If a cylindrical tank of radius 3 m is filled with water at the rate of \(\frac{3}{2} \, m^3/sec \), then the rate of change of its water level in (m/sec) is:
Show Hint
When working with related rates in geometry, always identify the constant dimensions before differentiating the volume formula to simplify your calculations.
Step 1: Recall the cylinder volume.
A cylinder of fixed radius $r$ and height $h$ has volume $V=\pi r^2 h$. Here only $h$ changes as water fills in. Step 2: Note the given data.
The radius is $r=3$ m (constant) and the fill rate is $\dfrac{dV}{dt}=\dfrac32\ \text{m}^3/\text{sec}$. Step 3: Differentiate with respect to time.
Since $r$ is constant,
\[ \frac{dV}{dt}=\pi r^2\frac{dh}{dt}. \] Step 4: Substitute the radius.
\[ \frac{3}{2}=\pi(3)^2\frac{dh}{dt}=9\pi\frac{dh}{dt}. \] Step 5: Solve for the level rate.
\[ \frac{dh}{dt}=\frac{3}{2\cdot9\pi}=\frac{3}{18\pi}=\frac{1}{6\pi}. \] Step 6: State the answer.
The water level rises at $\dfrac{1}{6\pi}$ m/sec.
\[ \boxed{\dfrac{1}{6\pi}} \]