Question

The volume of spherical balloon is increasing at the rate of 25 c.c./s. Find the rate of change of its surface area at the instant when its radius is 5cm.

Hint:

Notice that \( \frac{dV}{dt} = (\text{Surface Area}) \times \frac{dr}{dt} \). This is a useful shortcut for spherical rate problems.

Answer 1

Step 1: Understanding the Concept:
This problem involves related rates. We use the formulas for the volume and surface area of a sphere and relate their rates of change through the common variable, the radius.
Step 2: Key Formula or Approach:
1. Volume \( V = \frac{4}{3}\pi r^3 \).
2. Surface Area \( S = 4\pi r^2 \).
3. Given \( \frac{dV}{dt} = 25 \text{ cm}^3/\text{s} \). Find \( \frac{dS}{dt} \) when \( r = 5 \).
Step 3: Detailed Explanation:
Differentiate volume with respect to time \( t \):
\[ \frac{dV}{dt} = \frac{d}{dt} \left( \frac{4}{3}\pi r^3 \right) = \frac{4}{3}\pi (3r^2) \frac{dr}{dt} = 4\pi r^2 \frac{dr}{dt} \] Substitute the given values to find \( \frac{dr}{dt} \):
\[ 25 = 4\pi (5)^2 \frac{dr}{dt} \] \[ 25 = 100\pi \frac{dr}{dt} \implies \frac{dr}{dt} = \frac{25}{100\pi} = \frac{1}{4\pi} \text{ cm/s} \] Now, differentiate surface area \( S \) with respect to time \( t \):
\[ \frac{dS}{dt} = \frac{d}{dt} (4\pi r^2) = 8\pi r \frac{dr}{dt} \] Substitute \( r = 5 \) and \( \frac{dr}{dt} = \frac{1}{4\pi} \):
\[ \frac{dS}{dt} = 8\pi (5) \left( \frac{1}{4\pi} \right) \] \[ \frac{dS}{dt} = \frac{40\pi}{4\pi} = 10 \text{ cm}^2/\text{s} \] Step 4: Final Answer:
The rate of change of surface area is \( 10 \text{ cm}^2/\text{s} \).