Question

A spherical snow ball is forming so that its volume is increasing at the rate of $8\ \text{cm}^3/\text{sec}$. Find the rate of increase of its radius when the radius is $2\ \text{cm}$.

• A. $\pi\ \text{cm/sec}$
• B. $\frac{1}{8\pi}\ \text{cm/sec}$
• C. $2\pi\ \text{cm/sec}$
• D. $\frac{1}{2\pi}\ \text{cm/sec}$

Hint:

Notice that the rate of change of volume is exactly equal to the surface area multiplied by the rate of change of the radius: $\frac{dV}{dt} = \text{Surface Area} \times \frac{dr}{dt}$. Since the surface area of a sphere is $4\pi r^2$, at $r=2$ it is $16\pi$. Thus, $\frac{dr}{dt} = \frac{\text{Rate of Volume}}{\text{Surface Area}} = \frac{8}{16\pi} = \frac{1}{2\pi}$ instantly!

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The volume of a sphere increases at 8 cm³/s. Find the rate of radius increase when r = 2 cm.

Step 2: Key Formula or Approach:
V = (4/3)πr³. Differentiate with respect to time: dV/dt = 4πr²(dr/dt). Solve for dr/dt.

Step 3: Detailed Explanation:
Given dV/dt = 8 and r = 2: 8 = 4π(2)²(dr/dt) = 4π(4)(dr/dt) = 16π(dr/dt). Thus dr/dt = 8/(16π) = 1/(2π) cm/s.

Step 4: Final Answer:
The rate is 1/(2π) cm/s, option (D).