Chapter 6.6, Problem 15E

a.

To determine

Find the amplitude of sinusoidal function.

Answer to Problem 15E

  $5.685feet$

Explanation of Solution

Given information:

The Burncoat Head first high tide rose to $13.25feetat4.30A.M$ low tide at $1.88feetat10.51A.M$ . The second high tide was recorded at $4.53P.M$ .

Calculation:

Consider the high and low tide, $13.25feetat4.30A.M$ , $1.88feetat10.51A.M$ .

The amplitude is half the difference between high and low tide,

  $\begin{array}{l}A=\frac{13.25-1.88}{2}\\ A=5.685feet\end{array}$

Hence, amplitude is, $5.685feet$ .

b.

To determine

Find the vertical shift of sinusoidal function.

Answer to Problem 15E

  $7.565feet$

Explanation of Solution

Given information:

The Burn coat Head first high tide rose to $13.25feetat4.30A.M$ low tide at

  $1.88feetat10.51A.M$ . The second high tide was recorded at $4.53P.M$ .

Calculation:

Consider the high and low tide, $13.25feetat4.30A.M$ , $1.88feetat10.51A.M$ .

The vertical shift is the average of the high and low tide,

  $\begin{array}{l}h=\frac{13.25+1.88}{2}\\ h=\frac{15.13}{2}\\ 7.565feet\end{array}$

Hence, vertical shift is, $7.565feet$ .

c.

To determine

Find the period of sinusoidal function.

Answer to Problem 15E

  $12.4h$

Explanation of Solution

Given information:

The Burn coat Head first high tide rose to $13.25feetat4.30A.M$ low tide at

  $1.88feetat10.51A.M$ . The second high tide was recorded at $4.53P.M$ .

Calculation:

Consider the high and low tide, $13.25feetat4.30A.M$ , $1.88feetat10.51A.M$ .

The period of tide is

  $\begin{array}{l}12:00+4:53-4:30=12:23\\ p=12+\frac{23}{60}\\ p=12+0.4\\ p=12.4h\end{array}$

Hence, period is, $12.4h$ .

d.

To determine

Write sinusoidal function for tide.

Answer to Problem 15E

  $y=5.685\sin \left(\frac{\pi }{6.2}t-0.71\right)+7.565$

Explanation of Solution

Given information:

The Burn coat Head first high tide rose to $13.25feetat4.30A.M$ low tide at

  $1.88feetat10.51A.M$ . The second high tide was recorded at $4.53P.M$ .

Calculation:

Consider the high and low tide, $13.25feetat4.30A.M$ , $1.88feetat10.51A.M$ .

The standard form of sinusoidal function is,

  $y=A\sin \left(kt+c\right)+h$

Substitute the given values,

  $\begin{array}{l}y=5.685\sin \left(\frac{2\pi }{12.4}t+c\right)+7.565\\ putt=4.5andy=13.25tocalculatec,\\ 13.25=5.685\sin \left(\frac{2\pi }{12.4}\left(4.5\right)+c\right)+7.565\\ \frac{13.25-7.565}{5.685}=\sin \left(\frac{2\pi }{12.4}\left(4.5\right)+c\right)\\ 1=\sin \left(2.2811+c\right)\\ 1.5714=2.2811+c\\ c=-0.71\\ y=5.685\sin \left(\frac{\pi }{6.2}t-0.71\right)+7.565\end{array}$

Hence, sinusoidal function is $y=5.685\sin \left(\frac{\pi }{6.2}t-0.71\right)+7.565$ .

e.

To determine

Find the height of the water at $7:30P.M.$

Answer to Problem 15E

  $8.99feet$

Explanation of Solution

Given information:

The Burn coat Head first high tide rose to $13.25feetat4.30A.M$ low tide at

  $1.88feetat10.51A.M$ . The second high tide was recorded at $4.53P.M$ .

Calculation:

Consider the,

  $y=5.685\sin \left(\frac{\pi }{6.2}t-0.71\right)+7.565$

Put $t=19.5$ in above function,

  $\begin{array}{l}y=5.685\sin \left(\frac{\pi }{6.2}\left(19.5\right)-0.71\right)+7.565\\ y=8.99feet\end{array}$

Hence, the height of water is $8.99feet$ .