Question

A water tank has a square base of area 8 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

Step-by-step explanation:

Given

base of water tank
`A=8\ m^2`

Initial volume of water
`V=70\ m^3`

Water leaves the tank at the rate of

`\frac{\mathrm{d} V}{\mathrm{d} t}=-(2+4t^3)`

Integrating rate to get desired volume

at
`t=0,V=70\ m^3`

at
`t=3\ hr,V=V`

therefore

`\int_(70)^(V)dV=\int_(0)^(3)-\left ( 2+4t\right )dt`

`V-70=2t+2t^2|_0^3`

`V-70=-24`

`V=46\ m^3`

and
`volume=Area* h`

`h=(V)/(A)`

`h=(46)/(8)`

`h=5.75\ m`

Thus height of water remaining is
`5.75\ m`