Chapter 6.8, Problem 43E

a.

To determine

Find the time of next high tide.

Answer to Problem 43E

  $6.42P.M.$

Explanation of Solution

Given information:

The Hilton Head first high tide rose to $6:18A.M$ be $7.05feet$ .low tide $-0.30feet$ . The period for oscillation of the tides is $12$ hours and $24$ minutes.

Calculation:

Consider the $6:18A.M$ be $7.05feet$ .low tide $-0.30feet$ The period for oscillation of the tides is $12$ hours and $24$ minutes.

  $\begin{array}{l}6:18+12:24=18:42\\ or\\ 6.42P.M.\end{array}$

Hence, next high tide will occur at $6.42P.M.$

b.

To determine

Find the period of oscillation of tide.

Answer to Problem 43E

  $12.4h$

Explanation of Solution

Given information:

The Hilton Head first high tide rose to $6:18A.M$ be $7.05feet$ .low tide $-0.30feet$ . The period for oscillation of the tides is $12$ hours and $24$ minutes.

Calculation:

Consider the $6:18A.M$ be $7.05feet$ .low tide $-0.30feet$ the period for oscillation of the tides is $12$ hours and $24$ minutes.

  $\begin{array}{l}12:24=12+\frac{24}{60}\\ 12+0.4\\ 12.4h\end{array}$

Hence, the period of oscillation is $12.4h$

c.

To determine

Find the amplitude of tide.

Answer to Problem 43E

  $3.675ft.$

Explanation of Solution

Given information:

The Hilton Head first high tide rose to $6:18A.M$ be $7.05feet$ .low tide $-0.30feet$ . The period for oscillation of the tides is $12$ hours and $24$ minutes.

Calculation:

Consider the $6:18A.M$ be $7.05feet$ .low tide $-0.30feet$ the period for oscillation of the tides is $12$ hours and $24$ minutes.

  $\begin{array}{l}A=\frac{7.05-\left(-0.3\right)}{2}\\ A=\frac{7.05+0.3}{2}\\ A=\frac{7.35}{2}\\ A=3.675ft.\end{array}$

Hence, the amplitude is $3.675ft.$

d.

To determine

Write the sinusoidal function for $t=0$ .

Answer to Problem 43E

  $y=3.675\sin \left(\frac{\pi }{6.2}t-1.620\right)+3.375$

Explanation of Solution

Given information:

The Hilton Head first high tide rose to $6:18A.M$ be $7.05feet$ .low tide $-0.30feet$ . The period for oscillation of the tides is $12$ hours and $24$ minutes.

Calculation:

Consider the $6:18A.M$ be $7.05feet$ .low tide $-0.30feet$ the period for oscillation of the tides is $12$ hours and $24$ minutes.

The general expression for sinusoidal function is,

  $y=A\sin \left(kt+c\right)+h$

The vertical shift $h$ will be average of high and low tide.

  $\begin{array}{l}h=\frac{7.05+\left(-0.3\right)}{2}\\ h=\frac{7.05-0.3}{2}\\ h=\frac{6.75}{2}\\ h=3.375\\ k=2\pi /12.4\\ k=\frac{\pi }{6.2}\end{array}$

Substitute all values in standard equation,

  $\begin{array}{l}y=3.675\sin \left(\frac{\pi }{6.2}t+c\right)+3.375\\ putt=6.3,y=7.05tocalculatec,\\ 7.05=3.675\sin \left(\frac{\pi }{6.2}\left(6.3\right)+c\right)+3.375\\ \frac{7.05-3.375}{3.675}=\sin \left(\frac{\pi }{6.2}\left(6.3\right)+c\right)\\ \frac{7.05-3.375}{3.675}=\sin \left(3.19+c\right)\\ 3.19+c=arcsin\left(1\right)\\ 3.19+c=1.5708\\ c=-1.62\\ Now,the\sin usoidalfunctionbe,\\ y=3.675\sin \left(\frac{\pi }{6.2}t-1.620\right)+3.375\end{array}$

Hence, the sinusoidal function for $t=0$ is $y=3.675\sin \left(\frac{\pi }{6.2}t-1.620\right)+3.375$

e.

To determine

Find the time for tide height $6feet$ first time that day.

Answer to Problem 43E

  $4.767AM$

Explanation of Solution

Given information:

The Hilton Head first high tide rose to $6:18A.M$ be $7.05feet$ .low tide $-0.30feet$ . The period for oscillation of the tides is $12$ hours and $24$ minutes.

Calculation:

Consider the $6:18A.M$ be $7.05feet$ .low tide $-0.30feet$ the period for oscillation of the tides is $12$ hours and $24$ minutes.

The general expression for sinusoidal function is,

  $y=A\sin \left(kt+c\right)+h$

  $y=3.675\sin \left(\frac{\pi }{6.2}t-1.620\right)+3.375$

  $\begin{array}{l}puty=6\\ 6=3.675\sin \left(\frac{\pi }{6.2}t-1.620\right)+3.375\\ \frac{6-3.375}{3.675}=\sin \left(\frac{\pi }{6.2}t-1.620\right)\\ 0.71428=\sin \left(\frac{\pi }{6.2}t-1.620\right)\\ arc\sin 0.71428=\left(\frac{\pi }{6.2}t-1.620\right)\\ t=\frac{0.7955+1.62}{\pi /6.2}\\ t=4.767AM\end{array}$

Hence, the time for tide height $6feet$ first time that day $4.767AM$ .