Question

we need to determine how fast the height of the water is increasing. in other words, we are looking for a rate of change. in this problem, the volume and the height are both functions of the time t, where t is measured in minutes.
the rate of increase of the volume with respect to time is the derivative $\frac{dv}{dt}$. the rate of increase of the height with respect to time is the derivative $\frac{dh}{dt}$.
therefore, we need to differentiate each side of our area equation with respect to t. doing so gives the following result where $\frac{dv}{dt}$ is measured in m³/min and $\frac{dh}{dt}$ is measured in m/min.
$\frac{dv}{dt}=\frac{d(36\pi h)}{dt}$
$\frac{dv}{dt}=(\square)\frac{dh}{dt}$

Explanation:

Step1: Differentiate the right - hand side

Using the constant multiple rule of differentiation, if $y = cf(x)$ where $c$ is a constant and $f(x)$ is a function of $x$, then $\frac{dy}{dx}=c\frac{df(x)}{dx}$. Here, $V = 36\pi h$ and we are differentiating with respect to $t$. The derivative of $36\pi h$ with respect to $t$ is $36\pi\frac{dh}{dt}$ since $36\pi$ is a constant.
$\frac{d(36\pi h)}{dt}=36\pi\frac{dh}{dt}$

Step2: Substitute into the equation

We know that $\frac{dV}{dt}=\frac{d(36\pi h)}{dt}$, so $\frac{dV}{dt}=36\pi\frac{dh}{dt}$

Answer:

$36\pi$