Chapter 6.7, Problem 49E

a.

To determine

Find the amplitude of sinusoidal function.

Answer to Problem 49E

  $1.72feet$

Explanation of Solution

Given information:

The Daytona Beach first high tide rose to $3.99feetat12:03A.M$ low tide at $0.55feetat6:24A.M$ . The second high tide was recorded at $12:19P.M.$

Calculation:

Consider the high and low tide, $3.99feetat12:03A.M$ , $0.55feetat6:24A.M$ .

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

  $\begin{array}{l}A=\frac{3.99-0.55}{2}\\ A=1.72feet\end{array}$

Hence, amplitude is, $1.72feet$ .

b.

To determine

Find the vertical shift of sinusoidal function.

Answer to Problem 49E

  $2.27feet$

Explanation of Solution

Given information:

The Daytona Beach first high tide rose to $3.99feetat12:03A.M$ low tide at $0.55feetat6:24A.M$ . The second high tide was recorded at $12:19P.M.$

Calculation:

Consider the high and low tide, $3.99feetat12:03A.M$ , $0.55feetat6:24A.M$ .

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

  $\begin{array}{l}h=\frac{3.99+0.55}{2}\\ h=\frac{4.54}{2}\\ h=2.27feet\end{array}$

Hence, vertical shift is, $2.27feet$ .

c.

To determine

Find the period of sinusoidal function.

Answer to Problem 49E

  $12.3h$

Explanation of Solution

Given information:

The Daytona Beach first high tide rose to $3.99feetat12:03A.M$ low tide at $0.55feetat6:24A.M$ . The second high tide was recorded at $12:19P.M.$

Calculation:

Consider the high and low tide, $3.99feetat12:03A.M$ , $0.55feetat6:24A.M$ .

The period of tide is

  $\begin{array}{l}12.19+12-12.03=12.16\\ =12+\frac{16}{60}\\ =12+0.3\\ =12.3h\end{array}$

  $\begin{array}{l}hencek,\\ \frac{2\pi }{k}=12\left(60\right)+23\\ k=\frac{2\pi }{743}\end{array}$

Hence, period is, $12.3h$ .

d.

To determine

Write sinusoidal function for tide.

Answer to Problem 49E

  $y=1.72\sin \left(\frac{2\pi }{12.3}t+1.55\right)+2.27$

Explanation of Solution

Given information:

The Daytona Beach first high tide rose to $3.99feetat12:03A.M$ low tide at $0.55feetat6:24A.M$ . The second high tide was recorded at $12:19P.M.$

Calculation:

Consider the high and low tide, $3.99feetat12:03A.M$ , $0.55feetat6:24A.M$ .

The standard form of sinusoidal function is,

  $y=A\sin \left(kt+c\right)+h$

Substitute the given values,

  $\begin{array}{l}y=1.72\sin \left(\frac{2\pi }{12.3}t+c\right)+2.27\\ putt=0.05andy=3.99tocalculatec,\\ 3.99=1.72\sin \left(\frac{2\pi }{12.3}\left(0.05\right)+c\right)+2.27\\ \frac{3.99-2.27}{1.72}=\sin \left(0.0255+c\right)\\ 1.5708=0.0255+c\\ c=1.55\end{array}$

  $y=1.72\sin \left(\frac{2\pi }{12.3}t+1.55\right)+2.27$

Hence, sinusoidal function is $y=1.72\sin \left(\frac{2\pi }{12.3}t+1.55\right)+2.27$ .

e.

To determine

Find the height of the water at noon.

Answer to Problem 49E

  $3.96feet$

Explanation of Solution

Given information:

The Daytona Beach first high tide rose to $3.99feetat12:03A.M$ low tide at $0.55feetat6:24A.M$ . The second high tide was recorded at $12:19P.M.$

Calculation:

Consider the,

  $y=1.72\sin \left(\frac{2\pi }{12.3}t+1.55\right)+2.27$

Put $t=12$ in above function,

  $\begin{array}{l}y=1.72\sin \left(\frac{2\pi }{12.3}\left(12\right)+1.55\right)+2.27\\ y=1.72\sin \left(\frac{24\pi }{12.3}+1.55\right)+2.27\\ y=3.96feet\end{array}$

Hence, the height of water is $3.96feet$ .