Question

A water tank has a square base of area 5 square meters. Initially the tank contains 70 cubic meters. Water leaves the tank, starting at t=0, at the rate of 2 + 4 t cubic meters per hour. Here t is the time in hours. What is the depth of water remaining in the tank after 3 hours

Answer 1

Answer:

$9.2\ \text{m}$

Step-by-step explanation:

$b$ = Surface area of base = 5 square meters

Volume of water in tank = 70 cubic meters

The rate at which the volume is reducing is

$\dfrac{dV}{dt}=2+4t\\\Rightarrow dV=2+4tdt$

Integrating from $t=0$ to $t=3$

$V=\int^3_0(2+4t)dt\\\Rightarrow V=2t+2t^2|_0^3\\\Rightarrow V=2\times 3+2\times 3^2-0\\\Rightarrow V=24$

Volume of water remaining in the tank is $70-24=46\ \text{m}^3$

Suface area of base $\times$ depth = Volume

$5\times d=46\\\Rightarrow d=\dfrac{46}{5}\\\Rightarrow d=9.2\ \text{m}$

The depth of the water remaining in the tank is $9.2\ \text{m}$.