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:
This problem deals with rates of change involving geometric related rates. We are given the rate of change of the volume of a sphere over time ($\frac{dV}{dt}$) and need to calculate the rate of change of its radius ($\frac{dr}{dt}$) at the exact instant when the radius $r = 2\ \text{cm}$.

Step 2: Key Formula or Approach:
The formula for the volume $V$ of a perfect sphere in terms of its radius $r$ is: $$V = \frac{4}{3}\pi r^3$$ Differentiate both sides with respect to time $t$ using the chain rule: $$\frac{dV}{dt} = \frac{4}{3}\pi \cdot \left(3r^2 \frac{dr}{dt}\right) = 4\pi r^2 \frac{dr}{dt}$$

Step 3: Detailed Explanation:
From the problem description, we have: $$\frac{dV}{dt} = 8\ \text{cm}^3/\text{sec}$$ $$r = 2\ \text{cm}$$ Let's plug these values directly into our differentiated related rates equation: $$8 = 4\pi (2)^2 \cdot \frac{dr}{dt}$$ $$8 = 4\pi (4) \cdot \frac{dr}{dt}$$ $$8 = 16\pi \cdot \frac{dr}{dt}$$ Isolating the rate of change of the radius $\frac{dr}{dt}$: $$\frac{dr}{dt} = \frac{8}{16\pi} = \frac{1}{2\pi}\ \text{cm/sec}$$

Step 4: Final Answer:
The rate of increase of the radius is $\frac{1}{2\pi}\ \text{cm/sec}$, corresponding to option (D).