Question

A cone whose height is always equal to its diameter, is increasing in volume at the rate of \(40 \text{ cm}^3/\text{s}\). At what rate is the radius increasing when its circular base area is \(1 \text{ m}^2\)?

• A. 1 mm/s
• B. 0.001 cm/s
• C. 2 mm/s
• D. 0.002 cm/s

Hint:

Convert all units to consistent system (here cm).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept: h = 2r, \quad V = \frac{1}{3}\pi r^2 h = \frac{2}{3}\pi r^3
Step 2: Detailed Explanation: \frac{dV}{dt} = 2\pi r^2 \frac{dr}{dt} = 40
A = \pi r^2 = 1 \text{ m}^2 = 10^4 \text{ cm}^2 \Rightarrow r^2 = \frac{10^4}{\pi}
40 = 2\pi \left(\frac{10^4}{\pi}\right) \frac{dr}{dt} = 2 \times 10^4 \frac{dr}{dt}
\frac{dr}{dt} = \frac{40}{2 \times 10^4} = 0.002 \text{ cm/s}
Step 3: Final Answer: 0.002 \text{ cm/s}