Answer:
Explanation:
Given
base of water tank [tex]A=8\ m^2[/tex]
Initial volume of water [tex]V=70\ m^3[/tex]
Water leaves the tank at the rate of
[tex]\frac{\mathrm{d} V}{\mathrm{d} t}=-(2+4t^3)[/tex]
Integrating rate to get desired volume
at [tex]t=0,V=70\ m^3[/tex]
at [tex]t=3\ hr,V=V[/tex]
therefore
[tex]\int_{70}^{V}dV=\int_{0}^{3}-\left ( 2+4t\right )dt[/tex]
[tex]V-70=2t+2t^2|_0^3[/tex]
[tex]V-70=-24[/tex]
[tex]V=46\ m^3[/tex]
and [tex]volume=Area\times h[/tex]
[tex]h=\frac{V}{A}[/tex]
[tex]h=\frac{46}{8}[/tex]
[tex]h=5.75\ m[/tex]
Thus height of water remaining is [tex]5.75\ m[/tex]
