Question

The surface area of a spherical balloon is increasing at the rate of \( 2 \, \text{cm}^2/\text{sec} \). Then the rate of increase in the volume of the balloon, when the radius of the balloon is \( 6 \, \text{cm} \), is:

• A. \( 4 \, \text{cm}^3/\text{sec} \)
• B. \( 16 \, \text{cm}^3/\text{sec} \)
• C. \( 36 \, \text{cm}^3/\text{sec} \)
• D. \( 6 \, \text{cm}^3/\text{sec} \)

Hint:

To solve such problems, relate \( \frac{dV}{dt} \), \( \frac{dS}{dt} \), and \( \frac{dr}{dt} \) using their respective equations and carefully substitute the given values.

Answer 1

The Correct Option is D

The surface area \( S \) of a sphere is given by: \[ S = 4\pi r^2, \] where \( r \) is the radius of the sphere.
The volume \( V \) of the sphere is given by: \[ V = \frac{4}{3} \pi r^3. \]
We are given: \[ \frac{dS}{dt} = 2 \, \text{cm}^2/\text{sec}, \quad r = 6 \, \text{cm}. \]
We need to find \( \frac{dV}{dt} \), the rate of increase of volume.
Step 1: Relating \( \frac{dS}{dt} \) and \( \frac{dr}{dt} \)
Differentiate \( S = 4\pi r^2 \) with respect to \( t \): \[ \frac{dS}{dt} = 8\pi r \frac{dr}{dt}. \]
Rearrange to solve for \( \frac{dr}{dt} \): \[ \frac{dr}{dt} = \frac{\frac{dS}{dt}}{8\pi r}. \]
Substitute \( \frac{dS}{dt} = 2 \) and \( r = 6 \): \[ \frac{dr}{dt} = \frac{2}{8\pi \cdot 6} = \frac{1}{24\pi}. \]
Step 2: Relating \( \frac{dV}{dt} \) and \( \frac{dr}{dt} \)
Differentiate \( V = \frac{4}{3}\pi r^3 \) with respect to \( t \): \[ \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}. \]
Substitute \( r = 6 \) and \( \frac{dr}{dt} = \frac{1}{24\pi} \): \[ \frac{dV}{dt} = 4\pi (6)^2 \cdot \frac{1}{24\pi}. \]
Simplify: \[ \frac{dV}{dt} = 4\pi \cdot 36 \cdot \frac{1}{24\pi} = \frac{144}{24} = 6 \, \text{cm}^3/\text{sec}. \]
Final Answer: \[ \boxed{6 \, \text{cm}^3/\text{sec}} \]