Chapter 5, Problem 11PS

a.

To determine

Algebraically find the times at which the high and low tides occur.

Answer to Problem 11PS

  $\boxed{t=12.4n+6.2}$

  $\boxed{t=12.4n}$

Explanation of Solution

Given information:

The tide, or depth of the ocean near the shore, changes throughout the day. The water depth $d$ (in feet) of a bay can be modelled by

  $d=35-28\cos \frac{\pi }{6.2}t$

where $t$ represents the time in hours, with $t=0$ corresponding to $\text{12}:00\text{ a}.\text{m}.$

Calculation:

Let us consider the tide, or depth of the ocean near the shore, changes throughout the day. The water depth $d$ (in feet) of a bay can be modelled by

  $d=35-28\cos \frac{\pi }{6.2}t$

The low tides occur when the depth of the water is least. The value of $t$ for which the above expression has the least value can be calculated if we know the values of $t$ for $\cos \frac{\pi }{6.2}t$ which is maximum.

The maximum value of $\cos \frac{\pi }{6.2}t$ is $\text{1}$ .from this, we can determine the times when low tides occur.

  $\cos \frac{\pi }{6.2}t=1$

  $\frac{\pi }{6.2}t=2n\pi$

  $t=2n\times 6.2$

  $\boxed{t=12.4n}$

Similarly we can calculate the high tides occur when the depth of the water is greatest. The value of $t$ for which the above expression has the greatest value can be calculated if we know the values of $t$ for $\cos \frac{\pi }{6.2}t$ which is minimum.

  $\cos \frac{\pi }{6.2}t=-1$

  $\frac{\pi }{6.2}t=2(n+1)\pi$

  $t=2(n+1)\times 6.2$

  $\boxed{t=12.4n+6.2}$

Hence, the times at which the high and low tides occur are $\boxed{t=12.4n+6.2}$ and $\boxed{t=12.4n}$ respectively.

b.

To determine

If possible, algebraically find the time(s) at which the water depth is $\text{3}.\text{5 feet}$ .

Answer to Problem 11PS

There is no value of $t$ for which the depth will be $\text{3}.\text{5 feet}$ .

Explanation of Solution

Given information: The tide, or depth of the ocean near the shore, changes throughout the day. The water depth $d$ (in feet) of a bay can be modelled by

  $d=35-28\cos \frac{\pi }{6.2}t$

where $t$ represents the time in hours, with $t=0$ corresponding to $\text{12}:00\text{ a}.\text{m}.$

Calculation:

Since, the maximum value of $\cos \frac{\pi }{6.2}t$ is $\text{1}$ . So, the minimum value for depth of bay is

  $35-28\cos \frac{\pi }{6.2}t=35-28\left(1\right)$

  $=7$

Hence, there is no value of $t$ for which the depth will be $\text{3}.\text{5 feet}$ .

c.

To determine

Use a graphing utility to verify your results from parts (a) and (b).

Answer to Problem 11PS

  

Explanation of Solution

Given information: The tide, or depth of the ocean near the shore, changes throughout the day. The water depth $d$ (in feet) of a bay can be modelled by

  $d=35-28\cos \frac{\pi }{6.2}t$

where $t$ represents the time in hours, with $t=0$ corresponding to $\text{12}:00\text{ a}.\text{m}.$

Calculation:

The above two parts can be verified from the plot given below for the depth of the water in the bay at different points of the time of the day.

  

The minimum of the graph represents the time instants where low tides occur and maximum of the graph represents the time instants where high tides occur.

From the given plot, we can see that the minimum value of the depth is $7feet$ .

Hence, there are no time instants where the depth of the water is $\text{3}.\text{5 feet}$ .