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

Answer:

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 $volume=Area\times h$

$h=\frac{V}{A}$

$h=\frac{46}{8}$

$h=5.75\ m$

Thus height of water remaining is $5.75\ m$