Chapter 4.8, Problem 41E

Solution

To find: The time interval when water is $10\text{ft}$ from the top of the pier.

Option $D$ is correct.

Given information:

At high tide at $8:15\text{P}\text{.M}\text{.}$ , the water level on the side of a pier is $9\text{ft}$ from the top. At low tide $6\text{hr}12\min$ later, the water level is $13\text{ft}$ from the top. At which of the following times in that interval is the water $10\text{ft}$ from the top of the pier.

  $\begin{array}{l}\left(A\right)\text{9:15}\text{P}\text{.M}\text{.}\\ \left(B\right)\text{9:48}\text{P}\text{.M}\text{.}\\ \left(C\right)\text{9:52}\text{P}\text{.M}\text{.}\\ \left(D\right)10\text{:19}\text{P}\text{.M}\text{.}\\ \left(E\right)10\text{:21}\text{P}\text{.M}\text{.}\end{array}$

Explanation:

Let $t$ be the amount of time past $8:15\text{P}\text{.M}\text{.}$ in minutes. Let distance $d$ be the distance the water level is from the top of the pier. The period $T$ of the oscillation is $\text{12}\text{hours}\text{24}\text{minutes}$ . The maximum value of the oscillation $M$ is $\text{13}\text{feet}$ . The minimum value of the oscillation $m$ is $9\text{feet}$ .

The relationship between $d$ and $t$ in term of negative cosine on setting $t=0$ to be the minimum value of $d$ is:

  $\begin{array}{l}d\left(t\right)=\frac{M-m}{2}\cos \left(\frac{2\Pi t}{T}\right)+\frac{M+m}{2}\\ =-2\cos \left(\frac{\Pi t}{372}\right)+11\end{array}$

Now, find the value of $t$ such that $d\left(t\right)=10$ :

  $\begin{array}{l}10=-2\cos \left(\frac{\Pi t}{372}\right)+11\\ -1=-2\cos \left(\frac{\Pi t}{372}\right)\\ \frac{\Pi t}{372}=\arccos \left(\frac{1}{2}\right)\\ t=\frac{372}{\Pi }\arccos \left(\frac{1}{2}\right)\\ t=125\text{minutes}\end{array}$

This corresponds to time two hours and five minutes past $8:15\text{P}\text{.M}\text{.}$ The time is closest to $10:20\text{P}\text{.M}\text{.}$

Option $D$ is correct.