Chapter 1, Problem 62RE

Solution

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

The volume of oil remaining in the tank $t$ seconds later is equal to $4000\pi -2t$ 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 $r=10,h=40$ and calculate the volume below:

  $\begin{array}{c}V=\pi {\left(10\right)}^2\left(40\right)\\ =4000\pi \end{array}$

If the oil drains at a rate of $2$ cube ft per second, then after $t$ seconds, it will lose $2t$ cube feet oil. Therefore, the volume of oil remaining after $t$ seconds is below:

  $V=4000\pi -2t$

Hence, the volume of oil remaining in the tank $t$ seconds later is $4000\pi -2t$ cube ft.