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: Understanding the Question:
We are given the rate of change of the volume of water in a cylinder and the constant radius of the cylinder's base. We need to find the rate of change of the water level (height).

Step 2: Key Formula or Approach:
The volume $V$ of a cylinder is given by $V = \pi r^2 h$.
Since the radius $r$ is constant, we can differentiate both sides with respect to time $t$ to relate the rate of volume change ($\frac{dV}{dt}$) to the rate of height change ($\frac{dh}{dt}$).

Step 3: Detailed Explanation:
Given parameters:
Rate of volume increase: $\frac{dV}{dt} = 36\ \text{m}^3/\text{min}$
Radius of the base: $r = 3\ \text{m}$
Substitute $r = 3$ into the volume formula:
$$V = \pi (3)^2 h = 9\pi h$$ Differentiate both sides with respect to time $t$:
$$\frac{dV}{dt} = 9\pi \frac{dh}{dt}$$ Substitute the given value for $\frac{dV}{dt}$:
$$36 = 9\pi \frac{dh}{dt}$$ Solve for the rate of change of height $\frac{dh}{dt}$:
$$\frac{dh}{dt} = \frac{36}{9\pi}$$ $$\frac{dh}{dt} = \frac{4}{\pi}\ \text{m/min}$$

Step 4: Final Answer:
The water level is rising at a rate of $\frac{4}{\pi}\ \text{m/min}$, matching option (B).