Question

Water is being poured at the rate of $36\ \text{m}^3/\text{min}$ into a cylindrical vessel, whose circular base is of radius $3\ \text{m}$. Then the water level in the cylinder is rising at the rate of

• A. $4\pi\ \text{m/min}$
• B. $\frac{4}{\pi}\ \text{m/min}$
• C. $\frac{1}{4\pi}\ \text{m/min}$
• D. $2\pi\ \text{m/min}$

Hint:

In problems involving related rates for standard cylinders or prisms, remember that the cross-sectional area (like $\pi r^2$) is constant. Therefore, the rate of volume change is just the base area multiplied by the rate of height change!

Answer 1

The Correct Option is B

Step 1: Picture the situation.
Water is poured into a cylinder whose base radius stays fixed at $3$ m. As water fills, only the height of the water rises. We must find how fast the height rises.

Step 2: Note the given rates.
Volume is being added at $\frac{dV}{dt}=36\ \text{m}^3/\text{min}$. The radius is constant, $r=3$ m. We want $\frac{dh}{dt}$.

Step 3: Write the volume formula.
For a cylinder, $V=\pi r^2 h$. Since the radius does not change, only $h$ changes with time.

Step 4: Put the fixed radius in.
\[ V=\pi(3)^2 h=9\pi h. \] Now $V$ depends only on $h$, which keeps things simple.

Step 5: Differentiate with respect to time.
Because both volume and height change with time, we take $\frac{d}{dt}$ of both sides. The number $9\pi$ stays put. \[ \frac{dV}{dt}=9\pi\,\frac{dh}{dt}. \]

Step 6: Plug in the known rate.
\[ 36=9\pi\,\frac{dh}{dt}. \]

Step 7: Solve for the height rate.
\[ \frac{dh}{dt}=\frac{36}{9\pi}=\frac{4}{\pi}\ \text{m/min}. \] This matches option (2).
\[ \boxed{\dfrac{4}{\pi}\ \text{m/min}} \]