Question

At the surface of the ocean, the water pressure is the same as the air pressure above the water,\({\bf{15}}\;{\bf{lb/i}}{{\bf{n}}^{\bf{2}}}\). Below the surface, the water pressure increases by\({\bf{4}}{\bf{.34}}\;{\bf{lb/i}}{{\bf{n}}^{\bf{2}}}\)for every 10 ft of descent.

(a) Express the water pressure as a function of the depth below the ocean surface.

(b) At what depth is the pressure\({\bf{100}}\;{\bf{lb/i}}{{\bf{n}}^{\bf{2}}}\)?

Short answer

(a) The equation of the linear function is\(P = 0.434\;d + 15\).

(b) At the depth\(196\;ft\), the pressure is\(100\;{\rm{lb/i}}{{\rm{n}}^{\rm{2}}}\).

Step-by-step solution

Given data

Let d be the depth and P be the pressure.

We have, when\(d = 0,\;P = 15\)

So, the points are\(\left( {0,15} \right)\).\(\)

(a)Determine the water pressure as a function

Now pressure increases by\(4.34\,lb/i{n^2}\)for every\(10\;ft\).

So, pressure increases\(\frac{{4.34}}{{10}} = 0.434\,lb/i{n^2}\)for every\(1\;ft\).

So, the slope of the graph is\(0.434\)and\(y - {\rm{intercept}}\)is\(15\).

Hence, the equation of the linear function is.\(\)\(\)

(b) Determine the depth where pressure is \({\bf{100}}\;{\bf{lb/i}}{{\bf{n}}^{\bf{2}}}\)\(\)

Now \(P = 100\;lb/i{n^2}\)

Then,

\begin{aligned} & 100=0.434\ d+15 \\ & 0.434d=85 \\ & d=\frac{85}{0.434} \\& \approx 195.85\;feet \\ & \approx 196\;feet \end{aligned}

Hence, the depth is\(196\;ft\).