Chapter 6.6, Problem 4CFU

a.

To determine

To find: the location of buoy at $t=0$ .

Answer to Problem 4CFU

  $y=-5$

Explanation of Solution

Given information:

The function that represents a buy bobbing up and down in water is shown below.

  $y=-5\cos \left(\frac{\pi }{6}t\right)$

Calculation:

Substitute $t=0$ in the given function and solve $y$ as follows:

  $\begin{array}{c}y=-5\cos \left(\frac{\pi }{6}\left(0\right)\right)\\ =-5\cos \left(0\right)\\ =-5\left(1\right)\\ =-5\end{array}$

Thus the location of buoy at $t=0$ is $y=-5$ .

b.

To determine

To find: the maximum height of buoy.

Answer to Problem 4CFU

  $y=5$

Explanation of Solution

Given information:

The function that represents a buy bobbing up and down in water is shown below.

  $y=-5\cos \left(\frac{\pi }{6}t\right)$

Calculation:

Since the amplitude is negative, so in order to get the maximum height $\left(y\right)$ the value of cosine function should be negative. Substitute $t=6$ in the given function and solve $y$ as follows:

  $\begin{array}{c}y=-5\cos \left(\frac{\pi }{6}\left(6\right)\right)\\ =-5\cos \left(\pi \right)\\ =-5\left(-1\right)\\ =5\end{array}$

Thus the maximum height of buoy is $y=5$ .

c.

To determine

To find: the location of buoy at $t=7$ .

Answer to Problem 4CFU

  $y=4.33$

Explanation of Solution

Given information:

The function that represents a buy bobbing up and down in water is shown below.

  $y=-5\cos \left(\frac{\pi }{6}t\right)$

Calculation:

Substitute $t=7$ in the given function and solve $y$ as follows:

  $\begin{array}{c}y=-5\cos \left(\frac{\pi }{6}\left(7\right)\right)\\ =-5\cos \left(\frac{7\pi }{6}\right)\\ =-5\left(-\frac{\sqrt{3}}{2}\right)\\ =\frac{5\sqrt{3}}{2}\\ =4.33\end{array}$

Thus the location of buoy at $t=7$ is $y=4.33$ .