Question

during high tide around 11:00 p.m., the water level is 3 feet above a marker on a pier. the following day, during low tide around 5:12 a.m., the water level has dropped 3 feet below the marker. the height of the water is modeled by the function $y = 3cos(\frac{24pi}{149}x)$, where $x$ is the time in hours since 11:00 p.m. what is the height of the water at 2:15 a.m.? 0.22 feet below the marker 0.07 feet below the marker 1.26 feet above the marker 1.39 feet above the marker

Explanation:

Step1: Calculate time elapsed

The time from 11:00 p.m. to 2:15 a.m. is 3 hours and 15 minutes. Since 15 minutes is $\frac{15}{60}=\frac{1}{4}= 0.25$ hours, $x = 3.25$ hours.

Step2: Substitute $x$ into function

Substitute $x = 3.25$ into $y = 3\cos(\frac{24\pi}{149}x)$. So $y=3\cos(\frac{24\pi}{149}\times3.25)$. First, calculate $\frac{24\pi}{149}\times3.25=\frac{24\pi\times3.25}{149}=\frac{78\pi}{149}\approx1.65$. Then $y = 3\cos(1.65)$. Using a calculator, $\cos(1.65)\approx - 0.073$. So $y=3\times(- 0.073)=-0.219\approx - 0.22$.

Answer:

A. 0.22 feet below the marker