A) = (3.7 - 1.1) / 2 = 1.3 m (half of the difference between the maximum and minimum values)

Vertical shift (D) = (3.7 + 1.1) / 2 = 2.4 m (average of maximum and minimum values)

To find the frequency (B) and phase shift (C), we need to determine the period of the tidal cycle. In general, the period is the time it takes for one complete cycle. In this case, the period is 12 hours (between t = 3 hours and t = 9 hours).

The formula for the period (T) is given by T = 2π/B.

Substituting the given values, we have:

12 = 2π/B

Solving for B, we obtain:

B = 2π/12 = π/6

Now we can write the equation expressing the depth (y) in terms of time (t) as follows:

y = 1.3 * sin((π/6)(t - C)) + 2.4

We still need to find the value of C. We know that at t = 3 hours, the depth is a minimum of 1.1 m. Substituting these values into the equation:

1.1 = 1.3 * sin((π/6)(3 - C)) + 2.4

From here, you can solve for C by rearranging the equation and using trigonometric identities or by graphically plotting the function and finding the value of C where the minimum occurs at t = 3 hours.