Question

Water pours into a fish tank at a rate of 0.3 cubic meters per minute. How fast is the water level rising if the base of the fish tank is a 2 meter by 3 meter rectangle?

Answer 1

Answer:

$\frac{dh}{dt} = 0.05 m/min$

Explanation:

As we know that the dimensions of the base of the tank is given as

$L = 2 m$

$W = 3 m$

now the base area is given as

$A = 2m \times 3 m$

$A = 6 m^2$

now we know that volume of the liquid filled in the tank is given as

$V = Area \times height$

$V = 6 \times h$

now we will differentiate it with respect to time both sides

$\frac{dV}{dt} = 6 \times \frac{dh}{dt}$

here we know that

$\frac{dV}{dt} = 0.3 m^3/min$

now we have

$0.3 m^3/min = (6 m^2) \frac{dh}{dt}$

$\frac{dh}{dt} = 0.05 m/min$