Chapter 6.6, Problem 11E

To determine

To find: the equation for the sine function that represents the oscillation of tides in Georgia.

Answer to Problem 11E

  $y=3.55\sin \left(\frac{\pi }{6.2}t+\frac{2.34\pi }{3.1}\right)+4.24$

Explanation of Solution

Given information:

Equilibrium point = 4.24 feet

Amplitude = 3.55 feet

Phase shift = $-$ 4.68 hours

Period = 12.40 hours

Calculation:

The general equation for sine function is as follows:

  $y=A\sin \left(k\theta +c\right)+h$

Here, $A$ is coefficient of absolute value of amplitude, $k$ is absolute value of coefficient of $\theta$ and $h$ is the horizontal shift.

The value of $A$ can be evaluated as follows:

  $\begin{array}{c}\text{Amplitude}=\left|\pm A\right|\\ 3.55=\left|\pm A\right|\\ A=\pm 3.55\end{array}$

The value of $k$ can be evaluated as follows:

  $\begin{array}{c}\text{Period}=\frac{2\pi }{k}\\ 12.40=\frac{2\pi }{k}\\ k=\frac{0.2\pi }{12.40}\\ =\frac{\pi }{6.2}\end{array}$

The value of $c$ can be evaluated as follows:

  $\begin{array}{c}\text{Phase shift}=\frac{-c}{k}\\ -4.68=\frac{-c}{\frac{\pi }{6.2}}\\ c=\frac{\pi }{6.2}\left(4.68\right)\\ =\frac{2.34\pi }{3.1}\end{array}$

Substitute the value of $A,k,c\text{ and }h$ in the general equation of sine as shown below.

  $\begin{array}{l}y=A\sin \left(k\theta +c\right)+h\\ y=3.55\sin \left(\frac{\pi }{6.2}t+\frac{2.34\pi }{3.1}\right)+4.24\left[\theta =t\right]\end{array}$

Thus, the equation for the sine function that represents the oscillation of tides in Georgiais $y=3.55\sin \left(\frac{\pi }{6.2}t+\frac{2.34\pi }{3.1}\right)+4.24$ .