Question

at the surface of the ocean, the water pressure is the same as the air pressure above the water, 15 lb/in². below the surface, the water pressure increases by 4.34 lb/in² for every 10 ft of descent (a) express the water pressure in lb/in² as a function of the depth d in feet below the ocean surface. p = (b) at what depth is the pressure 110 lb/in²? (round your answer to the nearest whole number.) ft

Explanation:

Part (a)

Step1: Identify initial pressure and rate

The initial pressure at the surface (\(d = 0\)) is \(15\) lb/in². The rate of pressure increase is \(\frac{4.34}{10}\) lb/in² per foot (since it increases by \(4.34\) lb/in² for every \(10\) ft descent). Simplifying the rate, \(\frac{4.34}{10}= 0.434\) lb/in² per foot.

Step2: Formulate the linear function

A linear function has the form \(P(d)=m\times d + b\), where \(m\) is the slope (rate) and \(b\) is the y - intercept (initial value). Here, \(m = 0.434\) and \(b = 15\). So the pressure function is \(P(d)=0.434d + 15\).

Step1: Set up the equation

We know that \(P = 110\) lb/in², and we use the function \(P(d)=0.434d + 15\). So we set up the equation \(110=0.434d + 15\).

Step2: Solve for \(d\)

First, subtract \(15\) from both sides of the equation: \(110 - 15=0.434d\), which simplifies to \(95 = 0.434d\). Then, divide both sides by \(0.434\) to solve for \(d\): \(d=\frac{95}{0.434}\approx219\) (rounded to the nearest whole number).

Answer:

\(P = 0.434d+15\)

Part (b)