Chapter 10.5, Problem 28PS

Solution

(a)

To calculate: The sinusoidal model for tide depth $d$ as a function of time $t$ .

The sinusoidal model for the given data is $d=-6.5\cos \frac{\pi }{6}t+10$ .

Given information: The low tide is $3.5$ feet and after $6$ hours the high tide is $16.5$ feet.

Formula used: The general equation of the sinusoid function is given as:

  $y=a\sin b\left(x-h\right)+k$ or $y=a\cos b\left(x-h\right)+k$

Here, $a$ is amplitude, $h$ is the horizontal shift, $k$ is the vertical shift and $\frac{2\pi }{b}$ is the period.

Calculation:

Let $t=0$ represents midnight, so the minimum point is $\left(0,3.5\right)$ . The high tide occurs after $6$ hours, so the maximum point is $\left(6,16.5\right)$ .

The maximum value is $M=16.5$ and the minimum value is $m=3.5$ .

The minimum of the function is the $y$ intercept, so the model is a cosine function with no horizontal shift that is $h=0$ .

Calculated the vertical shift of the graph by the mean of the maximum and minimum values.

  $\begin{array}{c}k=\frac{16.5+3.5}{2}\\ =\frac{20}{2}\\ =10\end{array}$

The amplitude can be calculated by the half of the difference between maximum and minimum.

  $\begin{array}{c}a=\frac{3.5-16.5}{2}\\ =\frac{-13}{2}\\ =-6.5\end{array}$

Calculate the period of the function.

  $\begin{array}{c}P=2\left(6-0\right)\\ =12\end{array}$

From the graph it can be observed that the period is $12$ . Calculate the value of $b$ .

  $\begin{array}{c}P=\frac{2\pi }{b}\\ 12=\frac{2\pi }{b}\\ b=\frac{\pi }{6}\end{array}$

Substitute $-6.5$ for $a$ , $10$ for $k$ , $0$ for $h$ and $\frac{\pi }{6}$ for $b$ in the general equation of cosine.

  $\begin{array}{l}d=-6.5\cos \frac{\pi }{6}\left(t-0\right)+10\\ d=-6.5\cos \frac{\pi }{6}t+10\end{array}$

Therefore, the sinusoidal model for the given data is $d=-6.5\cos \frac{\pi }{6}t+10$ .

(b)

To calculate: The time when low and high tides occur in $24$ hour period.

The high tide occurs at $6:00$ and $18:00$ , and the low tides occurs at $0:00$ , $12:00$ and $24:00$ in $24$ hours.

Given information: The low tide is $3.5$ feet and after $6$ hours the high tide is $16.5$ feet.

Calculation:

From part (a), the sinusoidal model for given data is $d=-6.5\cos \frac{\pi }{6}t+10$ .

The high tide occurs for the depth $d=16.5$ . Substitute $16.5$ for $d$ in the above model.

  $\begin{array}{c}16.5=-6.5\cos \frac{\pi }{6}t+10\\ 6.5=-6.5\cos \frac{\pi }{6}t\\ -1=\cos \frac{\pi }{6}t\\ {\cos }^{-1}\left(-1\right)=\frac{\pi }{6}t\end{array}$

Further simplify to find the value of $t$ .

  $\begin{array}{c}\pi +2\pi n=\frac{\pi }{6}t\\ \left(\pi +2\pi n\right)\frac{6}{\pi }=t\\ t=6+12n\end{array}$

Here, $n$ is an integer. Substitute $0$ for $n$ .

  $\begin{array}{c}t=6+12\left(0\right)\\ =6\end{array}$

Substitute $1$ for $n$ .

  $\begin{array}{c}t=6+12\left(1\right)\\ =18\end{array}$

So, high tides occur at $6:00$ and $18:00$ in $24$ hours.

The low tide occurs for the depth $d=3.5$ . Substitute $3.5$ for $d$ in the above model.

  $\begin{array}{c}3.5=-6.5\cos \frac{\pi }{6}t+10\\ -6.5=-6.5\cos \frac{\pi }{6}t\\ 1=\cos \frac{\pi }{6}t\\ {\cos }^{-1}\left(1\right)=\frac{\pi }{6}t\end{array}$

Further simplify to find the value of $t$ .

  $\begin{array}{c}2\pi n=\frac{\pi }{6}t\\ \left(2\pi n\right)\frac{6}{\pi }=t\\ t=12n\end{array}$

Here, $n$ is an integer. Substitute $0$ for $n$ .

  $\begin{array}{c}t=12\left(0\right)\\ =6\end{array}$

Substitute $1$ for $n$ .

  $\begin{array}{c}t=12\left(1\right)\\ =12\end{array}$

Substitute $2$ for $n$ .

  $\begin{array}{c}t=12\left(2\right)\\ =24\end{array}$

So, low tides occur at $0:00$ , $12:00$ and $24:00$ .

Thus, the high tide occurs at $6:00$ and $18:00$ , and the low tides occurs at $0:00$ , $12:00$ and $24:00$ in $24$ hours.

(c)

To explain: The relation of the graph of the function in part (a) with the graph that shows the tide depth $d$ at Eastport $t$ hours after $3:00$ A.M.

The graph of the function in part (a) is shifted $3$ units to the left to form the graph of the tide depth $d$ at Eastport $t$ hours after $3:00$ A.M.

Given information: The low tide is $3.5$ feet and after $6$ hours the high tide is $16.5$ feet.

Calculation:

In part (a), it is given that $t=0$ represent midnight. So, the graph of the function started at $t=0$ .

Now, consider $3:00$ A.M. as the basis or start the graph at $t=3$ . So, move the graph $3$ units to the left.

Thus, the graph of the function in part (a) is shifted $3$ units to the left to form the graph of the tide depth $d$ at Eastport $t$ hours after $3:00$ A.M.