A tank in the shape of a rectangular parallel | vedclass

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(C) Let the volume of the rectangular parallelepiped be $V = 27 \ m^3$. Let $A$ be the area of the base and $h$ be the height of the tank. Then, $V =

Vedclass · Accepted Answer

$81$

Vedclass · Answer

(C) Let the volume of the rectangular parallelepiped be $V = 27 \ m^3$. Let $A$ be the area of the base and $h$ be the height of the tank. Then, $V = A 	imes h$. Differentiating with respect to time $t$, we get $\frac{dV}{dt} = A \frac{dh}{dt}$. According to the problem, the rate of change of the water level $\frac{dh}{dt}$ is thrice the rate of change of the water quantity $\frac{dV}{dt}$, i.e.,$\frac{dh}{dt} = 3 \frac{dV}{dt}$. Substituting this into the derivative equation: $\frac{dV}{dt} = A 	imes (3 \frac{dV}{dt})$. This implies $1 = 3A$, so $A = \frac{1}{3} \ m^2$. Since $V = A 	imes h$, we have $27 = \frac{1}{3} 	imes h$. Therefore, $h = 27 	imes 3 = 81 \ m$.

$A$ tank in the shape of a rectangular parallelepiped has a volume of $27 \ m^3$. This tank is filled with water such that the rate of change of the water level is thrice the rate of change of the water quantity falling into the tank. The height of the tank (in meters) is:

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