Chapter 14.5, Problem 49PPS

a.

To determine

To find: the maximum depth of the River T on that day.

Answer to Problem 49PPS

Maximum depth is $y_{\max }=11meters$ .

Explanation of Solution

Given information: Rivers due to ocean tides, the depth y in meters of the River T in

London varies as a sine function of x , the hour of the day. On a certain day the function was $y=3\sin \left[\frac{\pi }{6}(x-4)\right]+8$

, where x = 0, 1, 2…………..24 corresponding to 12:00 midnight, 1:00 A.M., 2:00 A.M. …12:00 midnight the next night.

Calculation:

Given equation is:

  $y=3\sin \left[\frac{\pi }{6}(x-4)\right]+8$ , where x = 0, 1, 2…………..24.

The depth reaches its maximum when $\sin \left[\frac{\pi }{6}(x-4)\right]$ reaches its maximum. Since $-1\le \sin \theta \le 1$

, the maximum of the sine will be 1:

  $\begin{array}{l}\sin {\left[\frac{\pi }{6}(x-4)\right]}_{\max }=1\\ y_{\max }=3\times 1+8=3+8\\ y_{\max }=11meters.\end{array}$

Therefore, the maximum depth is $y_{\max }=11meters$ .

b.

To determine

To find: at what times does the maximum depth occur.

Answer to Problem 49PPS

Maximum depth occur at 7 A.M. and 7 P.M.

Explanation of Solution

Given information: Rivers due to ocean tides, the depth y in meters of the River T in

London varies as a sine function of x , the hour of the day. On a certain day the function was $y=3\sin \left[\frac{\pi }{6}(x-4)\right]+8$

, where x = 0, 1, 2…………..24 corresponding to 12:00 midnight, 1:00 A.M., 2:00 A.M. …12:00 midnight the next night.

Calculation:

In order to find at what time the maximum of y is reaches must solve the equation:

  $\begin{array}{c}\sin {\left[\frac{\pi }{6}(x-4)\right]}_{\max }=1\\ \frac{\pi }{6}(x-4)=\frac{\pi }{2}+2k\pi \\ \frac{\pi }{6}(x-4)=\frac{(4k+1)\pi }{2}\\ x-4=3(4k+1)\\ x-4=12k+3\\ x=12k+3+4\\ x=12k+7\\ k=0\Rightarrow x=7\\ k=1\Rightarrow x=12\times 1+7=19\end{array}$

Therefore, maximum depth occur at 7 A.M. and 19=7 A.M.+12 =7 P.M.