Chapter 1, Problem 64RE

Solution

To write: The volume of oil remaining in the tank $t$ hours later as a function of $t$ .

The volume of oil remaining in the tank $t$ hours later is equal to $4000\pi -200\pi t$ cube ft.

Given information:

A cylindrical tank with diameter $20$ ft is filled with oil to depth of $40$ feet.

Calculation:

The diameter of the cylindrical tank is $20$ ft.

Then the radius of the cylindrical tank is $10$ ft.

The depth is $40$ ft.

Formula used:

The volume of the tank is given by

  $V=\pi r^{2}h$

Substitute $r=10$ and calculate the volume below:

  $\begin{array}{c}V=\pi {\left(10\right)}^{2}h\\ =100\pi h\end{array}$

If the oil drains at a rate of $2$ cube ft per hour, then after $t$ hours, it will decrease $2t$ feet height. Therefore, the depth of the oil remaining after $t$ hours is below:

  $h=40-2t$

Substitute $h=40-2t$ in the volume function of $h$ :

  $\begin{array}{c}V=100\pi \left(40-2t\right)\\ =4000\pi -200\pi t\end{array}$

Hence, the volume of oil remaining in the tank $t$ hours later is $4000\pi -200\pi t$ cube ft.