Chapter 9.1, Problem 54SR

(a)

To determine

To find: The direct variation equation that represents the situation.

Answer to Problem 54SR

The equation of the pressure is $P=0.43d$ .

Explanation of Solution

Given:

The pressure on the year of the swimmer varies directly with the depth at which the swimmer is swimming.

Calculation:

Consider the expression for the pressure on the ear is directly proportional to the depth d with the constant k as,

  $P=kd$

For the depth of 10 feet the pressure is of 4.3 psi. Substitute the values as,

  $\begin{array}{l}4.3=10k\\ k=0.43\end{array}$

Thus, the equation of the pressure is,

  $P=0.43d$

(b)

To determine

To find: The pressure at the depth at 60 feet.

Answer to Problem 54SR

The pressure is of $25.\text{8}\text{psi}$ .

Explanation of Solution

Given:

The depth is of 60 feet.

Calculation:

Consider the equation of the pressure is,

  $P=0.43d$

Then the pressure at 60 feet is,

  $\begin{array}{c}P=0.43\left(60\right)\\ =25.\text{8}\text{psi}\end{array}$

(c)

To determine

To find: The depth required for the pressure of 65 pounds.

Answer to Problem 54SR

The depth is of $151.16\text{feet}$ .

Explanation of Solution

Given:

The pressure is of 65 pounds.

Calculation:

Consider the pressure is of 65 pounds then the depth is,

  $\begin{array}{l}65=0.43d\\ d=\frac{65}{0.43}\\ d=151.16\text{feet}\end{array}$

(d)

To determine

To find: The table for the number of pounds of pressure at various depth of the water and draw the pressure versus depth graph.

Answer to Problem 54SR

The table is shown in Table 1 and the graph is shown in Figure 1

Explanation of Solution

Calculation:

Consider the table for the number of pounds of pressure at various depth is shown in Table 1

Table 1

Depth	Pressure
10	4.3
20	8.6
30	12.9
40	17.2

From the above points the graph for the function is shown in Figure 1

  

Figure 1