Question

A right circular cone is melting such that its volume is decreasing at the rate of \(12\pi\) \(\text{cm}^{3}/\text{sec}\). The radius of the base is always equal to half of its height. Then the rate of change of the radius when the radius is \(2\) cm is:

• A. radius is increasing at the rate of \(1\) cm/sec
• B. radius is decreasing at the rate of \(1\) cm/sec
• C. radius is increasing at the rate of \(1.5\) cm/sec
• D. radius is decreasing at the rate of \(1.5\) cm/sec

Hint:

Express the volume in terms of the radius only using h equal to 2r, then differentiate with respect to time.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
The radius is half the height, so \(h=2r\).

Step 2: Volume in Terms of r
\(V=\dfrac13\pi r^{2}h=\dfrac13\pi r^{2}(2r)=\dfrac23\pi r^{3}\).

Step 3: Differentiate
\(\dfrac{dV}{dt}=2\pi r^{2}\dfrac{dr}{dt}\).

Step 4: Calculation
Given \(\dfrac{dV}{dt}=-12\pi\) and \(r=2\): \(-12\pi=2\pi(4)\dfrac{dr}{dt}\), so \(\dfrac{dr}{dt}=-\dfrac{3}{2}\).
The negative sign means the radius is decreasing.

Final Answer:
The radius is decreasing at 1.5 cm per second. \[ \boxed{\text{(D) }\text{radius is decreasing at the rate of }1.5\text{ cm/sec}} \]