I NEED HELP UNDERSTANDING THIS PLEASE

Question

Throughout the day the depth of water at the end of a pier varies with the tides. High tide occurs at 4:00 a.m. with a depth of 6 meters. Low tide occurs at 10:00 a.m. with a depth of 2 meters.  

1.  Model the problem by using the given trigonometric equation to show the depth (y) of the water x hours after midnight, showing all your work. y = A cos(Bx + C) + D

Start by sketching a graph of the situation – sketch 2 cycles. (Pick appropriate intervals for the x- and y-axes and make the horizontal axis in time, not radians. Hint: What time should x = 0 be?)

1. Use the above graph and any extra work needed to determine the amplitude, period, and horizontal shift, and vertical shift to model the equation. Period and phase shift must be in radians.

Amplitude = _________ 

Period (in time) = ________ convert period to radians: ___________________________

Horizontal shift (in time) = ________ convert phase shift to radians: _______________________

(To find the phase shift use: -CB=x, where x is the horizontal shift in time.)

Vertical shift = _________

Equation: __________________________________________________________

2.  A large boat coming in at noon needs at least 4 meters of water to dock at the end of the pier. Will the boat be able to safely dock? Solve the problem by using the equation to find the exact depth of the water at noon. Explain your reasoning.

Show work below: (Hint: how much time after x=0 is noon?)

Answer 1

We can start by calculating A and D, the amplitude and vertical shift, as follows:

D, the vertical shift, also called the sinusoidal axis, or the average value, can be calculated by averaging the y-value of the high point and the y-value of the low point: D = (6 + 2) / 2 = 4.

A, the amplitude, will be the difference of the high and low point y-values divided by 2: (6 - 2) / 2 = 2. We should also think of A as how far away the high and low points are from the average value, D.

Next, we calculate B, the parameter controlling the length of the period by using the formula: B = 2π / period. We are told the 1st high tide is at 4 am, while the 1st low tide is at 10 am. These are 6 hrs apart, which means the highs are 12 hrs apart. Thus, B = 2π / 12 = π/6.

Lastly, an unshifted cosine function will begin at a peak, when x = 0. This cosine function has a peak when time = 4, meaning it is shifted right by 4 from a normal cosine curve. So C = - 4.

Putting all together, we get the following:

x: time elapsed since midnight (in hrs)

y: depth of water (in ft)

y = 2cos(π/6(x - 4)) + 4.

If you wanted to give the answer in the form your teacher requested, distribute the B-value across (x - 4), though the form above is more commonly used.