Chapter 4.4, Problem 75E

Solution

a.

To find: The time at which the first low tide occurs on Labor Day.

The time at which the first low tide occur on labor day is $1.00\text{ AM}$ .

Given information:

The sinusoidal function of time $=\frac{1}{2}$a lunar day.

Calculation:

One lunar day is equal to $12\text{ hr }24\min$ .

The time difference between the low and high is given by,

  $t=\frac{1}{2}\text{ Lunar day}$

So, half of the time is $6\text{ hr }12\min$ .

Given that the high point is at $7:12\text{ AM}$ .

Then subtract $6\text{ hr }12\min$ from the high point to get the time $1:00\text{ AM}$ on the same day.

Therefore, the time at which the first low tide occur on labor day is $1.00\text{ AM}$ .

b.

To find: The approximate depth of the water at $4:00\text{ AM}$ and at $\text{9}\text{.00}\text{PM}$ .

The approximate depth of the water at $4:00\text{ AM}$ is $8.5\text{ feet}$ .

The approximate depth of the water at $\text{9}\text{.00}\text{PM}$ is $10.7\text{ feet}$ .

Given information:

The sinusoidal function of time = $=\frac{1}{2}$a lunar day.

Calculation:

To find the depth $h$ , by generating the sinusoidal function of time.

  $h\left(t\right)=a\sin \left(b\left(t+c\right)\right)+d$

The amplitude a is given by,

  $a\Rightarrow \max -\left(\frac{\max +\min }{2}\right)=\left(\frac{\max +\min }{2}\right)-\min$

By solving to get,

  $a=2$

The period $\left|\frac{2\pi }{b}\right|\text{=744 minutes}$

  $\left|b\right|=\frac{2\pi }{744}$

The vertical offset $d$ is given by,

  $\begin{array}{c}d=\left(\frac{\max +\min }{2}\right)\\ =9\end{array}$

Substituting the values of $a,b,d$ in the above equation to get,

  $h\left(t\right)=2\sin \left(\frac{2\pi }{744}\left(t+c\right)\right)+9$

The temporal offset is $c$ , which is found by plugging in a value for $\left(t,h\left(t\right)\right)$ .

Here, let’s take the low tide at $1:24\text{ PM}$ or $\left(84,7\right)$

  $\begin{array}{l}c=\frac{744}{2\pi }\arcsin \left(\frac{7-9}{2}\right)-84\\ c=-270\end{array}$

Thus,

  $h\left(t\right)=2\sin \left(\frac{2\pi }{744}\left(t-270\right)\right)+9$

At $4:00\text{ PM}$ ,

  $\text{h}\left(240\right)\approx 8.5$

At $9:00\text{ PM}$ ,

  $\text{h}\left(1260\right)\approx 10.7$

Therefore,

The approximate depth of the water at $4:00\text{ AM}$ is $8.5\text{ feet}$_.

The approximate depth of the water at $4:00\text{ AM}$ is $10.7\text{ feet}$_.

c.

To find: The first time on the Labor Day that the water is $9\text{ feet}$ deep.

The time at which the water is $9\text{ feet}$ deep on the Labor Day is $4:30\text{ AM}$ .

Given information:

The sinusoidal function of time $=\frac{1}{2}$ a lunar day.

Calculation:

To find the depth $h$ , by generating the sinusoidal function of time

  $h\left(t\right)=a\sin \left(b\left(t+c\right)\right)+d$

The amplitude a is given by,

  $a\Rightarrow \max -\left(\frac{\max +\min }{2}\right)=\left(\frac{\max +\min }{2}\right)-\min$

By solving to get,

  $a=2$

The period $\left|\frac{2\pi }{b}\right|=744\text{ minutes}$

  $\left|b\right|=\frac{2\pi }{744}$

The vertical offset $d$ is given by,

  $\begin{array}{c}d=\left(\frac{\max +\min }{2}\right)\\ =9\end{array}$

Substituting the values of $a,b,d$ in the above equation to get,

  $h\left(t\right)=2\sin \left(\frac{2\pi }{744}\left(t+c\right)\right)+9$

The temporal offset is $c$ , which is found by plugging in a value for $\left(t,h\left(t\right)\right)$ .

Here, let’s take the low tide at $1:24\text{ PM}$ or $\left(84,7\right)$

  $\begin{array}{l}c=\frac{744}{2\pi }\arcsin \left(\frac{7-9}{2}\right)-84\\ c=-270\end{array}$

Thus,

  $h\left(t\right)=2\sin \left(\frac{2\pi }{744}\left(t-270\right)\right)+9$

Now to solve $h\left(t\right)=9$

  $\begin{array}{c}h\left(t\right)=9\\ =2\sin \left(\frac{2\pi }{744}\left(t-270\right)\right)+9\\ t=\frac{744}{2\pi }\arcsin \left(0\right)+270\\ t=270\end{array}$

Therefore, the time at which the water is $9\text{ feet}$ deep on the labor day is $4:30\text{ AM}$ .