Question

At the surface of the ocean the water pressure is the same as the air pressure above the water, $15\mathrm{lb}/{\mathrm{in}}^2$. Below the surface the water pressure increases by $4.34\mathrm{lb}/{\mathrm{in}}^2$for every 10 ft of descent.

(a)Find an equation for the relationship between pressure and depth below the ocean surface.

(b)Sketch a graph of this linear equation.

(c)What do the slope and y-intercept of the graph represent?

(d)At what depth is the pressure$100\mathrm{lb}/{\mathrm{in}}^2$?

Short answer

(a) The linear equation is $y=-0.94x+\mathrm{14.07.}$

(b) The answer is the graph of $y=-0.94x+\mathrm{14.07.}$

(c) The slope is $-0.94$and y-intercept is 14.07.

(d) At 108.07 ft, the pressure is$100\mathrm{lb}/{\mathrm{in}}^2$

Step-by-step solution

Part a. Step 1. Concept of linear equation.

A linear equation is an algebraic equation in which each term has an exponent of one and the graphing of the equation results in a straight line.

Part a. Step 2. Two point form of linear equation.

If two points $A(x_1,y_1)$and $B(x_2,y_2)$are given, the equation of $AB$is $\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right).$

Part a. Step 3. Forming an equation.

Let x describe the pressure of the ocean and y describe the depth of the ocean.

When $x_1=15lb/i{n}^2,y=0ft$

$x_2=4.34lb/i{n}^2,y=10ft$

Then the linear equation that relates the temperature y and x is

$$\begin{gathered} y-0=\frac{10-0}{4.34-15}\left(x-15\right) \\ y=\frac{10}{-10.66}\left(x-15\right) \\ y=-0.94x+14.07 \end{gathered}$$

Part b. Step 1. Concept of graph.

Graph is a mathematical representation of a network which describes the relationship between lines and points.

Part b. Step 2. Concept of graphing a linear equation.

The graph of linear equation in two variables is a line.

Part b. Step 3. Sketching the graph.

The graph is as follows,

Part c. Step 1. Concept of Slope.

Suppose a non-vertical line has a slope m and y-intercept b. This means that the line intersects the y-axis at the point $\left(0,b\right),$so the point slope form of the equation of the line with $x=0$and $y=b$becomes

$\begin{array}{c}(y-b)=m(x-0)\\ y=mx+b\end{array}$

Part c. Step 2. Concept of intercept.

The y-coordinate is a point where a line curve or a surface intersects the y-axis.

Part c. Step 3. Determining the slope and the intercept.

The linear equation is

$y=-0.94x+14.07$

Hence, the slope is $-0.94$and y-intercept is 14.07.

Part d. Step 1. Concept of linear equation.

A linear equation is an algebraic equation in which each term has an exponent of one and the graphing of the equation results in a straight line.

Part d. Step 2. Concept to determine two points.

If the value of x is given, then we have to evaluate the y by substituting the value in the equation and solve for y.

Part d. Step 3. Evaluating the depth at the given pressure.

The linear equation is

$y=-0.94x+14.07$

Substituting $x=100lb/i{n}^2$, we get

$\begin{array}{l}y=0.94\left(100\right)+14.07\\ y=94+14.07\\ y=108.07\end{array}$

Hence, at 108.07 ft, the pressure is $100\mathrm{lb}/{\mathrm{in}}^2$.