Question

Given y=Acos(Bx+C) + D. Throughout the day the depth of water at the end of a pier varies with the tides. High tide occurs at 4 am with a depth of 6 meters. Low tide occurs at 10 am with a depth of 2 meters. Model the problem by using the given trigonometric equation to show the depth of the water t hours after midnight showing all your work.

Solve the problem by finding the depth of the water at noon, explaining, the reasoning.

Answer 1

Given $y = A\cos(Bx + C) + D$. Throughout the day the depth of water at the end of a pier varies with the tides. High tide occurs at 4 am with a depth of 6 meters. Low tide occurs at 10 am with a depth of 2 meters. Model the problem by using the given trigonometric equation to show the depth of the water $t$ hours after midnight showing all your work.

Solve the problem by finding the depth of the water at noon, explaining the reasoning.

Solution

y varies from 2 to 6, which is $4 \pm 2$, so $A = 2$, $D = 4$

period is 12 hours, so $B = \frac{2\pi}{12} = \frac{\pi}{6}$

high tide is at $x = 4$, not $x = 0$

$$C = - \frac {2 \pi}{3}$$ $$\begin{array}{l} y = 2 \cos \left(\frac {\pi}{6} (x - 4)\right) + 4 \\ = 2 \cos \left(\frac {\pi}{6} x - \frac {2 \pi}{3}\right) + 4 \\ \end{array}$$

Equation should be

$$h = 2 \cos \left(\frac {\pi}{6} x - \frac {2 \pi}{3}\right) + 4$$

where $h$ is the height of the water and $t$ is the time (number of hours after midnight)