Question

we are given that the tank is being filled at a rate of 4 m³/min. in other words, the volume of water in the tank is increasing at a rate of 4 m³/min. therefore, we have the following.
$\frac{dv}{dt}=square$
substituting this value into the derivative $\frac{dv}{dt}=(36pi)\frac{dh}{dt}$ and then solving for $\frac{dh}{dt}$ gives the following result.
$\frac{dh}{dt}=square$
therefore, the height of the water in the tank is increasing at the following rate (in m/min).
$square$ m/min

Explanation:

Step1: Identify the given rate

The rate of change of volume $\frac{dV}{dt}$ is given as $4\ m^{3}/min$. So, $\frac{dV}{dt}=4$.

Step2: Solve for $\frac{dh}{dt}$

We know that $\frac{dV}{dt}=(36\pi)\frac{dh}{dt}$. Substitute $\frac{dV}{dt} = 4$ into the equation: $4=(36\pi)\frac{dh}{dt}$. Then, $\frac{dh}{dt}=\frac{4}{36\pi}=\frac{1}{9\pi}$.

Answer:

$\frac{dV}{dt}=4$
$\frac{dh}{dt}=\frac{1}{9\pi}$
$\frac{1}{9\pi}$ m/min