Question

the depth of the water at the end of a pier changes periodically along with the movement of tides. on a particular day, low tides occur at 12:00 am and 12:30 pm, with a depth of 2.5 m, while high tides occur at 6:15 am and 6:45 pm, with a depth of 5.5 m. let t = 0 be 12:00 am. write a cosine model, d = a cos(bt) + k, for the depth as a function of time. this amplitude is 1.5 meters. a = -1.5. the period is 4 hours. b =

Explanation:

Step1: Find the amplitude

The amplitude \(a\) of a cosine function \(y = a\cos(bx)+k\) is half the distance between the maximum and minimum values. The maximum depth is \(5.5\)m and the minimum is \(2.5\)m. So, \(a=\frac{5.5 - 2.5}{2}=\frac{3}{2}=1.5\)

Step2: Find the period

The time between two consecutive low - tides (or high - tides) gives the period. The low - tides occur at \(6:15\) am and \(6:45\) pm. The time difference between \(6:15\) am and \(6:45\) pm is \(12.5\) hours. But for a cosine function, if we consider the general tide - cycle (which is a sinusoidal function), the period \(T\) of a tide - cycle (from one low - tide to the next low - tide) is \(12\) hours. The formula for the period of \(y = a\cos(bx)+k\) is \(T=\frac{2\pi}{b}\). Since \(T = 12\), then \(b=\frac{2\pi}{T}=\frac{2\pi}{12}=\frac{\pi}{6}\)

Step3: Find the vertical shift \(k\)

The vertical shift \(k\) is the mid - value of the maximum and minimum. \(k=\frac{5.5 + 2.5}{2}=4\)

Answer:

The amplitude \(a = 1.5\), \(b=\frac{\pi}{6}\) and the model is \(d = 1.5\cos(\frac{\pi}{6}t)+4\)