Chapter 8.5, Problem 9PPS

(a)

To determine

To write:equation that relates the depth and the water pressure.

Answer to Problem 9PPS

The equation is $y=kx$ , that is, $y=0.4x$

Explanation of Solution

Given:

Depth(ft)	Pressure(psi)
$x$	$y$
$33$	$14.7$
$66$	$29.4$
$99$	$44.1$
$132$	$58.8$

Calculation:

Consider the table that shows the relationship between depth and pressure of the water.

Depth(ft)	Pressure(psi)
$x$	$y$
$33$	$14.7$
$66$	$29.4$
$99$	$44.1$
$132$	$58.8$

In order to find the equation that relates to the depth and water pressure, first find constant rate of change.

So, take any two points and use the formula for rate of change.

  $Rateofchange=\frac{changeiny}{changeinx}$

It means rate of change is change in pressure with respect to change in depth.

So, consider two points as : $(33,14,7),(66,29,4)$

Thus, constant rate of change is:

  $\begin{array}{l}Rateofchange=\frac{29.4-14.7}{66-33}\\ =\frac{14.7}{33}\\ =0.4\end{array}$

Hence, the equation is $y=kx$ , that is, $y=0.4x$

Conclusion:

Hence, the equation is $y=kx$ , that is, $y=0.4x$

(b)

To determine

To find: the water pressure at the given depth

Answer to Problem 9PPS

The water pressure at the depth of $900$ feet is $360psi$ .

Explanation of Solution

Calculation:

To find the pressure of the water when depth is $900$ feet, put the value of $x=900$ in equation.

  $\begin{array}{l}y=0.4(900)\\ =360\end{array}$

Therefore, the water pressure at the depth of $900$ feet is $360psi$

Conclusion:

Therefore, te water pressure at the depth of $900$ feet is $360psi$ .